Extensible prime generator

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Revision as of 20:23, 20 June 2022 by Alexei Kopylov (talk | contribs) (→‎Using formula for primes: Replace with another library which turns out to be much more efficient. Also use simpler formula (the old formula could be a little more efficient).)
Task
Extensible prime generator
You are encouraged to solve this task according to the task description, using any language you may know.
Task

Write a generator of prime numbers, in order, that will automatically adjust to accommodate the generation of any reasonably high prime.

The routine should demonstrably rely on either:

  1. Being based on an open-ended counter set to count without upper limit other than system or programming language limits. In this case, explain where this counter is in the code.
  2. Being based on a limit that is extended automatically. In this case, choose a small limit that ensures the limit will be passed when generating some of the values to be asked for below.
  3. If other methods of creating an extensible prime generator are used, the algorithm's means of extensibility/lack of limits should be stated.


The routine should be used to:

  • Show the first twenty primes.
  • Show the primes between 100 and 150.
  • Show the number of primes between 7,700 and 8,000.
  • Show the 10,000th prime.


Show output on this page.

Note: You may reference code already on this site if it is written to be imported/included, then only the code necessary for import and the performance of this task need be shown. (It is also important to leave a forward link on the referenced tasks entry so that later editors know that the code is used for multiple tasks).

Note 2: If a languages in-built prime generator is extensible or is guaranteed to generate primes up to a system limit, (231 or memory overflow for example), then this may be used as long as an explanation of the limits of the prime generator is also given. (Which may include a link to/excerpt from, language documentation).

Note 3:The task is written so it may be useful in solving the task   Emirp primes   as well as others (depending on its efficiency).


Reference



Ada

The solution is based on an open-ended counter, named "Current" counting up to the limit from the Compiler, namely 2**63-1.

The solution uses the package Miller_Rabin from the Miller-Rabin primality test. When using the gnat Ada compiler, the largest integer we can deal with is 2**63-1. For anything larger, we could use a big-num package.

<lang Ada>with Ada.Text_IO, Miller_Rabin;

procedure Prime_Gen is

  type Num is range 0 .. 2**63-1; -- maximum for the gnat Ada compiler
  
  MR_Iterations: constant Positive := 25; 
    -- the probability Pr[Is_Prime(N, MR_Iterations) = Probably_Prime] 
    -- is 1 for prime N and < 4**(-MR_Iterations) for composed N
  
  function Next(P: Num) return Num is
     N: Num := P+1;
     package MR is new Miller_Rabin(Num); use MR;
  begin
     while not (Is_Prime(N, MR_Iterations) = Probably_Prime) loop

N := N + 1;

     end loop;
     return N;
  end Next;
  
  Current: Num;
  Count: Num := 0;
  

begin

  -- show the first twenty primes
  Ada.Text_IO.Put("First 20 primes:");
  Current := 1;
  for I in 1 .. 20 loop
     Current := Next(Current);
     Ada.Text_IO.Put(Num'Image(Current));
  end loop;
  Ada.Text_IO.New_Line;
  
  -- show the primes between 100 and 150
  Ada.Text_IO.Put("Primes between 100 and 150:");
  Current := 99;
  loop
     Current := Next(Current);
     exit when Current > 150;
     Ada.Text_IO.Put(Num'Image(Current));
  end loop;
  Ada.Text_IO.New_Line;
  
  -- count primes between 7700 and 8000
  Ada.Text_IO.Put("Number of primes between 7700 and 8000:");
  Current := 7699;
  loop
     Current := Next(Current);
     exit when Current > 8000;
     Count := Count + 1;
  end loop;
  Ada.Text_IO.Put_Line(Num'Image(Count));
  
  Count := 10;
  Ada.Text_IO.Put_Line("Print the K_i'th prime, for $K=10**i:");
  begin
     loop

Current := 1; for I in 1 .. Count loop Current := Next(Current); end loop; Ada.Text_IO.Put(Num'Image(Count) & "th prime:" & Num'Image(Current)); Count := Count * 10;

     end loop;
  exception
     when Constraint_Error => 

Ada.Text_IO.Put_Line(" can't compute the" & Num'Image(Count) & "th prime:");

  end;

end;</lang>

Output:
First 20 primes: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
Primes between 100 and 150: 101 103 107 109 113 127 131 137 139 149
Number of primes between 7700 and 8000: 30
Print the K_i'th prime, for $K=10**i:
 10th prime: 29 100th prime: 541 1000th prime: 7919 10000th prime: 104729 100000th prime: 1299709 1000000th prime: 15485863

(The program has been stopped after running several days.)

AutoHotkey

<lang AutoHotkey>SetBatchLines, -1 p := 1 ;p functions as the counter Loop, 10000 { p := NextPrime(p) if (A_Index < 21) a .= p ", " if (p < 151 && p > 99) b .= p ", " if (p < 8001 && p > 7699) c++ } MsgBox, % "First twenty primes: " RTrim(a, ", ") . "`nPrimes between 100 and 150: " RTrim(b, ", ") . "`nNumber of primes between 7,700 and 8,000: " RTrim(c, ", ") . "`nThe 10,000th prime: " p

NextPrime(n) { Loop if (IsPrime(++n)) return n }

IsPrime(n) { if (n < 2) return, 0 else if (n < 4) return, 1 else if (!Mod(n, 2)) return, 0 else if (n < 9) return 1 else if (!Mod(n, 3)) return, 0 else { r := Floor(Sqrt(n)) f := 5 while (f <= r) { if (!Mod(n, f)) return, 0 if (!Mod(n, (f + 2))) return, 0 f += 6 } return, 1 } }</lang>

Output:
First twenty primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71
Primes between 100 and 150: 101, 103, 107, 109, 113, 127, 131, 137, 139, 149
Number of primes between 7,700 and 8,000: 30
The 10,000th prime: 104729

BQN

This implementation uses a simple segmented sieve similar to the one in Sieve of Eratosthenes. In order to match the task most closely, it returns a generator function (implemented as a closure) that outputs another prime on each call, using an underlying primes array that's extended as necessary. Working with one prime at a time is inefficient in an array language, and the given tasks can be solved more quickly using functions from bqn-libs primes.bqn, which also uses a more complicated and faster underlying sieve. <lang bqn># Function that returns a new prime generator PrimeGen ← {𝕤

 i ← 0  # Counter: index of next prime to be output
 primes ← ↕0
 next ← 2
 Sieve ← { p 𝕊 i‿n:
   E ← {↕∘⌈⌾(((𝕩|-i)+𝕩×⊢)⁼)n-i}       # Indices of multiples of 𝕩
   i + / (1⥊˜n-i) E⊸{0¨⌾(𝕨⊸⊏)𝕩}´ p    # Primes in segment [i,n)
 }
 {𝕤
   { i=≠primes ?                      # Extend if required
     next ↩ ((2⋆24)⊸+ ⌊ ט) old←next  # Sieve at most 16M new entries
     primes ∾↩ (primes(⍋↑⊣)√next) Sieve old‿next
   ;@}
   (i+↩1) ⊢ i⊑primes
 }

} _w_←{𝔽⍟𝔾∘𝔽_𝕣_𝔾∘𝔽⍟𝔾𝕩} # Looping utility for the session below</lang>

Output:

<lang bqn> pg ← PrimeGen@ (function block)

  PG¨ ↕20

⟨ 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 ⟩

  {p←↕0 ⋄ PG∘{  p∾↩𝕩}_w_(<⟜ 150) PG _w_(<⟜ 100)0 ⋄ p}

⟨ 101 103 107 109 113 127 131 137 139 149 ⟩

  {p←0  ⋄ PG∘{𝕤⋄p+↩1}_w_(<⟜8000) PG _w_(<⟜7700)0 ⋄ p}

30

  (PrimeGen@)⍟1e4 @  # Reset the count with a new generator

104729</lang>

C

Extends the list of primes by sieving more chunks of integers. There's no serious optimizations. The code can calculate all 32-bit primes in some seconds, and will overflow beyond that. <lang c>#include <stdio.h>

  1. include <stdlib.h>
  2. include <string.h>
  3. include <math.h>
  1. define CHUNK_BYTES (32 << 8)
  2. define CHUNK_SIZE (CHUNK_BYTES << 6)

int field[CHUNK_BYTES];

  1. define GET(x) (field[(x)>>6] & 1<<((x)>>1&31))
  2. define SET(x) (field[(x)>>6] |= 1<<((x)>>1&31))

typedef unsigned uint; typedef struct {

       uint *e;
       uint cap, len;

} uarray; uarray primes, offset;

void push(uarray *a, uint n) {

       if (a->len >= a->cap) {
               if (!(a->cap *= 2)) a->cap = 16;
               a->e = realloc(a->e, sizeof(uint) * a->cap);
       }
       a->e[a->len++] = n;

}

uint low; void init(void) {

       uint p, q;
       unsigned char f[1<<16];
       memset(f, 0, sizeof(f));
       push(&primes, 2);
       push(&offset, 0);
       for (p = 3; p < 1<<16; p += 2) {
               if (f[p]) continue;
               for (q = p*p; q < 1<<16; q += 2*p) f[q] = 1;
               push(&primes, p);
               push(&offset, q);
       }
       low = 1<<16;

}

void sieve(void) {

       uint i, p, q, hi, ptop;
       if (!low) init();
       memset(field, 0, sizeof(field));
       hi = low + CHUNK_SIZE;
       ptop = sqrt(hi) * 2 + 1;
       for (i = 1; (p = primes.e[i]*2) < ptop; i++) {
               for (q = offset.e[i] - low; q < CHUNK_SIZE; q += p)
                       SET(q);
               offset.e[i] = q + low;
       }
       for (p = 1; p < CHUNK_SIZE; p += 2)
               if (!GET(p)) push(&primes, low + p);
       low = hi;

}

int main(void) {

       uint i, p, c;
       while (primes.len < 20) sieve();
       printf("First 20:");
       for (i = 0; i < 20; i++)
               printf(" %u", primes.e[i]);
       putchar('\n');
       while (primes.e[primes.len-1] < 150) sieve();
       printf("Between 100 and 150:");
       for (i = 0; i < primes.len; i++) {
               if ((p = primes.e[i]) >= 100 && p < 150)
                       printf(" %u", primes.e[i]);
       }
       putchar('\n');
       while (primes.e[primes.len-1] < 8000) sieve();
       for (i = c = 0; i < primes.len; i++)
               if ((p = primes.e[i]) >= 7700 && p < 8000) c++;
       printf("%u primes between 7700 and 8000\n", c);
       for (c = 10; c <= 100000000; c *= 10) {
               while (primes.len < c) sieve();
               printf("%uth prime: %u\n", c, primes.e[c-1]);
       }
       return 0;

}</lang>

Output:
First 20: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
Between 100 and 150: 101 103 107 109 113 127 131 137 139 149
30 primes between 7700 and 8000
10th prime: 29
100th prime: 541
1000th prime: 7919
10000th prime: 104729
100000th prime: 1299709
1000000th prime: 15485863
10000000th prime: 179424673
100000000th prime: 2038074743

Alternative version based on The Genuine Sieve of Eratosthenes by Melissa O'Neil

It uses the Pairing Heap w/ generic data types from task Priority queue#C. Some notes:

  • It uses a wheel to skip over multiples of 2, 3, 5, and 7.
  • since we store the square of every prime found, the use of long int (64 bit in GCC) is required for a sieve of any non-trivial size.
  • ~100,000,000 primes is about the limits of this algorithm. This version is not as idiomatic to C as a page-segmented sieve would be.

<lang C>

  1. include <stdio.h>
  2. include <stdlib.h>
  3. include "pairheap.h"

int wheel2357[48] = {

   10,  2,  4,  2,  4,  6,  2,  6,
    4,  2,  4,  6,  6,  2,  6,  4,
    2,  6,  4,  6,  8,  4,  2,  4,
    2,  4,  8,  6,  4,  6,  2,  4,
    6,  2,  6,  6,  4,  2,  4,  6,
    2,  6,  4,  2,  4,  2, 10,  2,
    };

typedef struct { // elements in the priority queue

   pq_node_t hd;
   int       offset;      // index to skip value in 2,3,5,7 wheel
   long int  base_prime;

} w2357_multiples;

typedef struct {

   int      start_ndx;
   int      offset;
   long int candidate;
   heap_t   composites;
   int      count;      // count of primes returned.

} primegen_t;

primegen_t make_pgen() {

   w2357_multiples *composites;
   primegen_t gen;
   gen.start_ndx = 0;  // primes 2, 3, 5, 7, 11
   NEW_PQ_ELE(composites, 121);
   gen.offset = composites->offset = 1;
   gen.candidate = composites->base_prime = 11;
   gen.composites = (heap_t) composites;
   gen.count = 0;
   return gen;

}

long int next_prime(primegen_t *gen) {

   static short upto11[] = {
       2, 3, 5, 7, 11
       };
   if (gen->start_ndx < 5) {
       ++gen->count;
       return upto11[gen->start_ndx++];
   } else {
       for (;;) {
           // advance to the next prime candidate.
           gen->candidate += wheel2357[gen->offset++];
           if (gen->offset == 48)
               gen->offset = 0;
           // See if the composite number on top of the heap matches
           // the candidate.
           //
           w2357_multiples *top = (w2357_multiples *) gen->composites;
           if (top->hd.key == gen->candidate) { // not prime
               do {
                   // advance the top of heap to the next prime multiple
                   // that is not a multiple of 2, 3, 5, 7.
                   //
                   gen->composites = heap_pop(gen->composites);
                   top->hd.next = top->hd.down = NULL;
                   top->hd.key += top->base_prime  * wheel2357[top->offset++];
                   if (top->offset == 48)
                       top->offset = 0;
                   gen->composites = heap_merge((heap_t) top, gen->composites);
                   top = (w2357_multiples *) gen->composites;
               } while (top->hd.key == gen->candidate);
           } else {
               // prime found, add the square and it's position on the wheel
               // to the heap.
               //
               w2357_multiples *new;
               HEAP_PUSH(
                   new,
                   gen->candidate * gen->candidate,
                   &gen->composites);
               new->offset = gen->offset;
               new->base_prime = gen->candidate;
               ++gen->count;
               return gen->candidate;
           }
       }
   }

}

int main() {

   primegen_t primes = make_pgen();
   printf("first 20: ");
   for (int i = 1; i <= 20; i++)
       printf("%ld ", next_prime(&primes));
   putchar('\n');
   printf("between 100 and 150: ");
   long int p = next_prime(&primes);
   while (p < 150) {
       if (p > 100)
           printf("%ld ", p);
       p = next_prime(&primes);
   }
   putchar('\n');
   int count = 0;
   while (p < 8000) {
       if (p > 7700)
           ++count;
       p = next_prime(&primes);
   }
   printf("%d primes between 7700 and 8000.\n", count);
   long c;
   for (c = 10000; c <= 10000000; c *= 10) {
       while (primes.count < c)
           p = next_prime(&primes);
       printf("%ldth prime is %ld\n", c, p);
   }


   return 0;

} </lang>

Output:
first 20: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 
between 100 and 150: 101 103 107 109 113 127 131 137 139 149 
30 primes between 7700 and 8000.
10000th prime is 104729
100000th prime is 1299709
1000000th prime is 15485863
10000000th prime is 179424673

C++

Based on "The Genuine Sieve of Eratosthenes" by Melissa E. O'Neill. UPDATE: Added wheel optimization to match its C counterpart. <lang cpp>#include <iostream>

  1. include <cstdint>
  2. include <queue>
  3. include <utility>
  4. include <vector>
  5. include <limits>

template<typename integer> class prime_generator { public:

   integer next_prime();
   integer count() const {
       return count_;
   }

private:

   struct queue_item {
       queue_item(integer prime, integer multiple, unsigned int wheel_index) :
           prime_(prime), multiple_(multiple), wheel_index_(wheel_index) {}
       integer prime_;
       integer multiple_;
       unsigned int wheel_index_;
   };
   struct cmp {
       bool operator()(const queue_item& a, const queue_item& b) const {
           return a.multiple_ > b.multiple_;
       }
   };
   static integer wheel_next(unsigned int& index) {
       integer offset = wheel_[index];
       ++index;
       if (index == std::size(wheel_))
           index = 0;
       return offset;
   }
   typedef std::priority_queue<queue_item, std::vector<queue_item>, cmp> queue;
   integer next_ = 11;
   integer count_ = 0;
   queue queue_;
   unsigned int wheel_index_ = 0;
   static const unsigned int wheel_[];
   static const integer primes_[];

};

template<typename integer> const unsigned int prime_generator<integer>::wheel_[] = {

   2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2,
   6, 4, 6, 8, 4, 2, 4, 2, 4, 8, 6, 4, 6, 2, 4, 6,
   2, 6, 6, 4, 2, 4, 6, 2, 6, 4, 2, 4, 2, 10, 2, 10

};

template<typename integer> const integer prime_generator<integer>::primes_[] = {

   2, 3, 5, 7

};

template<typename integer> integer prime_generator<integer>::next_prime() {

   if (count_ < std::size(primes_))
       return primes_[count_++];
   integer n = next_;
   integer prev = 0;
   while (!queue_.empty()) {
       queue_item item = queue_.top();
       if (prev != 0 && prev != item.multiple_)
           n += wheel_next(wheel_index_);
       if (item.multiple_ > n)
           break;
       else if (item.multiple_ == n) {
           queue_.pop();
           queue_item new_item(item);
           new_item.multiple_ += new_item.prime_ * wheel_next(new_item.wheel_index_);
           queue_.push(new_item);
       }
       else
           throw std::overflow_error("prime_generator: overflow!");
       prev = item.multiple_;
   }
   if (std::numeric_limits<integer>::max()/n > n)
       queue_.emplace(n, n * n, wheel_index_);
   next_ = n + wheel_next(wheel_index_);
   ++count_;
   return n;

}

int main() {

   typedef uint32_t integer;
   prime_generator<integer> pgen;
   std::cout << "First 20 primes:\n";
   for (int i = 0; i < 20; ++i) {
       integer p = pgen.next_prime();
       if (i != 0)
           std::cout << ", ";
       std::cout << p;
   }
   std::cout << "\nPrimes between 100 and 150:\n";
   for (int n = 0; ; ) {
       integer p = pgen.next_prime();
       if (p > 150)
           break;
       if (p >= 100) {
           if (n != 0)
               std::cout << ", ";
           std::cout << p;
           ++n;
       }
   }
   int count = 0;
   for (;;) {
       integer p = pgen.next_prime();
       if (p > 8000)
           break;
       if (p >= 7700)
           ++count;
   }
   std::cout << "\nNumber of primes between 7700 and 8000: " << count << '\n';
   for (integer n = 10000; n <= 10000000; n *= 10) {
       integer prime;
       while (pgen.count() != n)
           prime = pgen.next_prime();
       std::cout << n << "th prime: " << prime << '\n';
   }
   return 0;

}</lang>

Output:
First 20 primes:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71
Primes between 100 and 150:
101, 103, 107, 109, 113, 127, 131, 137, 139, 149
Number of primes between 7700 and 8000: 30
10000th prime: 104729
100000th prime: 1299709
1000000th prime: 15485863
10000000th prime: 179424673

Faster Alternative

This is a "segmented sieve" implementation inspired by the original C solution. Execution time is about 4 seconds on my system (macOS 10.15.4, 3.2GHz Quad-Core Intel Core i5). <lang cpp>#include <algorithm>

  1. include <iostream>
  2. include <cmath>
  3. include <cstdint>
  4. include <vector>
  5. include <limits>

template<typename integer> class prime_generator { public:

   explicit prime_generator(integer initial_limit = 100, integer increment = 100000);
   integer next_prime();
   integer count() const {
       return count_;
   }

private:

   void find_primes(integer);
   integer count_ = 0;
   integer limit_;
   integer index_ = 0;
   integer increment_;
   std::vector<integer> primes_;
   std::vector<bool> sieve_;
   integer sieve_limit_ = 0;

};

template<typename integer> integer next_odd_number(integer n) {

   return n % 2 == 0 ? n + 1 : n;

}

template<typename integer> prime_generator<integer>::prime_generator(integer initial_limit, integer increment)

   : limit_(next_odd_number(initial_limit)), increment_(increment) {
   primes_.push_back(2);
   find_primes(3);

}

template<typename integer> integer prime_generator<integer>::next_prime() {

   if (index_ == primes_.size()) {
       if (std::numeric_limits<integer>::max() - increment_ < limit_)
           return 0;
       int start = limit_ + 2;
       limit_ = next_odd_number(limit_ + increment_);
       primes_.clear();
       find_primes(start);
   }
   ++count_;
   return primes_[index_++];

}

template<typename integer> integer isqrt(integer n) {

   return next_odd_number(static_cast<integer>(std::sqrt(n)));

}

template<typename integer> void prime_generator<integer>::find_primes(integer start) {

   index_ = 0;
   integer new_limit = isqrt(limit_);
   sieve_.resize(new_limit/2);
   for (integer p = 3; p * p <= new_limit; p += 2) {
       if (sieve_[p/2 - 1])
           continue;
       integer q = p * std::max(p, next_odd_number((sieve_limit_ + p - 1)/p));
       for (; q <= new_limit; q += 2*p)
           sieve_[q/2 - 1] = true;
   }
   sieve_limit_ = new_limit;
   size_t count = (limit_ - start)/2 + 1;
   std::vector<bool> composite(count, false);
   for (integer p = 3; p <= new_limit; p += 2) {
       if (sieve_[p/2 - 1])
           continue;
       integer q = p * std::max(p, next_odd_number((start + p - 1)/p)) - start;
       q /= 2;
       for (; q < count; q += p)
           composite[q] = true;
   }
   for (integer p = 0; p < count; ++p) {
       if (!composite[p])
           primes_.push_back(p * 2 + start);
   }

}

int main() {

   typedef uint64_t integer;
   prime_generator<integer> pgen(100, 500000);
   std::cout << "First 20 primes:\n";
   for (int i = 0; i < 20; ++i) {
       integer p = pgen.next_prime();
       if (i != 0)
           std::cout << ", ";
       std::cout << p;
   }
   std::cout << "\nPrimes between 100 and 150:\n";
   for (int n = 0; ; ) {
       integer p = pgen.next_prime();
       if (p > 150)
           break;
       if (p >= 100) {
           if (n != 0)
               std::cout << ", ";
           std::cout << p;
           ++n;
       }
   }
   int count = 0;
   for (;;) {
       integer p = pgen.next_prime();
       if (p > 8000)
           break;
       if (p >= 7700)
           ++count;
   }
   std::cout << "\nNumber of primes between 7700 and 8000: " << count << '\n';
   for (integer n = 10000; n <= 100000000; n *= 10) {
       integer prime;
       while (pgen.count() != n)
           prime = pgen.next_prime();
       std::cout << n << "th prime: " << prime << '\n';
   }
   return 0;

}</lang>

Output:
First 20 primes:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71
Primes between 100 and 150:
101, 103, 107, 109, 113, 127, 131, 137, 139, 149
Number of primes between 7700 and 8000: 30
10000th prime: 104729
100000th prime: 1299709
1000000th prime: 15485863
10000000th prime: 179424673
100000000th prime: 2038074743

Clojure

<lang Clojure>ns test-project-intellij.core

 (:gen-class)
 (:require [clojure.string :as string]))

(def primes " The following routine produces a infinite sequence of primes

 (i.e. can be infinite since the evaluation is lazy in that it 
 only produces values as needed).  The method is from clojure primes.clj library
 which produces primes based upon O'Neill's paper:
 'The Genuine Sieve of Eratosthenes'.  
  Produces primes based upon trial division on previously found primes up to
  (sqrt number), and uses 'wheel' to avoid
  testing numbers which are divisors of 2, 3, 5, or 7.
  A full explanation of the method is available at:
  [1] "
 (concat
   [2 3 5 7]
   (lazy-seq
     (let [primes-from   ; generates primes by only checking if primes 
                         ; numbers which are not divisible by 2, 3, 5, or 7
           (fn primes-from [n [f & r]]
             (if (some #(zero? (rem n %))
                       (take-while #(<= (* % %) n) primes))
               (recur (+ n f) r)
               (lazy-seq (cons n (primes-from (+ n f) r)))))
           ; wheel provides offsets from previous number to insure we are not landing on a divisor of 2, 3, 5, 7
           wheel (cycle [2 4 2 4 6 2 6 4 2 4 6 6 2 6  4  2
                         6 4 6 8 4 2 4 2 4 8 6 4 6 2  4  6
                         2 6 6 4 2 4 6 2 6 4 2 4 2 10 2 10])]
       (primes-from 11 wheel)))))

(defn between [lo hi]

 "Primes between lo and hi value "
 (->> (take-while #(<= % hi) primes)
      (filter #(>= % lo))
      ))

(println "First twenty:" (take 20 primes))

(println "Between 100 and 150:" (between 100 150))

(println "Number between 7,7700 and 8,000:" (count (between 7700 8000)))

(println "10,000th prime:" (nth primes (dec 10000)))  ; decrement by one since nth starts counting from 0


}</lang>

Output:
First 20: (2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71)
Between 100 and 150: (101 103 107 109 113 127 131 137 139 149)
Number between 7,700 and 8,000: 30
10000th prime: 104729

Alternate version using deferred execution Co-Inductive Streams and a Wheel

The above version is adequate for ranges up to the low millions, so covers the task requirements of primes up to just over a hundred thousands easily. However, it has a O(n^(3/2)) performance which means that it gets slow quite quickly with range as compared to a true incremental Sieve of Eratosthenes, which has O(n (log n)) performance. The following code is about the same speed for ranges in the low millions but quickly passes the above code in speed for large ranges to where it only takes 10's of seconds for a range of a hundred million where the above code takes thousands of seconds. The code is based on the Richard Bird list based Sieve of Eratosthenes mentioned in the O'Neil article but has infinite tree folding added as well as wheel factorization so that it is about the same speed and performance as a Sieve of Eratosthenes based on a priority queue; the code is written here in purely functional form with no mutation, as follows: <lang clojure>(deftype CIS [v cont]

 clojure.lang.ISeq
   (first [_] v)
   (next [_] (if (nil? cont) nil (cont)))
   (more [this] (let [nv (.next this)] (if (nil? nv) (CIS. nil nil) nv)))
   (cons [this o] (clojure.core/cons o this))
   (empty [_] (if (and (nil? v) (nil? cont)) nil (CIS. nil nil)))
   (equiv [this o] (loop [cis1 this, cis2 o] (if (nil? cis1) (if (nil? cis2) true false)
                                               (if (or (not= (type cis1) (type cis2))
                                                       (not= (.v cis1) (.v ^CIS cis2))
                                                       (and (nil? (.cont cis1))
                                                             (not (nil? (.cont ^CIS cis2))))
                                                       (and (nil? (.cont ^CIS cis2))
                                                             (not (nil? (.cont cis1))))) false
                                                 (if (nil? (.cont cis1)) true
                                                   (recur ((.cont cis1)) ((.cont ^CIS cis2))))))))
   (count [this] (loop [cis this, cnt 0] (if (or (nil? cis) (nil? (.cont cis))) cnt
                                           (recur ((.cont cis)) (inc cnt)))))
 clojure.lang.Seqable
   (seq [this] (if (and (nil? v) (nil? cont)) nil this))
 clojure.lang.Sequential
 Object
   (toString [this] (if (and (nil? v) (nil? cont)) "()" (.toString (seq (map identity this))))))

(comment " the wheel could also be a pre-determined vector as for the 2/3/5/7 wheel below... (def wheel

 [ 2 4 2 4 6 2 6 4 2 4 6 6 2 6  4  2
   6 4 6 8 4 2 4 2 4 8 6 4 6 2  4  6
   2 6 6 4 2 4 6 2 6 4 2 4 2 10 2 10 ])

")

(def wheel-primes [2 3 5 7 11 13 17])

(def next-prime 19)

(def nextnext-prime 23)

calculates the vector for very large wheels such as the 92160 element version here
the disadvantage is that it takes some time to calculate before the work can start...

(def wheel

 (loop [p 2, len 1, ^bytes ptrn [1]]
   (if (>= p next-prime)
     ptrn
     (let [cptrn (cycle ptrn), [f & rcyc] cptrn,
           np (+ p f), nlen (* len (- p 1)),
           culls
             (map (fn f _ f)
               (iterate (fn [[c [g & r]]] [(+ c (* p g)) r]) [(* p p) cptrn])),
           gaps (drop 1
                   (for [[gp _ _ _ cnt]
                           (iterate (fn [[_ v cls [g & rgs] c]]
                                 (let [[cl & rcls] cls, tv (+ v g),
                                       [sg & srgs] rgs, nc (+ c 1)]
                                   (if (= cl tv)
                                     [(+ g sg) (+ tv sg) rcls srgs nc]
                                     [g tv cls rgs nc])))
                               [f np culls rcyc 0]) :while (<= cnt nlen)] gp))]
       (recur np nlen (vec gaps))))))

(def wheellmt (- (count wheel) 1))

(defn primes-treeFolding

 "Computes the unbounded sequence of primes using a Sieve of Eratosthenes algorithm modified from Bird."
 []
 (letfn [(mltpls p pi
           (letfn [(nxtmltpl [c ci]
                     (let [nci (if (< ci wheellmt) (+ ci 1) 0)]  
                       (->CIS c #(-> (nxtmltpl (+ c (* p (get wheel ci))) nci)))))]
             (nxtmltpl (* p p) pi))),
         (allmtpls [^CIS pxs]
           (->CIS (mltpls (.v pxs)) #(-> (allmtpls ((.cont pxs)))))),
         (union [^CIS xs ^CIS ys]
           (let [xv (.v xs), yv (.v ys)]
             (if (< xv yv) (->CIS xv #(-> (union ((.cont xs)) ys)))
               (if (< yv xv)
                 (->CIS yv #(-> (union xs ((.cont ys)))))
                 (->CIS xv #(-> (union (next xs) ((.cont ys))))))))),
         (pairs [^CIS mltplss] (let [^CIS tl ((.cont mltplss))]
                                 (->CIS (union (.v mltplss) (.v tl))
                                         #(-> (pairs ((.cont tl))))))),
         (mrgmltpls [^CIS mltplss]
           (->CIS (.v ^CIS (.v mltplss))
                   #(-> (union ((.cont ^CIS (.v mltplss)))
                               (mrgmltpls (pairs ((.cont mltplss)))))))),
         (minusStrtAt [n ni ^CIS cmpsts]
           (let [nn (+ n (get wheel ni)), nni (if (< ni wheellmt) (+ ni 1) 0)]
             (if (< n (.v cmpsts))
               (->CIS [n ni] #(-> (minusStrtAt nn nni cmpsts)))
               (recur nn nni ((.cont cmpsts)))))),
         (xtraprmsndxd []
           (->CIS [next-prime 0] #(-> (minusStrtAt nextnext-prime 1
                               (mrgmltpls (allmtpls (xtraprmsndxd))))))),
         (stripndxs [^CIS ndxd]
           (->CIS (get (.v ndxd) 0) #(-> (stripndxs ((.cont ndxd))))))]
   (loop [i (- (count wheel-primes) 1), ff (fn [] (stripndxs (xtraprmsndxd)))]
       (if (<= i 0)
         (->CIS (get wheel-primes 0) ff)
         (recur (- i 1) (fn [] (->CIS (get wheel-primes i) ff)))))))</lang>

Now these functional incremental sieves are of limited use if one requires ranges of billions as they are hundreds of times slower than a version of a bit-packed page-segmented mutable array Sieve of Eratosthenes, which for Clojure there is a version at the end of Clojure Sieve_of_Eratosthenes#Unbounded_Versions section on the Sieve of Eratosthenes task page; this version will handle ranges of a billion in seconds rather than hundreds of seconds.

CoffeeScript

This uses the prime number generation algorithm outlined in the paper, "Two Compact Incremental Prime Sieves" by Jonathon P. Sorenson. This algorithm is essentially a rolling segmented SoE, and is quite fast for languages that have good array processing. <lang CoffeeScript> primes = () ->

   yield 2
   yield 3
   sieve = ([] for i in [1..3])
   sieve[0].push 3
   [r, s] = [3, 9]
   pos = 1
   n = 5
   loop
       isPrime = true
       if sieve[pos].length > 0  # this entry has a list of factors
           isPrime = false
           sieve[(pos + m) % sieve.length].push m for m in sieve[pos]
           sieve[pos] = []
       if n is s  # n is the next square
           if isPrime
               isPrime = false  # r divides n, so not actually prime
               sieve[(pos + r) % sieve.length].push r  # however, r is prime
           r += 2
           s = r*r
       yield n if isPrime
       n += 2
       pos += 1
       if pos is sieve.length
           sieve.push []  # array size must exceed largest prime found
           sieve.push []  # adding two entries keeps size = O(sqrt n)
           pos = 0
   undefined  # prevent CoffeeScript from aggregating values

module.exports = {

   primes

} </lang> Driver code: <lang CoffeeScript> primes = require('sieve').primes

gen = primes() console.log "The first 20 primes: #{gen.next().value for _ in [1..20]}"

p100_150 = (while (p = gen.next().value) < 150 then p).filter (n) -> n > 100 console.log "The primes between 100 and 150: #{p100_150}"

while gen.next().value < 7700

   undefined

count = 1 while gen.next().value < 8000

   ++count

console.log "There are #{count} primes between 7,700 and 8,000."

n = 10 c = 0 gen = primes() loop

   p = gen.next().value
   c += 1
   if c is n
       console.log "The #{n}th prime is #{p}"
       break if n is 10_000_000
       n *= 10

</lang>

Output:
The first 20 primes: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71
The primes between 100 and 150: 101,103,107,109,113,127,131,137,139,149
There are 30 primes between 7,700 and 8,000.
The 10th prime is 29
The 100th prime is 541
The 1000th prime is 7919
The 10000th prime is 104729
The 100000th prime is 1299709
The 1000000th prime is 15485863

D

This uses a Prime struct defined in the third entry of the Sieve of Eratosthenes task. Prime keeps and extends a dynamic array instance member of uints. The Prime struct has a opCall that returns the n-th prime number. The opCall calls a grow() private method until the dynamic array of primes is long enough to contain the required answer. The function grow() just grows the dynamic array geometrically and performs a normal sieving. On a 64 bit system this program works up to the maximum prime number that can be represented in the 32 bits of an uint. This program is less efficient than the C entry, so it's better to not use it past some tens of millions of primes, but it's enough for more limited usages. <lang d>void main() {

   import std.stdio, std.range, std.algorithm, sieve_of_eratosthenes3;
   Prime prime;
   writeln("First twenty primes:\n", 20.iota.map!prime);
   writeln("Primes primes between 100 and 150:\n",
           uint.max.iota.map!prime.until!q{a > 150}.filter!q{a > 99});
   writeln("Number of primes between 7,700 and 8,000: ",
           uint.max.iota.map!prime.until!q{a > 8_000}
           .filter!q{a > 7_699}.walkLength);
   writeln("10,000th prime: ", prime(9_999));

}</lang>

Output:
First twenty primes:
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]
Primes primes between 100 and 150:
[101, 103, 107, 109, 113, 127, 131, 137, 139, 149]
Number of primes between 7,700 and 8,000: 30
10,000th prime: 104729

Faster Alternative Version

<lang d>/// Prime sieve based on: http://www.cs.hmc.edu/~oneill/papers/Sieve-JFP.pdf

import std.container: Array, BinaryHeap, RedBlackTree;

struct LazyPrimeSieve {

   @property bool empty() const pure nothrow @safe @nogc {
       return i > 203_280_221; // Pi(2 ^^ 32).
   }
   @property auto front() const pure nothrow @safe @nogc {
       return prime;
   }
   @property void popFront() pure nothrow /*@safe*/ {
       prime = sieveOne();
   }

private:

   static struct Wheel2357 {
       static immutable ubyte[48] holes = [2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6,
           2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4, 2, 4, 8, 6, 4, 6, 2, 4, 6, 2, 6, 6,
           4, 2, 4, 6, 2, 6, 4, 2, 4, 2, 10, 2, 10];
       static immutable ubyte[4] spokes = [2, 3, 5, 7];
       static immutable ubyte first = 11;
       uint i;
       auto spin() pure nothrow @safe @nogc {
           return holes[i++ % $];
       }
   }
   static struct CompositeIterator {
       uint prime;
       Wheel2357 wheel;
       ulong composite;
       this(uint p) pure nothrow @safe @nogc {
           prime = p;
           composite = p * wheel.first;
       }
       void next() pure nothrow @safe @nogc {
           composite += prime * wheel.spin;
       }
   }
   version (heap)  // Less memory but slower.
       BinaryHeap!(Array!CompositeIterator, "a.composite > b.composite") iterators;
   else            // Faster but is more GC intensive.
       RedBlackTree!(CompositeIterator, "a.composite < b.composite", true) iterators;
   uint prime = 2;
   uint i = 1;
   Wheel2357 wheel;
   uint candidate = wheel.first;
   uint sieveOne() pure nothrow /*@safe*/ {
       switch (i) {
           case 0: .. case wheel.spokes.length - 1:
               return wheel.spokes[i++];
           case wheel.spokes.length:
               i++;
               return candidate;
           case wheel.spokes.length + 1:
               version (heap) {}
               else
                   iterators = new typeof(iterators);
               goto default;
           default:
               goto POST_RETURN;
               while (true) {
                   candidate += wheel.spin;
                   while (iterators.front.composite < candidate) {
                       auto it = iterators.front;
                       iterators.removeFront;
                       it.next;
                       iterators.insert(it);
                   }
                   if (iterators.front.composite != candidate) {
                       i++;
                       return candidate;
       POST_RETURN:
                       // Only insert primes that are multiply
                       // occuring in [0, 2 ^^ 32).
                       if (candidate < 2 ^^ 16)
                           iterators.insert(CompositeIterator(candidate));
                   }
               }
       }
   }

}


void main() /*@safe*/ {

   import std.stdio, std.algorithm, std.range;
   writeln("Sum of first 100,000 primes: ", LazyPrimeSieve().take(100_000).sum(0uL));
   writeln("First twenty primes:\n", LazyPrimeSieve().take(20));
   writeln("Primes primes between 100 and 150:\n",
           LazyPrimeSieve().until!q{a > 150}.filter!q{a > 99});
   writeln("Number of primes between 7,700 and 8,000: ",
           LazyPrimeSieve().until!q{a > 8_000}.filter!q{a > 7_699}.walkLength);
   writeln("10,000th prime: ", LazyPrimeSieve().dropExactly(9999).front);

}</lang>

Output:
Sum of first 100,000 primes: 62260698721
First twenty primes:
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]
Primes primes between 100 and 150:
[101, 103, 107, 109, 113, 127, 131, 137, 139, 149]
Number of primes between 7,700 and 8,000: 30
10,000th prime: 104729

Dart

A version based on a (hashed) Map:

<lang dart>Iterable<int> primesMap() {

   Iterable<int> oddprms() sync* {
     yield(3); yield(5); // need at least 2 for initialization
     final Map<int, int> bpmap = {9: 6};
     final Iterator<int> bps = oddprms().iterator;
     bps.moveNext(); bps.moveNext(); // skip past 3 to 5
     int bp = bps.current;
     int n = bp;
     int q = bp * bp;
     while (true) {
       n += 2;
       while (n >= q || bpmap.containsKey(n)) {
         if (n >= q) {
           final int inc = bp << 1;
           bpmap[bp * bp + inc] = inc;
           bps.moveNext(); bp = bps.current; q = bp * bp;
         } else {
           final int inc = bpmap.remove(n);
           int next = n + inc;
           while (bpmap.containsKey(next)) {
             next += inc;
           }
           bpmap[next] = inc;
         }
         n += 2;
       }
       yield(n);
     }
   }
   return [2].followedBy(oddprms());

}

void main() {

 print("The first 20 primes:");
 String str = "( ";
 primesMap().take(20).forEach((p)=>str += "$p "); print(str + ")");
 print("Primes between 100 and 150:");
 str = "( ";
 primesMap().skipWhile((p)=>p<100).takeWhile((p)=>p<150)
   .forEach((p)=>str += "$p "); print(str + ")");
 print("Number of primes between 7700 and 8000: ${
   primesMap().skipWhile((p)=>p<7700).takeWhile((p)=>p<8000).length
 }");
 print("The 10,000th prime:  ${
   primesMap().skip(9999).first
 }");
 final start = DateTime.now().millisecondsSinceEpoch;
 final answer = primesMap().takeWhile((p)=>p<2000000).reduce((a,p)=>a+p);
 final elapsed = DateTime.now().millisecondsSinceEpoch - start;

}</lang>

Output:
The first 20 primes:
( 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 )
Primes between 100 and 150:
( 101 103 107 109 113 127 131 137 139 149 )
Number of primes between 7700 and 8000: 30
The 10,000th prime:  104729
The sum of the primes to two million:  142913828922
This test bench took 356 milliseconds.

This version has a O(n log (log n)) computational complexity due to hash map access being O(1) on average but is somewhat slow due to the constant execution overhead and therefore only somewhat useful for ranges of up to about the tens of millions.

As a bonus, it solves Euler Problem 10 of summing all the primes up to two million quite quickly.

A faster alternative version based on the infinite page segmented sieve

The unbounded page segmented bit-packed version from the Sieve_of_Eratosthenes#Unbounded_infinite_iterators.2Fgenerators_of_primes at the bottom of the section can do the same job tens of times faster just by substituting 'primesPaged()' for 'primesMap()' in the 'main' function above in all places used. As noted for the listing on the task page, the code is only limited in range by the integer size limit on 32-bit execution environments, and will be tens of times faster than the above version. It will otherwise have the same output as above.

It solves the Euler Problem 10 in almost too short a time to be measured, and it becomes useful for ranges of hundreds of thousands. It can count all the primes to a billion on the low end tablet CPU of an Intel x5-Z8350 at 1.92 Gigahertz used to develop this in 40 seconds but using a generator slows the performance and it can use the provided `countPrimesTo` function to do the job four times as fast by directly manipulating the provided iteration of sieved bit-packed arrays.

Delphi

See Pascal.

EchoLisp

Standard prime functions handle numbers < 2e+9. See [2] . The bigint library handles large numbers. See [3]. The only limitations are time, memory, and browser performances .. <lang lisp>

the first twenty primes

(primes 20)

   → { 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 }
a stream to generate primes from a

(define (primes-from a)

   (let ((p (next-prime a)))
    (stream-cons p (primes-from p))))
primes between 100,150

(for/list ((p (primes-from 100))) #:break (> p 150) p)

   → (101 103 107 109 113 127 131 137 139 149)
the built-in function (primes-pi )counts the number of primes < a
count in [7700 ... 8000]

(- (primes-pi 8000) (primes-pi 7700) → 30

nth-prime

(nth-prime 10000) → 104729

big ones

(lib 'bigint) (define (p-digits n)

 (printf "(next-prime  %d ! ) has %d digits" n 
 (number-length (next-prime (factorial n )))))

(next-prime 0! ) has 1 digits (next-prime 10! ) has 7 digits (next-prime 100! ) has 158 digits (next-prime 200! ) has 375 digits (next-prime 300! ) has 615 digits (next-prime 400! ) has 869 digits ;; 9400 msec (FireFox)

is prime (1 + 116!) ?

(prime? (1+ (factorial 116))) → #t

</lang>

Elixir

The Sieve_of_Eratosthenes#Elixir Task page lists two "infinite" extensible generators at the bottom. The first of those, using a (hash) Map is reproduced here along with the code to fulfill the required tasks: <lang elixir>defmodule PrimesSoEMap do

 @typep stt :: {integer, integer, integer, Enumerable.integer, %{integer => integer}}
 @spec advance(stt) :: stt
 defp advance {n, bp, q, bps?, map} do
   bps = if bps? === nil do Stream.drop(oddprms(), 1) else bps? end
   nn = n + 2
   if nn >= q do
     inc = bp + bp
     nbps = bps |> Stream.drop(1)
     [nbp] = nbps |> Enum.take(1)
     advance {nn, nbp, nbp * nbp, nbps, map |> Map.put(nn + inc, inc)}
   else if Map.has_key?(map, nn) do
     {inc, rmap} = Map.pop(map, nn)
     [next] =
       Stream.iterate(nn + inc, &(&1 + inc))
         |> Stream.drop_while(&(Map.has_key?(rmap, &1))) |> Enum.take(1)
     advance {nn, bp, q, bps, Map.put(rmap, next, inc)}
   else
     {nn, bp, q, bps, map}
   end end
 end
 @spec oddprms() :: Enumerable.integer
 defp oddprms do # put first base prime cull seq in Map so never empty
   # advance base odd primes to 5 when initialized
   init = {7, 5, 25, nil, %{9 => 6}}
   [3, 5] # to avoid race, preseed with the first 2 elements...
     |> Stream.concat(
           Stream.iterate(init, &(advance &1))
             |> Stream.map(fn {p,_,_,_,_} -> p end))
 end
 @spec primes() :: Enumerable.integer
 def primes do
   Stream.concat([2], oddprms())
 end

end

IO.write "The first 20 primes are:\n( " PrimesSoEMap.primes() |> Stream.take(20) |> Enum.each(&(IO.write "#{&1} ")) IO.puts ")" IO.write "The primes between 100 to 150 are:\n( " PrimesSoEMap.primes() |> Stream.drop_while(&(&1<100))

 |> Stream.take_while(&(&1<150)) |> Enum.each(&(IO.write "#{&1} "))

IO.puts ")" IO.write "The number of primes between 7700 and 8000 is: " PrimesSoEMap.primes() |> Stream.drop_while(&(&1<7700))

 |> Stream.take_while(&(&1<8000)) |> Enum.count |> IO.puts

IO.write "The 10,000th prime is: " PrimesSoEMap.primes() |> Stream.drop(9999)

 |> Enum.take(1) |> List.first |>IO.puts

IO.write "The sum of all the priems to two million is: " testfunc =

 fn () ->
   ans =
     PrimesSoEMap.primes() |> Stream.take_while(&(&1<=2000000))
       |> Enum.sum() |> IO.puts
   ans end
timer.tc(testfunc)
 |> (fn {t,_} ->
   IO.puts "This test bench took #{t} microseconds." end).()</lang>
Output:
The first 20 primes are:
( 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 )
The primes between 100 to 150 are:
( 101 103 107 109 113 127 131 137 139 149 )
The number of primes between 7700 and 8000 is:  30
The 10,000th prime is:  104729
The sum of all the primes to two million is:  142913828922
This test bench took 7827912 microseconds.

The code solves the (trivial) task requirements quickly, but being purely functional, running on a Virtual Machine, and using the multi-precision ("Big Integer") `integer` type is somewhat slower than as implemented in some other languages.

The code is an "infinite" generator as the `integer` type in Elixir is of multi-precision and thus will never run out of range.

As a bonus, the above code solves the Euler Problem 10 of summing the primes to two million in about 7.83 seconds, or about a hundred thousand CPU cycles per prime on a !.92 Gigahertz CPU, which at least is within the 30 second time limit for that problem.

Alternate somewhat faster version (over two times)

The last code on the Sieve of Eratosthenes Task page uses deferred execution Co-Inductive Streams (CIS's) to implement the incremental functional Sieve of Eratosthenes using an infinite tree folding structure with a final output as a lazy Stream just as for the above. It is about twice as fast as the above for reasonable ranges of a few millions. It con be used just by substituting calls to the different named module as in `PrimesSoETreeFolding.primes` rather than `PrimesSoEMap.primes()`. It has the same limitations as to being "infinite" as the above.

F#

This task uses Unbounded_Page-Segmented_Bit-Packed Odds-Only Mutable Array Sieve F#

The functions

<lang fsharp> let primeZ fN =primes()|>Seq.unfold(fun g-> Some(fN(g()), g)) let primesI() =primeZ bigint let primes64() =primeZ int64 let primes32() =primeZ int32 let pCache =Seq.cache(primes32()) let isPrime g=if g<2 then false else let mx=int(sqrt(float g)) in pCache|>Seq.takeWhile(fun n->n<=mx)|>Seq.forall(fun n->g%n>0) let isPrime64 g=if g<2L then false else let mx=int(sqrt(float g)) in pCache|>Seq.takeWhile(fun n->n<=mx)|>Seq.forall(fun n->g%(int64 n)>0L) </lang>

The Task

<lang fsharp> Seq.take 20 primes32()|> Seq.iter (fun n-> printf "%d " n) </lang>

Output:
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71

<lang fsharp> primes32() |> Seq.skipWhile (fun n->n<100) |> Seq.takeWhile (fun n->n<=150) |> Seq.iter (fun n -> printf "%d " n) </lang>

Output:
101 103 107 109 113 127 131 137 139 149

<lang fsharp> printfn "%d" (primes32() |> Seq.skipWhile (fun n->n<7700) |> Seq.takeWhile (fun n->n<=8000) |> Seq.length) </lang>

Output:
30

To demonstrate extensibility I find the 10000th prime. <lang fsharp> Seq.item 9999 pCache </lang>

Output:
Real: 00:00:00.185, CPU: 00:00:00.190, GC gen0: 21, gen1: 0
val it : int = 104729

I then find the 10001st prime which takes less time. <lang fsharp> Seq.item 10000 pCache </lang>

Output:
Real: 00:00:00.004, CPU: 00:00:00.010, GC gen0: 1, gen1: 0
val it : int = 104743

To indiccate speed I time the following: <lang fsharp> let strt = System.DateTime.Now.Ticks for i = 1 to 8 do

 let n = pown 10 i // the item index below is zero based!
 printfn "The %dth prime is:  %A" n (primeZ int |> Seq.item (n - 1))

let timed = (System.DateTime.Now.Ticks - strt) / 10000L printfn "All of the last took %d milliseconds." timed </lang>

Output:
The 10th prime is:  29
The 100th prime is:  541
The 1000th prime is:  7919
The 10000th prime is:  104729
The 100000th prime is:  1299709
The 1000000th prime is:  15485863
The 10000000th prime is:  179424673
The 100000000th prime is:  2038074743
All of the last took 7937 milliseconds.

<lang fsharp>printfn "The first 20 primes are: %s"

 ( primesSeq() |> Seq.take 20
     |> Seq.fold (fun s p -> s + string p + " ") "" )

printfn "The primes from 100 to 150 are: %s"

 ( primesSeq() |> Seq.skipWhile ((>) (prime 100))
     |> Seq.takeWhile ((>=) (prime 150))
     |> Seq.fold (fun s p -> s + string p + " ") "" )

printfn "The number of primes from 7700 to 8000 are: %d"

 ( primesSeq() |> Seq.skipWhile ((>) (prime 7700))
     |> Seq.takeWhile ((>=) (prime 8000)) |> Seq.length )

let strt = System.DateTime.Now.Ticks for i = 1 to 8 do

 let n = pown 10 i // the item index below is zero based!
 printfn "The %dth prime is:  %A" n (primesSeq() |> Seq.item (n - 1))

let timed = (System.DateTime.Now.Ticks - strt) / 10000L printfn "All of the last took %d milliseconds." timed</lang>

Output:
The first 20 primes are:  2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 
The primes from 150 to 150 are:  101 103 107 109 113 127 131 137 139 149 
The number of primes from 7700 to 8000 are:  30
The 10th prime is:  29
The 100th prime is:  541
The 1000th prime is:  7919
The 10000th prime is:  104729
The 100000th prime is:  1299709
The 1000000th prime is:  15485863
The 10000000th prime is:  179424673
The 100000000th prime is:  2038074743
All of the last took 16634 milliseconds.

Even at that, this is slow due to the time to enumerate and there are much faster ways to do this without using enumeration as in manipulating the found prime bit representations directly...

Factor

Factor's math.primes vocabulary provides an extensible primality test upon which its prime number generators are built. For values below nine million, it checks for primality using a Sieve of Eratosthenes with wheels. For values nine million and above, it uses a Miller-Rabin probabalistic primality test.

Factor's fixnums automatically promote to bignums when large enough, so there are no limits to its prime generator other than the capabilities of the machine it's running on. <lang factor>USING: io math.primes prettyprint sequences ;

"First 20 primes: " write 20 nprimes .

"Primes between 100 and 150: " write 100 150 primes-between .

"Number of primes between 7,700 and 8,000: " write 7,700 8,000 primes-between length .

"10,000th prime: " write 10,000 nprimes last .</lang>

Output:
First 20 primes: { 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 }
Primes between 100 and 150: V{ 101 103 107 109 113 127 131 137 139 149 }
Number of primes between 7,700 and 8,000: 30
10,000th prime: 104729

Fortran

The Plan

Over the years, the storage of boolean variables has been a steady source of vexation. Few systems offer operation codes that can work on individual bits, so the usual approach is to allocate a convenient storage unit to hold the value. In Fortran, the default size of a LOGICAL variable is the same as that of an INTEGER variable, and these days, that means thirty-two bits to store the state of one. However, with the increasing use of character manipulation rather than just numbers, there is often support in the cpu for single-character access and some later Fortran compilers will recognise LOGICAL*1 and so reserve only eight bits per boolean variable. Even so, any attempt to store a boolean variable in a single bit will for every access require code to isolate that bit from the rest of the storage unit where it resides, and these operation codes will require more storage than would be saved. Similarly with collections of variables: some might best be aligned to word boundaries (and for double-sized variables, perhaps to even word boundaries) so the storage plan may well involve adding "padding" to preserve such alignment. Some languages (such as pl/i) offer a word ALIGNED to ensure this, and others (such as Pascal) offer PACKED for cramming, of use when dealing with records for a disc file and intending to save space. So, ... if there is a large array of boolean variables, and, there are not so many references to those variables, there is still an opportunity.

And indeed, the array shall be large. Some simple investigations show that storing a collection of prime numbers in an array of integers occupies rather more storage than storing a simple bit array spanning the same range of numbers, and given the obvious scheme of storing bits only for odd integers, this advantage is still greater - see the schedule in the source file. One could argue that the array of successive prime numbers could be stored in various space-saving ways, but, so also can the bit array be compressed. For instance, have a span of a "primorial" size such as 2*3*5 and a reference span with those factors marked "off": that removes seven odd numbers from consideration, leaving eight candidates for each surge and so only eight bits are required to state "prime" or not instead of fifteen. With non-binary computers, "bit fiddling" is less convenient but still possible. One must use techniques similar to those needed to work with the year, month, and day parts of an integer such as 20161015 in binary.

So, the plan is to have a long array of bits, and, rather than commit a lot of memory to this, do so in a disc file with random access. To follow the "extensible" aspect, this disc file will not be initialised to its maximum extent on the first invocation of the routine, instead, it will be extended as provoked by requests for NEXTPRIME and so forth.

Initialisation

When arranging a sieve of Eratosthenes, one of the problems is that one wishes to step along only with steps of prime number size to avoid wasted effort, but, before the sieve process is completed, there is no ready source of known prime numbers. This is especially difficult when instead of one long sieve covering the whole span of interest, the process is to proceed in surges, repeatedly using some limited size span. For this reason, it is often convenient to prepare an initial array of prime numbers knowing for example that Prime(4792) = 46337, and that the square of the next prime exceeds the range of signed 32-bit integers. But such pre-emptive preparation conflicts with the "extensible" notion, and requires special code and storage for the array.

Because Fortran passes parameters by reference (i.e. by address of the original) a trick is possible. The array SCHARS is shared storage to hold a record from the disc file (as a "buffer") and when subroutine GRASPPRIMEBAG is invoked to gain access to its disc file it notes whether it must create the file. If so, the first record is to be written, and the call is PSURGE(SCHARS) to do so within the shared bitpad. PSURGE knows that its first stepper is with F = 3 (because even numbers are not being represented) and proceeds with that, adjusting array SCHARS. When it is ready for the next sieve pass, it invokes F = NEXTPRIME(F) to find the next stepper, which will be five, and NEXTPRIME scans the bit array in SCHARS to find it. This is the same bitpad that PSURGE is in the process of adjusting. To support this startup ploy, GETSREC (invoked by NEXTPRIME) returns at once when SLAST = 0, signifying that there are no records in the work file as yet. Later, if GETSREC determines that the bit array is to be extended, it invokes PSURGE with its local array BIT8 as the bitpad to be developed then written to disc, leaving the shared SCHAR array as a record buffer for the use of NEXTPRIME when invoked by PSURGE.

Supporting NextPrime(n)

Since the bit array has a simple linear relationship to the numbers it is associated with, function NEXTPRIME(n) (and PREVIOUSPRIME(n)) can easily calculate the index to access the appropriate bits; similarly, function ISPRIME(n) need merely check n = NEXTPRIME(n - 1) rather than slog through possibly all potential prime factors up to SQRT(n) - though this does mean that prime numbers up to n must be available rather than merely up to SQRT(n). [A later adjustment has ISPRIME(n) repeat the code to locate the bit for n rather than use NEXTPRIME, which has to scan the bit array to the next prime] But there would be no escape from such a slog for function FIRSTFACTOR(n), unless one abandoned the bit array for a FF array and modified the sieve process to record the first factor. Something like Bit(i) = .false. would be replaced by if (FF(i) <= 1) then FF(i) = F Were the assignment to be made unconditionally (for faster running, perhaps), the array should be renamed to MaximumPrimeFactor. Either way, ISPRIME(n) remains easy, but much more storage than one bit per entry would be required.

If instead of the next prime one desires to find the n 'th prime number, then there is a difficulty that would not exist if the prime numbers only were stored in an array - but if they were then NEXTPRIME(n) would have difficulty. Of course, if storage is abundant, both forms of storage could be used and each request could be handled via a simple linear index calculation into the appropriate array.

Supporting Prime(n)

To find the n'th prime number, obviously one could scan along the bit array from the start, keeping count. This will soon become tedious, so in order to support function PRIME(n), each record starts with a count of all the primes that have preceded that record's span and so to find the count for a prime fingered in that record, the scan need work only from the start of that record. So the problem reduces to determining which record is the one containing the n 'th prime. Since the counts are obviously strictly increasing, a binary search would be a possibility as would be an interpolating search and one could even prepare an array containing the counts so that the search could proceed without needing disc accesses - at the cost of additional storage and organisational complexity, perhaps involving Aitken's interpolation formula, except that polynomials do not provide a good fit to the required shape. Fortunately, mathematicians have considered this aspect of prime numbers also, and a rather intimidating formula is available to give an estimate of the value of the n 'th prime number. Equipped with this, the appropriate record can be read, the count inspected, and a scan started to find the actual n 'th prime. Function PRIME(n) can be invoked just as array PRIME(n) might be, but with something of a roil of activity in the background.

Preparing the count field involves another trick, because the first record's count field instead holds the count of records in the bit file. Now note that the sieve starts with SORG = 3 which means that before the first block there is a count of one prime number. Thus, if function PRIME is accessing the first block, it must know that the count of previous primes is one and not refer to the count field which instead holds the record count. When PSURGE is preparing the next batch of bits (for the second and subsequent records) it accesses the previous record to find the previous count and scans that record to count its primes so as to prepare the previous count for the new record. When the second record is being prepared by PSURGE, the previous record (the first record) has a record count of one, and, this is exactly the desired count of previous primes for the first record. But only at this point, because in moments the first record will be rewritten with a record count of two. For all this to work, SORG = 3 must be the case: it is not a parameter but a constant.

Integer Overflow

The variables are all the default thirty-two bit two's complement integers and integer overflow is a possibility, especially because someone is sure to wonder what is the largest prime that can be represented in thirty-two bits - see the output. The code would be extensible in another way, if all appearances of INTEGER were to be replaced by INTEGER*8 though not all variables need be changed - such as C and B because they need only index a character in array SCHARS or a bit in a character. Using sixty-four bits for such variables is excessive even if the cpu uses a 64-bit data bus to memory. If such a change were to be made, then all would go well as cpu time and disc space were consumed up to the point when the count of prime numbers can no longer be fitted into the four character storage allowance in the record format. This will be when more than 4,294,967,295 primes have been counted (with 64-bit arithmetic its four bytes will manifest as an unsigned integer) in previous records, and Prime(4,294,967,295) = 104,484,802,043, so that the bit file would require a mere 6,530MB or so - which some may think is not too much. If so, expanding the allowance from four to five characters would be easy enough, and then 256 times as many primes could be counted. That would also expand the reach of the record counter, which otherwise would be limited to 4,294,967,295 records of 4096 bytes each, or a bit bag of 17,592,186,040,320 bytes - only seventeen terabytes...

Overflow is also a problem in many of the calculations. For instance, for a given (prime) number F, the marking of multiples of F via the sieve process starts with the bit corresponding to F² and if this exceeds the number corresponding to the last bit of the current sieve span, then the sieve process is complete for this span because all later values for F will be still further past the end. So, if LST is the number corresponding to the last bit of the current span, <lang Fortran> DO WHILE(F*F <= LST) !But, F*F might overflow the integer limit so instead,

     DO WHILE(F <= LST/F)                      !Except, LST might also overflow the integer limit, so 
     DO WHILE(F <= (IST + 2*(SBITS - 1))/F)    !Which becomes...
     DO WHILE(F <= IST/F + (MOD(IST,F) + 2*(SBITS - 1))/F) !Preserving the remainder from IST/F.</lang>

Except, IST might overflow the integer limit, in which case function PSURGE declares itself unable to proceed and returns false.

Overflow is detected by the sudden appearance of negative numbers, as is characteristic of two's complement integer arithmetic. This is not guaranteed to be used on all computers (notably, on a decimal computer such as the IBM1620 and others), and in its absence, the procedure will malfunction. Some systems detect integer overflow via hardware (a special "flag" register, or an interrupt) and there may be facilities for noticing such events. First Fortran (1957) offered special statements such as IF ACCUMULATOR OVERFLOW labelon,labeloff (yes, without brackets) and similarly for QUOTIENT OVERFLOW and DIVIDE CHECK but they were abandoned by the modernisers. The only general solution to this problem would be to convert to using multiple-precision (or "bignum") arithmetic, whereupon the code becomes extensible in another way simply by extending as needed the amount of storage allowed for variables.

These methods have been tested by converting INTEGER to INTEGER*2 and also by using a record size of sixteen bytes (because UltraEdit, when displaying in binary, shows that many bytes to a line), and it was a real pleasure for once to be able to read the 32-bit count field at the start of each record left-to-right in hexadecimal rather than in the crazed little-endian order that would otherwise have been used.

The Code

The source code employs the MODULE facility of F90 simply to avoid the tedium of setting up a COMMON storage area and having to declare the type of the functions in every routine that uses them. Otherwise, older style compilers will accept this, except for an occasional array facility (such as BIT8 = CHAR(255)) and the use of the $ format code to allow the next output to tag on to the same line. The rather more fearsome declaration RECURSIVE FUNCTION NEXTPRIME could be avoided if in subroutine PSURGE, the invocation of NEXTPRIME was replaced by in-line code. Similarly with subroutine GETSREC, though forgetting this didn't seem to make any difference. This is just convenience recursion, not structural recursion to some arbitrary depth.

Although recursion is now permissible if one utters the magic word RECURSIVE, this ability usually is not extended to the workings for formatted I/O so that if say a function is invoked in a WRITE statement's list, should that function attempt to use a WRITE statement, the run will be stopped. There can be slight dispensations if different types of WRITE statement are involved (say, formatted for one, and "free"-format for the other) but an ugly message is the likely result. The various functions are thus best invoked via an assignment statement to a scratch variable, which can then be printed. The functions are definitely not "pure" because although they are indeed functions only of their arguments, they all mess with shared storage, can produce error messages (and even a STOP), and can provoke I/O with a disc file, even creating such a file. For this reason, it would be unwise to attempt to invoke them via any sort of parallel processing. Similarly, the disc file is opened with exclusive use because of the possibility of writing to it. There are no facilities in standard Fortran to control the locking and unlocking of records of a disc file as would be needed when adding a new record and updating the record count. This would be needed if separate tasks could be accessing the bit file at the same time, and is prevented by exclusive use. If an interactive system were prepared to respond to requests for ISPRIME(n), etc. it should open the bit file only for its query then close it before waiting for the next request - which might be many milliseconds away.

The bit array is stored in an array of type CHARACTER*1 since this has been available longer and more widely than INTEGER*1. One hopes that the consequent genuflections to type checking via functions CHAR(i) and ICHAR(c) will not involve an overhead. <lang Fortran> MODULE PRIMEBAG !Need prime numbers? Plenty are available. C Creates and expands a disc file for a sieve of Eratoshenes, representing odd numbers only and starting with three. C Storage requirements: an array of N prime numbers in 16/32/64 bits vs. a bit array up to the 16/32/64 bit limit. C Word size N Prime N words in bits Bit array in bits. C 8 bit P(31) = 127 248 128 C P(54) = 251 432 256 C 16 bit P(3,512) = 32,749 56,192 32,768 C P(6,542) = 65,521 104,672 65,536 C 32 bit P(105,097,565) = 2,147,483,647 3,363,122,080 2,147,483,648 C P(203,280,221) = 4,294,967,291 6,504,967,072 4,294,967,296 C 64 bit 2.112E17  ? 1.352E19 9,223,372,036,854,775,808 ~ 9.22E18 C from n/Ln(n) 4.158E17  ? 2.661E19 18,446,744,073,709,551,616 ~ 1.84E19

      INTEGER MSG	!I/O unit number.
      INTEGER SSTASH	!For attachment to my stash file.
      INTEGER SRECLEN,SCHARS,SBITS	!Sizes.
      INTEGER SORG	!Where the sieve starts. This must be three.
      INTEGER SLAST	!Last record in my stash file.
      DATA SSTASH,SREC,SLAST/0,0,0/	!Prepared by PRIMEBAG.
      PARAMETER (SRECLEN = 1024)	!4K disc bloc size, but RECL (in OPEN) is in terms of four-byte integers.
      PARAMETER (SCHARS = (SRECLEN - 1)*4)	!Reserving space for one number at the start.
      PARAMETER (SBITS = SCHARS*8)	!Known size of a character.
      PARAMETER (SORG = 3)		!First odd number past two, which is not odd.
      CHARACTER*(*) SFILE		!A name is needed.
      PARAMETER (SFILE = "C:/Nicky/RosettaCode/Primes/PrimeSieve.bit")	!I don't have to count the characters.

Components of a buffered record for the stash.

      INTEGER SREC	!The record number.
      CHARACTER*1 C4(4)	!The start of the record - a counter.
      CHARACTER*1 SCHAR(0:SCHARS - 1)	!The majority of the record - a bit array, packed in 8-bit blobs...

Collect some bit twiddling assistants for AND and OR, rather than bit shifting.

      CHARACTER*1 BITON(0:7),BITOFF(0:7)	!Functions IBSET and IBCLR may not be available, and are little-endian anyway.
      PARAMETER (BITON =(/CHAR(2#10000000),CHAR(2#01000000),	!128,  64,	Reading strictly left-to-right.
    1                     CHAR(2#00100000),CHAR(2#00010000),	! 32,  16,	Uncompromising bigendery.
    1                     CHAR(2#00001000),CHAR(2#00000100),	!  8,   4,	Not just for bytes in words,
    3                     CHAR(2#00000010),CHAR(2#00000001)/))	!  2,   1.	But also bits in bytes.
      PARAMETER (BITOFF=(/CHAR(2#01111111),CHAR(2#10111111),	!127, 191,	BITON + BITOFF = 255.
    2                     CHAR(2#11011111),CHAR(2#11101111),	!223, 239,
    1                     CHAR(2#11110111),CHAR(2#11111011),	!247, 251,
    3                     CHAR(2#11111101),CHAR(2#11111110)/))	!253, 254.
      CONTAINS
       INTEGER FUNCTION I4UNPACK(C4)	!Convert four successive characters into an integer.
        CHARACTER*1 C4(4)	!The characters.
         I4UNPACK = ((ICHAR(C4(1))*256 + ICHAR(C4(2)))*256	!Convert the first four bytes
    1               + ICHAR(C4(3)))*256 + ICHAR(C4(4))		!To a four-byte integer.
       END FUNCTION I4UNPACK	!Big-endian style, irrespective of cpu endianness.
       SUBROUTINE C4PACK(I4)	!Convert an integer into successive bytes.

Could return the result via a fancy function, but for now a global variable will do.

        INTEGER I4,N	!The integer, and a copy to damage.
        INTEGER I	!A stepper.
         N = I4	!Keep the original safe.
         DO I = 4,1,-1	!Know that four characters will do. Fixed format makes this easy.
           C4(I) = CHAR(MOD(N,256))	!Grab the low-order eight bits.
           N = N/256			!And shift right eight.
         END DO			!Do it again.
       END SUBROUTINE C4PACK	!Stored big-endianly, irrespective of cpu endianness.
       LOGICAL FUNCTION GRASPPRIMEBAG(F)
        INTEGER F		!The I/O unit number to use.
        LOGICAL EXIST		!Use the keyword as a name
        INTEGER IOSTAT		!And don't worry over assignment direction.
        CHARACTER*3 STYLE	!One way or another.
         SSTASH = F		!I shall use it.
         INQUIRE (FILE = SFILE,EXIST = EXIST)	!Trouble with a missing "path" may arise.
         IF (EXIST) THEN	!If the file exists,
           STYLE = "OLD"	!I shall read it.
          ELSE			!But if it doesn't,
           STYLE = "NEW"	!I shall create it.
         END IF		!Enough prevarication.
         OPEN(SSTASH,FILE = SFILE, STATUS = STYLE,	!Go for the file.
    &     ACCESS = "DIRECT", RECL = SRECLEN, FORM = "UNFORMATTED",	!I have plans.
    &     ERR = 666, IOSTAT = IOSTAT)					!Which may be thwarted.
         IF (EXIST) THEN	!If there is one...
           CALL READSCHAR(1)	!The first record is also a header.
           SLAST = I4UNPACK(C4)	!The number of records stored.
          ELSE			!Otherwise, start from scratch.
           SLAST = 0			!No saved records.
           CALL PSURGE(SCHAR)		!During preparation of the first batch of bits.
         END IF		!All should now be in readiness.
         GRASPPRIMEBAG = .TRUE.!So, feel confidence.
        RETURN			!And escape.
 666     WRITE (*,667) IOSTAT,SFILE	!But, something may have gone wrong.
 667     FORMAT ("Pox! Error code ",I0,	!A "hole" in the directory path?
    1     " when attempting to open file ",A)	!Read-only access allowed when I want "update"?
         GRASPPRIMEBAG = .FALSE.		!Whatever, it didn't work.
       END FUNCTION GRASPPRIMEBAG	!So much for that.
       SUBROUTINE READSCHAR(R)	!Get record R into SCHAR, which may already hold it.
        INTEGER R		!The record number desired.
         IF (R.EQ.SREC) RETURN	!Perhaps it is already to hand.
         SREC = R		!If not, move attention to it.
         READ (SSTASH,REC = SREC) C4,SCHAR	!And read the record.
       END SUBROUTINE READSCHAR!Thus, I have a buffer too.
       LOGICAL FUNCTION PSURGE(BIT8)	!Add another record to the stash.

C Surges forward into the next batch of primes, to be stored via a bit array in the file. C Each record starts with a count of the number of primes that have gone before. C Except that for the first record, this is the record counter for the stash file. C Except that when starting the second record, one is also the number of primes before SORG.

        CHARACTER*1 BIT8(0:SCHARS - 1)	!Watch out! This may be SCHAR itself!
        INTEGER IST,LST	!The numbers spanned by the surge.
        INTEGER F		!A factor.
        INTEGER I		!Another factor and a stepper.
        INTEGER C		!Index for array BIT8.
        INTEGER NP		!Number of primes.

Carry forward the count of previous primes to start the following record..

  10     IF (SLAST.GT.0) THEN	!Is there a previous record?
           CALL READSCHAR(SLAST)	!Yes. Grab it. A good chance this is already in C4,SCHAR.
           NP = I4UNPACK(C4)		!Its count of the primes accumulated before it.
           DO I = 0,SCHARS - 1		!Find out how namy primes it fingered by scanning its bits.
             NP = NP + COUNT(IAND(ICHAR(SCHAR(I)),ICHAR(BITON)).NE.0)	!Whee! Eight at a go!
           END DO			!On to the next byte.
         END IF		!When creating a new record, its follower may not be sought in this run.

Concoct the next batch of bits. Contorted calculations avoid integer overflow.

  20     BIT8 = CHAR(255)		!All bits are aligned with numbers that might prove to be prime.
         IST = SORG + SLAST*(2*SBITS)	!Bit(0) of BIT8(0) corresponds to IST.
         LST = IST + 2*(SBITS - 1)	!Bit(last) to this number. Remember, only odd numbers have bits.
         IF (IST.LE.0) THEN		!Humm. I'd better check.
           WRITE (MSG,21) SLAST,IST,LST	!This works only with two's complement integers.
  21       FORMAT (/,"Integer overflow in the sieve of Eratosthenes!",	!Oh dear.
    1      /,"Advancing from surge ",I0," to span ",I0," to ",I0)	!These numbers will look odd.
           PSURGE = .FALSE.			!But it is better than no indication of what went wrong.
          RETURN			!Give in.
         END IF		!Enough worrying.
         F = 3			!The first possible factor. Zapping will start at F²

c DO WHILE(F.LE.LST/F) !If F² is past the end, so will be still larger F: enough.

         DO WHILE(F.LE.IST/F + (MOD(IST,F) + 2*(SBITS - 1))/F)	!"Synthetic division" avoiding overflow.
           I = (IST - 1)/F + 1		!I want the first multiple of F in IST:LST. F may be a factor of IST.
           IF (MOD(I,2).EQ.0) I = I + 1!If even, advance to the next odd multiple. Even numbers are omitted by design.
           IF (I.LT.F) I = F		!Less than F is superfluous: the position was zapped by earlier action.

c I = (I*F - IST)/2 !Current bit positions are for IST, IST+2, IST+4, etc.

           I = ((I - IST/F)*F - MOD(IST,F))/2	!Avoids overflow when calculating the start value, I*F.
           DO I = I,SBITS - 1,F	!Zap every F'th bit along. This is the sieve of Eratosthenes.
             C = I/8				!Eight bits per character.
             BIT8(C) = CHAR(IAND(ICHAR(BIT8(C)),	!For F = 3 and 5, characters will be hit more than once.
    1                            ICHAR(BITOFF(MOD(I,8)))))	!Whack a bit. All the above just for this!
           END DO			!On to the next bit.
  22       F = NEXTPRIME(F)		!So much for F. Next, please.
         END DO		!Are we there yet?

Correct the count in the header, if this is an added record.

  30     IF (SLAST.GT.0) THEN	!So, was there a pre-existing header record?
           CALL READSCHAR(1)		!Yes. Get the header record into C4,SCHAR.
           CALL C4PACK(SLAST + 1)	!This is the new record count.
           WRITE (SSTASH,REC = 1) C4,SCHAR	!Write it all back.
           SCHAR = BIT8	!Ensure that SCHAR and SREC will be agreed.
         END IF		!So much for the header's count.

Cast the bits into the stash by writing record SLAST + 1..

  40     IF (SLAST.EQ.0) THEN	!If we're writing the first record,
           CALL C4PACK(1)		!Then this is the record count.
          ELSE			!Otherwise,
           CALL C4PACK(NP)		!Place the previous primes count.
         END IF		!All this to help PRIME(i).
         SLAST = SLAST + 1	!This is now the last stashed record.
         WRITE (SSTASH,REC = SLAST) C4,BIT8	!I/O directly from the work area?
         SREC = SLAST		!This is where BIT8 was written.
         PSURGE = .TRUE.	!That assumes BIT8 is not SCHAR for SLAST > 1.
       END FUNCTION PSURGE	!That was fun!
       RECURSIVE SUBROUTINE GETSREC(R)	!Make present the bit array belonging to record R.
        INTEGER R		!The record number..
        CHARACTER*1 BIT8(0:SCHARS - 1)	!A scratchpad. Others may be relying on SCHAR.
         IF (SLAST.LE.0) RETURN!DANGER! The first record is being initialised!
         DO WHILE(SLAST.LT.R)	!If we haven't reached so far,
           IF (.NOT.PSURGE(BIT8)) THEN	!Slog forwards one record's worth.
             WRITE (MSG,1) R			!Or maybe not.
   1         FORMAT ("Cannot prepare surge ",I0)	!Explain.
             STOP "No bits, no go."			!And quit.
           END IF			!And having prepared the next block of bits,
         END DO		!Check afresh.
         CALL READSCHAR(R)	!Read the desired record's bits.
       END SUBROUTINE GETSREC	!Done.
       INTEGER FUNCTION PRIME(N)	!P(1) = 2, P(2) = 3, etc.

C Calculate P(n) ~ n.ln(n) C ~ n{ln(n) + ln(ln(n)) - 1 + (ln(ln(n)) - 2)/ln(n) - [ln(ln(n))**2 - 6*log(log(n)) + 11]/[2*(ln(n))**2] + ....} C J.B.Rosser's 1938 Theorem: n[ln(n) + ln(ln(n)) - 1] < P(n) < n[ln(n) + ln(ln(n))] C or, with E = ln(n) + ln(ln(n)), n[E - 1] < P(n) < n[E] C Experimentation shows that the undershoot of the first two terms involves many records worth of bits. C Including additional terms does much better, but can overshoot.

        INTEGER N		!The desired one.
        INTEGER R,NP		!Counts.
        INTEGER B,C		!Bit and character indices.
        DOUBLE PRECISION EST,LN,LLN	!Hope, if not actuality.
         IF (N.LE.0) STOP "Primes are counted positively!"	!Something must be wrong!
         IF (N.LE.1) THEN	!The start of the bit array being preempted.
           PRIME = 2			!So, no array access.
          ELSE			!Otherwise, the fun begins.
           LN = LOG(DFLOAT(N))	!Here we go.
           LLN = LOG(LN)	!A popular term.
           EST = N*(LN		!Estimate the value of the N'th prime.
    1               + LLN - 1		!Second term
    2               + (LLN - 2)/LN		!Third term.
    3               - (LLN**2 - 6*LLN + 11)/(2*LN**2))	!Fourth term.
           R = (EST - SORG)/(2*SBITS) + 1	!Thereby selecting a record to scan.
           IF (R.LE.0) R = 1			!And not making a mess with N < 6 or so.
   9       CALL GETSREC(R)	!Go for the record.
           IF (R.LE.1) THEN	!The first record starts with the record count.
             NP = 1		!And I know how many primes precede its start point
            ELSE		!While for all subsequent records,
             NP = I4UNPACK(C4)	!This counts the number of primes that precede record R's start number.
           END IF		!So now I'm ready to count onwards.
           IF (N.LE.NP) THEN	!Maybe not.
             R = R - 1			!The estimate took me too far ahead.
             GO TO 9			!Try again.
           END IF		!Could escalate to a binary search or even an interpolating search.

Commence scanning the bits.

           C = 0		!Start with the first character of SREC..
           B = -1		!Syncopation. The formula is known to always under-estimate.
  10       IF (NP.LT.N) THEN	!Are we there yet?
  11         B = B + 1			!No. Advance to the next bit.
             IF (B.GE.8) THEN		!Overflowed a character yet?
               B = 0			!Yes. Start afresh at the first bit.
               C = C + 1		!And advance one character.
               IF (C.GE.SCHARS) THEN	!Overflowed the record yet?
                 C = 0			!Yes. Start afresh at its first character.
                 R = R + 1		!And advance to the next record.
                 CALL GETSREC(R)	!Possibly, create it.
               END IF			!So much for records.
             END IF			!We're now ready to test bit B of character C of record R.
             IF (IAND(ICHAR(SCHAR(C)),ICHAR(BITON(B))).EQ.0) GO TO 11	!Not a prime. Search on.
             NP = NP + 1		!Count another prime.
             GO TO 10			!Pehaps this will be the one.
           END IF		!So much for the search.
           PRIME = SORG + (R - 1)*(2*SBITS) + (C*8 + B)*2	!The corresponding number.
           IF (PRIME.LE.0) WRITE (MSG,666) N,PRIME	!Or, possibly not.
 666       FORMAT ("Integer overflow! Prime(",I0,") gives ",I0,"!")	!Let us hope the caller notices.
         END IF		!So, all going well,
       END FUNCTION PRIME	!It is found.
       RECURSIVE INTEGER FUNCTION NEXTPRIME(N)	!Keep right on to the end of the road.

Can invoke GETSREC, which can invoke PSURGE, which ... invokes NEXTPRIME. Oh dear.

        INTEGER N	!Not necessarily itself a prime number.
        INTEGER NN	!A value to work with.
        INTEGER R	!A record number into the stash.
        INTEGER I,IST	!Number offsets.
        INTEGER C,B	!Character and bit index.
         IF (N.LE.1) THEN	!Suspicion prevails.
           NN = 2			!This is not represented in my bit array.
          ELSE			!Otherwise, the fun begins.
           NN = N + 1			!Advance, with a copy I can mess with.
           IF (MOD(NN,2).EQ.0) NN = NN + 1	!Thus, NN is now odd.
           IF (NN.LE.0) GO TO 666		!But perhaps not proper, due to overflow.
           R = (NN - SORG)/(2*SBITS)		!SORG is odd, so (NN - SORG) is even.
           CALL GETSREC(R + 1)		!The first record is numbered one, not zero.
           IST = SORG + R*(2*SBITS)	!The number for its first bit: even numbers are omitted..
           I = (NN - IST)/2		!Offset into the record. NN - IST is even.
           C = I/8			!Which character in SCHAR(0:SCHARS - 1)?
           B = MOD(I,8)		!Which bit in SCHAR(C)?
  10       IF (IAND(ICHAR(SCHAR(C)),ICHAR(BITON(B))).EQ.0) THEN	!On for a prime.
             NN = NN + 2	!Alas, it is off, so NN is not a prime. Perhaps this will be.
             B = B + 1		!Advance one bit. Each bit steps two.
             IF (B.GE.8) THEN	!Past the end of the character?
               B = 0			!Yes. Back to bit zero.
               C = C + 1		!And advance one chracter.
               IF (C.GE.SCHARS) THEN	!Past the end of the record?
                 IF (NN.LE.0) GO TO 666!Yes. If NN has overflowed, the end of the rope is reached.
                 C = 0			!Back to the start of a record.
                 R = R + 1		!Advance one record.
                 CALL GETSREC(R + 1)	!And read it. (Count is from 1, not 0).
               END IF		!So much for overflowing a record.
             END IF		!So much for overflowing a character.
             GO TO 10	!Try again.
           END IF		!So much for the bit array.
         END IF		!If there had been a scan.
         NEXTPRIME = NN	!The number for which the scan stopped.
         IF (NN.GT.0) RETURN	!All is well.
 666     WRITE (MSG,667) N,NN	!Or, maybe not. Careful: this won't appear if NEXTPRIME is invoked in a WRITE list.
 667     FORMAT ("Integer overflow! NextPrime(",I0,") gives ",I0,"!")	!The recipient could do a two's complement.
         NEXTPRIME = NN	!Prefer to return the bad value rather than fail to return anything.
       END FUNCTION NEXTPRIME	!No divisions, no sieving. Here, anyway
       INTEGER FUNCTION PREVIOUSPRIME(N)	!If N is good, this can't overflow.
        INTEGER N	!The number, not necessarily a prime.
        INTEGER NN	!A value to mess with.
        INTEGER R	!A record number.
        INTEGER I	!Offset.
        INTEGER C,B	!Character and bit fingers.
         IF (N.LE.3) THEN	!Suppress annoyances.
           NN = 2		!This is now called the first prime, not one.
          ELSE			!Otherwise, some work is to be done.
           NN = N - 1			!Step back one to ensure previousness.
           IF (MOD(NN,2).EQ.0) NN = NN - 1	!And here, oddness is a minimal requirement.
           R = (NN - SORG)/(2*SBITS)	!Finger the record containing the bit for NN.
           CALL GETSREC(R + 1)		!Record counting starts with one.
           I = (NN - (SORG + R*(2*SBITS)))/2	!Offset into that record.
           C = I/8			!Finger the character in SCHAR.
           B = MOD(I,8)		!And the bit within the character.
  10       IF (IAND(ICHAR(SCHAR(C)),ICHAR(BITON(B))).EQ.0) THEN	!On for a prime.
             NN = NN - 2	!Alas, it is off, so NN is not a prime. Perhaps this will be.
             B = B - 1		!Retreat one bit. Each bit steps two.
             IF (B.LT.0) THEN	!Past the start of the character?
               B = 7			!Yes. Back to the last bit.
               C = C - 1		!And retreat one chracter.
               IF (C.LT.0) THEN	!Past the start of the record?
                 C = SCHARS - 1	!Yes. Back to the end of a record.
                 R = R - 1		!Retreat one record.
                 CALL GETSREC(R + 1)	!And read it. (Count is from 1, not 0).
               END IF		!So much for overflowing a record.
             END IF		!So much for overflowing a character.
             GO TO 10		!Try again.
           END IF		!So much for the bit array.
         END IF		!Possibly, it was not needed.
         PREVIOUSPRIME = NN	!There.
       END FUNCTION PREVIOUSPRIME	!Doesn't overflow, either.
       LOGICAL FUNCTION ISPRIME(N)	!Could fool around explicity testing 2 and 3 and say 5,
        INTEGER N			!But that means also checking that N > 2, N > 3, and N > 5.

c ISPRIME = N .EQ. NEXTPRIME(N - 1) !This is so much easier, but involves scanning to reach the next prime.

        INTEGER R,IST,I,C,B		!Assistants for indexing the bit array.
         IF (N.LE.1) THEN	!First, preclude sillyness.
           ISPRIME = .FALSE.		!Not a prime.
         ELSE IF (N.EQ.2) THEN	!This is the only even number
           ISPRIME = .TRUE.		!That is a prime.
         ELSE IF (MOD(N,2).EQ.0) THEN	!Other even numbers
           ISPRIME = .FALSE.		!Are not prime numbers.
         ELSE			!Righto, now N is an odd number and there is a bit array for them.
           R = (N - SORG)/(2*SBITS)	!SORG is odd, so (N - SORG) is even.
           CALL GETSREC(R + 1)		!The first record is numbered one, not zero.
           IST = SORG + R*(2*SBITS)	!The number for its first bit: even numbers are omitted.
           I = (N - IST)/2		!Offset into the record. N - IST is even.
           C = I/8			!Which character in SCHAR(0:SCHARS - 1)?
           B = MOD(I,8)		!Which bit in SCHAR(C), indexing from zero?
           ISPRIME = IAND(ICHAR(SCHAR(C)),ICHAR(BITON(B))).GT.0	!The bit is on for a prime.
         END IF			!All that fuss to find a single bit.
       END FUNCTION ISPRIME		!But, no divisions up to SQRT(N) or the like.
     END MODULE PRIMEBAG	!Functions updating a disc file as a side effect...
     PROGRAM POKE
     USE PRIMEBAG
     INTEGER I,P,N,N1,N2	!Assorted assistants.
     INTEGER ORDER		!A collection of special values.
     PARAMETER (ORDER = 6)	!For one, two, and four byte integers.
     INTEGER EDGE(ORDER)	!Considered as two's complement and unsigned.
     PARAMETER (EDGE = (/31,54,3512,6542,105097565,203280221/))	!These primes are of interest.
     MSG = 6		!Standard output.
     IF (.NOT.GRASPPRIMEBAG(66)) STOP "Gan't grab my file!"	!Attempt in hope.

Case 1. C FORALL(I = 1:20) LIST(I) = PRIME(I) is rejected because function Prime(i) is rather impure.

  10 WRITE (MSG,11)
  11 FORMAT (19X,"First twenty primes: ", $)
     DO I = 1,20
       P = PRIME(I)
       WRITE (MSG,12) P
  12   FORMAT (I0,",",$)
     END DO

Case 2.

  20 WRITE (MSG,21)
  21 FORMAT (/,12X,"Primes between 100 and 150: ",$)
     P = 100
  22 P = NEXTPRIME(P)		!While (P:=NextPrime(P)) <= 150 do Print P;
     IF (P.LE.150) THEN	!But alas, no assignment within an expression.
       WRITE (MSG,23) P
  23   FORMAT (I0,",",$)
       GO TO 22
     END IF

Case 3.

  30 N1 = 7700	!Might as well parameterise this.
     N2 = 8000	!Rather than litter the source with explicit integers.
     N = 0
     P = N1
  31 P = NEXTPRIME(P)
     IF (P.LE.N2) THEN
       N = N + 1
       GO TO 31
     END IF
     WRITE (MSG,32) N1,N2,N
  32 FORMAT (/"Number of primes between ",I0," and ",I0,": ",I0)

Case 4.

  40 WRITE (MSG,41)
  41 FORMAT (/,"Tenfold steps...")
     N = 1
     DO I = 1,9	!This goes about as far as it can go.
       P = PRIME(N)
       WRITE (MSG,42) N,P
  42   FORMAT ("Prime(",I0,") = ",I0)
       N = N*10
     END DO

Cast forth some interesting values.

 100 WRITE (MSG,101)
 101 FORMAT (/,"Primes close to number sizes")
     DO N = 1,ORDER	!Step through the list.
       N1 = EDGE(N) - 1	!Syncopation for the special value.
       DO I = 1,2		!I want the prime on either side.
         N1 = N1 + 1			!So, there are two successive primes to finger.
         WRITE (MSG,102) N1			!Identify the index.
 102     FORMAT ("Prime(",I0,") = ",$)		!Piecemeal writing to the output,
         P = PRIME(N1)				!As this may fling forth a complaint.
         WRITE (MSG,103) P			!Show the value returned.
 103     FORMAT (I0,", ",$)			!Which may be unexpected.
       END DO			!On to the second.
       WRITE (MSG,*)		!End the line after the second result.
     END DO		!On to the next in the list.
     END	!Whee!</lang>

Although the structuralist talk up the merit of "structured" constructs in programming, there are often annoying obstacles. With a WHILE-loop one usually has to repeat the "next item" code to "prime" the loop, as in <lang Fortran> P = NEXTPRIME(100)

     DO WHILE (P.LE.150)
       ...stuff...
       P = NEXTPRIME(P)
     END DO</lang>

While this is not a large imposition in this example, if Fortran were to allow assignment within expressions as in Algol, the tedium of code replication and its risks could be avoided. <lang algol68> P:=100; WHILE (P:=NextPrime(P)) <= 150 DO stuff;</lang> If instead of NEXTPRIME the "next item" code was to read a record of a disc file, the repetition needed becomes tiresome. So, an IF-statement, and ... a GO TO...

The Results

When creating the bit file, everything appears in a blink up to the millionth prime, then a pause until the ten millionth prime, then about a minute to attain the hundred millionth prime. The blinking lights show that there is not much disc I/O in progress (actually, a solid-state unit) while the cpu is running at full speed. There is a further pause from there until overflow is reached. For a subsequent run with the disc file already prepared, all is completed in a blink. Pausing the Great Internet Mersenne Prime crunch (sixfold) increases the speed and I/O rate by about a third. During the file expansion, the rate of I/O slowly decreases at a decreasing rate - this is due to each successive sieve surge requiring more primes to step with, but the primes themselves thin out and being larger, require fewer steps to traverse the span. The effort for each surge is related to the sum of the reciprocals of the primes that will sieved with and this forms an interesting sequence in its own right. For a given surge width, fewer and fewer hits are made within that width by the larger prime numbers. However, 4092 bytes (four are reserved for the count, and on this windows XP system disc space is allocated in blocks of 4k) gives 32736 bits which with odd numbers only, spans 65472. With 32-bit two's complement, sqrt(2,147,483,647) = 46340·95 so even the largest stepper will land at least once within every span, except that only odd multiples are involved so already the hit rate has drifted below one per span and with extension to still larger numbers, the hit rate will fall further. Thus, if instead each pass were to be for the full width of the bit array in the disc file, the I/O system would suffer a thrashing during the multiple passes unless the entire file could be fitted into random-access memory. But such a scheme would not have the "extensible" aspect.

Output:

                   First twenty primes: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,
            Primes between 100 and 150: 101,103,107,109,113,127,131,137,139,149,
Number of primes between 7700 and 8000: 30

Tenfold steps...
Prime(1) = 2
Prime(10) = 29
Prime(100) = 541
Prime(1000) = 7919
Prime(10000) = 104729
Prime(100000) = 1299709
Prime(1000000) = 15485863
Prime(10000000) = 179424673
Prime(100000000) = 2038074743

Primes close to number sizes
Prime(31) = 127, Prime(32) = 131,
Prime(54) = 251, Prime(55) = 257,
Prime(3512) = 32749, Prime(3513) = 32771,
Prime(6542) = 65521, Prime(6543) = 65537,
Prime(105097565) = 2147483647, Prime(105097566) = Integer overflow! Prime(105097566) gives -2147483637!
-2147483637,
Prime(203280221) =
Integer overflow in the sieve of Eratosthenes!
Advancing from surge 32801 to span -2147420221 to -2147354751
Cannot prepare surge 65601
No bits, no go.

The disc file holding all primes up to the thirty-two bit limit occupies 134,352,896 bytes, or 128MB. Nothing much, these days. Activating 7-zip out of curiosity resulted in compressing the file by a factor of two. As the primes thin out there will be more and more characters with only one bit on (if not none) rather than a fuller selection. However, the existence of prime pairs shows that the bits will never be all lonely forever. For this run, the last record is number 32,801 and 105,097,477 (hex 643A905) primes have gone before.

 First value  Bit array...
           3: 11101101 10100110 01011010 01001100 10110010 10010001 01101101 00000010
              10011000 01100100 10100100 11000011 01100000 10000010 11010011 00001001
              00100110 01011000 01000000 10110100 00001001 00001101 00100010 01001010
              01000101 00010000 11000011 00101001 00010110 10000010 00101000 10100100 ...

  2147481603: 00000000 00000100 00000000 00000000 00010000 00000000 00000000 00000000
              00000000 00000000 00000000 00000001 01000000 10000000 10000000 00000000
              00100000 00001000 01001100 10000000 00000001 00000100 00000010 00000000
              00000100 00000000 01000000 00000000 00000010 00000001 00001100 00000000 ...

Would anyone prefer to see that bit array in the little-endian order within bytes?

The thinning out rather suggests an alternative encoding such as by counting the number of "off" bits until the next "on" bit, but this would produce a variable-length packing so that it would no longer be easy to find the bit associated with a given number by something so simple as R = (NN - SORG)/(2*SBITS) as in NEXTPRIME. A more accomplished data compression system might instead offer a reference to a library containing a table of prime numbers, or even store the code for a programme that will generate the data. File Pbag.for is 23,008 bytes long, and of course contains irrelevant commentary and flabby phrases such as "PARAMETER", "INTEGER", "GRASPPRIMEBAG", etc. As is the modern style, the code file is much larger at 548,923 bytes (containing kilobyte sequences of hex CC and of 00), but both are much smaller than the 134,352,896 bytes of file PrimeSieve.bit.

FreeBASIC

This program uses the Sieve Of Eratosthenes which is not very efficient for large primes but is quick enough for present purposes.

The size of the sieve array (of type Boolean) is calculated as 20 times the number of primes required which is big enough to compute up to 50 million primes (a sieve size of 1 billion bytes) which takes under 50 seconds on my i3 @ 2.13 GHz. I've limited the procedure to this but it should certainly be possible to use a much higher figure without running out of memory.

It would also be possible to use a more efficient algorithm to compute the optimal sieve size for smaller numbers of primes but this will suffice for now.

<lang freebasic>' FB 1.05.0

Enum SieveLimitType

 number 
 between
 countBetween

End Enum

Sub printPrimes(low As Integer, high As Integer, slt As SieveLimitType)

 If high < low OrElse low < 1 Then Return               ' too small
 If slt <> number AndAlso slt <> between AndAlso slt <> countBetween Then Return
 If slt <> number AndAlso (low < 2 OrElse high < 2) Then Return  
 If slt <> number AndAlso high > 1000000000 Then Return ' too big 
 If slt = number  AndAlso high > 50000000 Then Return   ' too big
 Dim As Integer n
 If slt = number Then
   n = 20 * high  big enough to accomodate 50 million primes to which this procedure is limited
 Else
   n = high
 End If  
 Dim a(2 To n) As Boolean  only uses 1 byte per element
 For i As Integer = 2 To n : a(i) = True : Next  set all elements to True to start with
 Dim As Integer p = 2, q
 ' mark non-prime numbers by setting the corresponding array element to False
 Do
   For j As Integer = p * p To n Step p
     a(j) = False
   Next j
   ' look for next True element in array after 'p'
   q = 0
   For j As Integer = p + 1 To Sqr(n)
     If a(j) Then
       q = j
       Exit For
     End If
   Next j    
   If q = 0 Then Exit Do
   p = q
 Loop
 Select Case As Const slt
    Case number
      Dim count As Integer = 0
      For i As Integer = 2 To n
        If a(i) Then
          count += 1
          If count >= low AndAlso count <= high Then
            Print i; " ";
          End If
          If count = high Then Exit Select
        End If
      Next
   Case between
      For i As Integer = low To high
        If a(i) Then
          Print i; " ";
        End if
      Next
   Case countBetween
      Dim count As Integer = 0
      For i As Integer = low To high
        If a(i) Then count += 1
      Next
      Print count;
 End Select
 Print

End Sub

Print "The first 20 primes are :" Print printPrimes(1, 20, number) Print Print "The primes between 100 and 150 are :" Print printPrimes(100, 150, between) Print Print "The number of primes between 7700 and 8000 is :"; printPrimes(7700, 8000, countBetween) Print Print "The 10000th prime is :"; Dim t As Double = timer printPrimes(10000, 10000, number) Print "Computed in "; CInt((timer - t) * 1000 + 0.5); " ms" Print Print "The 1000000th prime is :"; t = timer printPrimes(1000000, 1000000, number) Print "Computed in ";CInt((timer - t) * 1000 + 0.5); " ms" Print Print "The 50000000th prime is :"; t = timer printPrimes(50000000, 50000000, number) Print "Computed in ";CInt((timer - t) * 1000 + 0.5); " ms" Print Print "Press any key to quit" Sleep</lang>

Output:
The first 20 primes are :

 2  3  5  7  11  13  17  19  23  29  31  37  41  43  47  53  59  61  67  71

The primes between 100 and 150 are :

 101  103  107  109  113  127  131  137  139  149

The number of primes between 7700 and 8000 is : 30

The 10000th prime is : 104729
Computed in  8 ms

The 1000000th prime is : 15485863
Computed in  775 ms

The 50000000th prime is : 982451653
Computed in  46703 ms

Frink

Frink has built-in functions for efficiently enumerating through prime numbers, including primes, nextPrime, and previousPrime, and isPrime. These functions handle arbitrarily-large integers. <lang frink>println["The first 20 primes are: " + first[primes[], 20]] println["The primes between 100 and 150 are: " + primes[100,150]] println["The number of primes between 7700 and 8000 are: " + length[primes[7700,8000]]] println["The 10,000th prime is: " + nth[primes[], 10000-1]] // nth is zero-based</lang>

Output:
The first 20 primes are: [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]
The primes between 100 and 150 are: [101, 103, 107, 109, 113, 127, 131, 137, 139, 149]
The number of primes between 7700 and 8000 are: 30
The 10,000th prime is: 104729

Go

An implementation of "The Genuine Sieve of Eratosthenese" by Melissa E. O'Niell. This is the paper cited above in the "Faster Alternative Version" of D. The Go example here though strips away optimizations such as a wheel to show the central idea of storing prime multiples in a queue data structure. <lang go>package main

import (

   "container/heap"
   "fmt"

)

func main() {

   p := newP()
   fmt.Print("First twenty: ")
   for i := 0; i < 20; i++ {
       fmt.Print(p(), " ")
   }
   fmt.Print("\nBetween 100 and 150: ")
   n := p()
   for n <= 100 {
       n = p()
   }
   for ; n < 150; n = p() {
       fmt.Print(n, " ")
   }
   for n <= 7700 {
       n = p()
   }
   c := 0
   for ; n < 8000; n = p() {
       c++
   }
   fmt.Println("\nNumber beween 7,700 and 8,000:", c)
   p = newP()
   for i := 1; i < 10000; i++ {
       p()
   }
   fmt.Println("10,000th prime:", p())

}

func newP() func() int {

   n := 1
   var pq pQueue
   top := &pMult{2, 4, 0}
   return func() int {
       for {
           n++
           if n < top.pMult { // n is a new prime
               heap.Push(&pq, &pMult{prime: n, pMult: n * n})
               top = pq[0]
               return n
           }
           // n was next on the queue, it's a composite
           for top.pMult == n {
               top.pMult += top.prime
               heap.Fix(&pq, 0)
               top = pq[0]
           }
       }
   }

}

type pMult struct {

   prime int
   pMult int
   index int

}

type pQueue []*pMult

func (q pQueue) Len() int { return len(q) } func (q pQueue) Less(i, j int) bool { return q[i].pMult < q[j].pMult } func (q pQueue) Swap(i, j int) {

   q[i], q[j] = q[j], q[i]
   q[i].index = i
   q[j].index = j

} func (p *pQueue) Push(x interface{}) {

   q := *p
   e := x.(*pMult)
   e.index = len(q)
   *p = append(q, e)

} func (p *pQueue) Pop() interface{} {

   q := *p
   last := len(q) - 1
   e := q[last]
   *p = q[:last]
   return e

}</lang>

Output:
First twenty: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 
Between 100 and 150: 101 103 107 109 113 127 131 137 139 149 
Number beween 7,700 and 8,000: 30
10,000th prime: 104729

An alternative showing how to use a good and very fast open source Sieve of Atkin implementation via github.com/jbarham/primegen.go. Due to how Go's imports work, the bellow can be given directly to "go run" or "go build" and the latest version of the primegen package will be fetched and built if it's not already present on the system. (This example may not be exactly within the scope of this task, but it's a trivial to use and extremely fast prime generator probably worth considering whenever primes are needed in Go.) <lang go>package main

import ( "fmt" "github.com/jbarham/primegen.go" )

func main() { p := primegen.New()

fmt.Print("First twenty: ") for i := 0; i < 20; i++ { fmt.Print(p.Next(), " ") } fmt.Print("\nBetween 100 and 150: ") p.SkipTo(100) for n := p.Next(); n < 150; n = p.Next() { fmt.Print(n, " ") } p.SkipTo(7700) fmt.Println("\nNumber beween 7,700 and 8,000:", p.Count(8000)) p.Reset() for i := 1; i < 1e4; i++ { p.Next() } fmt.Println("10,000th prime:", p.Next()) }</lang>

Haskell

Library: primes
Works with: GHC version 7.8.3
Works with: primes version 0.2.1.0

This program uses the primes package, which uses a lazy wheel sieve to produce an infinite list of primes.

<lang haskell>#!/usr/bin/env runghc

import Data.List import Data.Numbers.Primes import System.IO

firstNPrimes :: Integer -> [Integer] firstNPrimes n = genericTake n primes

primesBetweenInclusive :: Integer -> Integer -> [Integer] primesBetweenInclusive lo hi =

 dropWhile (< lo) $ takeWhile (<= hi) primes

nthPrime :: Integer -> Integer nthPrime n = genericIndex primes (n - 1) -- beware 0-based indexing

main = do

 hSetBuffering stdout NoBuffering
 putStr "First 20 primes: "
 print $ firstNPrimes 20
 putStr "Primes between 100 and 150: "
 print $ primesBetweenInclusive 100 150
 putStr "Number of primes between 7700 and 8000: "
 print $ genericLength $ primesBetweenInclusive 7700 8000
 putStr "The 10000th prime: "
 print $ nthPrime 10000</lang>
Output:
First 20 primes: [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71]
Primes between 100 and 150: [101,103,107,109,113,127,131,137,139,149]
Number of primes between 7700 and 8000: 30
The 10000th prime: 104729

List based

Using list based unbounded sieve from here (runs instantly):

<lang haskell> λ> take 20 primesW [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71]

λ> takeWhile (< 150) . dropWhile (< 100) $ primesW

[101,103,107,109,113,127,131,137,139,149]

λ> length . takeWhile (< 8000) . dropWhile (< 7700) $ primesW

30

λ> (!! (10000-1)) primesW

104729</lang>

Using formula for primes

Works with: GHC version 8.10.7
Works with: primes version 0.2.1.0
Works with: exact-real version 0.12.5.1

There are closed formulas for generating primes. One of such formula is the following:

p(n) = [2 n (2n+1) { C (2*n-1)!} ]

here [x] is the integral part of x, {x} is the fractional part of x, and C is a real constant (similar to Mill's constant). We can prove that there is a constant C, such that p(n) is exactly the nth prime number for any natural number n. The first digits of c are c=0.359344964622775339841352348439200241924659634... Haskell have a library of constructive real numbers, where real numbers can be of an arbitrary precision. We can define this constant C exactly and use the above formula to calculate the nth prime. This is not the fastest method, but one of the coolest. And it is not as bad as you may think. It took about 0.3 seconds to calculate 10,000th prime using this formula.

<lang haskell> {-# LANGUAGE PostfixOperators #-} {-# LANGUAGE DataKinds #-} {-# LANGUAGE FlexibleInstances #-}

import Data.Numbers.Primes import Data.Array.Unboxed hiding ((!)) import qualified Data.Array.Unboxed as Array import Data.CReal import Data.CReal.Internal import GHC.TypeLits

instance KnownNat n => Enum (CReal n) where

 toEnum i         = fromIntegral i
 fromEnum _       = error "Cannot fromEnum CReal"
 enumFrom         = iterate (+ 1)
 enumFromTo n e   = takeWhile (<= e) $ iterate (+ 1)n
 enumFromThen n m = iterate (+(m-n)) n
 enumFromThenTo n m e = if m >= n then takeWhile (<= e) $ iterate (+(m-n)) n
                         else takeWhile (>= e) $ iterate (+(m-n)) n


-- partial_sum x y a b = (p,q) where -- p/q = sum_{a<i<=b} x(i) / poduct_{a<j<=j} y(j) -- The complexity of partial_sum x y 0 n is O(n log n) partial_sum x y = pq where

   pq a b = if a>=b then (0,1) 
        else if a==b-1 then (fromIntegral $ x b, fromIntegral $ y b ) 
        else (p_ab,q_ab) 
        where 
          c=(a+b) `div` 2
          (p_ac,q_ac) = pq a c 
          (p_cb,q_cb) = pq c b 
          p_ab = p_cb + q_cb*p_ac
          q_ab = q_ac*q_cb

-- c is the real constant that is used in the formula for primes c = crMemoize f where

 f n = 2^n * p `div` q  where
   n' = fromIntegral n
   u = head [ceiling (x) | x<-[(n' * log 2/ (log n'-1)/2 ) ..] , 2*x*log (2*x) - 2*x > n'*log 2] 
   -- Invariant: (2u+1)! > 2^n
   ar :: UArray Int  Int
   ar = listArray (1,u) $ primes  
   (p,q) = partial_sum (ar Array.!) (\n-> 2*n*(2*n+1) ) 0 u 


-- Fractorial part of x -- By definition it is in the interval [-0.5; 0.5] -- But it gurantes to work corectly if fractional part of x is in (-0.375; 0.375) fract x = x - fromIntegral (round (x :: CReal 3))

-- Factorial. -- The complexity of (n!) is O(n log n) (which is better than O(n^2) for product [1..n] ) (!) :: (RealFrac a, Num b) => a -> b (!) = fromIntegral . snd . partial_sum (const 0) id 0 . round

prime n = round( 2*n*(2*n+1) * fract ( c * ((2*n-1)!)))</lang>

Then you can use this function: <lang haskell> λ> :set +s λ> prime 10000 104729 (0.32 secs, 179,899,272 bytes) λ> length $ dropWhile (< 7700) $ takeWhile (< 8000) $ map prime [1..] 30 (3.09 secs, 3,418,225,920 bytes) λ> dropWhile (< 100) $ takeWhile (< 150) $ map prime [1..] [101,103,107,109,113,127,131,137,139,149] (0.02 secs, 20,239,464 bytes) λ> map prime [1..20] [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71] (0.01 secs, 10,485,208 bytes) </lang>

Icon and Unicon

Only works in Unicon (use of the Heap class). Brute force.

The expression: <lang unicon>![2,3,5,7] | (nc := 11) | (nc +:= |wheel2345)</lang> is an open-ended sequence generating potential primes.

<lang unicon>import Collections # to get the Heap class for use as a Priority Queue record filter(composite, prime) # next composite involving this prime

procedure main()

   every writes((primes()\20)||" " | "\n")
   every p := primes() do if 100 < p < 150 then writes(p," ") else if p >= 150 then break write()
   every (n := 0, p := primes()) do if 7700 < p < 8000 then n +:= 1 else if p >= 8000 then break write(n)
   every (i := 1, p := primes()) do if (i+:=1) >= 10000 then break write(p)

end

procedure primes()

   local wheel2357, nc
   wheel2357 := [2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2,
                 6, 4, 6, 8, 4, 2, 4, 2, 4, 8, 6, 4, 6, 2, 4, 6,
                 2, 6, 6, 4, 2, 4, 6, 2, 6, 4, 2, 4, 2, 10, 2, 10]
   suspend sieve(Heap(,getCompositeField), ![2,3,5.7] | (nc := 11) | (nc +:= |!wheel2357))

end

procedure sieve(pQueue, candidate)

   local nc
   if 0 = pQueue.size() then {   # 2 is prime
       pQueue.add(filter(candidate*candidate, candidate))
       return candidate
       }
   while candidate > (nc := pQueue.get()).composite do {
       nc.composite +:= nc.prime
       pQueue.add(nc)
       }
   pQueue.add(filter(nc.composite+nc.prime, nc.prime))
   if candidate < nc.composite then {   # new prime found!
       pQueue.add(filter(candidate*candidate, candidate))
       return candidate
       }

end

  1. Provide a function for comparing filters in the priority queue...

procedure getCompositeField(x); return x.composite; end</lang>

Output:
->ePrimes
2 3 5.7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 
101 103 107 109 113 127 131 137 139 149 
30
104729
->

J

Using the p: builtin, http://www.jsoftware.com/help/dictionary/dpco.htm reports "Currently, arguments larger than 2^31 are tested to be prime according to a probabilistic algorithm (Miller-Rabin)".

<lang J> p:i.20 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71

  (#~ >:&100)i.&.(p:inv) 150

101 103 107 109 113 127 131 137 139 149

  #(#~ >:&7700)i.&.(p:inv) 8000

30

  p:10000-1

104729</lang>

Note: p: gives the nth prime, where 0 is first, 1 is second, 2 (cardinal) is third (ordinal) and so on...

Note: 4&p: gives the next prime

<lang J> 4 p: 104729 104743</lang>

Java

Based on my second C++ solution, which in turn was based on the C solution. <lang java>import java.util.*;

public class PrimeGenerator {

   private int limit_;
   private int index_ = 0;
   private int increment_;
   private int count_ = 0;
   private List<Integer> primes_ = new ArrayList<>();
   private BitSet sieve_ = new BitSet();
   private int sieveLimit_ = 0;
   public PrimeGenerator(int initialLimit, int increment) {
       limit_ = nextOddNumber(initialLimit);
       increment_ = increment;
       primes_.add(2);
       findPrimes(3);
   }
   public int nextPrime() {
       if (index_ == primes_.size()) {
           if (Integer.MAX_VALUE - increment_ < limit_)
               return 0;
           int start = limit_ + 2;
           limit_ = nextOddNumber(limit_ + increment_);
           primes_.clear();
           findPrimes(start);
       }
       ++count_;
       return primes_.get(index_++);
   }
   public int count() {
       return count_;
   }
   private void findPrimes(int start) {
       index_ = 0;
       int newLimit = sqrt(limit_);
       for (int p = 3; p * p <= newLimit; p += 2) {
           if (sieve_.get(p/2 - 1))
               continue;
           int q = p * Math.max(p, nextOddNumber((sieveLimit_ + p - 1)/p));
           for (; q <= newLimit; q += 2*p)
               sieve_.set(q/2 - 1, true);
       }
       sieveLimit_ = newLimit;
       int count = (limit_ - start)/2 + 1;
       BitSet composite = new BitSet(count);
       for (int p = 3; p <= newLimit; p += 2) {
           if (sieve_.get(p/2 - 1))
               continue;
           int q = p * Math.max(p, nextOddNumber((start + p - 1)/p)) - start;
           q /= 2;
           for (; q >= 0 && q < count; q += p)
               composite.set(q, true);
       }
       for (int p = 0; p < count; ++p) {
           if (!composite.get(p))
               primes_.add(p * 2 + start);
       }
   }
   private static int sqrt(int n) {
       return nextOddNumber((int)Math.sqrt(n));
   }
   private static int nextOddNumber(int n) {
       return 1 + 2 * (n/2);
   }
   public static void main(String[] args) {
       PrimeGenerator pgen = new PrimeGenerator(20, 200000);
       System.out.println("First 20 primes:");
       for (int i = 0; i < 20; ++i) {
           if (i > 0)
               System.out.print(", ");
           System.out.print(pgen.nextPrime());
       }
       System.out.println();
       System.out.println("Primes between 100 and 150:");
       for (int i = 0; ; ) {
           int prime = pgen.nextPrime();
           if (prime > 150)
               break;
           if (prime >= 100) {
               if (i++ != 0)
                   System.out.print(", ");
               System.out.print(prime);
           }
       }
       System.out.println();
       int count = 0;
       for (;;) {
           int prime = pgen.nextPrime();
           if (prime > 8000)
               break;
           if (prime >= 7700)
               ++count;
       }
       System.out.println("Number of primes between 7700 and 8000: " + count);
       int n = 10000;
       for (;;) {
           int prime = pgen.nextPrime();
           if (prime == 0) {
               System.out.println("Can't generate any more primes.");
               break;
           }
           if (pgen.count() == n) {
               System.out.println(n + "th prime: " + prime);
               n *= 10;
           }
       }
   }

}</lang>

Output:
First 20 primes:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71
Primes between 100 and 150:
101, 103, 107, 109, 113, 127, 131, 137, 139, 149
Number of primes between 7700 and 8000: 30
10000th prime: 104729
100000th prime: 1299709
1000000th prime: 15485863
10000000th prime: 179424673
100000000th prime: 2038074743
Can't generate any more primes.

JavaScript

primeGenerator(num, showPrimes) This function takes two arguments:

num is either an integer as a limit, or an array of two integers to present a range;

showPrimes is a boolean to indicate whether the result should be a list (if true) or a single number (if false).

Sounds a bit weird, but I hope it will be intelligible by the testing examples below. First the code:

<lang JavaScript>function primeGenerator(num, showPrimes) {

 var i,
     arr = [];
 function isPrime(num) {
   // try primes <= 16
   if (num <= 16) return (
     num == 2 || num == 3 || num == 5 || num == 7 || num == 11 || num == 13
   );
   // cull multiples of 2, 3, 5 or 7
   if (num % 2 == 0 || num % 3 == 0 || num % 5 == 0 || num % 7 == 0)
     return false;
   // cull square numbers ending in 1, 3, 7 or 9
   for (var i = 10; i * i <= num; i += 10) {
     if (num % (i + 1) == 0) return false;
     if (num % (i + 3) == 0) return false;
     if (num % (i + 7) == 0) return false;
     if (num % (i + 9) == 0) return false;
   }
   return true;
 }
 if (typeof num == "number") {
   for (i = 0; arr.length < num; i++) if (isPrime(i)) arr.push(i);
   // first x primes
   if (showPrimes) return arr;
   // xth prime
   else return arr.pop();
 }
 if (Array.isArray(num)) {
   for (i = num[0]; i <= num[1]; i++) if (isPrime(i)) arr.push(i);
   // primes between x .. y
   if (showPrimes) return arr;
   // number of primes between x .. y
   else return arr.length;
 }
 // throw a default error if nothing returned yet
 // (surrogate for a quite long and detailed try-catch-block anywhere before)
 throw("Invalid arguments for primeGenerator()");

}</lang>

Test <lang>// first 20 primes console.log(primeGenerator(20, true));

// primes between 100 and 150 console.log(primeGenerator([100, 150], true));

// numbers of primes between 7700 and 8000 console.log(primeGenerator([7700, 8000], false));

// the 10,000th prime console.log(primeGenerator(10000, false));</lang>

Output <lang>Array [ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 51, 59, 61, 67, 71 ]

Array [ 101, 103, 107, 109, 113, 127, 131, 137, 139, 149 ]

30

104729</lang>

jq

Works with: jq version 1.4

Recent versions of jq include extensive support for unbounded stream generators, but in this section, we present a solution to the tasks that should work with any version of jq from 1.4 onwards. That is, instead of using an unbounded generator of a stream of primes, the core of the approach adopted here is a function, "extend_primes", which can be applied recursively to generate arbitrarily many, or indefinitely many, primes, as illustrated by the function named "primes" below.

Preliminaries: <lang jq># Recent versions of jq include the following definition:

  1. until/2 loops until cond is satisfied,
  2. and emits the value satisfying the condition:

def until(cond; next):

 def _until:
   if cond then . else (next|_until) end;
 _until;

def count(cond): reduce .[] as $x (0; if $x|cond then .+1 else . end);</lang>

Prime numbers: <lang jq># Is the input integer a prime?

  1. "previous" must be the array of sorted primes greater than 1 up to (.|sqrt)

def is_prime(previous):

 . as $in
 | (previous|length) as $plength
 | [false, 0]   # state: [found, ix]
 | until( .[0] or .[1] >= $plength;
          [ ($in % previous[.[1]]) == 0, .[1] + 1] )
 | .[0] | not ;
  1. extend_primes expects its input to be an array consisting of
  2. previously found primes, in order, and extends that array:

def extend_primes:

 if . == null or length == 0 then [2]
 else . as $previous
 | if . == [2] then [2,3]
   else . + [(2 + .[length-1]) | until( is_prime($previous) ; . + 2)]
   end
 end;
  1. If . is an integer > 0 then produce an array of . primes;
  2. otherwise emit an unbounded stream of primes:

def primes:

 . as $n
 | if type == "number" and $n > 0 then
     null | until( length == $n; extend_primes )
   else [2] | recurse(extend_primes) | .[length - 1]
   end;
  1. Primes up to and possibly including n:

def primes_upto(n):

 until( .[length-1] > n; extend_primes )
 | if .[length-1] > n then .[0:length-1] else . end;</lang>

The tasks: The tasks are completed separately here to highlight the fact that by using "extend_primes", each task can be readily completed without generating unnecessarily many primes. <lang jq>"First 20 primes:", (20 | primes), "",

"Primes between 100 and 150:",

  (primes_upto(150) | map(select( 100 < .))), "",

"The 10,000th prime is \( 10000 | primes | .[length - 1] )", "",

(( primes_upto(8000) | count( . > 7700) | length) as $length

   | "There are \($length) primes twixt 7700 and 8000.")</lang>
Output:

<lang sh>$ jq -r -c -n -f Extensible_prime_generator.jq First 20 primes: [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71]

Primes between 100 and 150: [101,103,107,109,113,127,131,137,139,149]

The 10,000th prime is 104729

There are 30 primes twixt 7700 and 8000.</lang>

Julia

Julia's Primes package, included in the distribution, is exact up to 2^64 = 18446744073709551616. After that, Primes can use the BigInt data type, and then may use a probabalistic prime determination algorithm for such integers of arbitrarily large size. The probabilistic formula is tune-able, and by default determines primes with Knuth's recommended level for cryptography of an error less than (0.25)^25 = 8.881784197001252e-16, or 1 in 1125899906842624. <lang julia>using Primes

sum = 2 currentprime = 2 for i in 2:100000

   currentprime = nextprime(currentprime + 1)
   sum += currentprime

end println("The sum of the first 100,000 primes is $sum")

curprime = 1 arr = zeros(Int, 20) for i in 1:20

   curprime = nextprime(curprime + 1)
   arr[i] = curprime

end println("The first 20 primes are ", arr)

println("the primes between 100 and 150 are ", primes(100,150)) println("The number of primes between 7,700 and 8,000 is ", length(primes(7700, 8000))) println("The 10,000th prime is ", prime(10000))</lang>

Output:
The sum of the first 100,000 primes is 62260698721
The first 20 primes are [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]
the primes between 100 and 150 are [101, 103, 107, 109, 113, 127, 131, 137, 139, 149]
The number of primes between 7,700 and 8,000 is 30
The 10,000th prime is 104729

Alternative True Generator

The above code just uses the "Primes" package as a set of tools to solve the tasks. The following code creates a very simple generator using `isprime` inside a iterator and then uses that to solve the tasks: <lang julia>using Primes: isprime

PrimesGen() = Iterators.filter(isprime, Iterators.countfrom(Int64(2)))

print("Sum of first 100,000 primes: ") println(Iterators.sum(Iterators.take(PrimesGen(), 100000))) print("First 20 primes: ( ") foreach((p->print(p," ")), Iterators.take(PrimesGen(), 20)) println(")") print("Primes between 100 and 150: ( ") for p in Iterators.filter((p->p>=100), PrimesGen()) p > 150 && break; print(p, " ") end println(")") let cnt = 0

   for p in PrimesGen()
       p > 8000 && break; if p > 7700 cnt += 1 end
   end; println("Number of primes between 7700 and 8000:  ", cnt)

end println("The 10,000th prime: ", Iterators.first(Iterators.drop(PrimesGen(), 9999))) println()</lang>

This outputs:

Output:
Sum of first 100,000 primes:  62260698721
First 20 primes:  ( 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 )
Primes between 100 and 150:  ( 101 103 107 109 113 127 131 137 139 149 )
Number of primes between 7700 and 8000:  30
The 10,000th prime:  104729

To show it's speed, lets use it to solve the Euler Problem 10 of calculating the sum of the primes to two million as follows: <lang julia>using Printf: @printf @time let sm = 0

         for p in Iterators.filter(isprime, Iterators.countfrom(UInt64(2)))
             p > 2000000 && break
             sm += p
         end; @printf("%d\n", sm) end</lang>

which outputs:

Output:
142913828922
  0.783845 seconds (328.02 k allocations: 5.042 MiB)

As shown, this is of adequate speed for this smallish range; however it wouldn't be adequate to do the same for a range of two billion.

This is an "infinite" iterator whose range is limited by the size of `Int64`, but as it will take about 300 thousand years to get there, it isn't much of a concern.

An "infinite" iterator based on a bit-packed page-segmented Sieve of Eratosthenes

The above code is more than adequate to solve the trivial tasks as required here, but is really too slow for "industrial strength" tasks for ranges of billions. The following code uses the Page Segmented Algorithm from Sieve_of_Eratosthenes#Julia to solve the task: <lang julia>using Printf: @printf

print("Sum of first 100,000 primes: ") println(Iterators.sum(Iterators.take(PrimesPaged(), 100000))) print("First 20 primes: ( ") foreach((p->@printf("%d ", p)), Iterators.take(PrimesPaged(), 20)) println(")") print("Primes between 100 and 150: ( ") for p in Iterators.filter((p->p>=100), PrimesPaged()) p > 150 && break; @printf("%d ", p)) end println(")") let cnt = 0

   for p in PrimesPaged()
       p > 8000 && break; if p > 7700 cnt += 1 end
   end; println("Number of primes between 7700 and 8000:  ", cnt)

end @printf("The 10,000th prime: %d\n", Iterators.first(Iterators.drop(PrimesPaged(), 9999)))</lang>

to produce the same output much faster.

To show how much faster it is, doing the same Euler Problem 10 as follows: <lang julia>using Printf: @printf @time let sm = 0

         for p in PrimesPaged()
             p > 2000000 && break
             sm += p
         end; @printf("%d\n", sm) end</lang>

produces:

Output:
142913828922
  0.016826 seconds (60 allocations: 23.891 KiB)

showing it is as over 40 times faster, but it will definitely get even relatively faster with increasing range as this range is relatively trivial for page segmentation and there are more optimizations one can make.

This generator is also "infinite" to the `UInt64` range, but now will "only" take hundreds of years to get there.

Kotlin

Although we could use the java.math.BigInteger type to generate arbitrarily large primes, there is no need to do so here as the primes to be generated are well within the limits of the 4-byte Int type. ((workwith|Kotlin|version 1.3}} <lang scala>fun isPrime(n: Int) : Boolean {

   if (n < 2) return false
   if (n % 2 == 0) return n == 2
   if (n % 3 == 0) return n == 3
   var d : Int = 5
   while (d * d <= n) {
       if (n % d == 0) return false
       d += 2
       if (n % d == 0) return false
       d += 4
   }
   return true

}

fun generatePrimes() = sequence {

       yield(2)
       var p = 3
       while (p <= Int.MAX_VALUE) {
          if (isPrime(p)) yield(p)
          p += 2
       }
   }

fun main(args: Array<String>) {

   val primes = generatePrimes().take(10000) // generate first 10,000 primes
   println("First 20 primes : ${primes.take(20).toList()}")
   println("Primes between 100 and 150 : ${primes.filter { it in 100..150 }.toList()}")
   println("Number of primes between 7700 and 8000 = ${primes.filter { it in 7700..8000 }.count()}")
   println("10,000th prime = ${primes.last()}")

}</lang>

Output:
First 20 primes : [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]
Primes between 100 and 150 : [101, 103, 107, 109, 113, 127, 131, 137, 139, 149]
Number of primes between 7700 and 8000 = 30
10,000th prime = 104729

Alternate with better performance

Of all the current submissions on the page, the above code has got to be the worst. While it is adequate to solve the trivial tasks required by the page, it is a Trial Division Sieve and has O(n^3/2)) asymptotic performance over `n`, the range, where even a purely functional incremental sieve has O(n log n) performance. There is no need to consider BigInteger at all for a prime generator starting from the lowest to the Long number range, as it will never get there in less than 100's of years. Even though the above code implements a rudimentary wheel factorization, it will still be extremely slow as that only provides constant factor gains.

The following odds-only incremental Sieve of Eratosthenes generator has O(n log (log n)) performance due to using a (mutable) HashMap: <lang scala>fun primesHM(): Sequence<Int> = sequence {

   yield(2)
   fun oddprms(): Sequence<Int> = sequence {
       yield(3); yield(5) // need at least 2 for initialization
       val hm = HashMap<Int,Int>()
       hm.put(9, 6)
       val bps = oddprms().iterator(); bps.next(); bps.next() // skip past 5
       yieldAll(generateSequence(SieveState(7, 5, 25)) {
           ss ->
               var n = ss.n; var q = ss.q
               n += 2
               while ( n >= q || hm.containsKey(n)) {
                   if (n >= q) {
                       val inc = ss.bp shl 1
                       hm.put(n + inc, inc)
                       val bp = bps.next(); ss.bp = bp; q = bp * bp
                   }
                   else {
                       val inc = hm.remove(n)!!
                       var next = n + inc
                       while (hm.containsKey(next)) {
                           next += inc
                       }
                       hm.put(next, inc)
                   }
                   n += 2
               }
               ss.n = n; ss.q = q
               ss
       }.map { it.n })
   }
   yieldAll(oddprms())

}</Lang>

it is faster than the first example even though not using wheel factorization (other than odds-only) and rapidly pulls far ahead of it with increasing range such that it is usable to a range of 100 million in the order of 10 seconds.

Alternate with "industrial strength" performance

For ranges of a billion and more, one needs a sieve based on Page Segmented mutable arrays. The last code on the Sieve of Eratosthenes Task page at: Sieve_of_Eratosthenes#Unbounded_Versions_2 can do the job. When called with the following same `main` as the first example with `primesPaged()` substituted for `generatePrimes()` or `primesHM()`, it produces the same output.

It can count the primes to one billion in about 15 seconds on a slow tablet CPU (Intel x5-Z8350 at 1.92 Gigahertz) with the following code: <lang kotlin>primesPaged().takeWhile { it <= 1_000_000_000 }.count()</lang>

Further speed-ups can be achieved of about a factor of four with maximum wheel factorization and by multi-threading by the factor of the effective number of cores used, but there is little point when most of the execution time as a generator is spend iterating over the results.

In order to take advantage of those optimizations, one needs to write functions that work directly with the provided sequence of culled bit pages such as the provided `countPrimesTo` function does, which counts the primes without the iteration about three times as fast.

Lua

The modest requirements of this task allow for naive implementations, such as what follows. This generator does not even use its own list of primes to help determine subsequent primes! (though easily fixed) So, it it sufficient, but not efficient. <lang lua>local primegen = {

 count_limit = 2,
 value_limit = 3,
 primelist = { 2, 3 },
 nextgenvalue = 5,
 nextgendelta = 2,
 tbd = function(n)
   if n < 2 then return false end
   if n % 2 == 0 then return n==2 end
   if n % 3 == 0 then return n==3 end
   local limit = math.sqrt(n)
   for f = 5, limit, 6 do
     if n % f == 0 or n % (f+2) == 0 then return false end
   end
   return true
 end,
 needmore = function(self)
   return (self.count_limit ~= nil and #self.primelist < self.count_limit)
       or (self.value_limit ~= nil and self.nextgenvalue < self.value_limit)
 end,
 generate = function(self, count_limit, value_limit)
   self.count_limit = count_limit
   self.value_limit = value_limit
   while self:needmore() do
     if (self.tbd(self.nextgenvalue)) then
       self.primelist[#self.primelist+1] = self.nextgenvalue
     end
     self.nextgenvalue = self.nextgenvalue + self.nextgendelta
     self.nextgendelta = 6 - self.nextgendelta
   end
 end,
 filter = function(self, f)
   local list = {}
   for k,v in ipairs(self.primelist) do
     if (f(v)) then list[#list+1] = v end
   end
   return list
 end,

}

primegen:generate(20, nil) print("First 20 primes: " .. table.concat(primegen.primelist, ", "))

primegen:generate(nil, 150) print("Primes between 100 and 150: " .. table.concat(primegen:filter(function(v) return v>=100 and v<=150 end), ", "))

primegen:generate(nil, 8000) print("Number of primes between 7700 and 8000: " .. #primegen:filter(function(v) return v>=7700 and v<=8000 end))

primegen:generate(10000, nil) print("The 10,000th prime: " .. primegen.primelist[#primegen.primelist])

primegen:generate(100000, nil) print("The 100,000th prime: " .. primegen.primelist[#primegen.primelist])</lang>

Output:
First 20 primes:  2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71
Primes between 100 and 150:  101, 103, 107, 109, 113, 127, 131, 137, 139, 149
Number of primes between 7700 and 8000:  30
The 10,000th prime:  104729
The 100,000th prime:  1299709

Lingo

The following script implements a Sieve of Eratosthenes that is automatically extended when a method call needs a higher upper limit.

<lang Lingo>-- parent script "sieve" property _sieve


-- @constructor


on new (me)

   me._sieve = []
   me._primeSieve(100) -- arbitrary initial size of sieve
   return me

end


-- Returns sorted list of first n primes p with p >= a (default: a=1)


on getNPrimes (me, n, a)

   if voidP(a) then a = 1
   i = a
   res = []
   repeat while TRUE
       if i>me._sieve.count then me._primeSieve(2*i)
       if me._sieve[i] then res.add(i)
       if res.count=n then return res
       i = i +1
   end repeat

end


-- Returns sorted list of primes p with a <= p <= b


on getPrimesInRange (me, a, b)

   if me._sieve.count<b then me._primeSieve(b)
   primes = []
   repeat with i = a to b
       if me._sieve[i] then primes.add(i)
   end repeat
   return primes

end


-- Returns nth prime


on getNthPrime (me, n)

   if me._sieve.count<2*n then me._primeSieve(2*n)
   i = 0
   found = 0
   repeat while TRUE
       i = i +1
       if i>me._sieve.count then me._primeSieve(2*i)
       if me._sieve[i] then found=found+1
       if found=n then return i
   end repeat

end


-- Sieve of Eratosthenes


on _primeSieve (me, limit)

   if me._sieve.count>=limit then
       return
   else if me._sieve.count>0 then
       return me._complementSieve(limit)
   end if
   me._sieve = [0]
   repeat with i = 2 to limit
       me._sieve[i] = 1
   end repeat
   c = sqrt(limit)
   repeat with i = 2 to c
       if (me._sieve[i]=0) then next repeat
       j = i*i
       repeat while (j<=limit)
           me._sieve[j] = 0
           j = j + i
       end repeat
   end repeat

end


-- Expands existing sieve to new limit


on _complementSieve (me, n)

   n1 = me._sieve.count
   repeat with i = n1+1 to n
       me._sieve[i] = 1
   end repeat
   c1 = sqrt(n1)
   repeat with i = 2 to c1
       if (me._sieve[i]=0) then next repeat
       j = n1 - (n1 mod i)
       repeat while (j<=n)
           me._sieve[j] = 0
           j = j + i
       end repeat
   end repeat
   c = sqrt(n)
   repeat with i = c1+1 to c
       if (me._sieve[i]=0) then next repeat
       j = i*i
       repeat while (j<=n)
           me._sieve[j] = 0
           j = j + i
       end repeat
   end repeat

end</lang>

<lang Lingo>sieve = script("sieve").new() put "First twenty primes: " & sieve.getNPrimes(20) put "Primes between 100 and 150: "& sieve.getPrimesInRange(100, 150) put "Number of primes between 7,700 and 8,000: " & sieve.getPrimesInRange(7700, 8000).count put "The 10,000th prime: " & sieve.getNthPrime(10000)</lang>

Output:
-- "First twenty primes: [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]"
-- "Primes between 100 and 150: [101, 103, 107, 109, 113, 127, 131, 137, 139, 149]"
-- "Number of primes between 7,700 and 8,000: 30"
-- "The 10,000th prime: 104729"

M2000 Interpreter

The fancy way, using two lambda which shares closures (they are pointers, so lanbdas get copies of pointers, which is by value passing, and by reference too). Also I use my loved GOTO. Change 200th to 10000th and wait... A 5@ is a literal for Decimals (Variables are Variant types, but when they get a value they hold that value type. Array items or in other containers they, get whatever we want, anytime)

Version 2 Now we can make a bigger computation using Fast! mode which eliminate Gui/console refresh to gain speed. We can make a refresh each 50primes, so we have a tiny delay on refreshing if we move a window above M2000 console.

Also I change IsPrime to not add to Inventory Known1, because we want this inventory to have all primes without any missing until the last one.

I provide another IsPrime2 which use PrimeNth to add to Known1.

Inventories are reference type. Lambda functions are value type, but closures which are reference type copied the reference by value, so we get then by reference. Inventories start now with 3 known primes, and PrimeNth works for odd numbers only (see x+=2 before the loop statement)

Loop statement check a flag in a block of code ({ }), so when the block ends restart again (resetting the loop flag)

<lang M2000 Interpreter> Module CheckPrimes {

     \\ Inventories are lists, Known and Known1 are pointers to Inventories
     Inventory Known=1:=2@,2:=3@,3:=5@
     Inventory Known1=2@, 3@, 5@
     \\ In a lambda all closures are copies
     \\ but Known and Know1 are copies of pointers
     \\ so are closures like by reference
     PrimeNth=lambda  Known, Known1  (n as long) -> {
           if n<1 then Error "Only >=1"
           if exist(known, n) then =eval(known) : exit
           if n>5 then {
                i=len(known1) 
                x=eval(known1, i-1)+2
           } else  x=5 : i=2
           { 
                 if i=n then  =known(n) : exit
                 ok=false
                 if frac(x) then 1000
                 if frac(x/2) else 1000
                 if frac(x/3) else 1000
                 x1=sqrt(x) : d=5@
                 Repeat 
                       if frac(x/d ) else exit
                       d += 2: if d>x1 then ok=true : exit
                       if frac(x/d) else exit
                       d += 4: if d<= x1 else ok=true: exit
                  Always
     1000    If ok then i++:Append Known, i:=x  : if not exist(Known1, x) then Append Known1, x
                  x+=2 : Loop }
     }
     \\ IsPrime has same closure, Known1
     IsPrime=lambda  Known1 (x as decimal) -> {
           if exist(Known1, x) then =true : exit
           if Eval(Known1, len(Known1)-1)>x then exit
           if frac(x/2) else exit
           if frac(x/3) else exit
           x1=sqrt(x):d = 5@
           {if frac(x/d ) else exit
                 d += 2: if d>x1 then =true : exit
                 if frac(x/d) else exit
                 d += 4: if d<= x1 else =true: exit
                 loop
            }
     }
     \\ fill Known1, PrimeNth is a closure here
     IsPrime2=lambda  Known1, PrimeNth (x as decimal) -> {
           if exist(Known1, x) then =true : exit
           i=len(Known1)
           if Eval(Known1, i-1)>x then exit
           {
                 z=PrimeNth(i)
                 if z<x then loop else.if z=x then =true :exit
                 i++
           }
     }
     Print "First twenty primes"
     n=PrimeNth(20)
     For i=1 to 20  : Print Known(i),: Next i
     Print
     Print "Primes between 100 and 150:"
     c=0     
     For i=100 to 150
           If IsPrime2(i) Then print i, : c++
     Next i
     Print
     Print "Count:", c
     Print "Primes between 7700 and 8000:"
     c=0
     For i=7700  to 8000
           If IsPrime(i) Then print i, : c++
     Next i
     Print
     Print "Count:", c
     Print "200th Prime:"
     Print PrimeNth(200)
     Print "List from 190th to 199th Prime:"
     For i=190 to 199 : Print Known(i), : Next i
     Print
     Print "Wait"
     Refresh  ' because refresh happen on next Print, which take time
     ' using set fast! we get no respond from GUI/M2000 Console
     ' also Esc, Break and Ctrl+C not work
     ' we have to use Refresh each 500 primes to have one Refresh
     Set fast !
     for i=500 to 10000 step 50: m=PrimeNth(i): Print "."; :Refresh:Next i
     Print
     Print "10000th Prime:", PrimeNth(10000)
     ' reset speed to fast (there are three levels: slow/fast/fast!)
     set fast
     Print
     Rem 1 : Print Known
     Rem  2: Print Known1

} CheckPrimes </lang>

Output:
First twenty primes
 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
Primes between 100 and 150:
 101 103 107 109 113 127 131 137 139 149
Count: 10
Primes between 7700 and 8000:
 7703 7717 7723 7727 7741 7753 7757 7759 7789 7793 7817 7823 7829 7841 7853 7867 7873 7877 7879 7883 7901 7907 7919 7927 7933 7937 7949 7951 7963 7993
Count: 30
200th Prime: 
 1223
List from 190th to 199th Prime:
 1151 1153 1163 1171 1181 1187 1193 1201 1213 1217
Wait
.... (truncate for output)
10000th Prime: 104729

Code Optimization

We can drop ok variable from PrimeNth, using a second label. Statement Restart, restart the block. Labels are hashed when first time used.

I use same indentation so you can copy it at same position as in example above. A loop statement mark once the current block for restart after then last statement on block.

<lang M2000 Interpreter> PrimeNth=lambda Known, Known1 (n as long) -> {

     if n<1 then Error "Only >=1"
     if exist(known, n) then =eval(known) : exit
     if n>5 then {
          i=len(known1) 
          x=eval(known1, i-1)+2
     } else  x=5 : i=2
     { 
           if i=n then  =known(n) : exit
           if  frac(x) then 999
           if frac(x/2) else 999
           if frac(x/3) else 999
           x1=sqrt(x) : d=5@
           {if frac(x/d ) else 999
                 d += 2: if d>x1 then  1000
                 if frac(x/d) else 999
                 d += 4: if d<= x1 else 1000
                 loop
            }
999     x++ : Restart

1000 i++:Append Known, i:=x  : if not exist(Known1, x) then Append Known1, x

        x++ : Loop }

} </lang>

Mathematica / Wolfram Language

Prime and PrimePi use sparse caching and sieving. For large values, the Lagarias–Miller–Odlyzko algorithm for PrimePi is used, based on asymptotic estimates of the density of primes, and is inverted to give Prime. PrimeQ first tests for divisibility using small primes, then uses the Miller–Rabin strong pseudoprime test base 2 and base 3, and then uses a Lucas test. <lang Mathematica>Prime[Range[20]] {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71} Select[Range[100,150], PrimeQ] {101, 103, 107, 109, 113, 127, 131, 137, 139, 149} PrimePi[8000] - PrimePi[7700] 30 Prime[10000] 104729</lang>

Nim

For such trivial ranges as the task requirements or solving the Euler Problem 10 of summing the primes to two million, a basic generator such as the hash table based version from the Sieve_of_Eratosthenes#Nim_Unbounded_Versions section of the Sieve of Eratosthenes task will suffice, as follows: <lang nim>import tables

type PrimeType = int

proc primesHashTable(): iterator(): PrimeType {.closure.} =

 iterator output(): PrimeType {.closure.} =
   # some initial values to avoid race and reduce initializations...
   yield 2.PrimeType; yield 3.PrimeType; yield 5.PrimeType; yield 7.PrimeType
   var h = initTable[PrimeType,PrimeType]()
   var n = 9.PrimeType
   let bps = primesHashTable()
   var bp = bps() # advance past 2
   bp = bps(); var q = bp * bp # to initialize with 3
   while true:
     if n >= q:
       let inc = bp + bp
       h[n + inc] = inc
       bp = bps(); q = bp * bp
     elif h.hasKey(n):
       var inc: PrimeType
       discard h.take(n, inc)
       var nxt = n + inc
       while h.hasKey(nxt): nxt += inc # ensure no duplicates
       h[nxt] = inc
     else: yield n
     n += 2.PrimeType
 output

var num = 0 stdout.write "The first 20 primes are: " var iter = primesHashTable() for p in iter():

 if num >= 20: break else: stdout.write(p, " "); num += 1

echo "" stdout.write "The primes between 100 and 150 are: " iter = primesHashTable() for p in iter():

 if p >= 150: break
 if p >= 100: stdout.write(p, " ")

echo "" num = 0 iter = primesHashTable() for p in iter():

 if p > 8000: break
 if p >= 7700: num += 1

echo "The number of primes between 7700 and 8000 is: ", num num = 1 iter = primesHashTable() for p in iter():

 if num >= 10000:
   echo "The 10,000th prime is:  ", p
   break
 num += 1

var sum = 0 iter = primesHashTable() for p in iter():

 if p >= 2_000_000:
   echo "The sum of the primes to two million is:  ", sum
   break
 sum += p</lang>
Output:
The first 20 primes are:  2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
The primes between 100 and 150 are:  101 103 107 109 113 127 131 137 139 149
The number of primes between 7700 and 8000 is:  30
The 10,000th prime is:  104729
The sum of the primes to two million is:  142913828922

The code isn't particularly fast but more than adequate to run this trivial series of tasks and capable for ranges of a few tens of millions. It is limited by the maximum size of the `int` numeric range, but is too slow to ever reach that in a reasonable time, as in many minutes to even reach the 32-bit integer number range of over two billion let alone to reach the integer limit for 64-bit machines.

Alternate much faster Page Segmented Mutable Bit-Packed Seq version

The same Sieve of Eratosthenes Nim Unbounded Versions section as above includes a Page Segmented version at the end that can be used the same way for ranges of a billion in only a second (or something approaching a hundred times faster); this can be used for "industrial strength ranges" as in many billions in a reasonable time. Its limit is just the numeric range used, so for 64-bit prime times is about 2e10 in years.

To use that sieve, just substitute the following in the code: <lang nim>for p in primesPaged():</lang>

wherever the following is in the code: <lang nim>for p in iter():</lang> and one doesn't need the lines including `iter` just above these lines at all.

As noted in that section, using an extensible generator isn't the best choice for huge ranges as much higher efficiency can be obtained by using functions that deal directly with the composite culled packed-bit array representations directly as the `countPrimesTo` function does. This makes further improvements to the culling efficiency pointless, as even if one were to reduce the culling speed to zero, it still would take several seconds per range of a billion to enumerate the found primes as a generator.

PARI/GP

PARI includes a nice prime generator which is quite extensible: <lang c>void showprimes(GEN lower, GEN upper) {

 forprime_t T;
 if (!forprime_init(&T, a,b)) return;
 while(forprime_next(&T))
 {
   pari_printf("%Ps\n", T.pp);
 }

}</lang>

Most of these functions are already built into GP: <lang parigp>primes(20) primes([100,150])

  1. primes([7700,8000]) /* or */

s=0; forprime(p=7700,8000,s++); s prime(10000)</lang>

Pascal

Works with: Free Pascal

there is something wrong

http://www.primos.mat.br/Ate100G.html -> 
75. de 16639648367 a 16875026921
76. de 16875026963 a 17110593779
77. de 17110593791 a 17346308407
...
my unit: 
 750000000 16875026921
 760000000 17110593779
 770000000 17346251243 <----Wrong

Limited to 2.7e14. Much faster than the other Version. 94s to 257 s for the primes 2..1E11 First the unit.Still a work in progress.

versus primesieve no chance at all, even single threaded 5x faster :-) ./primesieve -v primesieve 7.2, <https://primesieve.org> Copyright (C) 2010 - 2018 Kim Walisch ./primesieve -t1 100000071680 Sieve size = 256 KiB Threads = 1 100% Seconds: 19.891 Primes: 4118057696 <lang pascal> unit primsieve; //{$O+,R+} {$IFDEF FPC}

 {$MODE objFPC}
 {$CODEALIGN proc=32,loop=1}

{$IFEND} {segmented sieve of Erathostenes using only odd numbers} {using presieved sieve of small primes, to reduce the most time consuming} interface

 procedure InitPrime;
 procedure NextSieve;
 function SieveStart:Uint64;
 function SieveSize :LongInt;
 function Nextprime: Uint64;
 function TotalCount :Uint64;
 function PosOfPrime: Uint64;

implementation uses

 sysutils;

const

 smlPrimes :array [0..10] of Byte = (2,3,5,7,11,13,17,19,23,29,31);
 maxPreSievePrimeNum = 7;
 maxPreSievePrime = 17;//smlPrimes[maxPreSievePrimeNum];
 cSieveSize = 16384 * 4; //<= High(Word)+1 // Level I Data Cache

type

 tSievePrim = record
                svdeltaPrime:word;//diff between actual and new prime
                svSivOfs:word;    //Offset in sieve
                svSivNum:LongWord;//1 shl (1+16+32) = 5.6e14
              end;
 tpSievePrim = ^tSievePrim;

var //sieved with primes 3..maxPreSievePrime.here about 255255 Byte {$ALIGN 32}

 preSieve :array[0..3*5*7*11*13*17-1] of Byte;//must be > cSieveSize

{$ALIGN 32}

 Sieve :array[0..cSieveSize-1] of Byte;

{$ALIGN 32} //prime = FoundPrimesOffset + 2*FoundPrimes[0..FoundPrimesCnt]

 FoundPrimes : array[0..12252] of Word;

{$ALIGN 32}

 sievePrimes : array[0..78498] of tSievePrim;// 1e6^2 ->1e12

//sievePrimes : array[0..1077863] of tSievePrim;// maximum 1e14

 FoundPrimesOffset : Uint64;
 FoundPrimesCnt,
 FoundPrimesIdx,
 FoundPrimesTotal,
 SieveNum,
 SieveMaxIdx,
 preSieveOffset,
 LastInsertedSievePrime :NativeUInt;

procedure CopyPreSieveInSieve; forward; procedure CollectPrimes; forward; procedure sieveOneSieve; forward; procedure Init0Sieve; forward; procedure SieveOneBlock; forward;

//**************************************** procedure preSieveInit; var

 i,pr,j,umf : NativeInt;

Begin

 fillchar(preSieve[0],SizeOf(preSieve),#1);
 i := 1;
 pr := 3;// starts with pr = 3
 umf := 1;
 repeat
   IF preSieve[i] =1 then
   Begin
     pr := 2*i+1;
     j := i;
     repeat
       preSieve[j] := 0;
       inc(j,pr);
     until j> High(preSieve);
     umf := umf*pr;
   end;
   inc(i);
 until (pr = maxPreSievePrime)OR(umf>High(preSieve)) ;
 preSieveOffset := 0;

end;

function InsertSievePrimes(PrimPos:NativeInt):NativeInt; var

 j    :NativeUINt;
 i,pr : NativeUInt;

begin

 i := 0;
 //ignore first primes already sieved with
 if SieveNum = 0 then
   i := maxPreSievePrimeNum;
 pr :=0;
 j := Uint64(SieveNum)*cSieveSize*2-LastInsertedSievePrime;
 with sievePrimes[PrimPos] do
 Begin
   pr := FoundPrimes[i]*2+1;
   svdeltaPrime := pr+j;
   j := pr;
 end;
 inc(PrimPos);
 for i := i+1 to FoundPrimesCnt-1 do
 Begin
   IF PrimPos > High(sievePrimes) then
     BREAK;
   with sievePrimes[PrimPos] do
   Begin
     pr := FoundPrimes[i]*2+1;
     svdeltaPrime := (pr-j);
     j := pr;
   end;
   inc(PrimPos);
 end;
 LastInsertedSievePrime :=Uint64(SieveNum)*cSieveSize*2+pr;
 result := PrimPos;

end;

procedure CalcSievePrimOfs(lmt:NativeUint); //lmt High(sievePrimes) var

 i,pr : NativeUInt;
 sq : Uint64;

begin

 pr := 0;
 i := 0;
 repeat
   with sievePrimes[i] do
   Begin
     pr := pr+svdeltaPrime;
     IF sqr(pr)  < (cSieveSize*2) then
     Begin
       svSivNum := 0;
       svSivOfs := (pr*pr-1) DIV 2;
     end
     else
     Begin
       SieveMaxIdx := i;
       pr := pr-svdeltaPrime;
       BREAK;
     end;
   end;
   inc(i);
 until i > lmt;
 for i := i to lmt do
 begin
   with sievePrimes[i] do
   Begin
     pr := pr+svdeltaPrime;
     sq := sqr(pr);
     svSivNum := sq DIV (2*cSieveSize);
     svSivOfs := ( (sq - Uint64(svSivNum)*(2*cSieveSize))-1)DIV 2;
   end;
 end;

end;

procedure sievePrimesInit; var

 i,j,pr,PrimPos:NativeInt;

Begin

 LastInsertedSievePrime := 0;
 preSieveOffset := 0;
 SieveNum :=0;
 CopyPreSieveInSieve;
 //normal sieving of first sieve
 i := 1; // start with 3
 repeat
   while Sieve[i] = 0 do
     inc(i);
   pr := 2*i+1;
   inc(i);
   j := ((pr*pr)-1) DIV 2;
   if j > High(Sieve) then
     BREAK;
   repeat
     Sieve[j] := 0;
     inc(j,pr);
   until j > High(Sieve);
 until false;
 CollectPrimes;
 PrimPos := InsertSievePrimes(0);
 LastInsertedSievePrime := FoundPrimes[PrimPos]*2+1;
 IF PrimPos < High(sievePrimes) then
 Begin
   Init0Sieve;
   //Erste Sieb nochmals, aber ohne Eintrag
   sieveOneBlock;
   repeat
     sieveOneBlock;
     dec(SieveNum);
     PrimPos := InsertSievePrimes(PrimPos);
     inc(SieveNum);
  until PrimPos > High(sievePrimes);
 end;
 Init0Sieve;

end;

procedure Init0Sieve; begin

 FoundPrimesTotal :=0;
 preSieveOffset := 0;
 SieveNum :=0;
 CalcSievePrimOfs(High(sievePrimes));

end;

procedure CopyPreSieveInSieve; var

 lmt : NativeInt;

Begin

 lmt := preSieveOffset+cSieveSize;
 lmt := lmt-(High(preSieve)+1);
 IF lmt<= 0 then
 begin
   Move(preSieve[preSieveOffset],Sieve[0],cSieveSize);
   if lmt <> 0 then
     inc(preSieveOffset,cSieveSize)
   else
     preSieveOffset := 0;
 end
 else
 begin
   Move(preSieve[preSieveOffset],Sieve[0],cSieveSize-lmt);
   Move(preSieve[0],Sieve[cSieveSize-lmt],lmt);
   preSieveOffset := lmt
 end;

end;

procedure sieveOneSieve; var

 sp:tpSievePrim;
 pSieve :pByte;
 i,j,pr,sn,dSievNum :NativeUint;

Begin

 pr := 0;
 sn := sieveNum;
 sp := @sievePrimes[0];
 pSieve := @Sieve[0];
 For i := SieveMaxIdx downto 0 do
   with sp^ do
   begin
     pr := pr+svdeltaPrime;
     IF svSivNum = sn then
     Begin
       j := svSivOfs;
       repeat
         pSieve[j] := 0;
         inc(j,pr);
       until j > High(Sieve);
       dSievNum := j DIV cSieveSize;
       svSivOfs := j-dSievNum*cSieveSize;
       svSivNum := sn+dSievNum;

// svSivNum := svSivNum+dSievNum;

     end;
     inc(sp);
   end;
 i := SieveMaxIdx+1;
 repeat
   if i > High(SievePrimes) then
     BREAK;
   with sp^ do
   begin
     if svSivNum > sn then
     Begin
       SieveMaxIdx := I-1;
       Break;
     end;
     pr := pr+svdeltaPrime;
     j := svSivOfs;
     repeat
       Sieve[j] := 0;
       inc(j,pr);
     until j > High(Sieve);
     dSievNum := j DIV cSieveSize;
     svSivOfs := j-dSievNum*cSieveSize;
     svSivNum := sn+dSievNum;
   end;
   inc(i);
   inc(sp);
 until false;

end;

procedure CollectPrimes; //extract primes to FoundPrimes // var

  pSieve : pbyte;
  pFound : pWord;
  i,idx : NativeUint;

Begin

 FoundPrimesOffset := SieveNum*2*cSieveSize;
 FoundPrimesIdx := 0;
 pFound :=@FoundPrimes[0];
 i := 0;
 idx := 0;
 IF SieveNum = 0 then
 //include small primes used to pre-sieve
 Begin
   repeat
     pFound[idx]:= (smlPrimes[idx]-1) DIV 2;
     inc(idx);
   until smlPrimes[idx]>maxPreSievePrime;
   i := (smlPrimes[idx] -1) DIV 2;
 end;
 //grabbing the primes without if then -> reduces time extremly
 //primes are born to let branch-prediction fail.
 pSieve:= @Sieve[Low(Sieve)];
 repeat
   //store every value until a prime aka 1 is found
   pFound[idx]:= i;
   inc(idx,pSieve[i]);
   inc(i);
 until i>High(Sieve);
 FoundPrimesCnt:= idx;
 inc(FoundPrimesTotal,Idx);

end;

procedure SieveOneBlock; begin

 CopyPreSieveInSieve;
 sieveOneSieve;
 CollectPrimes;
 inc(SieveNum);

end;

procedure NextSieve; Begin

 SieveOneBlock;

end;

function Nextprime:Uint64; Begin

 result := FoundPrimes[FoundPrimesIdx]*2+1+FoundPrimesOffset;
 if (FoundPrimesIdx=0) AND (sievenum = 1) then
   inc(result);
 inc(FoundPrimesIdx);
 If FoundPrimesIdx>= FoundPrimesCnt then
   SieveOneBlock;

end;

function PosOfPrime: Uint64; Begin

 result := FoundPrimesTotal-FoundPrimesCnt+FoundPrimesIdx;

end;

function TotalCount :Uint64; begin

 result := FoundPrimesTotal;

end;

function SieveSize :LongInt; Begin

 result := 2*cSieveSize;

end;

function SieveStart:Uint64; Begin

 result := (SieveNum-1)*2*cSieveSize;

end;

procedure InitPrime; Begin

 Init0Sieve;
 SieveOneBlock;

end;

begin

 preSieveInit;
 sievePrimesInit;
 InitPrime;

end. {compiler: fpc/3.2.0/ppcx64 -Xs -O4 "%f" 50851378 in 1000079360 dauerte 529 ms 455052800 in 10000007168 dauerte 6297 ms 4118057696 in 100000071680 dauerte 93783 ms }</lang>

the test program

<lang pascal> program test; uses

 primsieve;

var

 i : NativeInt;
 cnt : Uint64;

Begin

 writeln('First 25 primes');
 For i := 1 to 25 do
   write(Nextprime:3);
 writeln;

Writeln;

 writeln('Primes betwenn 100 and 150');
 repeat
   i := NextPrime
 until i > 100;
 repeat
   write(i:4);
   i := NextPrime;
 until i>150;
 writeln;

Writeln;

 repeat
   i := NextPrime
 until i > 7700;
 cnt := 0;
 repeat
   inc(cnt);
   i := NextPrime;
 until i> 8000;
 writeln('between 7700 and 8000 are ',cnt,' primes');

Writeln;

 writeln('      i.th       prime');
 cnt := 10000;
 repeat
   while TotalCount < cnt do
     NextSieve;
   repeat
     i := NextPrime;
   until PosOfPrime = cnt;
   writeln(cnt:10,i:12);
   cnt := cnt*10;
 until cnt >100*1000*1000;

end.</lang>

output
First 25 primes
  2  3  5  7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

Primes betwenn 100 and 150
 101 103 107 109 113 127 131 137 139 149

between 7700 and 8000 are 30 primes

      i.th       prime
     10000      104729
    100000     1299709
   1000000    15485863
  10000000   179424673
 100000000  2038074743

real	0m1,121s


alternative

The main intention is the use in http://rosettacode.org/wiki/Emirp_primes. The speed is about 3x times slower than sieve of Atkin.About 13 secs for 10 billion/146 secs for 100 billion in 64-Bit. But i can hold all primes til 1e11 in 2.5 Gb memory.Test for isEmirp inserted. 32-bit is slow doing 64-Bit math.Using a dynamic array is slow too in NextPrime. <lang pascal>program emirp; {$IFDEF FPC}

 {$MODE DELPHI}
 {$OPTIMIZATION ON,REGVAR,PEEPHOLE,CSE,ASMCSE}
 {$CODEALIGN proc=8}

// {$R+,V+,O+} {$ELSE}

 {$APPLICATION CONSOLE}

{$ENDIF} uses

 sysutils;

type

 tSievenum      = NativeUint;

const

 cBitSize       = SizeOf(tSievenum)*8;
 cAndMask       = cBitSize-1;
 InitPrim      :array [0..9] of byte = (2,3,5,7,11,13,17,19,23,29);

(*

 {MAXANZAHL     =  2*3*5*7*11*13*17*19;*PRIM}
 MAXANZAHL     :array [0..8] of Longint =(2,6,30,210,2310,30030,
                                        510510,9699690,223092870);
 {WIFEMAXLAENGE =  1*2*4*6*10*12*16*18; *(PRIM-1)}
 WIFEMAXLAENGE :array [0..8] of longint =(1,2,8,48,480,5760,
                                        92160,1658880,36495360);
  • )

//Don't sieve with primes that are multiples of 2..InitPrim[BIS]

 BIS           =     5;
 MaxMulFac     =    22; {array [0..9] of byte= (2,4,6,10,14,22,26,34,40,50);}
 cMaxZahl      = 30030;
 cRepFldLen    =  5760;

 MaxUpperLimit =  100*1000*1000*1000-1;

 MAXIMUM       = ((MaxUpperLimit-1) DIV cMaxZahl+1)*cMaxZahl;
 MAXSUCHE      = (((MAXIMUM-1) div cMaxZahl+1)*cRepFldLen-1)
                   DIV cBitSize;

type

 tRpFldIdx  = 0..cRepFldLen-1;
 pNativeUint = ^ NativeUint;
 (* numberField as Bit array *)
 tsearchFld   = array of tSievenum;

 tSegment       = record
                    dOfs,
                    dSegment    :tSievenum;
                 end;
 tpSegment     = ^tSegment;
 tMulFeld    = array [0..MaxMulFac shr 1 -1] of tSegment;
 tnumberField= array [0..cMaxZahl-1] of word; //word->  0..cRepFldLen-1
 tRevIdx     = array [tRpFldIdx] of word;//word->  0..cMaxZahl-1
 tDiffFeld   = array [tRpFldIdx] of byte;
 tNewPosFeld = array [tRpFldIdx] of Uint64;

 tRecPrime   = record
                 rpPrime,
                 rpsvPos : Uint64;
                 rpOfs,
                 rpSeg   :LongWord;
               end;

var

 BitSet,
 BitClr : Array [0..cAndMask] Of NativeUint;
 deltaNewPos : tNewPosFeld;
 MulFeld   : tMulFeld;
 searchFld : tsearchFld;
 number    : tnumberField;
 DiffFld   : tDiffFeld;
 RevIdx    : tRevIdx;
 actSquare   : Uint64;
 NewStartPos,
 MaxPos    : Uint64;

const //K1 = $0101010101010101;

 K55 = $5555555555555555;
 K33 = $3333333333333333;
 KF1 = $0F0F0F0F0F0F0F0F;
 KF2 = $00FF00FF00FF00FF;
 KF4 = $0000FFFF0000FFFF;
 KF8 = $00000000FFFFFFFF;

function popcnt(n:Uint64):integer;overload;inline; var

 c,b,k : NativeUint;

begin

 b := n;
 k := NativeUint(K55);c :=  (b shr  1) AND k; b := (b AND k)+C;
 k := NativeUint(K33);c := ((b shr  2) AND k);b := (b AND k)+C;
 k := NativeUint(KF1);c := ((b shr  4) AND k);b := (b AND k)+c;
 k := NativeUint(KF2);c := ((b shr  8) AND k);b := (b AND k)+c;
 k := NativeUint(KF4);c := ((b shr 16) AND k);b := (b AND k)+c;
 k := NativeUint(KF8);c :=  (b shr 32)+(b AND k);
 result := c;

end;

function popcnt(n:LongWord):integer;overload; var

 c,k : LongWord;

begin

 result  := n;
 IF result = 0 then
   EXIT;
 k := LongWord(K55);c :=  (result  shr  1) AND k; result  := (result  AND k)+C;
 k := LongWord(K33);c := ((result  shr  2) AND k);result  := (result  AND k)+C;
 k := LongWord(KF1);c := ((result  shr  4) AND k);result  := (result  AND k)+c;
 k := LongWord(KF2);c := ((result  shr  8) AND k);result  := (result  AND k)+c;
 k := LongWord(KF4);
 result :=  (result  shr 16) AND k +(result  AND k);

end;

procedure Init; {simple sieve of erathosthenes only eliminating small primes} var

 pr,i,j,Ofs : NativeUint;

Begin

 //Init Bitmasks
 j := 1;
 For i := 0 to cAndMask do
 Begin
   BitSet[i] := J;
   BitClr[i] := NativeUint(NOT(J));
   j:= j+j;
 end;
 //building number wheel excluding multiples of small primes
 Fillchar(number,SizeOf(number),#0);
 For i := 0 to BIS do
 Begin
   pr := InitPrim[i];
   j := (High(number) div pr)*pr;
   repeat
     number[j] := 1;
     dec(j,pr);
   until j <= 0;
 end;

 // build reverse Index and save distances
 i := 1;
 j := 0;
 RevIdx[0]:= 1;
 repeat
   Ofs :=0;
   repeat
     inc(i);
     inc(ofs);
   until number[i] = 0;
   DiffFld[j] := ofs;
   inc(j);
   RevIdx[j] := i;
 until i = High(number);
 DiffFld[j] := 2;

 //calculate a bitnumber-index into cRepFldLen
 Fillchar(number,SizeOf(number),#0);
 Ofs := 1;
 for i := 0 to cRepFldLen-2 do
 begin
   inc(Ofs,DiffFld[i]);
   number[ofs] := i+1;
 end;

 //direct index into Mulfeld 2->0 ,4-> 1 ...
 For i := 0 to cRepFldLen-1 do
 Begin
   j := (DiffFld[i] shr 1) -1;
   DiffFld[i] := j;
 end;

end;

function CalcPos(m: Uint64): Uint64; {search right position of m} var

 i,res : NativeUint;

Begin

 res := m div cMaxZahl;
 i   := m-res* Uint64(cMaxzahl);//m mod cMaxZahl
 while (number[i]= 0) and (i <>1) do
 begin
   iF i = 0 THEN
   begin
     Dec(res,cRepFldLen);
     i := cMaxzahl;
   end;
   dec(i);
 end; {while}
 CalcPos := res *Uint64(cRepFldLen) +number[i];

end;

procedure CalcSqrOfs(out Segment,Ofs :Uint64); Begin

 Segment  := actSquare div cMaxZahl;
 Ofs      := actSquare-Segment*cMaxZahl; //ofs Mod cMaxZahl
 Segment  := Segment*cRepFldLen;

end;

procedure MulTab(sievePr:Nativeint); var

k,Segment,Segment0,Rest,Rest0: NativeUint;

Begin

 {multiplication-table of differences}
 {2* sievePr,4* ,6* ...MaxMulFac*sievePr }
 sievePr := sievePr+sievePr;
 Segment0 := sievePr div cMaxzahl;

 Rest0    := sievePr-Segment0*cMaxzahl;
 Segment0 := Segment0 * cRepFldLen;

 Segment := Segment0;
 Rest := Rest0;

 with MulFeld[0] do
 begin
   dOfs := Rest0;
   dSegment:= Segment0;
 end;

 for k := 1 to MaxMulFac shr 1-1 do
 begin
   Segment := Segment+Segment0;
   Rest    := Rest+Rest0;
   IF Rest >= cMaxzahl then
   Begin
     Rest:= Rest-cMaxzahl;
     Segment := Segment+cRepFldLen;
   end;
   with MulFeld[k] do
   begin
     dOfs := Rest;
     dSegment:= Segment;
   end;
 end;

end;

procedure CalcDeltaNewPos(sievePr,MulPos:NativeUint); var

 Ofs,Segment,prevPos,actPos : Uint64;
 i: NativeInt;

Begin

 MulTab(sievePr);
 //start at sqr sievePrime
 CalcSqrOfs(Segment,Ofs);
 NewStartPos := Segment+number[Ofs];
 prevPos := NewStartPos;
 deltaNewPos[0]:= prevPos;
 For i := 0 to cRepFldLen-2 do
 begin
   inc(mulpos);
   IF mulpos >= cRepFldLen then
     mulpos := 0;
   With MulFeld[DiffFld[mulpos]] do
   begin
     Ofs:= Ofs+dOfs;
     Segment := Segment+dSegment;
   end;
   If Ofs >= cMaxZahl then
   begin
     Ofs := Ofs-cMaxZahl;
     Segment := Segment+cRepFldLen;
   end;
   actPos := Segment+number[Ofs];
   deltaNewPos[i]:= actPos - prevPos;
   IF actPos> maxPos then
     BREAK;

   prevPos := actPos;
 end;
 deltaNewPos[cRepFldLen-1] := NewStartPos+cRepFldLen*sievePr-prevPos;

end;

procedure SieveByOnePrime(var sf:tsearchFld;sievePr:NativeUint); var

 pNewPos : ^Uint64;
 pSiev0,
 pSiev   : ^tSievenum;// dynamic arrays are slow
 Ofs      : Int64;
 Position : UINt64;
 i: NativeInt;

Begin

 pSiev0 := @sf[0];
 Ofs := MaxPos-sievePr *cRepFldLen;
 Position := NewStartPos;
 {unmark multiples of sieve prime}
 repeat
   IF Position < Ofs then
   Begin
     pNewPos:= @deltaNewPos[0];
     For i := Low(deltaNewPos) to High(deltaNewPos) do
     Begin
       pSiev := pSiev0;
       inc(pSiev,Position DIV cBitSize);
       //pSiev^ == @sf[Position DIV cBitSize]
       pSiev^ := pSiev^ AND BitCLR[Position AND cAndMask];
       inc(Position,pNewPos^);
       inc(pNewPos);
     end
   end
   else
   Begin
     pNewPos:= @deltaNewPos[0];
     For i := Low(deltaNewPos) to High(deltaNewPos) do
     Begin
       IF Position >= MaxPos then
         Break;
       pSiev := pSiev0;
       inc(pSiev,Position DIV cBitSize);
       pSiev^ := pSiev^ AND BitCLR[Position AND cAndMask];
       inc(Position,pNewPos^);
       inc(pNewPos);
     end
   end;
 until Position >= MaxPos;

end;

procedure SieveAll; var

 i,
 sievePr,
 PrimPos,
 srPrPos  : NativeUint;

Begin

 Init;
 MaxPos := CalcPos(MaxUpperLimit);
 {start of prime sieving}
 i := (MaxPos-1) DIV cBitSize+1;
 setlength(searchFld,i);
 IF Length(searchFld) <> i then
 Begin
   writeln('Not enough memory');
   Halt(-227);
 end;
 For i := High(searchFld) downto 0 do
    searchFld[i] := NativeUint(-1);
 {the first prime}
 srPrPos := 0;
 PrimPos := 0;
 sievePr := 1;
 actSquare := sievePr;
 repeat
   {next prime}
   inc(srPrPos);
   i := 2*(DiffFld[PrimPos]+1);
   //binom (a+b)^2; a^2 already known
   actSquare := actSquare+(2*sievePr+i)*i;
   inc(sievePr,i);

   IF actSquare > MaxUpperLimit THEN
     BREAK;
   {if sievePr == prime then sieve with sievePr}
   if BitSet[srPrPos AND cAndMask] AND
     searchFld[srPrPos DIV cBitSize] <> 0then
   Begin
     write(sievePr:8,#8#8#8#8#8#8#8#8);
     CalcDeltaNewPos(sievePr,PrimPos);
     SieveByOnePrime(searchFld,sievePr);
   end;
   inc(PrimPos);
   if PrimPos = cRepFldLen then
     dec(PrimPos,PrimPos);// := 0;
 until false;

end;

function InitRecPrime(pr: UInt64):tRecPrime; var

 svPos,sg : NativeUint;

Begin

 svPos := CalcPos(pr);
 sg := svPos DIV cRepFldLen;
 with result do
 Begin
   rpsvPos := svPos;
   rpSeg   := sg;
   rpOfs   := svPos - sg*cRepFldLen;
   rpPrime := RevIdx[rpOfs]+ sg*cMaxZahl;
 end;

end;

function InitPrimeSvPos(svPos: Uint64):tRecPrime; var

 sg : LongWord;

Begin

 sg := svPos DIV cRepFldLen;
 with result do
 Begin
   rpsvPos := svPos;
   rpSeg   := sg;
   rpOfs   := svPos - sg*cRepFldLen;
   rpPrime := RevIdx[rpOfs]+ sg*cMaxZahl;
 end;

end;

function NextPrime(var pr: tRecPrime):Boolean; var

 ofs : LongWord;
 svPos : Uint64;

Begin

 with pr do
 Begin
   svPos := rpsvPos;
   Ofs := rpOfs;
   repeat
     inc(svPos);
     if svPos > MaxPos then
     Begin
       result := false;
       EXIT;
     end;
     inc(Ofs);
     IF Ofs >= cRepFldLen then
     Begin
       ofs := 0;
       inc(rpSeg);
     end;
   until BitSet[svPos AND cAndMask] AND
     searchFld[svPos DIV cBitSize] <> 0;
   rpPrime := rpSeg*Uint64(cMaxZahl)+RevIdx[Ofs];
   rpSvPos := svPos;
   rpOfs := Ofs;
 end;
 result := true;

end;

function GetNthPrime(n: Uint64):tRecPrime; var

 i : longWord;
 cnt: Uint64;

Begin

 IF n > MaxPos then
   EXIT;

 i := 0;
 cnt := Bis;
 For i := 0 to n DIV cBitSize do
   inc(cnt,PopCnt(NativeUint(searchFld[i])));
 i := n DIV cBitSize+1;

 while cnt < n do
 Begin
   inc(cnt,PopCnt(NativeUint(searchFld[i])));
   inc(i);
 end;
 dec(i);

 dec(cnt,PopCnt(NativeUint(searchFld[i])));
 result := InitPrimeSvPos(i*Uint64(cBitSize)-1);
 while cnt < n do
   IF NextPrime(Result) then
     inc(cnt)
   else
     Break;

end;

procedure ShowPrimes(loLmt,HiLmt: NativeInt); var

 p1 :tRecPrime;

Begin

 IF HiLmt < loLmt then
   exit;
 p1 := InitRecPrime(loLmt);
 while p1.rpPrime < LoLmt do
   IF Not(NextPrime(p1)) Then
     EXIT;

 repeat
   write(p1.rpPrime,' ');
   IF Not(NextPrime(p1)) Then
     Break;
 until p1.rpPrime > HiLmt;
 writeln;

end;

function CountPrimes(loLmt,HiLmt: NativeInt):LongWord; var

 p1 :tRecPrime;

Begin

 result := 0;
 IF HiLmt < loLmt then
   exit;
 p1 := InitRecPrime(loLmt);
 while p1.rpPrime < LoLmt do
   IF Not(NextPrime(p1)) Then
     EXIT;
 repeat
   inc(result);
   IF Not(NextPrime(p1)) Then
     Break;
 until p1.rpPrime > HiLmt;

end;

procedure WriteCntSmallPrimes(n: NativeInt); var

 i, p,prPos,svPos : nativeUint;

Begin

 dec(n);
 IF n < 0 then
   EXIT;
 write('First ',n+1,' primes ');
 IF n < Bis then
 Begin
   For i := 0 to n do
     write(InitPrim[i]:3);
 end
 else
 Begin
   For i := 0 to BIS do
     write(InitPrim[i],' ');
   dec(n,Bis);

   svPos := 0;
   PrPos := 0;
   p     := 1;
   while n> 0 do
   Begin
     {next prime}
     inc(svPos);
     inc(p,2*(DiffFld[prPos]+1));
     if BitSet[svPos AND cAndMask] AND searchFld[svPos DIV cBitSize] <>0 then
     Begin
       write(p,' ');
       dec(n);
     end;
     inc(prPos);
     if prPos = cRepFldLen then
       dec(prPos,prPos);// := 0;
   end;
 end;
 writeln;

end;

function RvsNumL(var n: Uint64):Uint64; //reverse and last digit, most of the time n > base therefor repeat const

 base = 10;

var

 q, c: Int64;

Begin

 result := n;
 q := 0;
 repeat
   c:= result div Base;
   q := result+ (q-c)*Base;
   result := c;
 until result < Base;
 n := q*Base+result;

end;

function IsEmirp(n:Uint64):boolean; var

lastDgt:NativeUint;
ofs: NativeUint;
seg : Uint64;

Begin

 seg := n;
 lastDgt:= RvsNumL(n);
 result:= false;
 IF (seg = n) OR (n> MaxUpperLimit) then
   EXIT;

 IF lastDgt in [1,3,7,9] then
 Begin
   seg := n div cMaxZahl;
   ofs := n-seg* cMaxzahl;//m mod cMaxZahl
   IF (Number[ofs] <> 0) OR (ofs=1) then
   begin
     seg := seg *cRepFldLen+number[ofs];
     result := BitSet[seg AND cAndMask]  AND searchFld[seg DIV cBitSize] <> 0;
   end
 end;

end;

function GetEmirps(loLmt,HiLmt: Uint64):NativeInt; var

 p1 :tRecPrime;

Begin

 result := 0; 
 IF HiLmt < loLmt then
   exit;
 IF loLmt > MaxUpperLimit then
   Exit;
 IF HiLmt > MaxUpperLimit then
   HiLmt := MaxUpperLimit;
 p1 := InitRecPrime(loLmt);
 while p1.rpPrime < LoLmt do
   IF Not(NextPrime(p1)) Then
     EXIT;

 repeat
   if isEmirp(p1.rpPrime) then
     inc(result);
   iF not(NextPrime(p1)) then
     BREAK;
 until p1.rpPrime > HiLmt;

end;

var

 T1,T0: TDateTime;
 Anzahl :Uint64;
 i,j,dgtCnt,totalCnt : Uint64;
 n : LongInt;

Begin

 T0 := now;
 SieveAll;
 T1 := now;
 writeln('         ');
 Writeln('time for sieving ',FormatDateTime('NN:SS.ZZZ',T1-T0));
 Anzahl := BIS;
 For n := MaxPos DIV cBitSize-1 downto 0 do
   inc(Anzahl,PopCnt(NativeUint(searchFld[n])));
 n := MaxPos AND cAndMask;
 IF n >0 then
 Begin
   dec(n);
   repeat
     IF BitSet[n] AND searchFld[MaxPos DIV cBitSize] <> 0 then
       inc(Anzahl);
     dec(n);
   until n< 0;
 end;

 Writeln('there are ',Anzahl,' primes til ',MaxUpperLimit);
 WriteCntSmallPrimes(20);
 write('primes between 100 and 150: ');
 ShowPrimes(100,150);
 write('count of primes between 7700 and 8000 ');
 Writeln(CountPrimes(7700,8000));
 i := 100;
 repeat
   Writeln('the ',i, ' th prime ',GetNthPrime(i).rpPrime);
   i := i * 10;
 until i*25 > MaxUpperLimit;

 writeln;
 writeln('Count Emirps');
 writeln('             Emirp           Total');
 writeln('Decimals     Count           Count');
 totalCnt := 0;
 j := 10;
 i := 2;
 dgtCnt := 2;  // 13 is not present so 13<->31 isnt found
 repeat
   write(i:8);
   inc(dgtCnt,GetEmirps(  j,  j+j-1));//10..00->19..99
   inc(dgtCnt,GetEmirps(3*j,3*j+j-1));//30..00->39..99
   inc(dgtCnt,GetEmirps(7*j,7*j+j-1));//70..00->79..99
   inc(dgtCnt,GetEmirps(9*j,9*j+j-1));//90..00->99..99
   inc(TotalCnt,dgtCnt);
   writeln(dgtCnt:12,TotalCnt:14);
   j:=j*10;
   inc(i);
   dgtCnt := 0;    
 until j >= MaxUpperLimit;

end.</lang>

output
//64-Bit
time ./emirp
         
time for sieving 04:17.895
there are 4118054813 primes til 99999999999
First 20 primes 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 
primes between 100 and 150: 101 103 107 109 113 127 131 137 139 149 
count of primes between 7700 and 8000 30
the 100 th prime 541
the 1000 th prime 7919
the 10000 th prime 104729
the 100000 th prime 1299709
the 1000000 th prime 15485863
the 10000000 th prime 179424673
the 100000000 th prime 2038074743
the 1000000000 th prime 22801763489

Count Emirps
               Emirp         Total
Decimals       Count         Count
       2           8             8
       3          28            36
       4         204           240
       5        1406          1646
       6        9538         11184
       7       70474         81658
       8      535578        617236
       9     4192024       4809260
      10    33619380      38428640
      11   274890232     313318872

real	7m44.649s

Perl

Two examples of pure Perl extensible generators are shown in the Sieve of Eratosthenes#Extensible_sieves section.

The Math::Prime::Util module provides a highly performant, feature-rich library for generating, testing, and manipulating prime numbers in Perl. It offers full interoperability with Perl's bigint pragma.

Limits with a 64-bit Perl:

  • nth_prime takes about 20 seconds to return the 10^14th prime and should be fast for all results up to ~4e17. It will be impractically slow past that.
  • prime_count uses the LMO algorithm and takes about 35 seconds to return the count for primes to 10^16, and should have state of the art speed to 2^64-1. After that it will use a primality test in the interval so it still useful for large sizes with a small range.
  • fast approximations and upper/lower limits are available, which should be fast for any input size including bigints.
  • primes, next_prime, prev_prime, forprimes, prime_iterator, prime_iterator_object, and primality tests will work for practically any size input. The Math::Prime::Uti::GMP module is recommended for large inputs. With that module, these functions will work quickly for multi-thousand digit numbers.

<lang perl>use Math::Prime::Util qw(nth_prime prime_count primes);

  1. Direct solutions.
  2. primes([start],end) returns an array reference with all primes in the range
  3. prime_count([start],end) uses sieving or LMO to return fast prime counts
  4. nth_prime(n) does just that. It runs quite fast for native size inputs.

say "First 20: ", join(" ", @{primes(nth_prime(20))}); say "Between 100 and 150: ", join(" ", @{primes(100,150)}); say prime_count(7700,8000), " primes between 7700 and 8000"; say "${_}th prime: ", nth_prime($_) for map { 10**$_ } 1..8;</lang>

Output:
First 20: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
Between 100 and 150: 101 103 107 109 113 127 131 137 139 149
30 primes between 7700 and 8000
10th prime: 29
100th prime: 541
1000th prime: 7919
10000th prime: 104729
100000th prime: 1299709
1000000th prime: 15485863
10000000th prime: 179424673
100000000th prime: 2038074743

There are many other ways, including prev_prime / next_prime, a simple iterator, an OO style iterator, a tied array, and a Pari-like forprimes construct. For example, the above example using the OO iterator: <lang perl>use Math::Prime::Util "prime_iterator_object"; my $it = prime_iterator_object; say "First 20: ", join(" ", map { $it->iterate() } 1..20); $it->seek_to_value(100); print "Between 100 and 150:"; print " ", $it->iterate() while $it->value() <= 150; print "\n"; $it->seek_to_value(7700); my $c = 0; $c++ while $it->iterate() <= 8000; say "$c primes between 7700 and 8000"; say "${_}th prime: ", $it->ith($_) for map { 10**$_ } 1..8;</lang> Or using forprimes and a tied array: <lang perl>use Math::Prime::Util qw/forprimes/; use Math::Prime::Util::PrimeArray; tie my @primes, 'Math::Prime::Util::PrimeArray';

say "First 20: @primes[0..19]"; # Slice from the tied array print "Between 100 and 150: "; forprimes { print " $_"; } 100,150; print "\n";

  1. Count with forprimes

my $c = 0; forprimes { $c++ } 7700,8000; print "$c primes between 7700 and 8000\n";

  1. The tied array tries to do the right thing -- sieve a window if it sees
  2. forward or backward iteration, and nth_prime if it looks like random access.

say "${_}th prime: ", $primes[$_-1] for map { 10**$_ } 1..8;</lang>

Example showing bigints: <lang perl>use bigint; use Math::Prime::Util qw/forprimes prime_get_config/; warn "No GMP, expect slow results\n" unless prime_get_config->{gmp}; my $n = 10**200; forprimes { say $_-$n } $n,$n+1000;</lang>

Output:
357
627
799

Phix

Allows a new (and discard-able) sieve block to extend the list of already known primes. By starting with 1..10 done, we can then filter properly up to 100, so the next(/first) sieve cannot be larger than 90, then it goes 9900, with 99990000 logically next, but that blew 32-bit limits, besides capping it at 400k gave the best performance.

Unfortunately we lose the p2 check because we are not necessarily looking at that, on the plus side this completely avoids all marking/checking of even numbers.
This could almost certainly be further improved by halving the size of the sieve block.

I investigated the use of so-called "wheels", beguiled by the claim that "a 2-3-5-7 wheel saves 77%", until I realised the breakdown was 2: 50%, 3: 16%, 5: 7%, 7: 4% - it is unthinkable not to exclude even numbers, the added complexity (and subscripting) of a 30- or 210- element wheel does not seem worthwhile. While it would be trivial to unroll a 2-3 wheel, it seems far better just to avoid even numbers altogether (hence, I believe, this achieves 66 of those 77% savings). Some further discussion of this can be found on my talk page --Pete Lomax (talk)

with javascript_semantics
if platform()!=JS then free_console() end if
sequence primes = {2,3,5,7}
atom sieved = 10
 
procedure add_block()
integer N = min((sieved-1)*sieved,400000)
sequence sieve = repeat(1,N)    -- sieve[i] is really i+sieved
    for i=2 to length(primes) do -- (evens filtered on output)
        atom p = primes[i], p2 = p*p
        if p2>sieved+N then exit end if
        if p2<sieved+1 then
            p2 += ceil((sieved+1-p2)/p)*p
        end if
        p2 -= sieved
        if and_bits(p2,1)=0 then p2 += p end if
--      if sieve[p2] then           -- dang!
            for k=p2 to N by p*2 do
                sieve[k] = 0
            end for
--      end if
    end for
    for i=1 to N by 2 do
        if sieve[i] then
            primes &= i+sieved
        end if
    end for
    sieved += N
end procedure
 
function is_prime2(integer n)
    while sieved<n do
        add_block()
    end while
    return binary_search(n,primes)>0
end function
 
atom t0 = time()
while length(primes)<20 do add_block() end while
printf(1,"The first 20 primes are: ")   ?primes[1..20]
while sieved<150 do add_block() end while
sequence s = {}
for k=abs(binary_search(100,primes)) to length(primes) do
    integer p = primes[k]
    if p>150 then exit end if
    s &= p
end for
printf(1,"The primes between 100 and 150 are: ")    ?s
s = {}
for i=7700 to 8000 do
    if is_prime2(i) then s&=i end if
end for
printf(1,"There are %d primes between 7700 and 8000.\n",length(s))
for i=1 to iff(platform()=JS?7:8) do
    integer k = power(10,i)
    while length(primes)<k do
        add_block()
    end while
    printf(1,"The %,dth prime is : %d\n",{k,primes[k]})
end for
?time()-t0
if platform()!=JS then {} = wait_key() end if
Output:

(Takes 11s to reach 1e7 under pwa/p2js)

The first 20 primes are: {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71}
The primes between 100 and 150 are: {101,103,107,109,113,127,131,137,139,149}
There are 30 primes between 7700 and 8000.
The 10th prime is : 29
The 100th prime is : 541
The 1,000th prime is : 7919
The 10,000th prime is : 104729
The 100,000th prime is : 1299709
The 1,000,000th prime is : 15485863
The 10,000,000th prime is : 179424673
The 100,000,000th prime is : 2038074743
27.578

Using the builtins, for comparison only, output is identical though there is a marginal improvement in performance (for reasons lost in the mists of time!):

with javascript_semantics
atom t0 = time()
printf(1,"The first 20 primes are: %v\n",{get_primes(-20)})
sequence s = get_primes_le(150)[length(get_primes_le(100))+1..$]
printf(1,"The primes between 100 and 150 are: %v\n",{s})
integer n7700to8000 = length(get_primes_le(8000))-length(get_primes_le(7700))
printf(1,"There are %d primes between 7700 and 8000.\n",n7700to8000)
for i=1 to iff(platform()=JS?7:8) do
    integer k = power(10,i)
    printf(1,"The %,dth prime is : %d\n",{k,get_prime(k)})
end for
?time()-t0
if platform()!=JS then {} = wait_key() end if

PicoLisp

<lang PicoLisp>(de prime? (N Lst)

  (let S (sqrt N)
     (for D Lst
        (T (> D S) T)
        (T (=0 (% N D)) NIL) ) ) )

(de primeseq (A B)

  (let (I 1 R)
     (nth
        (make
           (link 2)
           (while (> A (inc 'I 2))
              (and (prime? I (made)) (link I)) )
           (setq R (length (made)))
           (while (> B I)
              (and (prime? I (made)) (link I))
              (inc 'I 2) ) )
        (inc R) ) ) )

(de take (N)

  (let I 1
     (make
        (link 2)
        (do (dec N)
           (until (prime? (inc 'I 2) (made)))
           (link I) ) ) ) )

(prin "First 20 primes: ") (println (take 20)) (prin "Primes between 100 and 150: ") (println (primeseq 100 150)) (prinl

  "Number of primes between 7700 and 8000: "
  (length (primeseq 7700 8000)) )

(for N (10 100 1000 10000 100000 1000000)

  (prinl
     N
     "th prime: "
     (last (take N)) ) )</lang>
Output:
First 20 primes: (2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71)
Primes between 100 and 150: (101 103 107 109 113 127 131 137 139 149)
Number of primes between 7700 and 8000: 30
10th prime: 29
100th prime: 541
1000th prime: 7919
10000th prime: 104729
100000th prime: 1299709
1000000th prime: 15485863

Faster version using a priority queue sieve and 2357 wheel

This version runs about twice as quickly by using the method above to find small primes as a base for populating the priority queues required for the lazy sieve of Eratosthenes algorithm by Melissa O'Neil. <lang PicoLisp> (load "plcommon/pairing-heap.l") # Pairing heap from RC task "Priority Queue"

(setq *WHEEL-2357*

  (10  2  4  2  4  6  2  6
    4  2  4  6  6  2  6  4
    2  6  4  6  8  4  2  4
    2  4  8  6  4  6  2  4
    6  2  6  6  4  2  4  6
    2  6  4  2  4  2 10  2 .))

(de "prime?" (N Primes)

  (let S (sqrt N)
     (for P Primes
        (T (> P S) T)
        (T (=0 (% N P)) NIL))))

(de "adjust-heap" (N H)

  (while (= (caar H) N)
     (let (
        Wheel (cadar H)
        P     (cddar H)
     )
     (setq H
        (heap-insert
           (cons
              (+ N (* P (car Wheel)))
              (cdr Wheel)
              P)
           (heap-rest H)))))
  H)

(de primes (Run?)

  (if (not Run?)
     (co 'primegen) # stop
     (co 'primegen
        (yield 2)
        (yield 3)
        (yield 5)
        (yield 7)
        (let (
           P     11
           Q     121
           Wp    (cdr *WHEEL-2357*)
           N     P
           Wn    (cdr *WHEEL-2357*)
           H     (heap-insert (cons Q Wp P) NIL)
        )
        (make
           (link P)
           (loop
              (cond
                 ((= N (caar H))
                    (setq H ("adjust-heap" N H))
                    (inc 'N (pop 'Wn)))
                 ((< N Q)
                    (yield N)
                    (inc 'N (pop 'Wn)))
                 (T
                    (loop
                       (inc 'P (pop 'Wp))
                       (T ("prime?" P (made))))
                    (link P)
                    (setq
                       Q  (* P P)
                       H  (heap-insert (cons Q Wp P) H))))))))))

</lang> Driver code: <lang PicoLisp> (prin "The first 20 primes: ") (do 20 (printsp (primes T))) (prinl)

(prin "between 100 and 150: ") (while (< (setq P (primes T)) 150)

  (when (> P 100)
     (printsp P)))

(prinl)

(setq Count 0) (while (< (setq P (primes T)) 8000)

  (when (> P 7700)
     (inc 'Count)))

(prinl "There are " Count " primes between 7700 and 8000.")

(de nthprime (N)

  (primes NIL)
  (do (dec N)
     (primes T))
  (primes T))

(de comma_fmt (N) (format N 0 "." ","))

(prinl "nth prime:") (for N (10 100 1000 10000 100000 1000000)

  (prinl (align 9 (comma_fmt N)) " " (align 12 (comma_fmt (nthprime N)))))

(bye) </lang>

Output:
The first 20 primes: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 
between 100 and 150: 101 103 107 109 113 127 131 137 139 149 
There are 30 primes between 7700 and 8000.
nth prime:
       10           29
      100          541
    1,000        7,919
   10,000      104,729
  100,000    1,299,709
1,000,000   15,485,863

PureBasic

<lang PureBasic>EnableExplicit DisableDebugger Define StartTime.i=ElapsedMilliseconds()

Procedure.b IsPrime(n.i)

 Define i.i=5
 If n<2 : ProcedureReturn #False : EndIf
 If n%2=0 : ProcedureReturn Bool(n=2) : EndIf
 If n%3=0 : ProcedureReturn Bool(n=3) : EndIf
 While i*i<=n
   If n%i=0 : ProcedureReturn #False : EndIf
   i+2
   If n%i=0 : ProcedureReturn #False : EndIf
   i+4
 Wend  
 ProcedureReturn #True

EndProcedure

If OpenConsole("Extensible prime generator")

 Define c.i=0, n.i=2
 Print("First twenty: ")
 While c<20
   If IsPrime(n)
     Print(Str(n)+" ")
     c+1
   EndIf
   n+1
 Wend
 
 Print(~"\nBetween 100 and 150: ")
 For n=100 To 150
   If IsPrime(n)
     Print(Str(n)+" ")
   EndIf
 Next
 
 Print(~"\nNumber beween 7'700 and 8'000: ")
 c=0
 For n=7700 To 8000
   c+IsPrime(n)
 Next
 Print(Str(c))
 
 Print(~"\n10'000th prime: ")
 c=0 : n=1
 While c<10000
   n+1
   c+IsPrime(n)    
 Wend
 Print(Str(n))  

EndIf Print(~"\nRuntime milliseconds: "+

     Str(ElapsedMilliseconds()-StartTime))

Input()</lang>

Output:
First twenty: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
Between 100 and 150: 101 103 107 109 113 127 131 137 139 149
Number beween 7'700 and 8'000: 30
10'000th prime: 104729
Runtime milliseconds: 75

Python

Python: Croft spiral

The Croft spiral sieve prime generator from the Prime decomposition task is used which contains the line <lang python>islice(count(7), 0, None, 2)</lang> The call to count(7) is to a generator of integers that counts from 7 upwards with no upper limit set.
The definition croft is a generator of primes and is used to generate as many primes as are asked for, in order.

<lang python>from __future__ import print_function from prime_decomposition import primes from itertools import islice


def p_range(lower_inclusive, upper_exclusive):

   'Primes in the range'
   for p in primes():
       if p >= upper_exclusive: break
       if p >= lower_inclusive: yield p

if __name__ == '__main__':

   print('The first twenty primes:\n  ', list(islice(primes(),20)))
   print('The primes between 100 and 150:\n  ', list(p_range(100, 150)))
   print('The number of primes between 7,700 and 8,000:\n  ', len(list(p_range(7700, 8000))))
   print('The 10,000th prime:\n  ', next(islice(primes(),10000-1, 10000)))</lang>
Output:
The first twenty primes:
   [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]
The primes between 100 and 150:
   [101, 103, 107, 109, 113, 127, 131, 137, 139, 149]
The number of primes between 7,700 and 8,000:
   30
The 10,000th prime:
   104729

Python: 210-wheel postponed incremental sieve

With more contemporary itertools use, 210-wheel incremental sieve with postponed primes processing and thus with radically reduced memory footprint which increases slower than square root of the number of primes produced:

<lang python>def wsieve(): # ideone.com/mqO25A

   wh11 = [ 2,4,2,4,6,2,6,4,2,4,6,6, 2,6,4,2,6,4,6,8,4,2,4,2,
            4,8,6,4,6,2,4,6,2,6,6,4, 2,4,6,2,6,4,2,4,2,10,2,10 ]
   cs = accumulate(chain([11], cycle(wh11)))
   yield next(cs)     # cf. ideone.com/WFv4f
   ps = wsieve()      #     codereview.stackexchange.com/q/92365/9064
   p = next(ps)       # 11         stackoverflow.com/q/30553925/849891
   psq = p*p          # 121
   D = dict(zip( accumulate(chain([0], wh11)), count(0) ))   # start from
   mults = {}
   for c in cs:
       if c in mults:
           wheel = mults.pop(c)
       elif c < psq:
           yield c
           continue
       else:          # c==psq:  map (p*) (roll wh from p) = roll (wh*p) from (p*p)
           x = [p*d for d in wh11]
           i = D[(p-11) % 210]
           wheel = accumulate(chain([psq+x[i]], cycle(x[i+1:] + x[:i+1])))
           p = next(ps)
           psq = p*p
       for m in wheel:
           if not m in mults:
               break
       mults[m] = wheel

def primes():

   yield from (2, 3, 5, 7)
   yield from wsieve()

print( list( islice( primes(), 0, 20))) print( list( takewhile( lambda x: x<150,

                  dropwhile( lambda x: x<100, primes()))))

print( len( list( takewhile( lambda x: x<8000,

                  dropwhile( lambda x: x<7700, primes())))))

print( next( islice( primes(), 10000-1, 10000)))</lang>

Output:
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]
[101, 103, 107, 109, 113, 127, 131, 137, 139, 149]
30
104729

Python: Iterative sieve on unbounded count from 2

After a [blog entry] on implementing a particular Haskell solution in Python.

<lang python>from itertools import count, takewhile, islice

def prime_sieve():

   sieved = count(2)
   prime = next(sieved)
   yield prime
   primes = [prime]
   for x in sieved:
       possible_prime_divs = takewhile(lambda p: p <= x**0.5, primes)
       if any(x % prime == 0 for prime in possible_prime_divs):
           continue
       yield x
       primes.append(x)

if __name__ == '__main__':

   def leq_150(x): return x <= 150
   def leq_8000(x): return x <= 8000
   
   print("Show the first twenty primes.\n   =",
       list(islice(prime_sieve(), 20)))
   print("Show the primes between 100 and 150\n   =",
       [x for x in takewhile(leq_150, prime_sieve()) if x >= 100])
   print("Show the number of primes between 7,700 and 8,000.\n   =",
       sum(1 for x in takewhile(leq_8000, prime_sieve()) if x >= 7700))
   print("Show the 10,000th prime.\n   =",
       next(islice(prime_sieve(), 10000-1, 10000)))</lang>
Output:
Show the first twenty primes.
   = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]
Show the primes between 100 and 150
   = [101, 103, 107, 109, 113, 127, 131, 137, 139, 149]
Show the number of primes between 7,700 and 8,000.
   = 30
Show the 10,000th prime.
   = 104729

Python: From partition function P

See Partition function P#Python: Mathloger video prime generator where a slight modification to the partition function creates an unbounded prime generator, (apart from memory and time limitations).

Racket

Personal note: unless I need the raw power of an application-specific primes generator/filter, I pretty well stick with the math/number-theory library. And even when I write an ASPG/F I question how it performs against the math/number-theory version!

The link referenced in the source: math/number-theory module documentation

<lang scheme>#lang racket

Using the prime functions from

(require math/number-theory)

(displayln "Show the first twenty primes.") (next-primes 1 20)

(displayln "Show the primes between 100 and 150.")

Note that in each of the in-range filters I "add1" to the stop value, so that (in this case) 150 is
considered. I'm pretty sure it's not prime... but technology moves so fast nowadays that things
might have changed!

(for/list ((i (sequence-filter prime? (in-range 100 (add1 150))))) i)

(displayln "Show the number of primes between 7,700 and 8,000.")

(for/sum (...) 1) counts the values in a sequence

(for/sum ((i (sequence-filter prime? (in-range 7700 (add1 8000))))) 1)

(displayln "Show the 10,000th prime.") (nth-prime (sub1 10000)) ; (nth-prime 0) => 2

If a languages in-built prime generator is extensible or is guaranteed to generate primes up to a
system limit, (2^31 or memory overflow for example), then this may be used as long as an
explanation of the limits of the prime generator is also given. (Which may include a link
to/excerpt from, language documentation).
Full details in
[[4]]
When reading the manual, note that "Integer" and "Natural" are unlimited (or bounded by whatever
big number representation there is (and the computational complexity of the work being asked).

(define 2^256 (expt 2 256)) 2^256 (next-prime 2^256)

(Oh, and this is a 64-bit laptop, I left my 256-bit PC in the office.)</lang>
Output:
Show the first twenty primes.
(2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71)
Show the primes between 100 and 150.
(101 103 107 109 113 127 131 137 139 149)
Show the number of primes between 7,700 and 8,000.
30
Show the 10,000th prime.
104729
115792089237316195423570985008687907853269984665640564039457584007913129639936
115792089237316195423570985008687907853269984665640564039457584007913129640233

Raku

(formerly Perl 6) Build a lazy infinite list of primes using the Raku builtin is-prime method. ( A lazy list will not bother to calculate the values until they are actually used. ) That is the first line. If you choose to calculate the entire list, it is fairly certain that you will run out of process / system memory before you finish... (patience too probably) but there aren't really any other constraints. The is-prime builtin uses a Miller-Rabin primality test with 100 iterations, so while it is not 100% absolutely guaranteed that every number it flags as prime actually is, the chances that it is wrong are about 4-100. Much less than the chance that a cosmic ray will cause an error in your computers CPU. Everything after the first line is just display code for the various task requirements.

<lang perl6>my @primes = lazy gather for 1 .. * { .take if .is-prime }

say "The first twenty primes:\n ", "[{@primes[^20].fmt("%d", ', ')}]"; say "The primes between 100 and 150:\n ", "[{@primes.&between(100, 150).fmt("%d", ', ')}]"; say "The number of primes between 7,700 and 8,000:\n ", +@primes.&between(7700, 8000); say "The 10,000th prime:\n ", @primes[9999];

sub between (@p, $l, $u) {

   gather for @p { .take if $l < $_ < $u; last if $_ >= $u }

}</lang>

Output:
The first twenty primes:
   [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]
The primes between 100 and 150:
   [101, 103, 107, 109, 113, 127, 131, 137, 139, 149]
The number of primes between 7,700 and 8,000:
   30
The 10,000th prime:
   104729

Red

<lang Red> Red [Description: "Prime checker/generator/counter"]

context [

   poke noprime: make bitset! 3 1 true
   top: 2
   noprimes: function [n [integer!] /extern top][
       either top < n [
           n: n + 100
           r: 2 
           while [r * r <= n][
               repeat q n / r - 1 [poke noprime q + 1 * r true] 
               until [not pick noprime r: r + 1]
           ]
           self/top: n
       ][top]
   ]
   set 'prime? func [
       "Check whether number is prime or return required prime"
       n [integer!] 
       /next "Return next closest prime to given number"
       /last "Return last closest prime to given number, or number itself if prime"
       /Nth  "Return Nth prime"
   ][
       noprimes case [
           Nth [to integer! n * 12 ]
           next [n + 100]
           true [n]
       ]
       case [
           next [until [not noprime/(n: n + 1)] n]
           last [while [noprime/:n][n: n - 1] n]
           Nth  [
               cnt: i: 0 
               while [cnt < n][
                   until [not noprime/(i: i + 1)] 
                   cnt: cnt + 1
               ] 
               i
           ]
           true [not noprime/:n]
       ]
   ]
   
   set 'primes function [
       "Return (number of) primes in given range"
       n [integer!] 
       /from     "Start considering primes from `start`"
           start "Default 1"
       /list     "First argument is interpreted as number of primes to list"
       /count    "Count primes from `start`"
   ][
       start: any [start 1]
       either list [
           noprimes start + (n * 12)
       ][
           set [start n] sort reduce [n start]
           noprimes start + n
       ]
       case [
           list [
               start: start - 1 
               collect [
                   loop n [
                       until [not noprime/(start: start + 1)] 
                       keep start
                   ]
               ]
           ]
           count [
               cnt: 0 
               repeat i n - start + 1 [
                   j: i - 1 
                   if not noprime/(j + start) [cnt: cnt + 1]
               ]
               cnt
           ]
           true [
               collect [
                   repeat i n - start + 1 [
                       j: i - 1 
                       if not noprime/(j: j + start) [keep j]
                   ]
               ]
           ]
       ]
   ]

] </lang>

Output:

>> primes/list 20 == [2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71] >> primes/from 150 100 == [101 103 107 109 113 127 131 137 139 149] >> primes/count/from 8000 7700 == 30 >> prime?/Nth 10000 == 104729

REXX

Programming note:   Most REXXes (of the 32-bit variety) run with an upper limit of roughly 2 Gbytes (for virtual storage), and that is the limit of this program (in building the stemmed array of prime numbers and prime number indicators, and available virtual storage will be the limiting factor of how many primes can be generated.

The method of extending primes (via the PRIMES subroutine) is of two kinds when invoking the PRIMES subroutine:

  • a positive number which will (possibly) generate primes up to that total amount, and
  • a negative number which will (possibly) generate primes up to   |number|.

Two arrays are available to the caller after invoking the PRIMES subroutine.   @.Nth   where this is the Nth prime.
Also, one can check if   1331   is a prime:       !.1331   has the value of   1   (is prime),     0   (isn't prime).

This REXX program uses some hardcoded divisions.   While the hardcoded divisions are not necessary, they do speed up the division process in validating if a number is prime (or not),   and the hardcoded divisions bypass the repetitive tests to see if a (low) prime that is being used for division is greater than the number being divided (tested).   The hardcoded divisions also bypass (or, at least, postpones) the computation of the integer square roots. <lang rexx>/*REXX program calculates and displays primes using an extendible prime number generator*/ parse arg f .; if f== then f= 20 /*allow specifying number for 1 ──► F.*/ _i= ' (inclusive) '; _b= 'between '; _tnp= 'the number of primes' _b; _tn= 'the primes' call primes f; do j=1 for f; $= $ @.j; end /*j*/

                                       say 'the first '    f    " primes are: "        $
                                       say

call primes -150; do j=100 to 150; if !.j==1 then $= $ j; end /*j*/

                                       say _tn  _b  '100  to  150'  _i  "are: "        $
                                       say

call primes -8000; do j=7700 to 8000; if !.j==1 then $= $ j; end /*j*/

                                       say _tnp  '7,700  and  8,000'  _i  "is: " words($)
                                       say

call primes 10000

                                       say 'the 10,000th prime is: '    @.10000

exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ primes: procedure expose !. @. $ #; parse arg H,,$; Hneg= H<0; H= abs(H)

       if symbol('#')=="LIT"  then call .primI  /*1st time here?  Then initialize stuff*/
       if Hneg  then  if  H<=@.#  then return   /*do we have a high enough  P  already?*/
                                  else nop      /*this is used to match the above THEN.*/
                else  if  H<=#    then return   /*are there enough primes currently ?  */
                                                /* [↓]  gen more primes within range.  */
         do j=@.#+2   by 2; parse var j  -1 _ /*find primes until have   H   Primes. */
         if     _==5  then iterate              /*is the right─most digit a 5  (five)? */
         if j// 3==0  then iterate              /*is  J  divisible by  three?  (& etc.)*/
         if j// 7==0  then iterate;  if j//11==0  then iterate; if j//13==0  then iterate
         if j//17==0  then iterate;  if j//19==0  then iterate; if j//23==0  then iterate
         if j//29==0  then iterate;  if j//31==0  then iterate; if j//37==0  then iterate
         if j//41==0  then iterate;  if j//43==0  then iterate; if j//47==0  then iterate
         if j//53==0  then iterate;  if j//59==0  then iterate; if j//61==0  then iterate
         if j//67==0  then iterate;  if j//71==0  then iterate; if j//73==0  then iterate
         if j//79==0  then iterate;  if j//83==0  then iterate; if j//89==0  then iterate
         if j//97==0  then iterate;  if j//101==0 then iterate; if j//103==0 then iterate
         x= j;          r= 0;  q= 1;   do while q<=x;  q= q*4;  end  /*R:  the sqrt(J).*/
                 do while q>1; q=q%4; _=x-r-q; r=r%2; if _>=0 then do;x=_;r=r+q; end; end
                   do k=@.lowP   while @.k<=r   /*÷ by the known odd primes (hardcoded)*/
                   if j//@.k==0  then iterate j /*J ÷ by a prime?  Then not prime.  ___*/
                   end   /*k*/                  /* [↑]  divide by odd primes up to √ J */
         #= # + 1                               /*bump the number of primes found.     */
         @.#= j;                       !.j= 1   /*assign to sparse array;  prime²;  P#.*/
         if Hneg  then if H<=@.#  then leave    /*is this a high enough prime?         */
                                  else nop      /*used to match the above  THEN.       */
                  else if H<=#    then leave    /*have enough primes been generated?   */
         end   /*j*/                            /* [↑]  keep generating until enough.  */
       return                                   /*return to invoker with more primes.  */

/*──────────────────────────────────────────────────────────────────────────────────────*/ .primI: !.=0; @.=0; /*!.x= a prime or not; @.n= Nth prime.*/

       L= 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103
               do #=1  for words(L);   p= word(L, #);   @.#= p;   !.p=1;    end   /*#*/
       #= # - 1;       @.lowP= #;      return   /*#:   # primes;  @.lowP:   start of ÷ */</lang>
output   when using the default input:
the first  20  primes are:   2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71

the primes between  100  to  150  (inclusive)  are:   101 103 107 109 113 127 131 137 139 149

the number of primes between  7,700  and  8,000  (inclusive)  is:  30

the 10,000th prime is:  104729

Ring

<lang ring> see "first twenty primes : " i = 1 nr = 0 while i <= 20

     nr += 1     
     if isPrime(nr) see " " + nr i += 1 ok

end

see "primes between 100 and 150 : " for nr = 100 to 150

   if isPrime(nr) see " " + nr ok

next see nl

see "primes between 7,700 and 8,000 : " i = 0 for nr = 7700 to 8000

   if isPrime(nr) i += 1 ok

next see i + nl

see "The 10,000th prime : " i = 1 nr = 0 while i <= 10000

     nr += 1     
     if isPrime(nr) i += 1 ok

end see nr + nl

func isPrime n

    if n <= 1 return false ok
    if n <= 3 return true ok
    if (n & 1) = 0 return false ok
    for t = 3 to sqrt(n) step 2
        if (n % t) = 0 return false ok
    next
    return true

</lang> Output:

first twenty primes :  2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
primes between 100 and 150 :  101 103 107 109 113 127 131 137 139 149
primes between 7,700 and 8,000 : 30
The 10,000th prime : 104729

Ruby

The prime library behaves like an enumerator. It has an "each" method, which takes an upper bound as argument. This argument is nil by default, which means no upper bound. <lang ruby>require "prime"

puts Prime.take(20).join(", ") puts Prime.each(150).drop_while{|pr| pr < 100}.join(", ") puts Prime.each(8000).drop_while{|pr| pr < 7700}.count puts Prime.take(10_000).last</lang>

Output:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71
101, 103, 107, 109, 113, 127, 131, 137, 139, 149
30
104729

Rust

Uses the code from Sieve_of_Eratosthenes#Unbounded_Page-Segmented_bit-packed_odds-only_version_with_Iterator; copy that code into a file named src/pagesieve.rs and prepend pub to all function declarations used in this code (count_primes_paged and primes_paged).

<lang rust>mod pagesieve;

use pagesieve::{count_primes_paged, primes_paged};

fn main() {

   println!("First 20 primes:\n {:?}",
            primes_paged().take(20).collect::<Vec<_>>());
   println!("Primes between 100 and 150:\n {:?}",
            primes_paged().skip_while(|&x| x < 100)
                          .take_while(|&x| x < 150)
                          .collect::<Vec<_>>());
   let diff = count_primes_paged(8000) - count_primes_paged(7700);
   println!("There are {} primes between 7,700 and 8,000", diff);
   // rust enumerations are zero base, so need to subtract 1!!!
   println!("The 10,000th prime is {}", primes_paged().nth(10_000 - 1).unwrap());

}</lang>

Output:
First 20 primes:
 [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]
Primes between 100 and 150:
 [101, 103, 107, 109, 113, 127, 131, 137, 139, 149]
There are 30 primes between 7,700 and 8,000
The 10,000th prime is 104729

Seed7

The sieve of eratosthenes cannot be used, because it needs a limit. Instead the function getPrime is used. GetPrime generates all primes in sequence. <lang seed7>$ include "seed7_05.s7i";

const func boolean: isPrime (in integer: number) is func

 result
   var boolean: prime is FALSE;
 local
   var integer: count is 2;
 begin
   if number = 2 then
     prime := TRUE;
   elsif number > 2 then
     while number rem count <> 0 and count * count <= number do
       incr(count);
     end while;
     prime := number rem count <> 0;
   end if;
 end func;

var integer: currentPrime is 1; var integer: primeNum is 0;

const func integer: getPrime is func

 result
   var integer: nextPrime is 0;
 begin
   repeat
     incr(currentPrime);
   until isPrime(currentPrime);
   nextPrime := currentPrime;
   incr(primeNum);
 end func;

const proc: main is func

 local
   var integer: aPrime is 0;
   var integer: count is 0;
 begin
   write("First twenty primes:");
   while primeNum < 20 do
     write(" " <& getPrime);
   end while;
   writeln;
   repeat
     aPrime := getPrime;
   until aPrime >= 100;
   write("Primes between 100 and 150:");
   while aPrime <= 150 do
     write(" " <& aPrime);
     aPrime := getPrime;
   end while;
   writeln;
   repeat
     aPrime := getPrime;
   until aPrime >= 7700;
   while aPrime <= 8000 do
     incr(count);
     aPrime := getPrime;
   end while;
   writeln("Number of primes between 7,700 and 8,000: " <& count);
   repeat
     aPrime := getPrime;
   until primeNum = 9999; # discard up to and including the 9,999 prime!
   writeln("The 10,000th prime: " <& getPrime);
 end func;</lang>
Output:
First twenty primes: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
Primes between 100 and 150: 101 103 107 109 113 127 131 137 139 149
Number of primes between 7,700 and 8,000: 30
The 10,000th prime: 104729

Sidef

<lang ruby>say ("First 20: ", 20.nth_prime.primes.join(' ')) say ("Between 100 and 150: ", primes(100,150).join(' ')) say (prime_count(7700,8000), " primes between 7700 and 8000") say ("10,000th prime: ", nth_prime(10_000))</lang>

Output:
First 20: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
Between 100 and 150: 101 103 107 109 113 127 131 137 139 149
30 primes between 7700 and 8000
10,000th prime: 104729

Swift

The Sieve_of_Eratosthenes#Unbounded_.28Odds-Only.29_Versions Swift examples contain several versions that are Extensible Prime Generators using the Sieve of Eratosthenese. One of the simplest is the Hashed Dictionary, reproduced here as follows: <lang swift>import Foundation

func soeDictOdds() -> UnfoldSequence<Int, Int> {

 var bp = 5; var q = 25
 var bps: UnfoldSequence<Int, Int>.Iterator? = nil
 var dict = [9: 6] // Dictionary<Int, Int>(9 => 6)
 return sequence(state: 2, next: { n in
   if n < 9 { if n < 3 { n = 3; return 2 }; defer {n += 2}; return n }
   while n >= q || dict[n] != nil {
     if n >= q {
       let inc = bp + bp
       dict[n + inc] = inc
       if bps == nil {
         bps = soeDictOdds().makeIterator()
         bp = (bps?.next())!; bp = (bps?.next())!; bp = (bps?.next())! // skip 2/3/5...
       }
       bp = (bps?.next())!; q = bp * bp // guaranteed never nil
     } else {
       let inc = dict[n] ?? 0
       dict[n] = nil
       var next = n + inc
       while dict[next] != nil { next += inc }
       dict[next] = inc
     }
     n += 2
   }
   defer { n += 2 }; return n
 })

}

print("The first 20 primes are: ", terminator: "") soeDictOdds().lazy.prefix(20).forEach { print($0, "", terminator: "") } print()

print("The primes between 100 and 150 are: ", terminator: "") soeDictOdds().lazy.drop(while: { $0 < Prime(100) }).lazy.prefix(while: { $0 <= 150 })

   .forEach { print($0, "", terminator: "") }

print()

print("The number of primes from 7700 to 8000 is  :", terminator: "") print(soeDictOdds().lazy.drop(while: { $0 < 7700 }).lazy.prefix(while: { $0 <= 8000 })

       .lazy.reduce(0, { a, _ in a + 1 }))

print("The 10,000th prime is: ", terminator: "") print((soeDictOdds().lazy.dropFirst(9999).first { $0 == $0 })!)

print("The sum of primes to 2 million is: ", terminator: "")

let start = NSDate() let answr = soeDictOdds().lazy.prefix(while: { $0 <= 2000000 })

             .reduce(0, { a, p in a + Int64(p) })

let elpsd = -start.timeIntervalSinceNow

print(answr) print(String(format: "This test took %.3f milliseconds.", elpsd * 1000))</lang>

Output:
The first 20 primes are:  2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
The primes between 100 and 150 are:  101 103 107 109 113 127 131 137 139 149
The number of primes from 7700 to 8000 is  :30
The 10,000th prime is:  104729
The sum of primes to 2 million is:  142913828922
This test took 862.497 milliseconds.

The above version is somewhat slow due to the overhead of computing the hash for the dictionary, but useful up to ranges of a few million as in solving the Euler Problem of summing all primes up to two million as shown above.

Alternate much faster version using Page Segmentation with bit-packed arrays

The Page Segmentation algorithm as per the the last version of the Sieve of Eratosthenes in the same Swift Unbounded section as the above code can use enumeration to do the same thing but is about 50 times faster, taking an almost imperceptible time to solve Euler Problem 10 and can count the primes to a billion (even by enumeration) in a few seconds. It can be used just by substituting `primesPaged()` for `soeDictOdds()` in the above testing code.

Tcl

Works with: Tcl version 8.6

<lang tcl>package require Tcl 8.6

  1. An iterative version of the Sieve of Eratosthenes.
  2. Effective limit is the size of memory.

coroutine primes apply {{} {

   yield
   while 1 {yield [coroutine primes_[incr p] apply {{} {

yield [info coroutine] set plist {} for {set n 2} true {incr n} { set found 0 foreach p $plist { if {$n%$p==0} { set found 1 break } } if {!$found} { lappend plist $n yield $n } }

   }}]}

}}

set p [primes] for {set primes {}} {[llength $primes] < 20} {} {

   lappend primes [$p]

} puts 1st20=[join $primes ,] rename $p {}

set p [primes] for {set primes {}} {[set n [$p]] <= 150} {} {

   if {$n >= 100 && $n <= 150} {

lappend primes $n

   }

} puts 100-150=[join $primes ,] rename $p {}

set p [primes] for {set count 0} {[set n [$p]] <= 8000} {} {

   incr count [expr {$n>=7700 && $n<=8000}]

} puts count7700-8000=$count rename $p {}

set p [primes] for {set count 0} {$count < 10000} {incr count} {

   set prime [$p]

} puts prime10000=$prime rename $p {}</lang>

Output:
1st20=2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71
100-150=101,103,107,109,113,127,131,137,139,149
count7700-8000=30
prime10000=104729

VBA

<lang vb>Option Explicit

Sub Main() Dim Primes() As Long, n As Long, temp$ Dim t As Single

   t = Timer
   
   n = 133218295 'limit for an Array of Longs with VBA on my computer
   Primes = ListPrimes(n)
   Debug.Print "For N = " & Format(n, "#,##0") & ", execution time : " & _
       Format(Timer - t, "0.000 s") & ", " & _
       Format(UBound(Primes) + 1, "#,##0") & " primes numbers."
   
   'First twenty primes
   For n = 0 To 19
       temp = temp & ", " & Primes(n)
   Next
   Debug.Print "First twenty primes : "; Mid(temp, 3)
   'Primes between 100 and 150
   n = 0: temp = vbNullString
   Do While Primes(n) < 100
       n = n + 1
   Loop
   Do While Primes(n) < 150
       temp = temp & ", " & Primes(n)
       n = n + 1
   Loop
   Debug.Print "Primes between 100 and 150 : " & Mid(temp, 3)
   'Number of primes between 7,700 and 8,000
   Dim ccount As Long
   n = 0
   Do While Primes(n) < 7700
       n = n + 1
   Loop
   Do While Primes(n) < 8000
       ccount = ccount + 1
       n = n + 1
   Loop
   Debug.Print "Number of primes between 7,700 and 8,000 : " & ccount
   'The 10 x Xth prime
   n = 1
   Do While n <= 100000
       n = n * 10
       Debug.Print "The " & n & "th prime: "; Format(Primes(n - 1), "#,##0")
   Loop
   Debug.Print "VBA has a limit in array's dim"
   Debug.Print "With my computer, the limit for an array of Long is : 133 218 295"
   Debug.Print "The last prime I could find is the : " & _
       Format(UBound(Primes), "#,##0") & "th, Value : " & _
       Format(Primes(UBound(Primes)), "#,##0")

End Sub

Function ListPrimes(MAX As Long) As Long() Dim t() As Boolean, L() As Long, c As Long, s As Long, i As Long, j As Long

   ReDim t(2 To MAX)
   ReDim L(MAX \ 2)
   s = Sqr(MAX)
   For i = 3 To s Step 2
       If t(i) = False Then
           For j = i * i To MAX Step i
               t(j) = True
           Next
       End If
   Next i
   L(0) = 2
   For i = 3 To MAX Step 2
       If t(i) = False Then
           c = c + 1
           L(c) = i
       End If
   Next i
   ReDim Preserve L(c)
   ListPrimes = L

End Function</lang>

Output:
For N = 133 218 295, execution time : 9,422 s, 7 550 284 primes numbers.
First twenty primes : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71
Primes between 100 and 150 : 101, 103, 107, 109, 113, 127, 131, 137, 139, 149
Number of primes between 7,700 and 8,000 : 30
The 10th prime: 29
The 100th prime: 541
The 1000th prime: 7 919
The 10000th prime: 104 729
The 100000th prime: 1 299 709
The 1000000th prime: 15 485 863
VBA has a limit in array's dim
With my computer, the limit for an array of Long is : 133 218 295
The last prime I could find is the : 7 550 283th, Value : 133 218 289

Wren

This uses a larger wheel than the sieve in the Wren-math module and is a bit faster if one is sieving beyond about 10^7. <lang ecmascript>import "/fmt" for Fmt

var primeSieve = Fn.new { |limit|

   if (limit < 2) return []
   var c = [false] * (limit + 1) // composite = true
   c[0] = true
   c[1] = true
   // no need to process the even numbers > 2
   var k = 9
   while (k <= limit) {
       c[k] = true
       k = k + 6
   }
   k = 25
   while (k <= limit) {
       c[k] = true
       k = k + 10
   }
   k = 49
   while (k <= limit) {
       c[k] = true
       k = k + 14
   }
   var p = 11
   var p2 = 121
   var inc = [2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,
       4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2,10]
   var w = 0
   while (p2 <= limit) {
       var i = p2
       while (i <= limit) {
           c[i] = true
           i = i + 2*p
       }
       var ok = true
       while (ok) {
           p = p + inc[w]
           w = (w + 1) % 48
           ok = c[p]
       }
       p2 = p * p
   }
   var primes = [2]
   var i = 3
   while (i <= limit) {
       if (!c[i]) primes.add(i)
       i = i + 2
   }
   return primes

}

var primes = primeSieve.call(1e8 * 2) System.print("The first 20 primes are: ") System.print(primes.take(20).toList) System.print("\nThe primes between 100 and 150 are:") System.print(primes.where { |p| p > 100 && p < 150 }.toList) System.write("\nThe number of primes between 7,700 and 8,000 is: ") Fmt.print("$,d", primes.count{ |p| p > 7700 && p < 8000 }) Fmt.print("\nThe 10,000th prime is: $,d", primes[9999]) Fmt.print("\nThe 100,000th prime is: $,d", primes[99999]) Fmt.print("\nThe 1,000,000th prime is: $,d", primes[999999])</lang>

Output:
The first 20 primes are: 
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]

The primes between 100 and 150 are:
[101, 103, 107, 109, 113, 127, 131, 137, 139, 149]

The number of primes between 7,700 and 8,000 is: 30

The 10,000th prime is: 104,729

The 100,000th prime is: 1,299,709

The 1,000,000th prime is: 15,485,863

Zig

Translation of: C

(Melissa O'Neil's Genuine Sieve version),

This version is based on the lazy sieve version written in C. However, it uses Zig's compile-time meta-programming logic to allow the programmer to specify the maximum number of bits for the primes to be generated. The advantage is that we can stop adding multiples to the priority queue once we pass the square root of the bit size. (e.g. for a u10, or 10 bit prime, it stops adding primes to the queue once a prime exceeds the maximum of a u5, or 31.) Therefore this version represents a spectrum of fixed or unbounded generators. PrimeGen(u32) (or PrimeGen(u64)) can be used as an unbounded sieve, whereas a PrimeGen(u10) represents a fixed sieve for all primes < 1024.

Implementation: <lang Zig> const std = @import("std"); const builtin = std.builtin; const heap = std.heap; const mem = std.mem; const meta = std.meta;

fn assertInt(comptime T: type) builtin.TypeInfo.Int {

   if (@typeInfo(T) != .Int)
       @compileError("data type must be an integer.");
   const int = @typeInfo(T).Int;
   if (int.is_signed == true or int.bits % 2 == 1 or int.bits < 4)
       @compileError("type must be an unsigned integer with even bit size (of at least 4 bits).");
   return int;

}

fn sqrtType(comptime T: type) type {

   const t = assertInt(T);
   return meta.Int(.unsigned, t.bits / 2);

}

// given an upper bound, max, return the most restrictive sieving data type. pub fn autoSieveType(comptime max: u64) type {

   if (max == 0)
       @compileError("The maximum sieving size must be non-zero.");
   var bit_len = 64 - @clz(u64, max);
   if (max & (max - 1) == 0) // power of two
       bit_len -= 1;
   if (bit_len % 2 == 1)
       bit_len += 1;
   if (bit_len < 4)
       bit_len = 4;
   return meta.Int(.unsigned, bit_len);

}

test "type meta functions" {

   const expect = std.testing.expect;
   expect(sqrtType(u20) == u10);
   expect(autoSieveType(8000) == u14);
   expect(autoSieveType(9000) == u14);
   expect(autoSieveType(16384) == u14);
   expect(autoSieveType(16385) == u16);
   expect(autoSieveType(32768) == u16);
   expect(autoSieveType(1000) == u10);
   expect(autoSieveType(10) == u4);
   expect(autoSieveType(4) == u4);
   expect(autoSieveType(std.math.maxInt(u32)) == u32);

}

const wheel2357 = [48]u8{

   10, 2, 4, 2, 4, 6, 2,  6,
   4,  2, 4, 6, 6, 2, 6,  4,
   2,  6, 4, 6, 8, 4, 2,  4,
   2,  4, 8, 6, 4, 6, 2,  4,
   6,  2, 6, 6, 4, 2, 4,  6,
   2,  6, 4, 2, 4, 2, 10, 2,

};

fn Wheel2357Multiple(comptime Int: type) type {

   _ = assertInt(Int);
   return struct {
       multiple: Int,
       base_prime: Int,
       offset: u6,
       fn less(self: Wheel2357Multiple(Int), other: Wheel2357Multiple(Int)) bool {
           return self.multiple < other.multiple;
       }
   };

}

pub fn PrimeGen(comptime Int: type) type {

   _ = assertInt(Int);
   return struct {
       const Self = @This();
       initial_primes: u16,
       offset: u6,
       candidate: Int,
       multiples: std.PriorityQueue(Wheel2357Multiple(Int)),
       allocator: *mem.Allocator,
       count: u32,
       pub fn init(alloc: *mem.Allocator) Self {
           return Self{
               .initial_primes = 0xAC, // primes 2, 3, 5, 7 in a bitmask
               .offset = 0,
               .candidate = 1,
               .count = 0,
               .allocator = alloc,
               .multiples = std.PriorityQueue(Wheel2357Multiple(Int)).init(alloc, Wheel2357Multiple(Int).less),
           };
       }
       pub fn deinit(self: *PrimeGen(Int)) void {
           self.multiples.deinit();
       }
       pub fn next(self: *PrimeGen(Int)) !?Int {
           if (self.initial_primes != 0) { // use the bitmask up first
               const p = @as(Int, @ctz(u16, self.initial_primes));
               self.initial_primes &= self.initial_primes - 1;
               self.count += 1;
               return p;
           } else {
               while (true) {
                   // advance to the next prime candidate.
                   if (@addWithOverflow(Int, self.candidate, wheel2357[self.offset], &self.candidate))
                       return null;
                   self.offset = (self.offset + 1) % @as(u6, wheel2357.len);
                   // See if the composite number on top of the heap matches
                   // the candidate.
                   //
                   var top = self.multiples.peek();
                   if (top == null or self.candidate < top.?.multiple) {
                       // prime found, add the square and it's position on the wheel
                       // to the heap.
                       //
                       if (self.candidate <= std.math.maxInt(sqrtType(Int)))
                           try self.multiples.add(Wheel2357Multiple(Int){
                               .multiple = self.candidate * self.candidate,
                               .base_prime = self.candidate,
                               .offset = self.offset,
                           });
                       self.count += 1;
                       return self.candidate;
                   } else {
                       while (true) {
                           // advance the top of heap to the next prime multiple
                           // that is not a multiple of 2, 3, 5, 7.
                           //
                           var mult = self.multiples.remove();
                           // If the multiple becomes too big (greater than the the maximum
                           // sieve size), then there's no reason to add it back to the queue.
                           //
                           var tmp: Int = undefined;
                           if (!@mulWithOverflow(Int, mult.base_prime, wheel2357[mult.offset], &tmp) and
                               !@addWithOverflow(Int, tmp, mult.multiple, &mult.multiple))
                           {
                               mult.offset = (mult.offset + 1) % @as(u6, wheel2357.len);
                               try self.multiples.add(mult);
                           }
                           top = self.multiples.peek();
                           if (top == null or self.candidate != top.?.multiple)
                               break;
                       }
                   }
               }
           }
       }
   };

} </lang> Driver code: <lang Zig> const std = @import("std"); const sieve = @import("sieve.zig"); const PrimeGen = sieve.PrimeGen; const heap = std.heap;

const stdout = std.io.getStdOut().writer(); pub fn main() !void {

   try part1();
   try part2();
   try part3();

}

// exercize 1: Print small primes fn part1() !void {

   var arena = heap.ArenaAllocator.init(heap.page_allocator);
   defer arena.deinit();
   var primes = PrimeGen(u8).init(&arena.allocator);
   defer primes.deinit();
   try stdout.print("The first 20 primes:", .{});
   while (try primes.next()) |p| {
       try stdout.print(" {}", .{p});
       if (primes.count == 20)
           break;
   }
   try stdout.print("\nThe primes between 100 and 150:", .{});
   while (try primes.next()) |p| if (p >= 100 and p <= 150)
       try stdout.print(" {}", .{p});
   try stdout.print("\n", .{});

}

// exercize 2: count medium primes fn part2() !void {

   var arena = heap.ArenaAllocator.init(heap.page_allocator);
   defer arena.deinit();
   var primes = PrimeGen(sieve.autoSieveType(8000)).init(&arena.allocator);
   defer primes.deinit();
   var count: i32 = 0;
   while (try primes.next()) |p| {
       if (p > 8000)
           break;
       if (p > 7700)
           count += 1;
   }
   try stdout.print("There are {} primes between 7700 and 8000.\n", .{count});

}

// exercize 3: find big primes fn part3() !void {

   var arena = heap.ArenaAllocator.init(heap.page_allocator);
   defer arena.deinit();
   var primes = PrimeGen(u32).init(&arena.allocator);
   defer primes.deinit();
   var c: u32 = 10;
   while (try primes.next()) |p| {
       if (primes.count == c) {
           try stdout.print("The {}th prime is {}\n", .{ c, p });
           if (c == 100_000_000)
               break;
           c *= 10;
       }
   }

}

</lang>

Output:
The first 20 primes: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
The primes between 100 and 150: 101 103 107 109 113 127 131 137 139 149
There are 30 primes between 7700 and 8000.
The 10th prime is 29
The 100th prime is 541
The 1000th prime is 7919
The 10000th prime is 104729
The 100000th prime is 1299709
The 1000000th prime is 15485863
The 10000000th prime is 179424673
The 100000000th prime is 2038074743

Alternative version based on "Two Compact Incremental Prime Sieves" by Jonathon P. Sorenson.

Like the version based on the O'Neil sieve, this version also uses metaprogramming, this time to properly calculate the sizes of the buckets required by the algorithm (the sieve is implemented as an array of buckets, each bucket containing a list of the unique prime factors of that entry.) It's quite memory efficient for a u32 and below (16 bytes per bucket), though it is not memory efficient for larger sieves. For example, a u36 requires a bucket size of 36 bytes.)

As with the O'Neil sieve, this version thus represents a spectrum of fixed or unbounded generators. PrimeGen(u32) (or PrimeGen(u64)) can be used as an unbounded sieve, whereas a PrimeGen(u10) represents a fixed sieve for all primes < 1024.

Since this version uses an array (O(1)) rather than an O(log n) priority queue, it is significantly faster. On my laptop, sieving up to 100,000,000 primes takes 2:18 seconds for the O'Neil sieve and 52 seconds for the Sorenson. <lang Zig> // Since in the common case (primes < 2^32 - 1), the stack only needs to be 8 16-bit words long // (only twice the size of a pointer) the required stacks are stored in each cell, rather // than using an indirection (e.g. linked list of integer cells) // const std = @import("std"); const builtin = std.builtin; const meta = std.meta; const mem = std.mem;

fn assertInt(comptime T: type) builtin.TypeInfo.Int {

   const Signedness = builtin.Signedness;
   if (@typeInfo(T) != .Int)
       @compileError("data type must be an integer.");
   const int = @typeInfo(T).Int;
   if (int.signedness == Signedness.signed or int.bits % 2 == 1 or int.bits < 4 or int.bits > 64)
       @compileError("type must be an unsigned integer with even bit size (of at least 4 bits).");
   return int;

}

// given a type, return the maximum stack size required by the algorthm. fn listSize(comptime T: type) usize {

   _ = assertInt(T);
   const primes = [_]u6{
       2,  3,  5,  7,  11, 13, 17, 19,
       23, 29, 31, 37, 41, 43, 47, 53,
   };
   // Find the first primorial that will overflow type T.
   // the size of the list is the primorial index minus one,
   // since the sieve doesn't include 2.
   //
   var i: usize = 0;
   var pi: T = 1;
   while (!@mulWithOverflow(T, pi, primes[i], &pi))
       i += 1;
   return i - 1;

}

fn sqrtType(comptime T: type) type {

   const t = assertInt(T);
   return meta.Int(.unsigned, t.bits / 2);

}

// stack type (actually just an array list) fn arrayList(comptime Int: type) type {

   return [listSize(Int)]sqrtType(Int);

}

// given an upper bound, max, return the most restrictive sieving data type. pub fn autoSieveType(comptime max: u64) type {

   if (max == 0)
       @compileError("The maximum sieving size must be non-zero.");
   var bit_len = 64 - @clz(u64, max);
   if (max & (max - 1) == 0) // power of two
       bit_len -= 1;
   if (bit_len % 2 == 1)
       bit_len += 1;
   if (bit_len < 4)
       bit_len = 4;
   return meta.Int(.unsigned, bit_len);

}

test "type meta functions" {

   const expect = std.testing.expect;
   try expect(sqrtType(u20) == u10);
   try expect(autoSieveType(8000) == u14);
   try expect(autoSieveType(9000) == u14);
   try expect(autoSieveType(16384) == u14);
   try expect(autoSieveType(16385) == u16);
   try expect(autoSieveType(32768) == u16);
   try expect(autoSieveType(1000) == u10);
   try expect(autoSieveType(10) == u4);
   try expect(autoSieveType(4) == u4);
   try expect(autoSieveType(std.math.maxInt(u32)) == u32);
   try expect(listSize(u64) == 14);
   try expect(listSize(u32) == 8);
   try expect(@sizeOf(arrayList(u32)) == 16);
   try expect(@sizeOf(arrayList(u36)) == 36);
   try expect(@sizeOf(arrayList(u64)) == 56);

}

pub fn PrimeGen(comptime Int: type) type {

   _ = assertInt(Int);
   return struct {
       const Self = @This();
       const Sieve = std.ArrayList(arrayList(Int));
       sieve: Sieve,
       count: usize,
       candidate: Int,
       rt: sqrtType(Int),
       sq: Int,
       pos: usize,
       // grow the sieve by a comptime fixed amount
       fn growBy(self: *Self, comptime n: usize) !void {
           var chunk: [n]arrayList(Int) = undefined;
           for (chunk) |*a|
               mem.set(sqrtType(Int), a, 0);
           try self.sieve.appendSlice(&chunk);
       }
       // add a known prime number to the sieve at postion k
       fn add(self: *Self, p: sqrtType(Int), k: usize) void {
           for (self.sieve.items[k]) |*x|
               if (x.* == 0) {
                   x.* = p;
                   return;
               };
           // each bucket is precalculated for the max size.
           // If we get here, there's been a mistake somewhere.
           unreachable;
       }
       pub fn init(alloc: *mem.Allocator) Self {
           return Self{
               .count = 0,
               .sieve = Sieve.init(alloc),
               .candidate = 3,
               .rt = 3,
               .sq = 9,
               .pos = 0,
           };
       }
       pub fn deinit(self: *Self) void {
           self.sieve.deinit();
       }
       pub fn next(self: *Self) !?Int {
           self.count += 1;
           if (self.count == 1) {
               try self.growBy(1); // prepare sieve
               return 2;
           } else {
               var is_prime = false;
               while (!is_prime) {
                   is_prime = true;
                   // Step 1: check the list at self.pos; if there are divisors then
                   // the candidate is not prime.  Move each divisor to its next multiple
                   // in the sieve.
                   //
                   if (self.sieve.items[self.pos][0] != 0) {
                       is_prime = false;
                       for (self.sieve.items[self.pos]) |*x| {
                           const p = x.*;
                           x.* = 0;
                           if (p == 0)
                               break;
                           self.add(p, (p + self.pos) % self.sieve.items.len);
                       }
                   }
                   // Step 2: If we've hit the next perfect square, and we thought the number
                   // was prime from step 1, note that it wasn't prime but rather was a non p-smooth
                   // number.  Add the square root to the sieve.  In any case, look ahead to the next
                   // square number.
                   //
                   if (self.candidate == self.sq) {
                       if (is_prime) {
                           is_prime = false;
                           self.add(self.rt, (self.pos + self.rt) % self.sieve.items.len);
                       }
                       // advance to the next root; if doing so would cause overflow then just ignore it,
                       // since we'll never see the next square.
                       //
                       var rt: sqrtType(Int) = undefined;
                       if (!@addWithOverflow(sqrtType(Int), self.rt, 2, &rt)) {
                           self.rt = rt;
                           self.sq = @as(Int, rt) * rt;
                       }
                   }
                   // advance the iterator; Note if we overflow, the candidate cannot be prime
                   // since the bit count must be even and all integers of the form 2^n - 1 with
                   // even n (except 2) are composite.
                   //
                   if (@addWithOverflow(Int, self.candidate, 2, &self.candidate)) {
                       std.debug.assert(!is_prime);
                       return null;
                   }
                   self.pos += 1;
                   if (self.pos == self.sieve.items.len) {
                       // expand the array by 2 to maintain the invariant: sieve.items.len > √candidate
                       try self.growBy(2);
                       self.pos = 0;
                   }
               }
               return self.candidate - 2;
           }
       }
   };

} </lang> The driver code is the same with this version, since the interface is unchanged from the O'Neil sieve.

Output:
The first 20 primes: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
The primes between 100 and 150: 101 103 107 109 113 127 131 137 139 149
There are 30 primes between 7700 and 8000.
The 10th prime is 29
The 100th prime is 541
The 1000th prime is 7919
The 10000th prime is 104729
The 100000th prime is 1299709
The 1000000th prime is 15485863
The 10000000th prime is 179424673
The 100000000th prime is 2038074743

zkl

<lang zkl>// http://stackoverflow.com/revisions/10733621/4

fcn postponed_sieve(){ # postponed sieve, by Will Ness

  vm.yield(2); vm.yield(3);	  # original code David Eppstein, 
  vm.yield(5); vm.yield(7);      #        ActiveState Recipe 2002
  D:=Dictionary();               
  ps:=Utils.Generator(postponed_sieve);  # a separate Primes Supply:
  p:=ps.pump(2,Void);            # (3) a Prime to add to dict
  q:=p*p;                        # (9) when its sQuare is 
  c:=9;                          # the next Candidate
  while(1){
     if (not D.holds(c)){        # not a multiple of any prime seen so far:
        if (c < q) vm.yield(c);  #   a prime, or

else{ # (c==q): # the next prime's square:

           add(D,c + 2*p,2*p);   #     (9+6,6 : 15,21,27,33,...)

p=ps.next(); # (5) q=p*p; # (25) }

     }else{                      # 'c' is a composite:

s := D.pop(c); # step of increment add(D,c + s,s); # next multiple, same step

     }
     c += 2;                     # next odd candidate
  }

}

fcn add(D,x,s){ # make no multiple keys in Dict

  while(D.holds(x)){ x += s }    # increment by the given step
  D[x] = s;

}</lang> <lang zkl>primes:=Utils.Generator(postponed_sieve); primes.walk(20).println(); // first 20 primes

primes.pump(List,fcn(p){ // the primes between 100 & 150

  if (p<100) Void.Skip else if(p>150) Void.Stop else p

}).println();

primes.reduce(fcn(n,p){ // count of primes between 7700 & 8000

  if (p<=7700) n else if(p>8000) Void.Stop else n+1

},0).println();

primes=Utils.Generator(postponed_sieve); // new Generator primes.drop(0d9_999); primes.next().println(); // 10,000th prime

  // or to carry on until the 100,000th:

primes.pump(Void,'wrap(p){ primes.n<=0d100_000 and p or Void.Stop }).println();</lang>

Output:
L(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71)
L(101,103,107,109,113,127,131,137,139,149)
30
104729
1299709

Using GMP (GNU Multiple Precision Arithmetic Library, probabilistic primes), this is a direct drop in for the above:

Library: GMP

<lang zkl>var [const] BN=Import.lib("zklBigNum"); // libGMP bigPrimes:=Walker(fcn(p){ p.nextPrime().copy(); }.fp(BN(1)));</lang> For example: <lang zkl>bigPrimes.walk(20).println(); // first 20 primes bigPrimes.pump(Void,'wrap(p){ bigPrimes.n<=0d10_000 and p or Void.Stop }).println();</lang>

Output:
L(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71)
104729