Chernick's Carmichael numbers: Difference between revisions

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(takes ~6.5 minutes)
(takes ~6.5 minutes)

=={{header|Nim}}==
Until a(9) a simple primality test using divisions by odd numbers is sufficient. But for a(10), it is necessary to improve the test. We have used here some optimizations found in other solutions:
:– eliminating multiples of 3, 5, 7, 11, 13, 17, 19, 23;
:– using a probability test which implies to use big integers; so, we have to convert the number to check to a big integer;
:– for n >= 5, checking only values of m which are multiple of 5 (in fact, we check only the multiples of 5 × 2^(n-4).

With these optimizations, the program executes in 4-5 minutes.

<lang Nim>import strutils, sequtils
import bignum

const
Max = 10
Factors: array[3..Max, int] = [1, 1, 2, 4, 8, 16, 32, 64] # 1 for n=3 then 2^(n-4).
FirstPrimes = [3, 5, 7, 11, 13, 17, 19, 23]

#---------------------------------------------------------------------------------------------------

iterator factors(n, m: Natural): Natural =
## Yield the factors of U(n, m).

yield 6 * m + 1
yield 12 * m + 1
var k = 2
for _ in 1..(n - 2):
yield 9 * k * m + 1
inc k, k

#---------------------------------------------------------------------------------------------------

proc mayBePrime(n: int): bool =
## First primality test.

if n < 23: return true

for p in FirstPrimes:
if n mod p == 0:
return false

result = true

#---------------------------------------------------------------------------------------------------

proc isChernick(n, m: Natural): bool =
## Check if U(N, m) if a Chernick-Carmichael number.

# Use the first and quick test.
for factor in factors(n, m):
if not factor.mayBePrime():
return false

# Use the slow probability test (need to use a big int).
for factor in factors(n, m):
if probablyPrime(newInt(factor), 25) == 0:
return false

result = true

#---------------------------------------------------------------------------------------------------

proc a(n: Natural): tuple[m: Natural, factors: seq[Natural]] =
## For a given "n", find the smallest Charnick-Carmichael number.

var m: Natural = 0
var incr = (if n >= 5: 5 else: 1) * Factors[n] # For n>= 5, a(n) is a multiple of 5.

while true:
inc m, incr
if isChernick(n, m):
return (m, toSeq(factors(n, m)))

#———————————————————————————————————————————————————————————————————————————————————————————————————

import strformat

for n in 3..Max:
let (m, factors) = a(n)

stdout.write fmt"a({n}) = U({n}, {m}) = "
var s = ""
for factor in factors:
s.addSep(" × ")
s.add($factor)
stdout.write s, '\n'</lang>

{{out}}
<pre>a(3) = U(3, 1) = 7 × 13 × 19
a(4) = U(4, 1) = 7 × 13 × 19 × 37
a(5) = U(5, 380) = 2281 × 4561 × 6841 × 13681 × 27361
a(6) = U(6, 380) = 2281 × 4561 × 6841 × 13681 × 27361 × 54721
a(7) = U(7, 780320) = 4681921 × 9363841 × 14045761 × 28091521 × 56183041 × 112366081 × 224732161
a(8) = U(8, 950560) = 5703361 × 11406721 × 17110081 × 34220161 × 68440321 × 136880641 × 273761281 × 547522561
a(9) = U(9, 950560) = 5703361 × 11406721 × 17110081 × 34220161 × 68440321 × 136880641 × 273761281 × 547522561 × 1095045121
a(10) = U(10, 3208386195840) = 19250317175041 × 38500634350081 × 57750951525121 × 115501903050241 × 231003806100481 × 462007612200961 × 924015224401921 × 1848030448803841 × 3696060897607681 × 7392121795215361</pre>


=={{header|PARI/GP}}==
=={{header|PARI/GP}}==

Revision as of 17:50, 3 November 2020

Task
Chernick's Carmichael numbers
You are encouraged to solve this task according to the task description, using any language you may know.

In 1939, Jack Chernick proved that, for n ≥ 3 and m ≥ 1: 

   U(n, m) = (6m + 1) * (12m + 1) * Product_{i=1..n-2} (2^i * 9m + 1)

is a Carmichael number if all the factors are primes and, for n > 4, m is a multiple of 2^(n-4).


Example
   U(3, m) = (6m + 1) * (12m + 1) * (18m + 1)
   U(4, m) = U(3, m) * (2^2 * 9m + 1)
   U(5, m) = U(4, m) * (2^3 * 9m + 1)
   ...
   U(n, m) = U(n-1, m) * (2^(n-2) * 9m + 1)
  • The smallest Chernick's Carmichael number with 3 prime factors, is: U(3, 1) = 1729.
  • The smallest Chernick's Carmichael number with 4 prime factors, is: U(4, 1) = 63973.
  • The smallest Chernick's Carmichael number with 5 prime factors, is: U(5, 380) = 26641259752490421121.


For n = 5, the smallest number m that satisfy Chernick's conditions, is m = 380, therefore U(5, 380) is the smallest Chernick's Carmichael number with 5 prime factors.

U(5, 380) is a Chernick's Carmichael number because m = 380 is a multiple of 2^(n-4), where n = 5, and the factors { (6*380 + 1), (12*380 + 1), (18*380 + 1), (36*380 + 1), (72*380 + 1) } are all prime numbers.


Task

For n ≥ 3, let a(n) be the smallest Chernick's Carmichael number with n prime factors.

  • Compute a(n) for n = 3..9.
  • Optional: find a(10).


Note: it's perfectly acceptable to show the terms in factorized form: 

 a(3) = 7 * 13 * 19
 a(4) = 7 * 13 * 19 * 37
 a(5) = 2281 * 4561 * 6841 * 13681 * 27361
 ...


See also


Related tasks



C

Library: GMP

<lang c>#include <stdio.h>

  1. include <stdlib.h>
  2. include <gmp.h>

typedef unsigned long long int u64;

  1. define TRUE 1
  2. define FALSE 0

int primality_pretest(u64 k) { 

   if (!(k %  3) || !(k %  5) || !(k %  7) || !(k % 11) || !(k % 13) || !(k % 17) || !(k % 19) || !(k % 23)) return (k <= 23);
   return TRUE;

}

int probprime(u64 k, mpz_t n) { 

   mpz_set_ui(n, k);
   return mpz_probab_prime_p(n, 0);

}

int is_chernick(int n, u64 m, mpz_t z) { 

   u64 t = 9 * m;
   if (primality_pretest(6 * m + 1) == FALSE) return FALSE;
   if (primality_pretest(12 * m + 1) == FALSE) return FALSE;
   for (int i = 1; i <= n - 2; i++) if (primality_pretest((t << i) + 1) == FALSE) return FALSE;
   if (probprime(6 * m + 1, z) == FALSE) return FALSE;
   if (probprime(12 * m + 1, z) == FALSE) return FALSE;
   for (int i = 1; i <= n - 2; i++) if (probprime((t << i) + 1, z) == FALSE) return FALSE;
   return TRUE;

}

int main(int argc, char const *argv[]) { 

   mpz_t z;
   mpz_inits(z, NULL);

   for (int n = 3; n <= 10; n ++) {
       u64 multiplier = (n > 4) ? (1 << (n - 4)) : 1;

       if (n > 5) multiplier *= 5;

       for (u64 k = 1; ; k++) {
           u64 m = k * multiplier;

           if (is_chernick(n, m, z) == TRUE) {
               printf("a(%d) has m = %llu\n", n, m);
               break;
           }
       }
   }

   return 0;

}</lang>

Output:
a(3) has m = 1
a(4) has m = 1
a(5) has m = 380
a(6) has m = 380
a(7) has m = 780320
a(8) has m = 950560
a(9) has m = 950560
a(10) has m = 3208386195840

C++

Library: GMP

<lang cpp>#include <gmp.h>

  1. include <iostream>

using namespace std;

typedef unsigned long long int u64;

bool primality_pretest(u64 k) { // for k > 23 

   if (!(k %  3) || !(k %  5) || !(k %  7) || !(k % 11) ||
       !(k % 13) || !(k % 17) || !(k % 19) || !(k % 23)
   ) {
       return (k <= 23);
   }

   return true;

}

bool probprime(u64 k, mpz_t n) { 

   mpz_set_ui(n, k);
   return mpz_probab_prime_p(n, 0);

}

bool is_chernick(int n, u64 m, mpz_t z) { 

   if (!primality_pretest(6 * m + 1)) {
       return false;
   }

   if (!primality_pretest(12 * m + 1)) {
       return false;
   }

   u64 t = 9 * m;

   for (int i = 1; i <= n - 2; i++) {
       if (!primality_pretest((t << i) + 1)) {
           return false;
       }
   }

   if (!probprime(6 * m + 1, z)) {
       return false;
   }

   if (!probprime(12 * m + 1, z)) {
       return false;
   }

   for (int i = 1; i <= n - 2; i++) {
       if (!probprime((t << i) + 1, z)) {
           return false;
       }
   }

   return true;

}

int main() { 

   mpz_t z;
   mpz_inits(z, NULL);

   for (int n = 3; n <= 10; n++) {

       // `m` is a multiple of 2^(n-4), for n > 4
       u64 multiplier = (n > 4) ? (1 << (n - 4)) : 1;

       // For n > 5, m is also a multiple of 5
       if (n > 5) {
           multiplier *= 5;
       }

       for (u64 k = 1; ; k++) {

           u64 m = k * multiplier;

           if (is_chernick(n, m, z)) {
               cout << "a(" << n << ") has m = " << m << endl;
               break;
           }
       }
   }

   return 0;

}</lang>

Output:
a(3) has m = 1
a(4) has m = 1
a(5) has m = 380
a(6) has m = 380
a(7) has m = 780320
a(8) has m = 950560
a(9) has m = 950560
a(10) has m = 3208386195840

(takes ~3.5 minutes)

F#

This task uses Extensible Prime Generator (F#) <lang fsharp> // Generate Chernick's Carmichael numbers. Nigel Galloway: June 1st., 2019 let fMk m k=isPrime(6*m+1) && isPrime(12*m+1) && [1..k-2]|>List.forall(fun n->isPrime(9*(pown 2 n)*m+1)) let fX k=Seq.initInfinite(fun n->(n+1)*(pown 2 (k-4))) |> Seq.filter(fun n->fMk n k ) let cherCar k=let m=Seq.head(fX k) in printfn "m=%d primes -> %A " m ([6*m+1;12*m+1]@List.init(k-2)(fun n->9*(pown 2 (n+1))*m+1)) [4..9] |> Seq.iter cherCar </lang>

Output:
cherCar(4): m=1 primes -> [7; 13; 19; 37] 
cherCar(5): m=380 primes -> [2281; 4561; 6841; 13681; 27361] 
cherCar(6): m=380 primes -> [2281; 4561; 6841; 13681; 27361; 54721] 
cherCar(7): m=780320 primes -> [4681921; 9363841; 14045761; 28091521; 56183041; 112366081; 224732161] 
cherCar(8): m=950560 primes -> [5703361; 11406721; 17110081; 34220161; 68440321; 136880641; 273761281; 547522561] 
cherCar(9): m=950560 primes -> [5703361; 11406721; 17110081; 34220161; 68440321; 136880641; 273761281; 547522561; 1095045121]

Go

Basic only

<lang go>package main

import ( 

   "fmt"
   "math/big"

)

var ( 

   zero = new(big.Int)
   prod = new(big.Int)
   fact = new(big.Int)

)

func ccFactors(n, m uint64) (*big.Int, bool) { 

   prod.SetUint64(6*m + 1)
   if !prod.ProbablyPrime(0) {
       return zero, false
   }
   fact.SetUint64(12*m + 1)
   if !fact.ProbablyPrime(0) { // 100% accurate up to 2 ^ 64
       return zero, false
   }
   prod.Mul(prod, fact)
   for i := uint64(1); i <= n-2; i++ {
       fact.SetUint64((1<<i)*9*m + 1)
       if !fact.ProbablyPrime(0) {
           return zero, false
       }
       prod.Mul(prod, fact)
   }
   return prod, true

}

func ccNumbers(start, end uint64) { 

   for n := start; n <= end; n++ {
       m := uint64(1)
       if n > 4 {
           m = 1 << (n - 4)
       }
       for {
           num, ok := ccFactors(n, m)
           if ok {
               fmt.Printf("a(%d) = %d\n", n, num)
               break
           }
           if n <= 4 {
               m++
           } else {
               m += 1 << (n - 4)
           }
       }
   }

}

func main() { 

   ccNumbers(3, 9)

}</lang>

Output:
a(3) = 1729
a(4) = 63973
a(5) = 26641259752490421121
a(6) = 1457836374916028334162241
a(7) = 24541683183872873851606952966798288052977151461406721
a(8) = 53487697914261966820654105730041031613370337776541835775672321
a(9) = 58571442634534443082821160508299574798027946748324125518533225605795841

Basic plus optional

Library: GMP(Go wrapper)


To reach a(10) in a reasonable time, a much more efficient approach is needed.

The following version takes account of the optimizations referred to in the Talk page and previewed in the C++ entry above.

It also uses a wrapper for the C library, GMP, which despite the overhead of cgo is still much faster than Go's native big.Int library.

The resulting executable is several hundred times faster than before and, even on my modest Celeron @1.6GHZ, reaches a(9) in under 10ms and a(10) in about 22 minutes. <lang go>package main

import ( 

   "fmt"
   big "github.com/ncw/gmp"

)

const ( 

   min = 3
   max = 10

)

var ( 

   prod       = new(big.Int)
   fact       = new(big.Int)
   factors    = [max]uint64{}
   bigFactors = [max]*big.Int{}

)

func init() { 

   for i := 0; i < max; i++ {
       bigFactors[i] = big.NewInt(0)
   }

}

func isPrimePretest(k uint64) bool { 

   if k%3 == 0 || k%5 == 0 || k%7 == 0 || k%11 == 0 ||
       k%13 == 0 || k%17 == 0 || k%19 == 0 || k%23 == 0 {
       return k <= 23
   }
   return true

}

func ccFactors(n, m uint64) bool { 

   if !isPrimePretest(6*m + 1) {
       return false
   }
   if !isPrimePretest(12*m + 1) {
       return false
   }
   factors[0] = 6*m + 1
   factors[1] = 12*m + 1
   t := 9 * m
   for i := uint64(1); i <= n-2; i++ {
       tt := (t << i) + 1
       if !isPrimePretest(tt) {
           return false
       }
       factors[i+1] = tt
   }

   for i := 0; i < int(n); i++ {
       fact.SetUint64(factors[i])
       if !fact.ProbablyPrime(0) {
           return false
       }
       bigFactors[i].Set(fact)
   }
   return true

}

func prodFactors(n uint64) *big.Int { 

   prod.Set(bigFactors[0])
   for i := 1; i < int(n); i++ {
       prod.Mul(prod, bigFactors[i])
   }
   return prod

}

func ccNumbers(start, end uint64) { 

   for n := start; n <= end; n++ {
       mult := uint64(1)
       if n > 4 {
           mult = 1 << (n - 4)
       }
       if n > 5 {
           mult *= 5
       }
       m := mult
       for {
           if ccFactors(n, m) {
               num := prodFactors(n)
               fmt.Printf("a(%d) = %d\n", n, num)
               fmt.Printf("m(%d) = %d\n", n, m)
               fmt.Println("Factors:", factors[:n], "\n")
               break
           }
           m += mult
       }
   }

}

func main() { 

   ccNumbers(min, max)

}</lang>

Output:
a(3) = 1729
m(3) = 1
Factors: [7 13 19] 

a(4) = 63973
m(4) = 1
Factors: [7 13 19 37] 

a(5) = 26641259752490421121
m(5) = 380
Factors: [2281 4561 6841 13681 27361] 

a(6) = 1457836374916028334162241
m(6) = 380
Factors: [2281 4561 6841 13681 27361 54721] 

a(7) = 24541683183872873851606952966798288052977151461406721
m(7) = 780320
Factors: [4681921 9363841 14045761 28091521 56183041 112366081 224732161] 

a(8) = 53487697914261966820654105730041031613370337776541835775672321
m(8) = 950560
Factors: [5703361 11406721 17110081 34220161 68440321 136880641 273761281 547522561] 

a(9) = 58571442634534443082821160508299574798027946748324125518533225605795841
m(9) = 950560
Factors: [5703361 11406721 17110081 34220161 68440321 136880641 273761281 547522561 1095045121] 

a(10) = 24616075028246330441656912428380582403261346369700917629170235674289719437963233744091978433592331048416482649086961226304033068172880278517841921
m(10) = 3208386195840
Factors: [19250317175041 38500634350081 57750951525121 115501903050241 231003806100481 462007612200961 924015224401921 1848030448803841 3696060897607681 7392121795215361]

Java

<lang java> import java.math.BigInteger; import java.util.ArrayList; import java.util.List;

public class ChernicksCarmichaelNumbers { 

   public static void main(String[] args) {
       for ( long n = 3 ; n < 10 ; n++ ) {
           long m = 0;
           boolean foundComposite = true;
           List<Long> factors = null;
           while ( foundComposite ) {
               m += (n <= 4 ? 1 : (long) Math.pow(2, n-4) * 5);
               factors = U(n, m);
               foundComposite = false;
               for ( long factor : factors ) {
                   if ( ! isPrime(factor) ) {
                       foundComposite = true;
                       break;
                   }
               }
           }
           System.out.printf("U(%d, %d) = %s = %s %n", n, m, display(factors), multiply(factors));
       }
   }
   
   private static String display(List<Long> factors) {
       return factors.toString().replace("[", "").replace("]", "").replaceAll(", ", " * ");
   }
   
   private static BigInteger multiply(List<Long> factors) {
       BigInteger result = BigInteger.ONE;
       for ( long factor : factors ) {
           result = result.multiply(BigInteger.valueOf(factor));
       }
       return result;
   }
   
   private static List<Long> U(long n, long m) {
       List<Long> factors = new ArrayList<>();
       factors.add(6*m + 1);
       factors.add(12*m + 1);
       for ( int i = 1 ; i <= n-2 ; i++ ) {
           factors.add(((long)Math.pow(2, i)) * 9 * m + 1);
       }
       return factors;
   }

   private static final int MAX = 100_000;
   private static final boolean[] primes = new boolean[MAX];
   private static boolean SIEVE_COMPLETE = false;
   
   private static final boolean isPrimeTrivial(long test) {
       if ( ! SIEVE_COMPLETE ) {
           sieve();
           SIEVE_COMPLETE = true;
       }
       return primes[(int) test];
   }
   
   private static final void sieve() {
       //  primes
       for ( int i = 2 ; i < MAX ; i++ ) {
           primes[i] = true;            
       }
       for ( int i = 2 ; i < MAX ; i++ ) {
           if ( primes[i] ) {
               for ( int j = 2*i ; j < MAX ; j += i ) {
                   primes[j] = false;
               }
           }
       }
   }

   //  See http://primes.utm.edu/glossary/page.php?sort=StrongPRP
   public static final boolean isPrime(long testValue) {
       if ( testValue == 2 ) return true;
       if ( testValue % 2 == 0 ) return false;
       if ( testValue <= MAX ) return isPrimeTrivial(testValue);
       long d = testValue-1;
       int s = 0;
       while ( d % 2 == 0 ) {
           s += 1;
           d /= 2;
       }
       if ( testValue < 1373565L ) {
           if ( ! aSrp(2, s, d, testValue) ) {
               return false;
           }
           if ( ! aSrp(3, s, d, testValue) ) {
               return false;
           }
           return true;
       }
       if ( testValue < 4759123141L ) {
           if ( ! aSrp(2, s, d, testValue) ) {
               return false;
           }
           if ( ! aSrp(7, s, d, testValue) ) {
               return false;
           }
           if ( ! aSrp(61, s, d, testValue) ) {
               return false;
           }
           return true;
       }
       if ( testValue < 10000000000000000L ) {
           if ( ! aSrp(3, s, d, testValue) ) {
               return false;
           }
           if ( ! aSrp(24251, s, d, testValue) ) {
               return false;
           }
           return true;
       }
       //  Try 5 "random" primes
       if ( ! aSrp(37, s, d, testValue) ) {
           return false;
       }
       if ( ! aSrp(47, s, d, testValue) ) {
           return false;
       }
       if ( ! aSrp(61, s, d, testValue) ) {
           return false;
       }
       if ( ! aSrp(73, s, d, testValue) ) {
           return false;
       }
       if ( ! aSrp(83, s, d, testValue) ) {
           return false;
       }
       //throw new RuntimeException("ERROR isPrime:  Value too large = "+testValue);
       return true;
   }

   private static final boolean aSrp(int a, int s, long d, long n) {
       long modPow = modPow(a, d, n);
       //System.out.println("a = "+a+", s = "+s+", d = "+d+", n = "+n+", modpow = "+modPow);
       if ( modPow == 1 ) {
           return true;
       }
       int twoExpR = 1;
       for ( int r = 0 ; r < s ; r++ ) {
           if ( modPow(modPow, twoExpR, n) == n-1 ) {
               return true;
           }
           twoExpR *= 2;
       }
       return false;
   }
   
   private static final long SQRT = (long) Math.sqrt(Long.MAX_VALUE);
   
   public static final long modPow(long base, long exponent, long modulus) {
       long result = 1;
       while ( exponent > 0 ) {
           if ( exponent % 2 == 1 ) {
               if ( result > SQRT || base > SQRT ) {
                   result = multiply(result, base, modulus);
               }
               else {
                   result = (result * base) % modulus;
               }
           }
           exponent >>= 1;
           if ( base > SQRT ) {
               base = multiply(base, base, modulus);
           }
           else {
               base = (base * base) % modulus;
           }
       }
       return result;
   }



   //  Result is a*b % mod, without overflow.
   public static final long multiply(long a, long b, long modulus) {
       long x = 0;
       long y = a % modulus;
       long t;
       while ( b > 0 ) {
           if ( b % 2 == 1 ) {
               t = x + y;
               x = (t > modulus ? t-modulus : t);
           }
           t = y << 1;
           y = (t > modulus ? t-modulus : t);
           b >>= 1;
       }
       return x % modulus;
   }

} </lang>

Output:
U(3, 1) = 7 * 13 * 19 = 1729 
U(4, 1) = 7 * 13 * 19 * 37 = 63973 
U(5, 380) = 2281 * 4561 * 6841 * 13681 * 27361 = 26641259752490421121 
U(6, 380) = 2281 * 4561 * 6841 * 13681 * 27361 * 54721 = 1457836374916028334162241 
U(7, 780320) = 4681921 * 9363841 * 14045761 * 28091521 * 56183041 * 112366081 * 224732161 = 24541683183872873851606952966798288052977151461406721 
U(8, 950560) = 5703361 * 11406721 * 17110081 * 34220161 * 68440321 * 136880641 * 273761281 * 547522561 = 53487697914261966820654105730041031613370337776541835775672321 
U(9, 950560) = 5703361 * 11406721 * 17110081 * 34220161 * 68440321 * 136880641 * 273761281 * 547522561 * 1095045121 = 58571442634534443082821160508299574798027946748324125518533225605795841

Julia

<lang julia>using Primes

function trial_pretest(k::UInt64) 

   if ((k %  3)==0 || (k %  5)==0 || (k %  7)==0 || (k % 11)==0 ||
       (k % 13)==0 || (k % 17)==0 || (k % 19)==0 || (k % 23)==0)
       return (k <= 23)
   end

   return true

end

function gcd_pretest(k::UInt64) 

   if (k <= 107)
       return true
   end

   gcd(29*31*37*41*43*47*53*59*61*67, k) == 1 &&
   gcd(71*73*79*83*89*97*101*103*107, k) == 1

end

function is_chernick(n::Int64, m::UInt64) 

   t = 9*m

   if (!trial_pretest(6*m + 1))
       return false
   end

   if (!trial_pretest(12*m + 1))
       return false
   end

   for i in 1:n-2
       if (!trial_pretest((t << i) + 1))
           return false
       end
   end

   if (!gcd_pretest(6*m + 1))
       return false
   end

   if (!gcd_pretest(12*m + 1))
       return false
   end

   for i in 1:n-2
       if (!gcd_pretest((t << i) + 1))
           return false
       end
   end

   if (!isprime(6*m + 1))
       return false
   end

   if (!isprime(12*m + 1))
       return false
   end

   for i in 1:n-2
       if (!isprime((t << i) + 1))
           return false
       end
   end

   return true

end

function chernick_carmichael(n::Int64, m::UInt64) 

   prod = big(1)

   prod *= 6*m + 1
   prod *= 12*m + 1

   for i in 1:n-2
       prod *= ((big(9)*m)<<i) + 1
   end

   prod

end

function cc_numbers(from, to) 

   for n in from:to

       multiplier = 1

       if (n > 4) multiplier = 1 << (n-4) end
       if (n > 5) multiplier *= 5 end

       m = UInt64(multiplier)

       while true

           if (is_chernick(n, m))
               println("a(", n, ") = ", chernick_carmichael(n, m))
               break
           end

           m += multiplier
       end
   end

end

cc_numbers(3, 10)</lang>

Output:
a(3) = 1729
a(4) = 63973
a(5) = 26641259752490421121
a(6) = 1457836374916028334162241
a(7) = 24541683183872873851606952966798288052977151461406721
a(8) = 53487697914261966820654105730041031613370337776541835775672321
a(9) = 58571442634534443082821160508299574798027946748324125518533225605795841
a(10) = 24616075028246330441656912428380582403261346369700917629170235674289719437963233744091978433592331048416482649086961226304033068172880278517841921

(takes ~6.5 minutes)

Nim

Until a(9) a simple primality test using divisions by odd numbers is sufficient. But for a(10), it is necessary to improve the test. We have used here some optimizations found in other solutions:

– eliminating multiples of 3, 5, 7, 11, 13, 17, 19, 23;
– using a probability test which implies to use big integers; so, we have to convert the number to check to a big integer;
– for n >= 5, checking only values of m which are multiple of 5 (in fact, we check only the multiples of 5 × 2^(n-4).

With these optimizations, the program executes in 4-5 minutes.

<lang Nim>import strutils, sequtils import bignum

const 

 Max = 10
 Factors: array[3..Max, int] = [1, 1, 2, 4, 8, 16, 32, 64]   # 1 for n=3 then 2^(n-4).
 FirstPrimes = [3, 5, 7, 11, 13, 17, 19, 23]
  1. ---------------------------------------------------------------------------------------------------

iterator factors(n, m: Natural): Natural = 

 ## Yield the factors of U(n, m).

 yield 6 * m + 1
 yield 12 * m + 1
 var k = 2
 for _ in 1..(n - 2):
   yield 9 * k * m + 1
   inc k, k
  1. ---------------------------------------------------------------------------------------------------

proc mayBePrime(n: int): bool = 

 ## First primality test.

 if n < 23: return true

 for p in FirstPrimes:
   if n mod p == 0:
     return false

 result = true
  1. ---------------------------------------------------------------------------------------------------

proc isChernick(n, m: Natural): bool = 

 ## Check if U(N, m) if a Chernick-Carmichael number.

 # Use the first and quick test.
 for factor in factors(n, m):
   if not factor.mayBePrime():
     return false

 # Use the slow probability test (need to use a big int).
 for factor in factors(n, m):
   if probablyPrime(newInt(factor), 25) == 0:
     return false

 result = true
  1. ---------------------------------------------------------------------------------------------------

proc a(n: Natural): tuple[m: Natural, factors: seq[Natural]] = 

 ## For a given "n", find the smallest Charnick-Carmichael number.

 var m: Natural = 0
 var incr = (if n >= 5: 5 else: 1) * Factors[n]  # For n>= 5, a(n) is a multiple of 5.

 while true:
   inc m, incr
   if isChernick(n, m):
     return (m, toSeq(factors(n, m)))
  1. ———————————————————————————————————————————————————————————————————————————————————————————————————

import strformat

for n in 3..Max: 

 let (m, factors) = a(n)

 stdout.write fmt"a({n}) = U({n}, {m}) = "
 var s = ""
 for factor in factors:
   s.addSep(" × ")
   s.add($factor)
 stdout.write s, '\n'</lang>
Output:
a(3) = U(3, 1) = 7 × 13 × 19
a(4) = U(4, 1) = 7 × 13 × 19 × 37
a(5) = U(5, 380) = 2281 × 4561 × 6841 × 13681 × 27361
a(6) = U(6, 380) = 2281 × 4561 × 6841 × 13681 × 27361 × 54721
a(7) = U(7, 780320) = 4681921 × 9363841 × 14045761 × 28091521 × 56183041 × 112366081 × 224732161
a(8) = U(8, 950560) = 5703361 × 11406721 × 17110081 × 34220161 × 68440321 × 136880641 × 273761281 × 547522561
a(9) = U(9, 950560) = 5703361 × 11406721 × 17110081 × 34220161 × 68440321 × 136880641 × 273761281 × 547522561 × 1095045121
a(10) = U(10, 3208386195840) = 19250317175041 × 38500634350081 × 57750951525121 × 115501903050241 × 231003806100481 × 462007612200961 × 924015224401921 × 1848030448803841 × 3696060897607681 × 7392121795215361

PARI/GP

<lang parigp> cherCar(n)={ 

 my(C=vector(n));C[1]=6; C[2]=12; for(g=3,n,C[g]=2^(g-2)*9);
 my(i=1); my(N(g)=while(i<=n&ispseudoprime(g*C[i]+1),i=i+1); return(i>n));
    i=1;  my(G(g)=while(i<=n&isprime(g*C[i]+1),i=i+1); return(i>n));
 i=1; if(n>4,i=2^(n-4)); if(n>5,i=i*5); my(m=i); while(!(N(m)&G(m)),m=m+i);
 printf("cherCar(%d): m = %d\n",n,m)}

for(x=3,9,cherCar(x)) </lang>

Output:
cherCar(3): m = 1
cherCar(4): m = 1
cherCar(5): m = 380
cherCar(6): m = 380
cherCar(7): m = 780320
cherCar(8): m = 950560
cherCar(9): m = 950560
cherCar(10): m = 3208386195840

Perl

Library: ntheory

<lang perl>use 5.020; use warnings; use ntheory qw/:all/; use experimental qw/signatures/;

sub chernick_carmichael_factors ($n, $m) { 

   (6*$m + 1, 12*$m + 1, (map { (1 << $_) * 9*$m + 1 } 1 .. $n-2));

}

sub chernick_carmichael_number ($n, $callback) { 

   my $multiplier = ($n > 4) ? (1 << ($n-4)) : 1;

   for (my $m = 1 ; ; ++$m) {
       my @f = chernick_carmichael_factors($n, $m * $multiplier);
       next if not vecall { is_prime($_) } @f;
       $callback->(@f);
       last;
   }

}

foreach my $n (3..9) { 

   chernick_carmichael_number($n, sub (@f) { say "a($n) = ", vecprod(@f) });

}</lang>

Output:
a(3) = 1729
a(4) = 63973
a(5) = 26641259752490421121
a(6) = 1457836374916028334162241
a(7) = 24541683183872873851606952966798288052977151461406721
a(8) = 53487697914261966820654105730041031613370337776541835775672321
a(9) = 58571442634534443082821160508299574798027946748324125518533225605795841

Phix

Library: Phix/mpfr
Translation of: Sidef

<lang Phix>function chernick_carmichael_factors(integer n, m) 

   sequence res = {6*m + 1, 12*m + 1}
   for i=1 to n-2 do
       res &= power(2,i) * 9*m + 1
   end for
   return res

end function

include mpfr.e mpz p = mpz_init() randstate state = gmp_randinit_mt()

function m_prime(atom a) 

   mpz_set_d(p,a)
   return mpz_probable_prime_p(p, state)

end function

function is_chernick_carmichael(integer n, m) 

   return iff(n==2 ? m_prime(6*m + 1) and m_prime(12*m + 1)
                   : m_prime(power(2,n-2) * 9*m + 1) and 
                     is_chernick_carmichael(n-1, m))

end function

function chernick_carmichael_number(integer n) 

   integer m = iff(n>4 ? power(2,n-4) : 1), mm = m
   while not is_chernick_carmichael(n, mm) do mm += m end while
   return {chernick_carmichael_factors(n, mm),mm}

end function

for n=3 to 9 do 

   {sequence f, integer m} = chernick_carmichael_number(n)
   mpz_set_si(p,1)
   for i=1 to length(f) do
       mpz_mul_d(p,p,f[i])
       f[i] = sprintf("%d",f[i])
   end for
   printf(1,"U(%d,%d): %s = %s\n",{n,m,mpz_get_str(p),join(f," * ")})

end for</lang>

Output:
U(3,1): 1729 = 7 * 13 * 19
U(4,1): 63973 = 7 * 13 * 19 * 37
U(5,380): 26641259752490421121 = 2281 * 4561 * 6841 * 13681 * 27361
U(6,380): 1457836374916028334162241 = 2281 * 4561 * 6841 * 13681 * 27361 * 54721
U(7,780320): 24541683183872873851606952966798288052977151461406721 = 4681921 * 9363841 * 14045761 * 28091521 * 56183041 * 112366081 * 224732161
U(8,950560): 53487697914261966820654105730041031613370337776541835775672321 = 5703361 * 11406721 * 17110081 * 34220161 * 68440321 * 136880641 * 273761281 * 547522561
U(9,950560): 58571442634534443082821160508299574798027946748324125518533225605795841 = 5703361 * 11406721 * 17110081 * 34220161 * 68440321 * 136880641 * 273761281 * 547522561 * 1095045121

Pleasingly fast, note however that a(10) remains well out of reach / would probably need a complete rewrite.

with cheat

Translation of: C

with added cheat for the a(10) case - I found a nice big prime factor of k and added that on each iteration instead of 1.

You could also use the sequence {1,1,1,1,19,19,4877,457,457,12564169}, if you know a way to build that, and then it wouldn't be cheating anymore... <lang Phix>include mpfr.e sequence ppp = {3,5,7,11,13,17,19,23} function primality_pretest(atom k) 

   for i=1 to length(ppp) do
       if remainder(k,ppp[i])=0 then return (k<=23) end if
   end for
   return true

end function

function probprime(atom k, mpz n) 

   mpz_set_d(n, k)
   return mpz_prime(n)

end function

function is_chernick(integer n, atom m, mpz z) 

   atom t = 9 * m;
   if primality_pretest(6 * m + 1) == false then return false end if
   if primality_pretest(12 * m + 1) == false then return false end if
   for i=1 to n-3 do
       if primality_pretest(t*power(2,i) + 1) == false then return false end if
   end for
   if probprime(6 * m + 1, z) == false then return false end if
   if probprime(12 * m + 1, z) == false then return false end if
   for i=1 to n-2 do
       if probprime(t*power(2,i) + 1, z) == false then return false end if
   end for
   return true

end function

procedure main() 

   atom t0 = time()
   mpz z = mpz_init(0)

   for n=3 to 10 do
       atom multiplier = iff(n>4 ? power(2,n-4) : 1), k = 1
       if n>5 then multiplier *= 5 end if

       while true do
           if n=10 then k += 12564168 end if   -- cheat!
           atom m = k * multiplier;
           if is_chernick(n, m, z) then
               printf(1,"a(%d) has m = %d\n", {n, m})
               exit
           end if
           k += 1
       end while
   end for
   ?elapsed(time()-t0)

end procedure main()</lang>

Output:
a(3) has m = 1
a(4) has m = 1
a(5) has m = 380
a(6) has m = 380
a(7) has m = 780320
a(8) has m = 950560
a(9) has m = 950560
a(10) has m = 3208386195840
"0.1s"

Raku

(formerly Perl 6)

Works with: Rakudo version 2019.03
Translation of: Perl

Use the ntheory library from Perl 5 for primality testing since it is much, much faster than Rakus built-in .is-prime method.

<lang perl6>use Inline::Perl5; use ntheory:from<Perl5> <:all>;

sub chernick-factors ($n, $m) { 

   6*$m + 1, 12*$m + 1, |((1 .. $n-2).map: { (1 +< $_) * 9*$m + 1 } )

}

sub chernick-carmichael-number ($n) { 

   my $multiplier = 1 +< (($n-4) max 0);
   my $iterator   = $n < 5 ?? (1 .. *) !! (1 .. *).map: * * 5;

   $multiplier * $iterator.first: -> $m {
       [&&] chernick-factors($n, $m * $multiplier).map: { is_prime($_) }
   }

}

for 3 .. 9 -> $n { 

   my $m = chernick-carmichael-number($n);
   my @f = chernick-factors($n, $m);
   say "U($n, $m): {[*] @f} = {@f.join(' ⨉ ')}";

}</lang>

Output:
U(3, 1): 1729 = 7 ⨉ 13 ⨉ 19
U(4, 1): 63973 = 7 ⨉ 13 ⨉ 19 ⨉ 37
U(5, 380): 26641259752490421121 = 2281 ⨉ 4561 ⨉ 6841 ⨉ 13681 ⨉ 27361
U(6, 380): 1457836374916028334162241 = 2281 ⨉ 4561 ⨉ 6841 ⨉ 13681 ⨉ 27361 ⨉ 54721
U(7, 780320): 24541683183872873851606952966798288052977151461406721 = 4681921 ⨉ 9363841 ⨉ 14045761 ⨉ 28091521 ⨉ 56183041 ⨉ 112366081 ⨉ 224732161
U(8, 950560): 53487697914261966820654105730041031613370337776541835775672321 = 5703361 ⨉ 11406721 ⨉ 17110081 ⨉ 34220161 ⨉ 68440321 ⨉ 136880641 ⨉ 273761281 ⨉ 547522561
U(9, 950560): 58571442634534443082821160508299574798027946748324125518533225605795841 = 5703361 ⨉ 11406721 ⨉ 17110081 ⨉ 34220161 ⨉ 68440321 ⨉ 136880641 ⨉ 273761281 ⨉ 547522561 ⨉ 1095045121

Sidef

<lang ruby>func chernick_carmichael_factors (n, m) { 

   [6*m + 1, 12*m + 1, {|i| 2**i * 9*m + 1 }.map(1 .. n-2)...]

}

func is_chernick_carmichael (n, m) { 

   (n == 2) ? (is_prime(6*m + 1) && is_prime(12*m + 1))
            : (is_prime(2**(n-2) * 9*m + 1) && __FUNC__(n-1, m))

}

func chernick_carmichael_number(n, callback) { 

   var multiplier = (n>4 ? 2**(n-4) : 1)
   var m = (1..Inf -> first {|m| is_chernick_carmichael(n, m * multiplier) })
   var f = chernick_carmichael_factors(n, m * multiplier)
   callback(f...)

}

for n in (3..9) { 

   chernick_carmichael_number(n, {|*f| say "a(#{n}) = #{f.join(' * ')}" })

}</lang>

Output:
a(3) = 7 * 13 * 19
a(4) = 7 * 13 * 19 * 37
a(5) = 2281 * 4561 * 6841 * 13681 * 27361
a(6) = 2281 * 4561 * 6841 * 13681 * 27361 * 54721
a(7) = 4681921 * 9363841 * 14045761 * 28091521 * 56183041 * 112366081 * 224732161
a(8) = 5703361 * 11406721 * 17110081 * 34220161 * 68440321 * 136880641 * 273761281 * 547522561
a(9) = 5703361 * 11406721 * 17110081 * 34220161 * 68440321 * 136880641 * 273761281 * 547522561 * 1095045121

zkl

Translation of: Go
Library: GMP

GNU Multiple Precision Arithmetic Library

Using GMP (probabilistic primes), because it is easy and fast to check primeness. <lang zkl>var [const] BI=Import("zklBigNum"); // libGMP

fcn ccFactors(n,m){ // not re-entrant 

  prod:=BI(6*m + 1);
  if(not prod.probablyPrime())    return(False);
  fact:=BI(12*m + 1);
  if(not fact.probablyPrime())    return(False);
  prod.mul(fact);
  foreach i in ([1..n-2]){
     fact.set((2).pow(i) *9*m + 1);
     if(not fact.probablyPrime()) return(False);
     prod.mul(fact);
  }
  prod

}

fcn ccNumbers(start,end){ 

  foreach n in ([start..end]){
     a,m := ( if(n<=4) 1  else (2).pow(n - 4) ), a;
     while(1){

if(num := ccFactors(n,m)){ println("a(%d) = %,d".fmt(n,num)); break; } m+=a; 

     }
  }

}</lang> <lang zkl>ccNumbers(3,9);</lang>

Output:
a(3) = 1,729
a(4) = 63,973
a(5) = 26,641,259,752,490,421,121
a(6) = 1,457,836,374,916,028,334,162,241
a(7) = 24,541,683,183,872,873,851,606,952,966,798,288,052,977,151,461,406,721
a(8) = 53,487,697,914,261,966,820,654,105,730,041,031,613,370,337,776,541,835,775,672,321
a(9) = 58,571,442,634,534,443,082,821,160,508,299,574,798,027,946,748,324,125,518,533,225,605,795,841
Retrieved from "https://rosettacode.org/wiki/Chernick%27s_Carmichael_numbers?oldid=290929"