Sequence: smallest number greater than previous term with exactly n divisors: Difference between revisions

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Calculate the sequence where each term a_n is the smallest natural number greater than the previous term, that has exactly n divisors.

Task

Show here, on this page, at least the first 15 terms of the sequence.

See also

OEIS:A069654

Related tasks

11l

Translation of: Python

F divisors(n)
   V divs = [1]
   L(ii) 2 .< Int(n ^ 0.5) + 3
      I n % ii == 0
         divs.append(ii)
         divs.append(Int(n / ii))
   divs.append(n)
   R Array(Set(divs))

F sequence(max_n)
   V previous = 0
   V n = 0
   [Int] r
   L
      n++
      V ii = previous
      I n > max_n
         L.break
      L
         ii++
         I divisors(ii).len == n
            r.append(ii)
            previous = ii
            L.break
   R r

L(item) sequence(15)
   print(item)

Output:

Action!

Calculations on a real Atari 8-bit computer take quite long time. It is recommended to use an emulator capable with increasing speed of Atari CPU.

CARD FUNC CountDivisors(CARD a)
  CARD i,count

  i=1 count=0
  WHILE i*i<=a
  DO
    IF a MOD i=0 THEN
      IF i=a/i THEN
        count==+1
      ELSE
        count==+2
      FI
    FI
    i==+1
  OD
RETURN (count)

PROC Main()
  CARD a
  BYTE i

  a=1
  FOR i=1 TO 15
  DO
    WHILE CountDivisors(a)#i
    DO
      a==+1
    OD
    IF i>1 THEN
      Print(", ")
    FI
    PrintC(a)
  OD
RETURN

Output:

Screenshot from Atari 8-bit computer

1, 2, 4, 6, 16, 18, 64, 66, 100, 112, 1024, 1035, 4096, 4288, 4624

Ada

Translation of: Go

with Ada.Text_IO;

procedure Show_Sequence is

   function Count_Divisors (N : in Natural) return Natural is
      Count : Natural := 0;
      I     : Natural;
   begin
      I := 1;
      while I**2 <= N loop
         if N mod I = 0 then
            if I = N / I then
               Count := Count + 1;
            else
               Count := Count + 2;
            end if;
         end if;
         I := I + 1;
      end loop;

      return Count;
   end Count_Divisors;

   procedure Show (Max : in Natural) is
      use Ada.Text_IO;
      N : Natural := 1;
   Begin
      Put_Line ("The first" & Max'Image & "terms of the sequence are:");
      for Divisors in 1 .. Max loop
         while Count_Divisors (N) /= Divisors loop
            N := N + 1;
         end loop;
         Put (N'Image);
      end loop;
      New_Line;
   end Show;

begin
   Show (15);
end Show_Sequence;

Output:

The first 15terms of the sequence are:
 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

ALGOL 68

Translation of: Go

with a small optimisation.

BEGIN

    PROC count divisors = ( INT n )INT:
         BEGIN
            INT i2, count := 0;
            FOR i WHILE ( i2 := i * i ) < n DO
                IF n MOD i = 0 THEN count +:= 2 FI
            OD;
            IF i2 = n THEN count + 1 ELSE count FI
         END # count divisors # ;
 
    INT max = 15;

    print( ( "The first ", whole( max, 0 ), " terms of the sequence are:" ) );
    INT next := 1;
    FOR i WHILE next <= max DO
        IF next = count divisors( i ) THEN
            print( ( " ", whole( i, 0 ) ) );
            next +:= 1
        FI
    OD

END

Output:

The first 15 terms of the sequence are: 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

ALGOL W

Translation of: Go

...via Algol 68 and with a small optimisation.

begin
    integer max, next, i;

    integer procedure countDivisors ( Integer value n ) ;
    begin
        integer count, i, i2;
        count := 0;
        i     := 1;
        while  begin i2 := i * i;
                     i2 < n
        end do begin
            if n rem i = 0 then count := count + 2;
            i := i + 1
        end;
        if i2 = n then count + 1 else count
    end countDivisors ;

    max := 15;
    write( i_w := 1, s_w := 0, "The first ", max, " terms of the sequence are: " );
    i := next := 1;
    while next <= max do begin
        if next = countDivisors( i ) then begin
            writeon( i_w := 1, s_w := 0, " ", i );
            next := next +  1
        end;
        i := i + 1
    end;
    write()
end.

Output:

The first 15 terms of the sequence are:  1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

Arturo

i: new 0
next: new 1
MAX: 15
while [next =< MAX][
    if next = size factors i [
        prints ~"|i| "
        inc 'next
    ]
    inc 'i
]
print ""

Output:

1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

AutoHotkey

Translation of: Go

MAX := 15
next := 1, i := 1
while (next <= MAX)
	if (next = countDivisors(A_Index)) 
		Res.= A_Index ", ", next++
MsgBox % "The first " MAX " terms of the sequence are:`n" Trim(res, ", ")
return

countDivisors(n){
	while (A_Index**2 <= n)
		if !Mod(n, A_Index)
			count += (A_Index = n/A_Index) ? 1 : 2
	return count
}

Outputs:

The first 15 terms of the sequence are:
1, 2, 4, 6, 16, 18, 64, 66, 100, 112, 1024, 1035, 4096, 4288, 4624

AWK

# syntax: GAWK -f SEQUENCE_SMALLEST_NUMBER_GREATER_THAN_PREVIOUS_TERM_WITH_EXACTLY_N_DIVISORS.AWK
# converted from Kotlin
BEGIN {
    limit = 15
    printf("first %d terms:",limit)
    n = 1
    while (n <= limit) {
      if (n == count_divisors(++i)) {
        printf(" %d",i)
        n++
      }
    }
    printf("\n")
    exit(0)
}
function count_divisors(n,  count,i) {
    for (i=1; i*i<=n; i++) {
      if (n % i == 0) {
        count += (i == n / i) ? 1 : 2
      }
    }
    return(count)
}

Output:

first 15 terms: 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

C

Translation of: Go

#include <stdio.h>

#define MAX 15

int count_divisors(int n) {
    int i, count = 0;
    for (i = 1; i * i <= n; ++i) {
        if (!(n % i)) {
            if (i == n / i)
                count++;
            else
                count += 2;
        }
    }
    return count;
}

int main() {
    int i, next = 1;
    printf("The first %d terms of the sequence are:\n", MAX);
    for (i = 1; next <= MAX; ++i) {
        if (next == count_divisors(i)) {           
            printf("%d ", i);
            next++;
        }
    }
    printf("\n");
    return 0;
}

Output:

The first 15 terms of the sequence are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

C++

Translation of: C

#include <iostream>

#define MAX 15

using namespace std;

int count_divisors(int n) {
    int count = 0;
    for (int i = 1; i * i <= n; ++i) {
        if (!(n % i)) {
            if (i == n / i)
                count++;
            else
                count += 2;
        }
    }
    return count;
}

int main() {
    cout << "The first " << MAX << " terms of the sequence are:" << endl;
    for (int i = 1, next = 1; next <= MAX; ++i) {
        if (next == count_divisors(i)) {           
            cout << i << " ";
            next++;
        }
    }
    cout << endl;
    return 0;
}

Output:

The first 15 terms of the sequence are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

Alternative

Translation of: Pascal

More terms and quicker than the straightforward C version above. Try It Online!

#include <cstdio>
#include <chrono>

using namespace std::chrono;

const int MAX = 32;
 
unsigned int getDividersCnt(unsigned int n) {
  unsigned int d = 3, q, dRes, res = 1;
  while (!(n & 1)) { n >>= 1; res++; }
  while ((d * d) <= n) { q = n / d; if (n % d == 0) { dRes = 0;
    do { dRes += res; n = q; q /= d; } while (n % d == 0);
      res += dRes; } d += 2; } return n != 1 ? res << 1 : res; }

int main() { unsigned int i, nxt, DivCnt;
  printf("The first %d anti-primes plus are: ", MAX);
  auto st = steady_clock::now(); i = nxt = 1; do {
    if ((DivCnt = getDividersCnt(i)) == nxt ) { printf("%d ", i);
      if ((++nxt > 4) && (getDividersCnt(nxt) == 2))
        i = (1 << (nxt - 1)) - 1; } i++; } while (nxt <= MAX);
  printf("%d ms", (int)(duration<double>(steady_clock::now() - st).count() * 1000));
}

Output:

The first 32 anti-primes plus are: 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624 4632 65536 65572 262144 262192 263169 269312 4194304 4194306 4477456 4493312 4498641 4498752 268435456 268437200 1073741824 1073741830 223 ms

Dyalect

Translation of: Go

func countDivisors(n) {
    var count = 0
    var i = 1
    while i * i <= n {
        if n % i == 0 {
            if i == n / i {
                count += 1
            } else {
                count += 2
            }
        }
        i += 1
    }
    return count
}
 
let max = 15
print("The first \(max) terms of the sequence are:")
var (i, next) = (1, 1)
while next <= max {
    if next == countDivisors(i) {
        print("\(i) ", terminator: "")
        next += 1
    }
    i += 1
}

print()

Output:

The first 15 terms of the sequence are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

F#

First 28 are easy with a Naive implementation

// Nigel Galloway: November 19th., 2017
let fN g=[1..(float>>sqrt>>int)g]|>List.fold(fun Σ n->if g%n>0 then Σ else if g/n=n then Σ+1 else Σ+2) 0
let A069654=let rec fG n g=seq{match g-fN n with 0->yield n; yield! fG(n+1)(g+1) |_->yield! fG(n+1)g} in fG 1 1

A069654 |> Seq.take 28|>Seq.iter(printf "%d "); printfn ""

Output:

1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624 4632 65536 65572 262144 262192 263169 269312 4194304 4194306 4477456 4493312 4498641 4498752

Factor

USING: io kernel math math.primes.factors prettyprint sequences ;

: next ( n num -- n' num' )
    [ 2dup divisors length = ] [ 1 + ] do until [ 1 + ] dip ;

: A069654 ( n -- seq )
    [ 2 1 ] dip [ [ next ] keep ] replicate 2nip ;

"The first 15 terms of the sequence are:" print 15 A069654 .

Output:

The first 15 terms of the sequence are:
{ 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624 }

FreeBASIC

#define UPTO 15

function divisors(byval n as ulongint) as uinteger
    'find the number of divisors of an integer
    dim as integer r = 2, i
    for i = 2 to n\2
        if n mod i = 0 then r += 1
    next i
    return r
end function

dim as ulongint i = 2
dim as integer n, nfound = 1

print 1;" ";    'special case

while nfound < UPTO
    n = divisors(i)
    if n = nfound + 1 then
        nfound += 1
        print i;" ";
    end if
    i+=1
wend
print
end

Output:

 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

Go

package main

import "fmt"

func countDivisors(n int) int {
    count := 0
    for i := 1; i*i <= n; i++ {
        if n%i == 0 {
            if i == n/i {
                count++
            } else {
                count += 2
            }
        }
    }
    return count
}

func main() {
    const max = 15
    fmt.Println("The first", max, "terms of the sequence are:")
    for i, next := 1, 1; next <= max; i++ {
        if next == countDivisors(i) {
            fmt.Printf("%d ", i)
            next++
        }
    }
    fmt.Println()
}

Output:

The first 15 terms of the sequence are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

Haskell

import Text.Printf (printf)

sequence_A069654 :: [(Int,Int)]
sequence_A069654 = go 1 $ (,) <*> countDivisors <$> [1..]
 where go t ((n,c):xs) | c == t    = (t,n):go (succ t) xs
                       | otherwise = go t xs
       countDivisors n = foldr f 0 [1..floor $ sqrt $ realToFrac n]
        where f x r | n `mod` x == 0 = if n `div` x == x then r+1 else r+2
                    | otherwise      = r

main :: IO ()
main = mapM_ (uncurry $ printf "a(%2d)=%5d\n") $ take 15 sequence_A069654

Output:

a( 1)=    1
a( 2)=    2
a( 3)=    4
a( 4)=    6
a( 5)=   16
a( 6)=   18
a( 7)=   64
a( 8)=   66
a( 9)=  100
a(10)=  112
a(11)= 1024
a(12)= 1035
a(13)= 4096
a(14)= 4288
a(15)= 4624

J

sieve=: 3 :0
 NB. sieve y  returns a vector of y boxes.
 NB. In each box is an array of 2 columns.
 NB. The first column is the factor tally
 NB. and the second column is a number with
 NB. that many factors.
 NB. Rather than factoring, the algorithm
 NB. counts prime seive cell hits.
 NB. The boxes are not ordered by factor tally.
 range=. <. + i.@:|@:-
 tally=. y#0
 for_i.#\tally do.
  j=. }:^:(y<:{:)i * 1 range >: <. y % i
  tally=. j >:@:{`[`]} tally
 end.
 (</.~ {."1) (,. i.@:#)tally
)

   A=: sieve 100000
   B=: /:~ A
   missing=: [: (-.~i.@:#) (<0 0)&{&>
   C=: ({.~ {.@:missing) B
   D=:{:"1 L:_1 C
   (] , [ {~ (I. {:))&.>/@:|. D
+-----------------------------------------------------------------------+
|0 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624 4632 65536 65572|
+-----------------------------------------------------------------------+

Intermediate steps and explanation for a smaller sieve:

   ] A=: sieve 60
+---+---+----+----+----+----+----+----+----+-----+
|0 0|1 1|2  2|3  4|4  6|6 12|5 16|8 24|9 36|10 48|
|   |   |2  3|3  9|4  8|6 18|    |8 30|    |     |
|   |   |2  5|3 25|4 10|6 20|    |8 40|    |     |
|   |   |2  7|3 49|4 14|6 28|    |8 42|    |     |
|   |   |2 11|    |4 15|6 32|    |8 54|    |     |
|   |   |2 13|    |4 21|6 44|    |8 56|    |     |
|   |   |2 17|    |4 22|6 45|    |    |    |     |
|   |   |2 19|    |4 26|6 50|    |    |    |     |
|   |   |2 23|    |4 27|6 52|    |    |    |     |
|   |   |2 29|    |4 33|    |    |    |    |     |
|   |   |2 31|    |4 34|    |    |    |    |     |
|   |   |2 37|    |4 35|    |    |    |    |     |
|   |   |2 41|    |4 38|    |    |    |    |     |
|   |   |2 43|    |4 39|    |    |    |    |     |
|   |   |2 47|    |4 46|    |    |    |    |     |
|   |   |2 53|    |4 51|    |    |    |    |     |
|   |   |2 59|    |4 55|    |    |    |    |     |
|   |   |    |    |4 57|    |    |    |    |     |
|   |   |    |    |4 58|    |    |    |    |     |
+---+---+----+----+----+----+----+----+----+-----+


   ] B=: /:~ A            NB. ascending sort by tally
+---+---+----+----+----+----+----+----+----+-----+
|0 0|1 1|2  2|3  4|4  6|5 16|6 12|8 24|9 36|10 48|
|   |   |2  3|3  9|4  8|    |6 18|8 30|    |     |
|   |   |2  5|3 25|4 10|    |6 20|8 40|    |     |
|   |   |2  7|3 49|4 14|    |6 28|8 42|    |     |
|   |   |2 11|    |4 15|    |6 32|8 54|    |     |
|   |   |2 13|    |4 21|    |6 44|8 56|    |     |
|   |   |2 17|    |4 22|    |6 45|    |    |     |
|   |   |2 19|    |4 26|    |6 50|    |    |     |
|   |   |2 23|    |4 27|    |6 52|    |    |     |
|   |   |2 29|    |4 33|    |    |    |    |     |
|   |   |2 31|    |4 34|    |    |    |    |     |
|   |   |2 37|    |4 35|    |    |    |    |     |
|   |   |2 41|    |4 38|    |    |    |    |     |
|   |   |2 43|    |4 39|    |    |    |    |     |
|   |   |2 47|    |4 46|    |    |    |    |     |
|   |   |2 53|    |4 51|    |    |    |    |     |
|   |   |2 59|    |4 55|    |    |    |    |     |
|   |   |    |    |4 57|    |    |    |    |     |
|   |   |    |    |4 58|    |    |    |    |     |
+---+---+----+----+----+----+----+----+----+-----+


   NB. truncate to the fully populated length

   NB. missing lists the unavailable factor tallies
   missing=: [: (-.~i.@:#) (<0 0)&{&>
   ] C=: ({.~ {.@:missing) B   NB. retain tally counts with no holes
+---+---+----+----+----+----+----+
|0 0|1 1|2  2|3  4|4  6|5 16|6 12|
|   |   |2  3|3  9|4  8|    |6 18|
|   |   |2  5|3 25|4 10|    |6 20|
|   |   |2  7|3 49|4 14|    |6 28|
|   |   |2 11|    |4 15|    |6 32|
|   |   |2 13|    |4 21|    |6 44|
|   |   |2 17|    |4 22|    |6 45|
|   |   |2 19|    |4 26|    |6 50|
|   |   |2 23|    |4 27|    |6 52|
|   |   |2 29|    |4 33|    |    |
|   |   |2 31|    |4 34|    |    |
|   |   |2 37|    |4 35|    |    |
|   |   |2 41|    |4 38|    |    |
|   |   |2 43|    |4 39|    |    |
|   |   |2 47|    |4 46|    |    |
|   |   |2 53|    |4 51|    |    |
|   |   |2 59|    |4 55|    |    |
|   |   |    |    |4 57|    |    |
|   |   |    |    |4 58|    |    |
+---+---+----+----+----+----+----+



   NB. Remove the tally column from each box (by retaining the values)
   ,.L:_1 D=:{:"1 L:_1 C
+-+-+--+--+--+--+--+
|0|1| 2| 4| 6|16|12|
| | | 3| 9| 8|  |18|
| | | 5|25|10|  |20|
| | | 7|49|14|  |28|
| | |11|  |15|  |32|
| | |13|  |21|  |44|
| | |17|  |22|  |45|
| | |19|  |26|  |50|
| | |23|  |27|  |52|
| | |29|  |33|  |  |
| | |31|  |34|  |  |
| | |37|  |35|  |  |
| | |41|  |38|  |  |
| | |43|  |39|  |  |
| | |47|  |46|  |  |
| | |53|  |51|  |  |
| | |59|  |55|  |  |
| | |  |  |57|  |  |
| | |  |  |58|  |  |
+-+-+--+--+--+--+--+


   NB. finally enlist successively larger values
   (] , [ {~ (I. {:))&.>/@:|. D
+---------------+
|0 1 2 4 6 16 18|
+---------------+

Java

Translation of: C

public class AntiPrimesPlus {

    static int count_divisors(int n) {
        int count = 0;
        for (int i = 1; i * i <= n; ++i) {
            if (n % i == 0) {
                if (i == n / i)
                    count++;
                else
                    count += 2;
            }
        }
        return count;
    }

    public static void main(String[] args) {
        final int max = 15;
        System.out.printf("The first %d terms of the sequence are:\n", max);
        for (int i = 1, next = 1; next <= max; ++i) {
            if (next == count_divisors(i)) {           
                System.out.printf("%d ", i);
                next++;
            }
        }
        System.out.println();
    }
}

Output:

The first 15 terms of the sequence are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

jq

Adapted from Julia

Works with: jq

Works with gojq, the Go implementation of jq

# The number of prime factors (as in prime factorization)
def numfactors:
  . as $num
  | reduce range(1; 1 + sqrt|floor) as $i (null;
       if ($num % $i) == 0 
       then ($num / $i) as $r
       | if $i == $r then .+1 else .+2 end
       else . 
       end );

# Output: a stream
def A06954:
  foreach range(1; infinite) as $i ({k: 0};
     .j = .k + 1
     | until( $i == (.j | numfactors); .j += 1)
     | .k = .j ;
     .j ) ;

"First 15 terms of OEIS sequence A069654: ",
[limit(15; A06954)]

Output:

First 15 terms of OEIS sequence A069654: 
[1,2,4,6,16,18,64,66,100,112,1024,1035,4096,4288,4624]

Julia

Translation of: Perl

using Primes

function numfactors(n)
    f = [one(n)]
    for (p,e) in factor(n)
        f = reduce(vcat, [f*p^j for j in 1:e], init=f)
    end
    length(f)
end

function A06954(N)
    println("First $N terms of OEIS sequence A069654: ")
    k = 0
    for i in 1:N
        j = k
        while (j += 1) > 0
            if i == numfactors(j)
                print("$j ")
                k = j
                break
            end
        end
    end
end

A06954(15)

Output:

First 15 terms of OEIS sequence A069654:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

Kotlin

Translation of: Go

// Version 1.3.21

const val MAX = 15

fun countDivisors(n: Int): Int {
    var count = 0
    var i = 1
    while (i * i <= n) {
        if (n % i == 0) {
            count += if (i == n / i) 1 else 2
        }
        i++
    }
    return count
}

fun main() {
    println("The first $MAX terms of the sequence are:")
    var i = 1
    var next = 1
    while (next <= MAX) {
        if (next == countDivisors(i)) {
            print("$i ")
            next++
        }
        i++
    }
    println()
}

Output:

The first 15 terms of the sequence are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

Mathematica / Wolfram Language

res = {#, DivisorSigma[0, #]} & /@ Range[100000];
highest = 0;
filter = {};
Do[
 If[r[[2]] == highest + 1,
  AppendTo[filter, r[[1]]];
  highest = r[[2]]
  ]
 ,
 {r, res}
 ]
Take[filter, 15]

Output:

{1, 2, 4, 6, 16, 18, 64, 66, 100, 112, 1024, 1035, 4096, 4288, 4624}

Nim

import strformat

const MAX = 15

func countDivisors(n: int): int =
  var i = 1 
  var count = 0
  while i * i <= n:
    if n mod i == 0:
      if i == n div i:
        inc count
      else:
        inc count, 2
    inc i
  count

var i, next = 1
echo fmt"The first {MAX} terms of the sequence are:"
while next <= MAX:
  if next == countDivisors(i):
    write(stdout, fmt"{i} ")
    inc next
  inc i
write(stdout, "\n")

Output:

The first 15 terms of the sequence are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

Counting divisors by prime factorisation.
If divCnt= Count of divisors is prime then the only candidate ist n = prime^(divCnt-1). There will be more rules. If divCnt is odd then the divisors of divCnt are a^(even_factor*i)*..*k^(even_factor*j). I think of next = 33 aka 11*3 with the solution 1031^2 * 2^10=1,088,472,064 with a big distance to next= 32 => 1073741830.
Try it online!

program AntiPrimesPlus;
{$IFDEF FPC}
  {$MODE Delphi}
{$ELSE}
  {$APPTYPE CONSOLE} // delphi
{$ENDIF}
uses
  sysutils,math;
const
  MAX =32;
 
function getDividersCnt(n:Uint32):Uint32;
// getDividersCnt by dividing n into its prime factors
// aka n = 2250 = 2^1*3^2*5^3 has (1+1)*(2+1)*(3+1)= 24 dividers
var
  divi,quot,deltaRes,rest : Uint32;
begin
  result := 1;
 
  //divi  := 2; //separat without division
  while Not(Odd(n)) do
  Begin
    n := n SHR 1;
    inc(result);
  end;
 
  //from now on only odd numbers
  divi  := 3;
  while (sqr(divi)<=n) do
  Begin
    DivMod(n,divi,quot,rest);
    if rest = 0 then
    Begin
      deltaRes := 0;
      repeat
        inc(deltaRes,result);      
        n := quot;      
        DivMod(n,divi,quot,rest);
      until rest <> 0;
      inc(result,deltaRes);
    end;
    inc(divi,2);
  end;
  //if last factor of n is prime
  IF n <> 1 then
    result := result*2;
end;
 
var
  T0 : Int64;
  i,next,DivCnt: Uint32;
begin
  writeln('The first ',MAX,' anti-primes plus are:');
  T0:= GetTickCount64;
  i := 1;
  next := 1;
  repeat
    DivCnt := getDividersCnt(i);
    IF DivCnt= next then
    Begin
      write(i,' ');
      inc(next);
      //if next is prime then only prime( => mostly 2 )^(next-1) is solution
      IF (next > 4) AND (getDividersCnt(next) = 2) then
        i := 1 shl (next-1) -1;// i is incremented afterwards
    end;
    inc(i);
  until Next > MAX;
  writeln;
  writeln(GetTickCount64-T0,' ms');
end.

Output:

The first 32 anti-primes plus are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624 4632 65536 65572 262144 262192 263169 269312 4194304 4194306 4477456 4493312 4498641 4498752 268435456 268437200 1073741824 1073741830 
525 ms

Perl

Library: ntheory

use strict;
use warnings;
use ntheory 'divisors';

print "First 15 terms of OEIS: A069654\n";
my $m = 0;
for my $n (1..15) {
    my $l = $m;
    while (++$l) {
        print("$l "), $m = $l, last if $n == divisors($l);
    }
}

Output:

First 15 terms of OEIS: A069654
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

Phix

Uses the optimisation trick from pascal, of n:=power(2,next-1) when next is a prime>4.

with javascript_semantics
constant limit = 15
sequence res = repeat(0,limit)
integer next = 1
atom n = 1
while next<=limit do
    integer k = length(factors(n,1))
    if k=next then
        res[k] = n
        next += 1
        if next>4 and is_prime(next) then
            n := power(2,next-1)-1 -- (-1 for +=1 next)
        end if
    end if
    n += 1
end while
printf(1,"The first %d terms are: %v\n",{limit,res})

Output:

The first 15 terms are: {1,2,4,6,16,18,64,66,100,112,1024,1035,4096,4288,4624}

You can raise the limit to 32, the 25th takes about 4s but the rest are all near-instant, however I lost patience waiting for the 33rd.
(The last two trailing .0 just mean "not a 31-bit integer" and don't appear when run on 64 bit.)

The first 32 terms are: {1,2,4,6,16,18,64,66,100,112,1024,1035,4096,4288,4624,
                         4632,65536,65572,262144,262192,263169,269312,4194304,
                         4194306,4477456,4493312,4498641,4498752,268435456,
                         268437200,1073741824.0,1073741830.0}

PL/I

See #Polyglot:PL/I and PL/M

PL/M

Translation of: Go

via Algol 68

100H: /* FIND THE SMALLEST NUMBER > THE PREVIOUS ONE WITH EXACTLY N DIVISORS */

   /* CP/M BDOS SYSTEM CALL */
   BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
   /* CONSOLE OUTPUT ROUTINES */
   PR$CHAR:   PROCEDURE( C ); DECLARE C BYTE;    CALL BDOS( 2, C );  END;
   PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S );  END;
   PR$NL:     PROCEDURE; CALL PR$STRING( .( 0DH, 0AH, '$' ) );       END;
   PR$NUMBER: PROCEDURE( N );
      DECLARE N ADDRESS;
      DECLARE V ADDRESS, N$STR( 6 ) BYTE INITIAL( '.....$' ), W BYTE;
      N$STR( W := LAST( N$STR ) - 1 ) = '0' + ( ( V := N ) MOD 10 );
      DO WHILE( ( V := V / 10 ) > 0 );
         N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
      END;
      CALL PR$STRING( .N$STR( W ) );
   END PR$NUMBER;

   /* TASK */

   /* RETURNS THE DIVISOR COUNT OF N */
   COUNT$DIVISORS: PROCEDURE( N )ADDRESS;
      DECLARE N ADDRESS;
      DECLARE ( I, I2, COUNT ) ADDRESS;
      COUNT = 0;
      I     = 1;
      DO WHILE( ( I2 := I * I ) < N );
         IF N MOD I = 0 THEN COUNT = COUNT + 2;
         I = I + 1;
      END;
      IF I2 = N THEN RETURN ( COUNT + 1 ); ELSE RETURN ( COUNT );
   END COUNT$DIVISORS ;
 
   DECLARE MAX LITERALLY '15';
   DECLARE ( I, NEXT ) ADDRESS;

   CALL PR$STRING( .'THE FIRST $' );
   CALL PR$NUMBER( MAX );
   CALL PR$STRING( .' TERMS OF THE SEQUENCE ARE:$' );
   NEXT = 1;
   I    = 1;
   DO WHILE( NEXT <= MAX );
      IF NEXT = COUNT$DIVISORS( I ) THEN DO;
         CALL PR$CHAR( ' ' );
         CALL PR$NUMBER( I );
         NEXT = NEXT + 1;
      END;
      I = I + 1;
   END;

EOF

Output:

THE FIRST 15 TERMS OF THE SEQUENCE ARE: 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

Polyglot:PL/I and PL/M

Works with: 8080 PL/M Compiler

... under CP/M (or an emulator)

Should work with many PL/I implementations.
The PL/I include file "pg.inc" can be found on the Polyglot:PL/I and PL/M page. Note the use of text in column 81 onwards to hide the PL/I specifics from the PL/M compiler.

Translation of: PL/M

 /* FIND THE SMALLEST NUMBER > THE PREVIOUS ONE WITH EXACTLY N DIVISORS */

sequence_100H: procedure options                                                (main);

/* PROGRAM-SPECIFIC %REPLACE STATEMENTS MUST APPEAR BEFORE THE %INCLUDE AS */
/* E.G. THE CP/M PL/I COMPILER DOESN'T LIKE THEM TO FOLLOW PROCEDURES      */
   /* PL/I                                                                      */
      %replace maxnumber by         15;
   /* PL/M */                                                                   /*
      DECLARE  MAXNUMBER LITERALLY '15';
   /* */

/* PL/I DEFINITIONS                                                             */
%include 'pg.inc';
/* PL/M DEFINITIONS: CP/M BDOS SYSTEM CALL AND CONSOLE I/O ROUTINES, ETC. */    /*
   DECLARE BINARY LITERALLY 'ADDRESS', CHARACTER LITERALLY 'BYTE';
   DECLARE FIXED  LITERALLY ' ',       BIT       LITERALLY 'BYTE';
   DECLARE STATIC LITERALLY ' ',       RETURNS   LITERALLY ' ';
   DECLARE FALSE  LITERALLY '0',       TRUE      LITERALLY '1';
   DECLARE HBOUND LITERALLY 'LAST',    SADDR     LITERALLY '.';
   BDOSF: PROCEDURE( FN, ARG )BYTE;
                               DECLARE FN BYTE, ARG ADDRESS; GOTO 5;   END; 
   BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5;   END;
   PRCHAR:   PROCEDURE( C );   DECLARE C BYTE;      CALL BDOS( 2, C ); END;
   PRSTRING: PROCEDURE( S );   DECLARE S ADDRESS;   CALL BDOS( 9, S ); END;
   PRNL:     PROCEDURE;        CALL PRCHAR( 0DH ); CALL PRCHAR( 0AH ); END;
   PRNUMBER: PROCEDURE( N );
      DECLARE N ADDRESS;
      DECLARE V ADDRESS, N$STR( 6 ) BYTE, W BYTE;
      N$STR( W := LAST( N$STR ) ) = '$';
      N$STR( W := W - 1 ) = '0' + ( ( V := N ) MOD 10 );
      DO WHILE( ( V := V / 10 ) > 0 );
         N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
      END; 
      CALL BDOS( 9, .N$STR( W ) );
   END PRNUMBER;
   MODF: PROCEDURE( A, B )ADDRESS;
      DECLARE ( A, B )ADDRESS;
      RETURN( A MOD B );
   END MODF;
   MIN: PROCEDURE( A, B ) ADDRESS; 
      DECLARE  ( A, B ) ADDRESS;
      IF A < B THEN RETURN( A ); ELSE RETURN( B );
   END MIN;
/* END LANGUAGE DEFINITIONS */

   /* TASK */

   COUNTDIVISORS: PROCEDURE( N )RETURNS /* THE DIVISOR COUNT OF N */            (
                                FIXED BINARY                                    )
                                ;
      DECLARE N                 FIXED BINARY;
      DECLARE ( I, I2, COUNT )  FIXED BINARY;
      COUNT = 0;
      I     = 1;
      I2    = 1;
      DO WHILE( I2 < N );
         IF MODF( N, I ) = 0 THEN COUNT = COUNT + 2;
         I  = I + 1;
         I2 = I * I;
      END;
      IF I2 = N THEN RETURN ( COUNT + 1 ); ELSE RETURN ( COUNT );
   END COUNTDIVISORS ;
 
   DECLARE ( I, NEXT ) FIXED BINARY;

   CALL PRSTRING( SADDR( 'THE FIRST $' ) );
   CALL PRNUMBER( MAXNUMBER );
   CALL PRSTRING( SADDR( ' TERMS OF THE SEQUENCE ARE:$' ) );
   NEXT = 1;
   I    = 1;
   DO WHILE( NEXT <= MAXNUMBER );
      IF NEXT = COUNTDIVISORS( I ) THEN DO;
         CALL PRCHAR( ' ' );
         CALL PRNUMBER( I );
         NEXT = NEXT + 1;
      END;
      I = I + 1;
   END;

EOF: end sequence_100H;

Output:

THE FIRST 15 TERMS OF THE SEQUENCE ARE: 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

Python

def divisors(n):
    divs = [1]
    for ii in range(2, int(n ** 0.5) + 3):
        if n % ii == 0:
            divs.append(ii)
            divs.append(int(n / ii))
    divs.append(n)
    return list(set(divs))


def sequence(max_n=None):
    previous = 0
    n = 0
    while True:
        n += 1
        ii = previous
        if max_n is not None:
            if n > max_n:
                break
        while True:
            ii += 1
            if len(divisors(ii)) == n:
                yield ii
                previous = ii
                break


if __name__ == '__main__':
    for item in sequence(15):
        print(item)

Output:

Quackery

factors is defined at Factors of an integer#Quackery.

  [ stack ]                is terms (   --> s )

  [ temp put
    [] terms put
    0 1
    [ dip 1+
      over factors size
      over = if
        [ over
          terms take
          swap join
          terms put
          1+ ]
      terms share size
      temp share = until ]
    terms take
    temp release
    dip 2drop ]            is task  ( n --> [ )

  15 task echo

Output:

[ 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624 ]

R

#Need to add 1 to account for skipping n. Not the most efficient way to count divisors, but quite clear.
divisorCount <- function(n) length(Filter(function(x) n %% x == 0, seq_len(n %/% 2))) + 1
A06954 <- function(terms)
{
  out <- 1
  while((resultCount <- length(out)) != terms)
  {
    n <- resultCount + 1
    out[n] <- out[resultCount]
    while(divisorCount(out[n]) != n) out[n] <- out[n] + 1
  }
  out
}
print(A06954(15))

Output:

[1]    1    2    4    6   16   18   64   66  100  112 1024 1035 4096 4288 4624

Raku

(formerly Perl 6)

Works with: Rakudo version 2019.03

sub div-count (\x) {
    return 2 if x.is-prime;
    +flat (1 .. x.sqrt.floor).map: -> \d {
        unless x % d { my \y = x div d; y == d ?? y !! (y, d) }
    }
}

my $limit = 15;

my $m = 1;
put "First $limit terms of OEIS:A069654";
put (1..$limit).map: -> $n { my $ = $m = first { $n == .&div-count }, $m..Inf };

Output:

First 15 terms of OEIS:A069654
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

REXX

Programming note: this Rosetta Code task (for 15 sequence numbers) doesn't require any optimization, but the code was optimized for listing higher numbers.

The method used is to find the number of proper divisors (up to the integer square root of X), and add one.

Optimization was included when examining even or odd index numbers (determine how much to increment the do loop).

/*REXX program finds and displays   N   numbers of the   "anti─primes plus"   sequence. */
parse arg N .                                    /*obtain optional argument from the CL.*/
if N=='' | N==","  then N= 15                    /*Not specified?  Then use the default.*/
idx= 1                                           /*the maximum number of divisors so far*/
say '──index──  ──anti─prime plus──'             /*display a title for the numbers shown*/
#= 0                                             /*the count of anti─primes found  "  " */
        do i=1  until #==N                       /*step through possible numbers by twos*/
        d= #divs(i);  if d\==idx  then iterate   /*get # divisors;  Is too small?  Skip.*/
        #= # + 1;     idx= idx + 1               /*found an anti─prime #;  set new minD.*/
        say center(#, 8)  right(i, 15)           /*display the index and the anti─prime.*/
        end   /*i*/
exit 0                                            /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
#divs: procedure; parse arg x 1 y                /*X and Y:  both set from 1st argument.*/
       if x<7  then do                           /*handle special cases for numbers < 7.*/
                    if x<3   then return x       /*   "      "      "    "  one and two.*/
                    if x<5   then return x - 1   /*   "      "      "    "  three & four*/
                    if x==5  then return 2       /*   "      "      "    "  five.       */
                    if x==6  then return 4       /*   "      "      "    "  six.        */
                    end
       odd= x // 2                               /*check if   X   is  odd  or not.      */
       if odd  then      #= 1;                   /*Odd?   Assume  Pdivisors  count of 1.*/
               else do;  #= 3;    y= x % 2       /*Even?     "        "        "    " 3.*/
                    end                          /* [↑]  Even,  so add 2 known divisors.*/
                                                 /* [↓] start with known number of Pdivs*/
          do k=3  for x%2-3  by 1+odd  while k<y /*for odd numbers, skip over the evens.*/
          if x//k==0  then do                    /*if no remainder, then found a divisor*/
                           #= # + 2;   y= x % k  /*bump the # Pdivs;  calculate limit Y.*/
                           if k>=y  then do;   #= # - 1;   leave
                                         end     /* [↑]  has the limit been reached?    */
                           end                   /*                         ___         */
                      else if k*k>x  then leave  /*only divide up to the   √ x          */
          end   /*k*/                            /* [↑]  this form of DO loop is faster.*/
       return # + 1                              /*bump "proper divisors" to "divisors".*/

output when using the default input:

──index──  ──anti─prime plus──
   1                   1
   2                   2
   3                   4
   4                   6
   5                  16
   6                  18
   7                  64
   8                  66
   9                 100
   10                112
   11               1024
   12               1035
   13               4096
   14               4288
   15               4624

Ring

# Project : ANti-primes

see "working..." + nl
see "wait for done..." + nl + nl
see "the first 15 Anti-primes Plus are:" + nl + nl
num = 1
n = 0
result = list(15)
while num < 16
      n = n + 1
      div = factors(n)
      if div = num
         result[num] = n
         num = num + 1
      ok
end
see "["
for n = 1 to len(result)
    if n < len(result)
       see string(result[n]) + ","
    else
       see string(result[n]) + "]" + nl + nl
    ok
next
see "done..." + nl

func factors(an)
     ansum = 2
     if an < 2
        return(1)
     ok
     for nr = 2 to an/2
         if an%nr = 0
            ansum = ansum+1
         ok
     next
     return ansum

Output:

working...
wait for done...

the first 15 Anti-primes Plus are:

[1,2,4,6,16,18,64,66,100,112,1024,1035,4096,4288,4624]

done...

Ruby

require 'prime'
 
def num_divisors(n)
  n.prime_division.inject(1){|prod, (_p,n)| prod *= (n + 1) } 
end

seq = Enumerator.new do |y|
  cur = 0
  (1..).each do |i|
    if num_divisors(i) == cur + 1 then
      y << i
      cur += 1
    end
  end
end

p seq.take(15)

Output:

[1, 2, 4, 6, 16, 18, 64, 66, 100, 112, 1024, 1035, 4096, 4288, 4624]

Sidef

func n_divisors(n, from=1) {
    from..Inf -> first_by { .sigma0 == n }
}

with (1) { |from|
    say 15.of { from = n_divisors(_+1, from) }
}

Output:

[1, 2, 4, 6, 16, 18, 64, 66, 100, 112, 1024, 1035, 4096, 4288, 4624]

Swift

Includes an optimization borrowed from the Pascal example above.

// See https://en.wikipedia.org/wiki/Divisor_function
func divisorCount(number: Int) -> Int {
    var n = number
    var total = 1
    // Deal with powers of 2 first
    while n % 2 == 0 {
        total += 1
        n /= 2
    }
    // Odd prime factors up to the square root
    var p = 3
    while p * p <= n {
        var count = 1
        while n % p == 0 {
            count += 1
            n /= p
        }
        total *= count
        p += 2
    }
    // If n > 1 then it's prime
    if n > 1 {
        total *= 2
    }
    return total
}

let limit = 32
var n = 1
var next = 1
while next <= limit {
    if next == divisorCount(number: n) {
        print(n, terminator: " ")
        next += 1
        if next > 4 && divisorCount(number: next) == 2 {
            n = 1 << (next - 1) - 1;
        }
    }
    n += 1
}
print()

Output:

1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624 4632 65536 65572 262144 262192 263169 269312 4194304 4194306 4477456 4493312 4498641 4498752 268435456 268437200 1073741824 1073741830

Wren

Translation of: Phix

Library: Wren-math

import "/math" for Int

var limit = 24
var res = List.filled(limit, 0)
var next = 1
var n = 1
while (next <= limit) {
    var k = Int.divisors(n).count
    if (k == next) {
        res[k-1] = n        
        next = next + 1
        if (next > 4 && Int.isPrime(next)) n = 2.pow(next-1) - 1
    }
    n = n + 1 
}
System.print("The first %(limit) terms are:")
System.print(res)

Output:

The first 24 terms are:
[1, 2, 4, 6, 16, 18, 64, 66, 100, 112, 1024, 1035, 4096, 4288, 4624, 4632, 65536, 65572, 262144, 262192, 263169, 269312, 4194304, 4194306]

XPL0

func Divs(N);   \Count divisors of N
int  N, D, C;
[C:= 0;
if N > 1 then
    [D:= 1;
    repeat  if rem(N/D) = 0 then C:= C+1;
            D:= D+1;
    until   D >= N;
    ];
return C;
];

int An, N;
[An:= 1;  N:= 0;
loop [if Divs(An) = N then
        [IntOut(0, An);  ChOut(0, ^ );
        N:= N+1;
        if N >= 15 then quit;
        ];
     An:= An+1;
     ];
]

Output:

1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624

zkl

fcn countDivisors(n)
   { [1..(n).toFloat().sqrt()] .reduce('wrap(s,i){ s + (if(0==n%i) 1 + (i!=n/i)) },0) }

n:=15;
println("The first %d anti-primes plus are:".fmt(n));
(1).walker(*).tweak(
   fcn(n,rn){ if(rn.value==countDivisors(n)){ rn.inc(); n } else Void.Skip }.fp1(Ref(1)))
.walk(n).concat(" ").println();

Output:

The first 15 anti-primes plus are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624