# Sequence: smallest number greater than previous term with exactly n divisors

Sequence: smallest number greater than previous term with exactly n divisors is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.
 NOTE: Task has been split into multiple tasks. Please check that the entries are correct with the changed task description and fix / move accordingly

Calculate the sequence where each term an 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

Related tasks

## ALGOL 68

Translation of: Go
`BEGIN     PROC count divisors = ( INT n )INT:         BEGIN            INT count := 0;            FOR i WHILE i*i <= n DO                IF n MOD i = 0 THEN                    count +:= IF i = n OVER i THEN 1 ELSE 2 FI                FI            OD;            count         END # count divisors # ;     INT max = 15;     print( ( "The first ", whole( max, 0 ), " terms of the sequence are:", newline ) );    INT next := 1;    FOR i WHILE next <= max DO        IF next = count divisors( i ) THEN            print( ( whole( i, 0 ), " " ) );            next +:= 1        FI    OD;    print( ( newline, newline ) ) 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
```

## 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
```

## 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
```

## 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
```

## 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
```

## Pascal

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 dividersvar  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```

## 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
```

## Phix

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

`constant limit = 32sequence res = repeat(0,limit)integer next = 1atom n = 1while next<=limit do    integer k = length(factors(n,1))    if k=next then        res[k] = n        next += 1        if next>4 and length(factors(next,1))=2 then            n := power(2,next-1)-1 -- n is incremented afterwards        end if    end if    n += 1end whileprintf(1,"The first %d terms are:\n",limit)pp(res,{pp_Pause,0,pp_StrFmt,1})`
Output:
```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,
4498752,268435456,268437200,1073741824,1073741830}
```

## 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                                             /*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 do;  #= 1;             end   /*Odd?   Assume  Pdivisors  count of 1.*/               else do;  #= 3;    y= x%2;  end   /*Even?     "        "        "    " 3.*/                                                 /* [↑]   start with known num of Pdivs.*/                  do k=3  for x%2-3  by 1+odd  while k<y  /*for odd numbers, skip evens.*/                  if x//k==0  then do            /*if no remainder, then found a divisor*/                                   #=#+2;  y=x%k /*bump  #  Pdivs,  calculate limit  Y. */                                   if k>=y  then do;  #= #-1;  leave;  end      /*limit?*/                                   end                                          /*  ___ */                              else if k*k>x  then leave        /*only divide up to √ 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..." + nlsee "wait for done..." + nl + nlsee "the first 15 Anti-primes Plus are:" + nl + nlnum = 1n = 0result = list(15)while num < 16      n = n + 1      div = factors(n)      if div = num         result[num] = n         num = num + 1      okendsee "["for n = 1 to len(result)    if n < len(result)       see string(result[n]) + ","    else       see string(result[n]) + "]" + nl + nl    oknextsee "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...
```

## 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]
```

## 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
```