Factors of an integer

From Rosetta Code
Task
Factors of an integer
You are encouraged to solve this task according to the task description, using any language you may know.

Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.

You may see other such operations in the Basic Data Operations category, or:

Integer Operations
Arithmetic | Comparison

Boolean Operations
Bitwise | Logical

String Operations
Concatenation | Interpolation | Comparison | Matching

Memory Operations
Pointers & references | Addresses

Compute the factors of a positive integer. These factors are the positive integers by which the number being factored can be divided to yield a positive integer result (though the concepts function correctly for zero and negative integers, the set of factors of zero has countably infinite members, and the factors of negative integers can be obtained from the factors of related positive numbers without difficulty; this task does not require handling of either of these cases). Note that even prime numbers will have at least two factors; ‘1’ and themselves.

See also:

0815

<lang 0815> <:1:~>|~#:end:>~x}:str:/={^:wei:~%x<:a:x=$~ =}:wei:x<:1:+{>~>x=-#:fin:^:str:}:fin:{{~% </lang>

ACL2

<lang Lisp>(defun factors-r (n i)

  (declare (xargs :measure (nfix (- n i))))
  (cond ((zp (- n i))
         (list n))
        ((= (mod n i) 0)
         (cons i (factors-r n (1+ i))))
        (t (factors-r n (1+ i)))))

(defun factors (n)

  (factors-r n 1))</lang>

ActionScript

<lang ActionScript>function factor(n:uint):Vector.<uint> { var factors:Vector.<uint> = new Vector.<uint>(); for(var i:uint = 1; i <= n; i++) if(n % i == 0)factors.push(i); return factors; }</lang>

Ada

<lang Ada>with Ada.Text_IO; with Ada.Command_Line; procedure Factors is

  Number  : Positive;
  Test_Nr : Positive := 1;

begin

  if Ada.Command_Line.Argument_Count /= 1 then
     Ada.Text_IO.Put (Ada.Text_IO.Standard_Error, "Missing argument!");
     Ada.Command_Line.Set_Exit_Status (Ada.Command_Line.Failure);
     return;
  end if;
  Number := Positive'Value (Ada.Command_Line.Argument (1));
  Ada.Text_IO.Put ("Factors of" & Positive'Image (Number) & ": ");
  loop
     if Number mod Test_Nr = 0 then
        Ada.Text_IO.Put (Positive'Image (Test_Nr) & ",");
     end if;
     exit when Test_Nr ** 2 >= Number;
     Test_Nr := Test_Nr + 1;
  end loop;
  Ada.Text_IO.Put_Line (Positive'Image (Number) & ".");

end Factors;</lang>

Aikido

<lang aikido>import math

function factor (n:int) {

   var result = []
   function append (v) {
       if (!(v in result)) {
           result.append (v)
       }
   }
   var sqrt = cast<int>(Math.sqrt (n))
   append (1)
   for (var i = n-1 ; i >= sqrt ; i--) {
       if ((n % i) == 0) {
           append (i)
           append (n/i)
       }
   }
   append (n)
   return result.sort()

}

function printvec (vec) {

   var comma = ""
   print ("[")
   foreach v vec {
       print (comma + v)
       comma = ", "
   }
   println ("]")

}

printvec (factor (45)) printvec (factor (25)) printvec (factor (100))</lang>

ALGOL 68

Works with: ALGOL 68 version Revision 1 - no extensions to language used
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8-8d

Note: The following implements generators, eliminating the need of declaring arbitrarily long int arrays for caching. <lang algol68>MODE YIELDINT = PROC(INT)VOID;

PROC gen factors = (INT n, YIELDINT yield)VOID: (

 FOR i FROM 1 TO ENTIER sqrt(n) DO
   IF n MOD i = 0 THEN
     yield(i); 
     INT other = n OVER i;
     IF i NE other THEN yield(n OVER i) FI
   FI
 OD

);

[]INT nums2factor = (45, 53, 64);

FOR i TO UPB nums2factor DO

 INT num = nums2factor[i];
 STRING sep := ": ";
 print(num);
  1. FOR INT j IN # gen factors(num, # ) DO ( #
    1. (INT j)VOID:(
      print((sep,whole(j,0))); 
      sep:=", "
  1. OD # ));
 print(new line)

OD</lang> Output:

        +45: 1, 45, 3, 15, 5, 9
        +53: 1, 53
        +64: 1, 64, 2, 32, 4, 16, 8

APL

<lang APL> factors←{(0=(⍳⍵)|⍵)/⍳⍵}

     factors 12345

1 3 5 15 823 2469 4115 12345

     factors 720

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 30 36 40 45 48 60 72 80 90 120 144 180 240 360 720</lang>

AutoHotkey

<lang AutoHotkey>msgbox, % factors(45) "`n" factors(53) "`n" factors(64)

Factors(n) { Loop, % floor(sqrt(n))

  {  v := A_Index = 1 ? 1 "," n : mod(n,A_Index) ? v : v "," A_Index "," n//A_Index
  }
  Sort, v, N U D,
  Return, v

}</lang>

Output:

1,3,5,9,15,45
1,53
1,2,4,8,16,32,64

AutoIt

<lang AutoIt>;AutoIt Version: 3.2.10.0 $num = 45 MsgBox (0,"Factors", "Factors of " & $num & " are: " & factors($num)) consolewrite ("Factors of " & $num & " are: " & factors($num)) Func factors($intg)

  $ls_factors=""
  For $i = 1 to $intg/2
     if ($intg/$i - int($intg/$i))=0 Then

$ls_factors=$ls_factors&$i &", "

     EndIf
  Next
  Return $ls_factors&$intg

EndFunc</lang>

Output:

Factors of 45 are: 1, 3, 5, 9, 15, 45

AWK

<lang AWK>

  1. syntax: GAWK -f FACTORS_OF_AN_INTEGER.AWK

BEGIN {

   print("enter a number or C/R to exit")

} { if ($0 == "") { exit(0) }

   if ($0 !~ /^[0-9]+$/) {
     printf("invalid: %s\n",$0)
     next
   }
   n = $0
   printf("factors of %s:",n)
   for (i=1; i<=n; i++) {
     if (n % i == 0) {
       printf(" %d",i)
     }
   }
   printf("\n")

} </lang>

output:

enter a number or C/R to exit
invalid: -1
factors of 0:
factors of 1: 1
factors of 2: 1 2
factors of 11: 1 11
factors of 64: 1 2 4 8 16 32 64
factors of 100: 1 2 4 5 10 20 25 50 100
factors of 32766: 1 2 3 6 43 86 127 129 254 258 381 762 5461 10922 16383 32766
factors of 32767: 1 7 31 151 217 1057 4681 32767

BASIC

Works with: QBasic

This example stores the factors in a shared array (with the original number as the last element) for later retrieval.

Note that this will error out if you pass 32767 (or higher). <lang qbasic>DECLARE SUB factor (what AS INTEGER)

REDIM SHARED factors(0) AS INTEGER

DIM i AS INTEGER, L AS INTEGER

INPUT "Gimme a number"; i

factor i

PRINT factors(0); FOR L = 1 TO UBOUND(factors)

   PRINT ","; factors(L);

NEXT PRINT

SUB factor (what AS INTEGER)

   DIM tmpint1 AS INTEGER
   DIM L0 AS INTEGER, L1 AS INTEGER
   REDIM tmp(0) AS INTEGER
   REDIM factors(0) AS INTEGER
   factors(0) = 1
   FOR L0 = 2 TO what
       IF (0 = (what MOD L0)) THEN
           'all this REDIMing and copying can be replaced with:
           'REDIM PRESERVE factors(UBOUND(factors)+1)
           'in languages that support the PRESERVE keyword
           REDIM tmp(UBOUND(factors)) AS INTEGER
           FOR L1 = 0 TO UBOUND(factors)
               tmp(L1) = factors(L1)
           NEXT
           REDIM factors(UBOUND(factors) + 1)
           FOR L1 = 0 TO UBOUND(factors) - 1
               factors(L1) = tmp(L1)
           NEXT
           factors(UBOUND(factors)) = L0
       END IF
   NEXT

END SUB</lang>

Sample outputs:

Gimme a number? 17
 1 , 17
Gimme a number? 12345
 1 , 3 , 5 , 15 , 823 , 2469 , 4115 , 12345
Gimme a number? 32765
 1 , 5 , 6553 , 32765
Gimme a number? 32766
 1 , 2 , 3 , 6 , 43 , 86 , 127 , 129 , 254 , 258 , 381 , 762 , 5461 , 10922 ,
 16383 , 32766

Batch File

Command line version: <lang dos>@echo off set res=Factors of %1: for /L %%i in (1,1,%1) do call :fac %1 %%i echo %res% goto :eof

fac

set /a test = %1 %% %2 if %test% equ 0 set res=%res% %2</lang>

Outputs:

>factors 32767
Factors of 32767: 1 7 31 151 217 1057 4681 32767

>factors 45
Factors of 45: 1 3 5 9 15 45

>factors 53
Factors of 53: 1 53

>factors 64
Factors of 64: 1 2 4 8 16 32 64

>factors 100
Factors of 100: 1 2 4 5 10 20 25 50 100

Interactive version: <lang dos>@echo off set /p limit=Gimme a number: set res=Factors of %limit%: for /L %%i in (1,1,%limit%) do call :fac %limit% %%i echo %res% goto :eof

fac

set /a test = %1 %% %2 if %test% equ 0 set res=%res% %2</lang>

Outputs:

>factors
Gimme a number:27
Factors of 27: 1 3 9 27

>factors
Gimme a number:102
Factors of 102: 1 2 3 6 17 34 51 102

BBC BASIC

<lang bbcbasic> INSTALL @lib$+"SORTLIB"

     sort% = FN_sortinit(0, 0)
     
     PRINT "The factors of 45 are " FNfactorlist(45)
     PRINT "The factors of 12345 are " FNfactorlist(12345)
     END
     
     DEF FNfactorlist(N%)
     LOCAL C%, I%, L%(), L$
     DIM L%(32)
     FOR I% = 1 TO SQR(N%)
       IF (N% MOD I% = 0) THEN
         L%(C%) = I%
         C% += 1
         IF (N% <> I%^2) THEN
           L%(C%) = (N% DIV I%)
           C% += 1
         ENDIF
       ENDIF
     NEXT I%
     CALL sort%, L%(0)
     FOR I% = 0 TO C%-1
       L$ += STR$(L%(I%)) + ", "
     NEXT
     = LEFT$(LEFT$(L$))</lang>

Output:

The factors of 45 are 1, 3, 5, 9, 15, 45
The factors of 12345 are 1, 3, 5, 15, 823, 2469, 4115, 12345

bc

<lang bc>/* Calculate the factors of n and return their count.

* This function mutates the global array f[] which will
* contain all factors of n in ascending order after the call!
*/

define f(n) {

   auto i, d, h, h[], l, o
   /* Local variables:
    * i: Loop variable.
    * d: Complementary (higher) factor to i.
    * h: Will always point to the last element of h[].
    * h[]: Array to hold the greater factor of the pair (x, y), where 
    *      x * y == n. The factors are stored in descending order.
    * l: Will always point to the next free spot in f[].
    * o: For saving the value of scale.
    */
   /* Use integer arithmetic */
   o = scale
   scale = 0
   /* Two factors are 1 and n (if n != 1) */
   f[l++] = 1
   if (n == 1) return(1)
   h[0] = n
   /* Main loop */
   for (i = 2; i < h[h]; i++) {
       if (n % i == 0) {
           d = n / i
           if (d != i) {
               h[++h] = d
           }
           f[l++] = i
       }
   }
   /* Append the values in h[] to f[] */
   while (h >= 0) {
       f[l++] = h[h--]
   }
   scale = o
   return(l)

}</lang>

Befunge

<lang Befunge>10:p&v: >:0:g%#v_0:g\:0:g/\v

    >:0:g:*`|      >           >0:g1+0:p
            >:0:g:*-#v_0:g\>$>:!#@_.v
                     >     ^ ^  ," "<</lang>

C

<lang c>#include <stdio.h>

  1. include <stdlib.h>

typedef struct {

   int *list;
   short count; 

} Factors;

void xferFactors( Factors *fctrs, int *flist, int flix ) {

   int ix, ij;
   int newSize = fctrs->count + flix;
   if (newSize > flix)  {
       fctrs->list = realloc( fctrs->list, newSize * sizeof(int));
   }
   else {
       fctrs->list = malloc(  newSize * sizeof(int));
   }
   for (ij=0,ix=fctrs->count; ix<newSize; ij++,ix++) {
       fctrs->list[ix] = flist[ij];
   }
   fctrs->count = newSize;

}

Factors *factor( int num, Factors *fctrs) {

   int flist[301], flix;
   int dvsr;
   flix = 0;
   fctrs->count = 0;
   free(fctrs->list);
   fctrs->list = NULL;
   for (dvsr=1; dvsr*dvsr < num; dvsr++) {
       if (num % dvsr != 0) continue;
       if ( flix == 300) {
           xferFactors( fctrs, flist, flix );
           flix = 0;
       }
       flist[flix++] = dvsr;
       flist[flix++] = num/dvsr;
   }
   if (dvsr*dvsr == num) 
       flist[flix++] = dvsr;
   if (flix > 0)
       xferFactors( fctrs, flist, flix );
   return fctrs;

}

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

   int nums2factor[] = { 2059, 223092870, 3135, 45 };
   Factors ftors = { NULL, 0};
   char sep;
   int i,j;
   for (i=0; i<4; i++) {
       factor( nums2factor[i], &ftors );
       printf("\nfactors of %d are:\n  ", nums2factor[i]);
       sep = ' ';
       for (j=0; j<ftors.count; j++) {
           printf("%c %d", sep, ftors.list[j]);
           sep = ',';
       }
       printf("\n");
   }
   return 0;

}</lang>

Prime factoring

<lang C>#include <stdio.h>

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

/* 65536 = 2^16, so we can factor all 32 bit ints */ char bits[65536];

typedef unsigned long ulong; ulong primes[7000], n_primes;

typedef struct { ulong p, e; } prime_factor; /* prime, exponent */

void sieve() { int i, j; memset(bits, 1, 65536); bits[0] = bits[1] = 0; for (i = 0; i < 256; i++) if (bits[i]) for (j = i * i; j < 65536; j += i) bits[j] = 0;

/* collect primes into a list. slightly faster this way if dealing with large numbers */ for (i = j = 0; i < 65536; i++) if (bits[i]) primes[j++] = i;

n_primes = j; }

int get_prime_factors(ulong n, prime_factor *lst) { ulong i, e, p; int len = 0;

for (i = 0; i < n_primes; i++) { p = primes[i]; if (p * p > n) break; for (e = 0; !(n % p); n /= p, e++); if (e) { lst[len].p = p; lst[len++].e = e; } }

return n == 1 ? len : (lst[len].p = n, lst[len].e = 1, ++len); }

int ulong_cmp(const void *a, const void *b) { return *(const ulong*)a < *(const ulong*)b ? -1 : *(const ulong*)a > *(const ulong*)b; }

int get_factors(ulong n, ulong *lst) { int n_f, len, len2, i, j, k, p; prime_factor f[100];

n_f = get_prime_factors(n, f);

len2 = len = lst[0] = 1; /* L = (1); L = (L, L * p**(1 .. e)) forall((p, e)) */ for (i = 0; i < n_f; i++, len2 = len) for (j = 0, p = f[i].p; j < f[i].e; j++, p *= f[i].p) for (k = 0; k < len2; k++) lst[len++] = lst[k] * p;

qsort(lst, len, sizeof(ulong), ulong_cmp); return len; }

int main() { ulong fac[10000]; int len, i, j; ulong nums[] = {3, 120, 1024, 2UL*2*2*2*3*3*3*5*5*7*11*13*17*19 };

sieve();

for (i = 0; i < 4; i++) { len = get_factors(nums[i], fac); printf("%lu:", nums[i]); for (j = 0; j < len; j++) printf(" %lu", fac[j]); printf("\n"); }

return 0; }</lang>output

3: 1 3
120: 1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120
1024: 1 2 4 8 16 32 64 128 256 512 1024
3491888400: 1 2 3 4 5 6 7 8 9 10 11 ...(>1900 numbers)... 1163962800 1745944200 3491888400

C++

<lang Cpp>#include <iostream>

  1. include <vector>
  2. include <algorithm>
  3. include <iterator>

std::vector<int> GenerateFactors(int n) {

   std::vector<int> factors;
   factors.push_back(1);
   factors.push_back(n);
   for(int i = 2; i * i <= n; ++i)
   {
       if(n % i == 0)
       {
           factors.push_back(i);
           if(i * i != n)
               factors.push_back(n / i);
       }
   }
   std::sort(factors.begin(), factors.end());
   return factors;

}

int main() {

   const int SampleNumbers[] = {3135, 45, 60, 81};
   for(size_t i = 0; i < sizeof(SampleNumbers) / sizeof(int); ++i)
   {
       std::vector<int> factors = GenerateFactors(SampleNumbers[i]);
       std::cout << "Factors of " << SampleNumbers[i] << " are:\n";
       std::copy(factors.begin(), factors.end(), std::ostream_iterator<int>(std::cout, "\n"));
       std::cout << std::endl;
   }

}</lang>

C#

C# 3.0 <lang csharp>using System; using System.Linq; using System.Collections.Generic;

public static class Extension {

   public static List<int> Factors(this int me)
   {
       return Enumerable.Range(1, me).Where(x => me % x == 0).ToList();
   }

}

class Program {

   static void Main(string[] args)
   {
       Console.WriteLine(String.Join(", ", 45.Factors()));        
   }

}</lang>

C# 1.0 <lang csharp>static void Main(string[] args) { do { Console.WriteLine("Number:"); Int64 p = 0; do { try { p = Convert.ToInt64(Console.ReadLine()); break; } catch (Exception) { }

} while (true);

Console.WriteLine("For 1 through " + ((int)Math.Sqrt(p)).ToString() + ""); for (int x = 1; x <= (int)Math.Sqrt(p); x++) { if (p % x == 0) Console.WriteLine("Found: " + x.ToString() + ". " + p.ToString() + " / " + x.ToString() + " = " + (p / x).ToString()); }

Console.WriteLine("Done."); } while (true); }</lang>

Example output:

Number:
32434243
For 1 through 5695
Found: 1. 32434243 / 1 = 32434243
Found: 307. 32434243 / 307 = 105649
Done.

Chapel

Inspired by the Clojure solution: <lang chapel>iter factors(n) { for i in 1..floor(sqrt(n)):int { if n % i == 0 then { yield i; yield n / i; } } }</lang>

Clojure

<lang lisp>(defn factors [n] (filter #(zero? (rem n %)) (range 1 (inc n))))

(print (factors 45))</lang>

(1 3 5 9 15 45)

Improved version. Considers small factors from 1 up to (sqrt n) -- we increment it because range does not include the end point. Pair each small factor with its co-factor, flattening the results, and put them into a sorted set to get the factors in order. <lang lisp>(defn factors [n]

 (into (sorted-set)
   (mapcat (fn [x] [x (/ n x)])
     (filter #(zero? (rem n %)) (range 1 (inc (Math/sqrt n)))) )))</lang>

Same idea, using for comprehensions. <lang lisp>(defn factors [n]

 (into (sorted-set)
   (reduce concat
     (for [x (range 1 (inc (Math/sqrt n))) :when (zero? (rem n x))]
       [x (/ n x)]))))</lang>

CoffeeScript

<lang coffeescript># Reference implementation for finding factors is slow, but hopefully

  1. robust--we'll use it to verify the more complicated (but hopefully faster)
  2. algorithm.

slow_factors = (n) ->

 (i for i in [1..n] when n % i == 0)
 
  1. The rest of this code does two optimizations:
  2. 1) When you find a prime factor, divide it out of n (smallest_prime_factor).
  3. 2) Find the prime factorization first, then compute composite factors from those.

smallest_prime_factor = (n) ->

 for i in [2..n]
   return n if i*i > n
   return i if n % i == 0

prime_factors = (n) ->

 return {} if n == 1
 spf = smallest_prime_factor n
 result = prime_factors(n / spf)
 result[spf] or= 0
 result[spf] += 1
 result

fast_factors = (n) ->

 prime_hash = prime_factors n
 exponents = []
 for p of prime_hash
   exponents.push
     p: p
     exp: 0
 result = []
 while true
   factor = 1
   for obj in exponents
     factor *= Math.pow obj.p, obj.exp
   result.push factor
   break if factor == n
   # roll the odometer
   for obj, i in exponents
     if obj.exp < prime_hash[obj.p]
       obj.exp += 1
       break
     else
       obj.exp = 0
 
 return result.sort (a, b) -> a - b
   

verify_factors = (factors, n) ->

 expected_result = slow_factors n
 throw Error("wrong length") if factors.length != expected_result.length
 for factor, i in expected_result
   console.log Error("wrong value") if factors[i] != factor     
   
 

for n in [1, 3, 4, 8, 24, 37, 1001, 11111111111, 99999999999]

 factors = fast_factors n
 console.log n, factors
 if n < 1000000
   verify_factors factors, n</lang>

output <lang coffeescript>> coffee factors.coffee 1 [ 1 ] 3 [ 1, 3 ] 4 [ 1, 2, 4 ] 8 [ 1, 2, 4, 8 ] 24 [ 1, 2, 3, 4, 6, 8, 12, 24 ] 37 [ 1, 37 ] 1001 [ 1, 7, 11, 13, 77, 91, 143, 1001 ] 11111111111 [ 1, 21649, 513239, 11111111111 ] 99999999999 [ 1,

 3,
 9,
 21649,
 64947,
 194841,
 513239,
 1539717,
 4619151,
 11111111111,
 33333333333,
 99999999999 ]</lang>

Common Lisp

We iterate in the range 1..sqrt(n) collecting ‘low’ factors and corresponding ‘high’ factors, and combine at the end to produce an ordered list of factors. <lang lisp>(defun factors (n &aux (lows '()) (highs '()))

 (do ((limit (isqrt n)) (factor 1 (1+ factor)))
     ((= factor limit)
      (when (= n (* limit limit))
        (push limit highs))
      (nreconc lows highs))
   (multiple-value-bind (quotient remainder) (floor n factor)
     (when (zerop remainder)
       (push factor lows)
       (push quotient highs)))))</lang>

D

Procedural Style

<lang d>import std.stdio, std.math, std.algorithm;

T[] factors(T)(in T n) pure nothrow {

   if (n == 1)
       return [n];
   T[] res = [1, n];
   T limit = cast(T)real(n).sqrt + 1;
   for (T i = 2; i < limit; i++) {
       if (n % i == 0) {
           res ~= i;
           immutable q = n / i;
           if (q > i)
               res ~= q;
       }
   }
   return res.sort().release;

}

void main() {

   writefln("%(%s\n%)", [45, 53, 64, 1111111].map!factors);

}</lang>

Output:
[1, 3, 5, 9, 15, 45]
[1, 53]
[1, 2, 4, 8, 16, 32, 64]
[1, 239, 4649, 1111111]

Functional Style

<lang d>import std.stdio, std.algorithm, std.range;

auto factors(I)(I n) {

   return iota(1, n + 1).filter!(i => n % i == 0);

}

void main() {

   36.factors.writeln;

}</lang>

Output:
[1, 2, 3, 4, 6, 9, 12, 18, 36]

E

This example is in need of improvement:
Use a cleverer algorithm such as in the Common Lisp example.

<lang e>def factors(x :(int > 0)) {

   var xfactors := []
   for f ? (x % f <=> 0) in 1..x {
     xfactors with= f
   }
   return xfactors

}</lang>

Ela

Using higher-order function

<lang ela>open list

factors m = filter (\x -> m % x == 0) [1..m]</lang>

Using comprehension

<lang ela>factors m = [x \\ x <- [1..m] | m % x == 0]</lang>

Erlang

<lang erlang>factors(N) ->

   [I || I <- lists:seq(1,trunc(N/2)), N rem I == 0]++[N].</lang>

F#

If number % divisor = 0 then both divisor AND number / divisor are factors.

So, we only have to search till sqrt(number).

Also, this is lazily evaluated. <lang fsharp>let factors number = seq {

   for divisor in 1 .. (float >> sqrt >> int) number do
   if number % divisor = 0 then
       yield divisor
       yield number / divisor

}</lang>

Factor

   USE: math.primes.factors
   ( scratchpad ) 24 divisors .
   { 1 2 3 4 6 8 12 24 }

FALSE

<lang false>[1[\$@$@-][\$@$@$@$@\/*=[$." "]?1+]#.%]f: 45f;! 53f;! 64f;!</lang>

Forth

This is a slightly optimized algorithm, since it realizes there are no factors between n/2 and n. The values are saved on the stack and - in true Forth fashion - printed in descending order. <lang Forth>: factors dup 2/ 1+ 1 do dup i mod 0= if i swap then loop ;

.factors factors begin dup dup . 1 <> while drop repeat drop cr ;

45 .factors 53 .factors 64 .factors 100 .factors</lang>

Fortran

Works with: Fortran version 90 and later

<lang fortran>program Factors

 implicit none
 integer :: i, number
 
 write(*,*) "Enter a number between 1 and 2147483647"
 read*, number
 do i = 1, int(sqrt(real(number))) - 1
   if (mod(number, i) == 0) write (*,*) i, number/i
 end do
 
 ! Check to see if number is a square
 i = int(sqrt(real(number))) 
 if (i*i == number) then
    write (*,*) i
 else if (mod(number, i) == 0) then
    write (*,*) i, number/i
 end if
   

end program</lang>

Frink

Frink has built-in factoring functions which use wheel factoring, trial division, Pollard p-1 factoring, and Pollard rho factoring. It also recognizes some special forms (e.g. Mersenne numbers) and handles them efficiently. Integers can either be decomposed into prime factors or all factors.

The factors[n] function will return the prime decomposition of n.

The allFactors[n, include1=true, includeN=true, sort=true, onlyToSqrt=false] function will return all factors of n. The optional arguments include1 and includeN indicate if the numbers 1 and n are to be included in the results. If the optional argument sort is true, the results will be sorted. If the optional argument onlyToSqrt=true, then only the factors less than or equal to the square root of the number will be produced.

The following produces all factors of n, including 1 and n:

<lang frink>allFactors[n]</lang>

FunL

Function to compute set of factors: <lang funl>def factors( n ) = {d | d <- 1..n if d|n}</lang>

Test: <lang funl>for x <- [103, 316, 519, 639, 760]

 println( 'The set of factors of ' + x + ' is ' + factors(x) )</lang>
 
Output:
The set of factors of 103 is {1, 103}
The set of factors of 316 is {158, 4, 79, 1, 2, 316}
The set of factors of 519 is {1, 3, 173, 519}
The set of factors of 639 is {9, 639, 71, 213, 1, 3}
The set of factors of 760 is {8, 19, 4, 40, 152, 5, 10, 76, 1, 95, 190, 760, 20, 2, 38, 380}

GAP

<lang gap># Built-in function DivisorsInt(Factorial(5));

  1. [ 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 ]
  1. A possible implementation, not suitable to large n

div := n -> Filtered([1 .. n], k -> n mod k = 0);

div(Factorial(5));

  1. [ 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 ]
  1. Another implementation, usable for large n (if n can be factored quickly)

div2 := function(n)

 local f, p;
 f := Collected(FactorsInt(n));
 p := List(f, v -> List([0 .. v[2]], k -> v[1]^k));
 return SortedList(List(Cartesian(p), Product));

end;

div2(Factorial(5));

  1. [ 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 ]</lang>

Go

Trial division, no prime number generator, but with some optimizations. It's good enough to factor any 64 bit integer, with large primes taking several seconds. <lang go>package main

import "fmt"

func main() {

   printFactors(-1)
   printFactors(0)
   printFactors(1)
   printFactors(2)
   printFactors(3)
   printFactors(53)
   printFactors(45)
   printFactors(64)
   printFactors(600851475143)
   printFactors(999999999999999989)

}

func printFactors(nr int64) {

   if nr < 1 {
       fmt.Println("\nFactors of", nr, "not computed")
       return
   }
   fmt.Printf("\nFactors of %d: ", nr)
   fs := make([]int64, 1)
   fs[0] = 1
   apf := func(p int64, e int) {
       n := len(fs)
       for i, pp := 0, p; i < e; i, pp = i+1, pp*p {
           for j := 0; j < n; j++ {
               fs = append(fs, fs[j]*pp)
           }
       }
   }
   e := 0
   for ; nr & 1 == 0; e++ {
       nr >>= 1
   }
   apf(2, e)
   for d := int64(3); nr > 1; d += 2 {
       if d*d > nr {
           d = nr
       }
       for e = 0; nr%d == 0; e++ {
           nr /= d
       }
       if e > 0 {
           apf(d, e)
       }
   }
   fmt.Println(fs)
   fmt.Println("Number of factors =", len(fs))

}</lang> Output:

Factors of -1 not computed

Factors of 0 not computed

Factors of 1: [1]
Number of factors = 1

Factors of 2: [1 2]
Number of factors = 2

Factors of 3: [1 3]
Number of factors = 2

Factors of 53: [1 53]
Number of factors = 2

Factors of 45: [1 3 9 5 15 45]
Number of factors = 6

Factors of 64: [1 2 4 8 16 32 64]
Number of factors = 7

Factors of 600851475143: [1 71 839 59569 1471 104441 1234169 87625999 6857 486847 5753023 408464633 10086647 716151937 8462696833 600851475143]
Number of factors = 16

Factors of 999999999999999989: [1 999999999999999989]
Number of factors = 2

Groovy

A straight brute force approach up to the square root of N: <lang groovy>def factorize = { long target ->

   if (target == 1) return [1L]
   if (target < 4) return [1L, target]
   def targetSqrt = Math.sqrt(target)
   def lowfactors = (2L..targetSqrt).grep { (target % it) == 0 }
   if (lowfactors == []) return [1L, target]
   def nhalf = lowfactors.size() - ((lowfactors[-1] == targetSqrt) ? 1 : 0)
   
   [1] + lowfactors + (0..<nhalf).collect { target.intdiv(lowfactors[it]) }.reverse() + [target]

}</lang>

Test: <lang groovy>((1..30) + [333333]).each { println ([number:it, factors:factorize(it)]) }</lang> Output:

[number:1, factors:[1]]
[number:2, factors:[1, 2]]
[number:3, factors:[1, 3]]
[number:4, factors:[1, 2, 4]]
[number:5, factors:[1, 5]]
[number:6, factors:[1, 2, 3, 6]]
[number:7, factors:[1, 7]]
[number:8, factors:[1, 2, 4, 8]]
[number:9, factors:[1, 3, 9]]
[number:10, factors:[1, 2, 5, 10]]
[number:11, factors:[1, 11]]
[number:12, factors:[1, 2, 3, 4, 6, 12]]
[number:13, factors:[1, 13]]
[number:14, factors:[1, 2, 7, 14]]
[number:15, factors:[1, 3, 5, 15]]
[number:16, factors:[1, 2, 4, 8, 16]]
[number:17, factors:[1, 17]]
[number:18, factors:[1, 2, 3, 6, 9, 18]]
[number:19, factors:[1, 19]]
[number:20, factors:[1, 2, 4, 5, 10, 20]]
[number:21, factors:[1, 3, 7, 21]]
[number:22, factors:[1, 2, 11, 22]]
[number:23, factors:[1, 23]]
[number:24, factors:[1, 2, 3, 4, 6, 8, 12, 24]]
[number:25, factors:[1, 5, 25]]
[number:26, factors:[1, 2, 13, 26]]
[number:27, factors:[1, 3, 9, 27]]
[number:28, factors:[1, 2, 4, 7, 14, 28]]
[number:29, factors:[1, 29]]
[number:30, factors:[1, 2, 3, 5, 6, 10, 15, 30]]
[number:333333, factors:[1, 3, 7, 9, 11, 13, 21, 33, 37, 39, 63, 77, 91, 99, 111, 117, 143, 231, 259, 273, 333, 407, 429, 481, 693, 777, 819, 1001, 1221, 1287, 1443, 2331, 2849, 3003, 3367, 3663, 4329, 5291, 8547, 9009, 10101, 15873, 25641, 30303, 37037, 47619, 111111, 333333]]

Haskell

Using D. Amos module Primes [1] for finding prime factors <lang Haskell>import HFM.Primes(primePowerFactors) import Data.List

factors = map product.

         mapM (uncurry((. enumFromTo 0) . map .(^) )) . primePowerFactors</lang>

List comprehension

Naive, functional, no import <lang Haskell>factors_naive n = [i | i <-[1..n], (mod n i) == 0]</lang> <lang Haskell>factors_naive 6 [1,2,3,6] </lang>

Factor, cofactor. Rearrange a list of tuples to a sorted list <lang Haskell>import Data.List tuple_to_list lt = (fst lt) ++ (snd lt) factors_co n = sort (tuple_to_list(unzip

       [ (j, (div n j)) | j <- 
               [i | i <- 
                       [1..truncate (sqrt (fromIntegral n))]
                       , (mod n i) == 0]] ))

</lang> <lang Haskell>factors_co 6 [1,2,3,6] </lang>

A cleaner, simplified version of the code above, without the sorting nor the tuples, increasing speed and making it possible to see results in real time (if using GHCi) <lang Haskell>import Data.List factors n = lows ++ (reverse $ map (div n) lows)

   where lows = filter ((== 0) . mod n) [1..truncate . sqrt $ fromIntegral n]

</lang> <lang Haskell>*Main> :set +s

  • Main> factors 120

[1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120] (0.01 secs, 7578656 bytes)</lang>

HicEst

<lang hicest> DLG(NameEdit=N, TItle='Enter an integer')

DO i = 1, N^0.5
  IF( MOD(N,i) == 0) WRITE() i, N/i
ENDDO

END</lang>

Icon and Unicon

<lang Icon>procedure main(arglist) numbers := arglist ||| [ 32767, 45, 53, 64, 100] # combine command line provided and default set of values every writes(lf,"factors of ",i := !numbers,"=") & writes(divisors(i)," ") do lf := "\n" end

link factors</lang> Sample Output:

factors of 32767=1 7 31 151 217 1057 4681 32767
factors of 45=1 3 5 9 15 45
factors of 53=1 53
factors of 64=1 2 4 8 16 32 64
factors of 100=1 2 4 5 10 20 25 50 100
divisors

J

J has a primitive, q: which returns its prime factors. <lang J>q: 40

2 2 2 5</lang>

Alternatively, q: can produce provide a table of the exponents of the unique relevant prime factors <lang J> __ q: 420 2 3 5 7 2 1 1 1</lang>

With this, we can form lists of each of the potential relevant powers of each of these prime factors <lang J> ((^ i.@>:)&.>/) __ q: 420 ┌─────┬───┬───┬───┐ │1 2 4│1 3│1 5│1 7│ └─────┴───┴───┴───┘</lang>

From here, it's a simple matter (*/&>@{) to compute all possible factors of the original number <lang J>factrs=: */&>@{@((^ i.@>:)&.>/)@q:~&__

  factrs 40
1  5
2 10
4 20
8 40</lang>

However, a data structure which is organized around the prime decomposition of the argument can be hard to read. So, for reader convenience, we should probably arrange them in a monotonically increasing list:

<lang J> factors=: [: /:~@, */&>@{@((^ i.@>:)&.>/)@q:~&__

  factors 420

1 2 3 4 5 6 7 10 12 14 15 20 21 28 30 35 42 60 70 84 105 140 210 420</lang>

A less efficient, but concise variation on this theme:

<lang J> ~.,*/&> { 1 ,&.> q: 40 1 5 2 10 4 20 8 40</lang>

This computes 2^n intermediate values where n is the number of prime factors of the original number.

Another less efficient approach, in which remainders are examined up to the square root, larger factors obtained as fractions, and the combined list nubbed and sorted might be: <lang J>factorsOfNumber=: monad define

 Y=. y"_
 /:~ ~. ( , Y%]) ( #~ 0=]|Y) 1+i.>.%:y

)

  factorsOfNumber 40

1 2 4 5 8 10 20 40</lang>

Another approach:

<lang J>odometer =: #: i.@(*/) factors=: (*/@:^"1 odometer@:>:)/@q:~&__</lang>

See http://www.jsoftware.com/jwiki/Essays/Odometer

Java

Works with: Java version 5+

<lang java5>public static TreeSet<Long> factors(long n) {

TreeSet<Long> factors = new TreeSet<Long>();
factors.add(n);
factors.add(1L);
for(long test = n - 1; test >= Math.sqrt(n); test--)
 if(n % test == 0)
 {
  factors.add(test);
  factors.add(n / test);
 }
return factors;

}</lang>

JavaScript

<lang javascript>function factors(num) {

var
 n_factors = [],
 i;
for (i = 1; i <= Math.floor(Math.sqrt(num)); i += 1)
 if (num % i === 0)
 {
  n_factors.push(i);
  if (num / i !== i)
   n_factors.push(num / i);
 }
n_factors.sort(function(a, b){return a - b;});  // numeric sort
return n_factors;

}

factors(45); // [1,3,5,9,15,45] factors(53); // [1,53] factors(64); // [1,2,4,8,16,32,64]</lang>

jq

Works with: jq version 1.4

<lang jq># This implementation uses "sort" for tidiness def factors:

 . as $num
 | reduce range(1; 1 + sqrt|floor) as $i
     ([];
      if ($num % $i) == 0 then
        ($num / $i) as $r
        | if $i == $r then . + [$i] else . + [$i, $r] end
      else . 
      end )
 | sort;

def task:

 (45, 53, 64) | "\(.): \(factors)" ;

task</lang>

Output:
$ jq -n -M -r -c -f factors.jq
45: [1,3,5,9,15,45]
53: [1,53]
64: [1,2,4,8,16,32,64]

Julia

<lang julia>function factors(n)

   f = [one(n)]
   for (p,e) in factor(n)
       f = reduce(vcat, f, [f*p^j for j in 1:e])
   end
   return length(f) == 1 ? [one(n), n] : sort!(f)

end</lang> Example output:

julia> factors(45)
6-element Array{Int64,1}:
  1
  3
  5
  9
 15
 45

K

<lang K> f:{d:&~x!'!1+_sqrt x;?d,_ x%|d}

  f 1

1

  f 3

1 3

  f 120

1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120

  f 1024

1 2 4 8 16 32 64 128 256 512 1024

  f 600851475143

1 71 839 1471 6857 59569 104441 486847 1234169 5753023 10086647 87625999 408464633 716151937 8462696833 600851475143

  #f 3491888400 / has 1920 factors

1920

  / Number of factors for 3491888400 .. 3491888409
  #:'f' 3491888400+!10

1920 16 4 4 12 16 32 16 8 24</lang>

LFE

Using List Comprehensions

This following function is elegant looking and concise. However, it will not handle large numbers well: it will consume a great deal of memory (on one large number, the function consumed 4.3GB of memory on my desktop machine): <lang lisp> (defun factors (n)

 (list-comp
   ((<- i (when (== 0 (rem n i))) (lists:seq 1 (trunc (/ n 2)))))
   i))

</lang>

Non-Stack-Consuming

This version will not consume the stack (this function only used 18MB of memory on my machine with a ridiculously large number): <lang lisp> (defun factors (n)

 "Tail-recursive prime factors function."
 (factors n 2 '()))

(defun factors

 ((1 _ acc) (++ acc '(1)))
 ((n _ acc) (when (=< n 0))
   #(error undefined))
 ((n k acc) (when (== 0 (rem n k)))
   (factors (div n k) k (cons k acc)))
 ((n k acc)
   (factors n (+ k 1) acc)))

</lang>

Output in the REPL: <lang lisp> > (factors 10677106534462215678539721403561279) (104729 104729 104729 98731 98731 32579 29269 1) </lang>

Liberty BASIC

<lang lb>num = 10677106534462215678539721403561279 maxnFactors = 1000 dim primeFactors(maxnFactors), nPrimeFactors(maxnFactors) global nDifferentPrimeNumbersFound, nFactors, iFactor


print "Start finding all factors of ";num; ":"

nDifferentPrimeNumbersFound=0 dummy = factorize(num,2) nFactors = showPrimeFactors(num) dim factors(nFactors) dummy = generateFactors(1,1) sort factors(), 0, nFactors-1 for i=1 to nFactors

  print i;"     ";factors(i-1)

next i

print "done"

wait


function factorize(iNum,offset)

   factorFound=0
   i = offset
   do
       if (iNum MOD i)=0 _
       then
           if primeFactors(nDifferentPrimeNumbersFound) = i _
           then
              nPrimeFactors(nDifferentPrimeNumbersFound) = nPrimeFactors(nDifferentPrimeNumbersFound) + 1
           else
              nDifferentPrimeNumbersFound = nDifferentPrimeNumbersFound + 1
              primeFactors(nDifferentPrimeNumbersFound) = i
              nPrimeFactors(nDifferentPrimeNumbersFound) = 1
           end if
           if iNum/i<>1 then dummy = factorize(iNum/i,i)
           factorFound=1
        end if
        i=i+1
   loop while factorFound=0 and i<=sqr(iNum)
   if factorFound=0 _
   then
      nDifferentPrimeNumbersFound = nDifferentPrimeNumbersFound + 1
      primeFactors(nDifferentPrimeNumbersFound) = iNum
      nPrimeFactors(nDifferentPrimeNumbersFound) = 1
   end if

end function


function showPrimeFactors(iNum)

  showPrimeFactors=1
  print iNum;" = ";
  for i=1 to nDifferentPrimeNumbersFound
     print primeFactors(i);"^";nPrimeFactors(i);
     if i<nDifferentPrimeNumbersFound then print " * "; else print ""
     showPrimeFactors = showPrimeFactors*(nPrimeFactors(i)+1)
  next i
  end function


function generateFactors(product,pIndex)

  if pIndex>nDifferentPrimeNumbersFound _
  then
     factors(iFactor) = product
     iFactor=iFactor+1
  else
     for i=0 to nPrimeFactors(pIndex)
        dummy = generateFactors(product*primeFactors(pIndex)^i,pIndex+1)
     next i
  end if
  end function</lang>

Outcome: <lang lb>Start finding all factors of 10677106534462215678539721403561279: 10677106534462215678539721403561279 = 29269^1 * 32579^1 * 98731^2 * 104729^3 1 1 2 29269 3 32579 4 98731 5 104729 6 953554751 7 2889757639 8 3065313101 9 3216557249 10 3411966091 11 9747810361 12 10339998899 13 10968163441 14 94145414120981 15 99864835517479 16 285308661456109 17 302641427774831 18 317573913751019 19 321027175754629 20 336866824130521 21 357331796744339 22 1020878431297169 23 1082897744693371 24 1148684789012489 25 9295070881578575111 26 9859755075476219149 27 10458744358910058191 28 29880090805636839461 29 31695334089430275799 30 33259198413230468851 31 33620855089606540541 32 35279725624365333809 33 37423001741237879131 34 106915577231321212201 35 113410797903992051459 36 973463478356842592799919 37 1032602289299548955255621 38 1095333837964291484285239 39 3129312029983540559911069 40 3319420643851943354153471 41 3483202590619213772296379 42 3694810384914157044482761 43 11197161487859039232598529 44 101949856624833767901342716951 45 108143405156052462534965931709 46 327729719588146219298926345301 47 364792324112959639158827476291 48 10677106534462215678539721403561279 done</lang>

<lang logo>to factors :n

 output filter [equal? 0 modulo :n ?] iseq 1 :n

end

show factors 28  ; [1 2 4 7 14 28]</lang>

Lua

<lang lua>function Factors( n )

   local f = {}
   
   for i = 1, n/2 do
       if n % i == 0 then 
           f[#f+1] = i
       end
   end
   f[#f+1] = n
   
   return f

end</lang>


Maple

<lang Maple> numtheory:-divisors(n); </lang>

Mathematica / Wolfram Language

<lang Mathematica>Factorize[n_Integer] := Divisors[n]</lang>

MATLAB / Octave

<lang Matlab> function fact(n);

   f = factor(n);	% prime decomposition
   K = dec2bin(0:2^length(f)-1)-'0';   % generate all possible permutations
   F = ones(1,2^length(f));	
   for k = 1:size(K)
     F(k) = prod(f(~K(k,:)));		% and compute products 
   end; 
   F = unique(F);	% eliminate duplicates
   printf('There are %i factors for %i.\n',length(F),n);
   disp(F);
 end;
</lang>

Output:

>> fact(12)
There are 6 factors for 12.
    1    2    3    4    6   12
>> fact(28)
There are 6 factors for 28.
    1    2    4    7   14   28
>> fact(64)
There are 7 factors for 64.
    1    2    4    8   16   32   64
>>fact(53)
There are 2 factors for 53.
    1   53

Maxima

The builtin divisors function does this. <lang maxima>(%i96) divisors(100); (%o96) {1,2,4,5,10,20,25,50,100}</lang>

Such a function could be implemented like so: <lang maxima>divisors2(n) := map( lambda([l], lreduce("*", l)),

   apply( cartesian_product,
   map( lambda([fac],
       setify(makelist(fac[1]^i, i, 0, fac[2]))),
   ifactors(n))));</lang>

MAXScript

<lang MAXScript> fn factors n = ( return (for i = 1 to n+1 where mod n i == 0 collect i) ) </lang> Output: <lang MAXScript> factors 3

  1. (1, 3)

factors 7

  1. (1, 7)

factors 14

  1. (1, 2, 7, 14)

factors 60

  1. (1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60)

factors 54

  1. (1, 2, 3, 6, 9, 18, 27, 54)

</lang>

Mercury

Mercury is both a logic language and a functional language. As such there are two possible interfaces for calculating the factors of an integer. This code shows both styles of implementation. Note that much of the code here is ceremony put in place to have this be something which can actually compile. The actual factoring is contained in the predicate factor/2 and in the function factor/1. The function form is implemented in terms of the predicate form rather than duplicating all of the predicate code.

The predicates main/2 and factor/2 are shown with the combined type and mode statement (e.g. int::in) as is the usual case for simple predicates with only one mode. This makes the code more immediately understandable. The predicate factor/5, however, has its mode broken out onto a separate line both to show Mercury's mode statement (useful for predicates which can have varying instantiation of parameters) and to stop the code from extending too far to the right. Finally the function factor/1 has its mode statements removed (shown underneath in a comment for illustration purposes) because good coding style (and the default of the compiler!) has all parameters "in"-moded and the return value "out"-moded.

This implementation of factoring works as follows:

  1. The input number itself and 1 are both considered factors.
  2. The numbers between 2 and the square root of the input number are checked for even division.
  3. If the incremental number divides evenly into the input number, both the incremental number and the quotient are added to the list of factors.

This implementation makes use of Mercury's "state variable notation" to keep a pair of variables for accumulation, thus allowing the implementation to be tail recursive. !Accumulator is syntax sugar for a *pair* of variables. One of them is an "in"-moded variable and the other is an "out"-moded variable.  !:Accumulator is the "out" portion and !.Accumulator is the "in" portion in the ensuing code.

Using the state variable notation avoids having to keep track of strings of variables unified in the code named things like Acc0, Acc1, Acc2, Acc3, etc.

fac.m

<lang Mercury>:- module fac.

- interface.
- import_module io.
- pred main(io::di, io::uo) is det.
- implementation.
- import_module float, int, list, math, string.

main(!IO) :-

   io.command_line_arguments(Args, !IO),
   list.filter_map(string.to_int, Args, CleanArgs),
   list.foldl((pred(Arg::in, !.IO::di, !:IO::uo) is det :-
                   factor(Arg, X),
                   io.format("factor(%d, [", [i(Arg)], !IO),
                   io.write_list(X, ",", io.write_int, !IO),
                   io.write_string("])\n", !IO)
              ), CleanArgs, !IO).
- pred factor(int::in, list(int)::out) is det.

factor(N, Factors) :-

   Limit = float.truncate_to_int(math.sqrt(float(N))),

factor(N, 2, Limit, [], Unsorted),

   list.sort_and_remove_dups([1, N | Unsorted], Factors).

- pred factor(int, int, int, list(int), list(int)).
- mode factor(in, in, in, in, out) is det.

factor(N, X, Limit, !Accumulator) :-

   ( if X  > Limit 
         then true
         else ( if 0 = N mod X 
                    then !:Accumulator = [X, N / X | !.Accumulator]
                    else true ),
              factor(N, X + 1, Limit, !Accumulator) ).
- func factor(int) = list(int).

%:- mode factor(in) = out is det. factor(N) = Factors :- factor(N, Factors).

- end_module fac.</lang>

Use and output

Use of the code looks like this:

$ mmc fac.m && ./fac 100 999 12345678 booger
factor(100, [1,2,4,5,10,20,25,50,100])
factor(999, [1,3,9,27,37,111,333,999])
factor(12345678, [1,2,3,6,9,18,47,94,141,282,423,846,14593,29186,43779,87558,131337,262674,685871,1371742,2057613,4115226,6172839,12345678])

МК-61/52

П9	1	П6	КИП6	ИП9	ИП6	/	П8	^	[x]
x#0	21	-	x=0	03	ИП6	С/П	ИП8	П9	БП
04	1	С/П	БП	21

MUMPS

<lang MUMPS>factors(num) New fctr,list,sep,sqrt If num<1 Quit "Too small a number" If num["." Quit "Not an integer" Set sqrt=num**0.5\1 For fctr=1:1:sqrt Set:num/fctr'["." list(fctr)=1,list(num/fctr)=1 Set (list,fctr)="",sep="[" For Set fctr=$Order(list(fctr)) Quit:fctr="" Set list=list_sep_fctr,sep="," Quit list_"]"

w $$factors(45) ; [1,3,5,9,15,45] w $$factors(53) ; [1,53] w $$factors(64) ; [1,2,4,8,16,32,64]</lang>

NetRexx

Translation of: REXX

<lang NetRexx>/* NetRexx ***********************************************************

  • 21.04.2013 Walter Pachl
  • 21.04.2013 add method main to accept argument(s)
                                                                                                                                          • /

options replace format comments java crossref symbols nobinary class divl

 method main(argwords=String[]) static
   arg=Rexx(argwords)
   Parse arg a b
   Say a b
   If a= Then Do
     help='java divl low [high] shows'
     help=help||' divisors of all numbers between low and high'
     Say help
     Return
     End
   If b= Then b=a
   loop x=a To b
     say x '->' divs(x)
     End

method divs(x) public static returns Rexx

 if x==1 then return 1               /*handle special case of 1     */
 lo=1
 hi=x
 odd=x//2                            /* 1 if x is odd               */
 loop j=2+odd By 1+odd While j*j<x   /*divide by numbers<sqrt(x)    */
   if x//j==0 then Do                /*Divisible?  Add two divisors:*/
     lo=lo j                         /* list low divisors           */
     hi=x%j hi                       /* list high divisors          */
     End
   End
 If j*j=x Then                       /*for a square number as input */
   lo=lo j                           /* add its square root         */
 return lo hi                        /* return both lists           */</lang>

Output:

java divl 1 10
1 -> 1
2 -> 1 2
3 -> 1 3
4 -> 1 2 4
5 -> 1 5
6 -> 1 2 3 6
7 -> 1 7
8 -> 1 2 4 8
9 -> 1 3 9
10 -> 1 2 5 10

Nimrod

<lang nimrod>import intsets, math, algorithm

proc factors(n): seq[int] =

 var fs = initIntSet()
 for x in 1 .. int(sqrt(float(n))):
   if n mod x == 0:
     fs.incl(x)
     fs.incl(n div x)
 result = @[]
 for x in fs:
   result.add(x)
 sort(result, system.cmp[int])

echo factors(45)</lang>

Niue

<lang Niue>[ 'n ; [ negative-or-zero [ , ] if

      [ n not-factor [ , ] when ] else ] n times n ] 'factors ;

[ dup 0 <= ] 'negative-or-zero ; [ swap dup rot swap mod 0 = not ] 'not-factor ;

( tests ) 100 factors .s .clr ( => 1 2 4 5 10 20 25 50 100 ) newline 53 factors .s .clr ( => 1 53 ) newline 64 factors .s .clr ( => 1 2 4 8 16 32 64 ) newline 12 factors .s .clr ( => 1 2 3 4 6 12 ) </lang>

Oberon-2

Oxford Oberon-2 <lang oberon2> MODULE Factors; IMPORT Out,SYSTEM; TYPE LIPool = POINTER TO ARRAY OF LONGINT; LIVector= POINTER TO LIVectorDesc; LIVectorDesc = RECORD cap: INTEGER; len: INTEGER; LIPool: LIPool; END;

PROCEDURE New(cap: INTEGER): LIVector; VAR v: LIVector; BEGIN NEW(v); v.cap := cap; v.len := 0; NEW(v.LIPool,cap); RETURN v END New;

PROCEDURE (v: LIVector) Add(x: LONGINT); VAR newLIPool: LIPool; BEGIN IF v.len = LEN(v.LIPool^) THEN (* run out of space *) v.cap := v.cap + (v.cap DIV 2); NEW(newLIPool,v.cap); SYSTEM.MOVE(SYSTEM.ADR(v.LIPool^),SYSTEM.ADR(newLIPool^),v.cap * SIZE(LONGINT)); v.LIPool := newLIPool END; v.LIPool[v.len] := x; INC(v.len) END Add;

PROCEDURE (v: LIVector) At(idx: INTEGER): LONGINT; BEGIN RETURN v.LIPool[idx]; END At;


PROCEDURE Factors(n:LONGINT): LIVector; VAR j: LONGINT; v: LIVector; BEGIN v := New(16); FOR j := 1 TO n DO IF (n MOD j) = 0 THEN v.Add(j) END; END; RETURN v END Factors;

VAR v: LIVector; j: INTEGER; BEGIN v := Factors(123); FOR j := 0 TO v.len - 1 DO Out.LongInt(v.At(j),4);Out.Ln END; Out.Int(v.len,6);Out.String(" factors");Out.Ln END Factors. </lang> Output:

   
   1
   3
  41
 123
     4 factors

Objeck

<lang objeck>use IO; use Structure;

bundle Default {

 class Basic {
   function : native : GenerateFactors(n : Int)  ~ IntVector {
     factors := IntVector->New();
     factors-> AddBack(1);
     factors->AddBack(n);
     for(i := 2; i * i <= n; i += 1;) {
       if(n % i = 0) {
         factors->AddBack(i);
         if(i * i <> n) {
           factors->AddBack(n / i);
         };
       };
     };
     factors->Sort();


     return factors;
   }
    
   function : Main(args : String[]) ~ Nil {
     numbers := [3135, 45, 60, 81];
     for(i := 0; i < numbers->Size(); i += 1;) {
       factors := GenerateFactors(numbers[i]);
       
       Console->GetInstance()->Print("Factors of ")->Print(numbers[i])->PrintLine(" are:");
       each(i : factors) {
         Console->GetInstance()->Print(factors->Get(i))->Print(", ");
       };
       "\n\n"->Print();
     };
   }
 }

}</lang>

OCaml

<lang ocaml>let rec range = function 0 -> [] | n -> range(n-1) @ [n]

let factors n =

 List.filter (fun v -> (n mod v) = 0) (range n)</lang>

Oz

<lang oz>declare

 fun {Factors N}
    Sqr = {Float.toInt {Sqrt {Int.toFloat N}}}

    Fs = for X in 1..Sqr append:App do
            if N mod X == 0 then
               CoFactor = N div X
            in
               if CoFactor == X then %% avoid duplicate factor
                  {App [X]}          %% when N is a square number
               else
                  {App [X CoFactor]}
               end
            end
         end
 in
    {Sort Fs Value.'<'}
 end

in

 {Show {Factors 53}}</lang>

PARI/GP

<lang parigp>divisors(n)</lang>

Pascal

Translation of: Fortran
Works with: Free Pascal version 2.6.2

<lang pascal>program Factors; var

 i, number: integer;

begin

 write('Enter a number between 1 and 2147483647: ');
 readln(number);

 for i := 1 to round(sqrt(number)) - 1 do
   if number mod i = 0 then
     write (i, ' ',  number div i, ' ');

 // Check to see if number is a square
 i := round(sqrt(number));
 if i*i = number then
    write(i)
 else if number mod i = 0 then
    write(i, number/i);
 writeln;

end.</lang> Output:

Enter a number between 1 and 2147483647: 49
1 49 7

Enter a number between 1 and 2147483647: 353435
1 25755 3 8585 5 5151 15 1717 17 1515 51 505 85 303 101 255 

small improvement

the factors are in ascending order.

Works with: Free Pascal

<lang pascal>program factors; {Looking for extreme composite numbers: http://wwwhomes.uni-bielefeld.de/achim/highly.txt}

const

 MAXFACTORCNT = 1920; //number := 3491888400;

var

 FaktorList : array[0..MAXFACTORCNT] of LongWord;
 i, number,quot,cnt: LongWord;

begin

 writeln('Enter a number between 1 and 4294967295: ');
 write('3491888400 is a nice choice ');
 readln(number);
 cnt := 0;
 i := 1;
 repeat
   quot := number div i;
   if quot *i-number = 0 then
   begin
     FaktorList[cnt] := i;
     FaktorList[MAXFACTORCNT-cnt] := quot;
     inc(cnt);
   end;
   inc(i);
 until i> quot;
 writeln(number,' has ',2*cnt,' factors');
 dec(cnt);
 For i := 0 to cnt do
   write(FaktorList[i],' ,');
 For i := cnt downto 1 do
   write(FaktorList[MAXFACTORCNT-i],' ,');

{ the last without ','}

 writeln(FaktorList[MAXFACTORCNT]);

end.</lang>

output
Enter a number between 1 and 4294967295: 
3491888400 is a nice choice 120
120 has 16 factors
1 ,2 ,3 ,4 ,5 ,6 ,8 ,10 ,12 ,15 ,20 ,24 ,30 ,40 ,60 ,120

Perl

<lang perl>sub factors {

       my($n) = @_;
       return grep { $n % $_ == 0 }(1 .. $n);

} print join ' ',factors(64), "\n";</lang>

Or more intelligently:

<lang perl>sub factors {

 my $n = shift;
 $n = -$n if $n < 0;
 my @divisors;
 for (1 .. int(sqrt($n))) {  # faster and less memory than map/grep
   push @divisors, $_ unless $n % $_;
 }
 # Return divisors including top half, without duplicating a square
 @divisors, map { $_*$_ == $n ? () : int($n/$_) } reverse @divisors;

} print join " ", factors(64), "\n";</lang>

One could also use a module, e.g.:

Library: ntheory

<lang perl>use ntheory qw/divisors/; print join " ", divisors(12345678), "\n";

  1. Alternately something like: fordivisors { say } 12345678; </lang>

Perl 6

Works with: Rakudo Star version 2013-10

<lang perl6>sub factors (Int $n) {

   sort uniq 
   map { $^x, $n div $^x },
   grep { $n %% $^x },
   1 .. sqrt $n;

}</lang>

If we don't bother about performance at all we can make the code quite shorter by reviewing all integers from 1 to $n:

<lang perl6>sub factors (Int $n) { grep $n %% *, 1 .. $n }</lang>

PHP

<lang PHP>function GetFactors($n){

  $factors = array(1, $n);
  for($i = 2; $i * $i <= $n; $i++){
     if($n % $i == 0){
        $factors[] = $i;
        if($i * $i != $n)
           $factors[] = $n/$i;
     }
  }
  sort($factors);
  return $factors;

}</lang>

PicoLisp

<lang PicoLisp>(de factors (N)

  (filter
     '((D) (=0 (% N D)))
     (range 1 N) ) )</lang>

PL/I

<lang PL/I>do i = 1 to n;

  if mod(n, i) = 0 then put skip list (i);

end;</lang>

PowerShell

Straightforward but slow

<lang powershell>function Get-Factor ($a) {

   1..$a | Where-Object { $a % $_ -eq 0 }

}</lang> This one uses a range of integers up to the target number and just filters it using the Where-Object cmdlet. It's very slow though, so it is not very usable for larger numbers.

A little more clever

<lang powershell>function Get-Factor ($a) {

   1..[Math]::Sqrt($a) `
       | Where-Object { $a % $_ -eq 0 } `
       | ForEach-Object { $_; $a / $_ } `
       | Sort-Object -Unique

}</lang> Here the range of integers is only taken up to the square root of the number, the same filtering applies. Afterwards the corresponding larger factors are calculated and sent down the pipeline along with the small ones found earlier.

ProDOS

Uses the math module: <lang ProDOS>editvar /newvar /value=a /userinput=1 /title=Enter an integer: do /delimspaces %% -a- >b printline Factors of -a-: -b- </lang>

Prolog

Simple Brute Force Implementation <lang Prolog> brute_force_factors( N , Fs ) :-

 integer(N) ,
 N > 0 ,  
 setof( F , ( between(1,N,F) , N mod F =:= 0 ) , Fs )
 .

</lang>

A Slightly Smarter Implementation <lang Prolog> smart_factors(N,Fs) :-

 integer(N) ,
 N > 0 ,
 setof( F , factor(N,F) , Fs )
 .

factor(N,F) :-

 L is floor(sqrt(N)) ,
 between(1,L,X) ,
 0 =:= N mod X ,
 ( F = X ; F is N // X )
 .

</lang>

Not every Prolog has between/3: you might need this:

<lang Prolog>

between(X,Y,Z) :-

 integer(X) ,
 integer(Y) ,
 X =< Z ,
 between1(X,Y,Z)
 .

between1(X,Y,X) :-

 X =< Y
 .

between1(X,Y,Z) :-

 X < Y ,
 X1 is X+1 ,
 between1(X1,Y,Z)
 .  

</lang>

Output:

?- N=36 ,( brute_force_factors(N,Factors) ; smart_factors(N,Factors) ).
N = 36, Factors = [1, 2, 3, 4, 6, 9, 12, 18, 36] ;
N = 36, Factors = [1, 2, 3, 4, 6, 9, 12, 18, 36] .

?- N=53,( brute_force_factors(N,Factors) ; smart_factors(N,Factors) ).
N = 53, Factors = [1, 53] ;
N = 53, Factors = [1, 53] .

?- N=100,( brute_force_factors(N,Factors);smart_factors(N,Factors) ).
N = 100, Factors = [1, 2, 4, 5, 10, 20, 25, 50, 100] ;
N = 100, Factors = [1, 2, 4, 5, 10, 20, 25, 50, 100] .

?- N=144,( brute_force_factors(N,Factors);smart_factors(N,Factors) ).
N = 144, Factors = [1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144] ;
N = 144, Factors = [1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144] .

?- N=32765,( brute_force_factors(N,Factors);smart_factors(N,Factors) ).
N = 32765, Factors = [1, 5, 6553, 32765] ;
N = 32765, Factors = [1, 5, 6553, 32765] .

?- N=32766,( brute_force_factors(N,Factors);smart_factors(N,Factors) ).
N = 32766, Factors = [1, 2, 3, 6, 43, 86, 127, 129, 254, 258, 381, 762, 5461, 10922, 16383, 32766] ;
N = 32766, Factors = [1, 2, 3, 6, 43, 86, 127, 129, 254, 258, 381, 762, 5461, 10922, 16383, 32766] .

38 ?- N=32767,( brute_force_factors(N,Factors);smart_factors(N,Factors) ).
N = 32767, Factors = [1, 7, 31, 151, 217, 1057, 4681, 32767] ;
N = 32767, Factors = [1, 7, 31, 151, 217, 1057, 4681, 32767] .

PureBasic

<lang PureBasic>Procedure PrintFactors(n)

 Protected i, lim=Round(sqr(n),#PB_Round_Up)
 NewList F.i()
 For i=1 To lim
   If n%i=0
     AddElement(F()): F()=i
     AddElement(F()): F()=n/i
   EndIf
 Next
 ;- Present the result
 SortList(F(),#PB_Sort_Ascending)
 ForEach F()
   Print(str(F())+" ")
 Next

EndProcedure

If OpenConsole()

 Print("Enter integer to factorize: ")
 PrintFactors(Val(Input()))
 Print(#CRLF$+#CRLF$+"Press ENTER to quit."): Input()

EndIf</lang> Output can look like

Enter integer to factorize: 96
1 2 3 4 6 8 12 16 24 32 48 96

Python

Naive and slow but simplest (check all numbers from 1 to n): <lang python>>>> def factors(n):

     return [i for i in range(1, n + 1) if not n%i]</lang>

Slightly better (realize that there are no factors between n/2 and n): <lang python>>>> def factors(n):

     return [i for i in range(1, n//2 + 1) if not n%i] + [n]

>>> factors(45) [1, 3, 5, 9, 15, 45]</lang>

Much better (realize that factors come in pairs, the smaller of which is no bigger than sqrt(n)): <lang python>>>> from math import sqrt >>> def factor(n):

     factors = set()
     for x in range(1, int(sqrt(n)) + 1):
       if n % x == 0:
         factors.add(x)
         factors.add(n//x)
     return sorted(factors)

>>> for i in (45, 53, 64): print( "%i: factors: %s" % (i, factor(i)) )

45: factors: [1, 3, 5, 9, 15, 45] 53: factors: [1, 53] 64: factors: [1, 2, 4, 8, 16, 32, 64]</lang>

More efficient when factoring many numbers: <lang python>from itertools import chain, cycle, accumulate # last of which is Python 3 only

def factors(n):

   def prime_powers(n):
       # c goes through 2, 3, 5, then the infinite (6n+1, 6n+5) series
       for c in accumulate(chain([2, 1, 2], cycle([2,4]))):
           if c*c > n: break
           if n%c: continue
           d,p = (), c
           while not n%c:
               n,p,d = n//c, p*c, d + (p,)
           yield(d)
       if n > 1: yield((n,))
   r = [1]
   for e in prime_powers(n):
       r += [a*b for a in r for b in e]
   return r</lang>

R

<lang R>factors <- function(n) {

  if(length(n) > 1) 
  {
     lapply(as.list(n), factors)
  } else
  {
     one.to.n <- seq_len(n)
     one.to.n[(n %% one.to.n) == 0]
  }

} factors(60)</lang>

1  2  3  4  5  6 10 12 15 20 30 60

<lang R>factors(c(45, 53, 64))</lang>

[[1]]
[1]  1  3  5  9 15 45
[[2]]
[1]  1 53
[[3]]
[1]  1  2  4  8 16 32 64

Racket

<lang Racket>

  1. lang racket
a naive version

(define (naive-factors n)

 (for/list ([i (in-range 1 (add1 n))] #:when (zero? (modulo n i))) i))

(naive-factors 120) ; -> '(1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120)

much better
use `factorize' to get prime factors and construct the
list of results from that

(require math) (define (factors n)

 (sort (for/fold ([l '(1)]) ([p (factorize n)])
         (append (for*/list ([e (in-range 1 (add1 (cadr p)))] [x l])
                   (* x (expt (car p) e)))
                 l))
       <))

(naive-factors 120) ; -> same

to see how fast it is

(define huge 1200034005600070000008900000000000000000) (time (length (factors huge)))

I get 42ms for getting a list of 7776 numbers
but actually the math library comes with a `divisors' function that
does the same, except even faster

(divisors 120) ; -> same

(time (length (divisors huge)))

And this one clocks at 17ms

</lang>

REALbasic

<lang vb>Function factors(num As UInt64) As UInt64()

 'This function accepts an unsigned 64 bit integer as input and returns an array of unsigned 64 bit integers
 Dim result() As UInt64
 Dim iFactor As UInt64 = 1
 While iFactor <= num/2 'Since a factor will never be larger than half of the number
   If num Mod iFactor = 0 Then
     result.Append(iFactor)
   End If
   iFactor = iFactor + 1
 Wend
 result.Append(num) 'Since a given number is always a factor of itself
 Return result

End Function</lang>

REXX

optimized version

This REXX version has no effective limits on the number of digits in the number to be factored   [by adjusting the number of digits (precision)].
This REXX version also supports negative integers and zero.
It also indicates primes in the output as well as the number of factors. <lang rexx>/*REXX pgm displays divisors of any (negative/zero/positive) integers.*/ @.=left(,7); @.1='{unity}'; @.2='[prime]' /*unity & prime tags.*/ parse arg low high inc . /*get optional args. */ high=word(high low 20,1); low=word(low 1,1); inc=word(inc 1,1) /*opts*/ w=length(high)+1; numeric digits max(9,w) /*'nuff digs for // */ say center('n',1+w) '#divisors' center('divisors',60) /*header. */ say copies('─',1+w) '─────────' copies('─' ,60) /*separator.*/

    do n=low  to high  by inc;   divs=divisors(n);  #=words(divs);  p=@.#
    if divs=='infinite'  then #='∞';  if n<1  then p=@..   /*handle N<1*/
    say  right(n,w)" "    center('['#"]",9)    "──► "    p    ' '    divs
    end   /*n*/                       /* [↑]   process a range of ints.*/

exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────DIVISORS subroutine─────────────────*/ divisors: procedure; parse arg x; x=abs(x); if x==1 then return 1; b=x if x==0 then return 'infinite'; odd=x//2 a=1 /* [↓] use only EVEN|ODD integers*/

  do j=2+odd  by 1+odd  while j*j<x   /*divide by all integers up to √x*/
  if x//j==0  then do; a=a j; b=x%j b; end /*add divs to α&ß lists if ÷*/
  end   /*j*/                         /* [↑]  %  is REXX integer divide*/
                                      /* [↓] adjust for square.      _ */

if j*j==x then return a j b /*Was X a square? If so, add √x.*/

                return  a   b         /*return divisors  (both lists). */</lang>

output when the input used is:   -6 200

  n   #divisors                           divisors
───── ───────── ────────────────────────────────────────────────────────────
  -6     [4]    ──►            1 2 3 6
  -5     [2]    ──►            1 5
  -4     [3]    ──►            1 2 4
  -3     [2]    ──►            1 3
  -2     [2]    ──►            1 2
  -1     [1]    ──►            1
   0     [∞]    ──►            infinite
   1     [1]    ──►  {unity}   1
   2     [2]    ──►  [prime]   1 2
   3     [2]    ──►  [prime]   1 3
   4     [3]    ──►            1 2 4
   5     [2]    ──►  [prime]   1 5
   6     [4]    ──►            1 2 3 6
   7     [2]    ──►  [prime]   1 7
   8     [4]    ──►            1 2 4 8
   9     [3]    ──►            1 3 9
  10     [4]    ──►            1 2 5 10
  11     [2]    ──►  [prime]   1 11
  12     [6]    ──►            1 2 3 4 6 12
  13     [2]    ──►  [prime]   1 13
  14     [4]    ──►            1 2 7 14
  15     [4]    ──►            1 3 5 15
  16     [5]    ──►            1 2 4 8 16
  17     [2]    ──►  [prime]   1 17
  18     [6]    ──►            1 2 3 6 9 18
  19     [2]    ──►  [prime]   1 19
  20     [6]    ──►            1 2 4 5 10 20
  21     [4]    ──►            1 3 7 21
  22     [4]    ──►            1 2 11 22
  23     [2]    ──►  [prime]   1 23
  24     [8]    ──►            1 2 3 4 6 8 12 24
  25     [3]    ──►            1 5 25
  26     [4]    ──►            1 2 13 26
  27     [4]    ──►            1 3 9 27
  28     [6]    ──►            1 2 4 7 14 28
  29     [2]    ──►  [prime]   1 29
  30     [8]    ──►            1 2 3 5 6 10 15 30
  31     [2]    ──►  [prime]   1 31
  32     [6]    ──►            1 2 4 8 16 32
  33     [4]    ──►            1 3 11 33
  34     [4]    ──►            1 2 17 34
  35     [4]    ──►            1 5 7 35
  36     [9]    ──►            1 2 3 4 6 9 12 18 36
  37     [2]    ──►  [prime]   1 37
  38     [4]    ──►            1 2 19 38
  39     [4]    ──►            1 3 13 39
  40     [8]    ──►            1 2 4 5 8 10 20 40
  41     [2]    ──►  [prime]   1 41
  42     [8]    ──►            1 2 3 6 7 14 21 42
  43     [2]    ──►  [prime]   1 43
  44     [6]    ──►            1 2 4 11 22 44
  45     [6]    ──►            1 3 5 9 15 45
  46     [4]    ──►            1 2 23 46
  47     [2]    ──►  [prime]   1 47
  48    [10]    ──►            1 2 3 4 6 8 12 16 24 48
  49     [3]    ──►            1 7 49
  50     [6]    ──►            1 2 5 10 25 50
  51     [4]    ──►            1 3 17 51
  52     [6]    ──►            1 2 4 13 26 52
  53     [2]    ──►  [prime]   1 53
  54     [8]    ──►            1 2 3 6 9 18 27 54
  55     [4]    ──►            1 5 11 55
  56     [8]    ──►            1 2 4 7 8 14 28 56
  57     [4]    ──►            1 3 19 57
  58     [4]    ──►            1 2 29 58
  59     [2]    ──►  [prime]   1 59
  60    [12]    ──►            1 2 3 4 5 6 10 12 15 20 30 60
  61     [2]    ──►  [prime]   1 61
  62     [4]    ──►            1 2 31 62
  63     [6]    ──►            1 3 7 9 21 63
  64     [7]    ──►            1 2 4 8 16 32 64
  65     [4]    ──►            1 5 13 65
  66     [8]    ──►            1 2 3 6 11 22 33 66
  67     [2]    ──►  [prime]   1 67
  68     [6]    ──►            1 2 4 17 34 68
  69     [4]    ──►            1 3 23 69
  70     [8]    ──►            1 2 5 7 10 14 35 70
  71     [2]    ──►  [prime]   1 71
  72    [12]    ──►            1 2 3 4 6 8 9 12 18 24 36 72
  73     [2]    ──►  [prime]   1 73
  74     [4]    ──►            1 2 37 74
  75     [6]    ──►            1 3 5 15 25 75
  76     [6]    ──►            1 2 4 19 38 76
  77     [4]    ──►            1 7 11 77
  78     [8]    ──►            1 2 3 6 13 26 39 78
  79     [2]    ──►  [prime]   1 79
  80    [10]    ──►            1 2 4 5 8 10 16 20 40 80
  81     [5]    ──►            1 3 9 27 81
  82     [4]    ──►            1 2 41 82
  83     [2]    ──►  [prime]   1 83
  84    [12]    ──►            1 2 3 4 6 7 12 14 21 28 42 84
  85     [4]    ──►            1 5 17 85
  86     [4]    ──►            1 2 43 86
  87     [4]    ──►            1 3 29 87
  88     [8]    ──►            1 2 4 8 11 22 44 88
  89     [2]    ──►  [prime]   1 89
  90    [12]    ──►            1 2 3 5 6 9 10 15 18 30 45 90
  91     [4]    ──►            1 7 13 91
  92     [6]    ──►            1 2 4 23 46 92
  93     [4]    ──►            1 3 31 93
  94     [4]    ──►            1 2 47 94
  95     [4]    ──►            1 5 19 95
  96    [12]    ──►            1 2 3 4 6 8 12 16 24 32 48 96
  97     [2]    ──►  [prime]   1 97
  98     [6]    ──►            1 2 7 14 49 98
  99     [6]    ──►            1 3 9 11 33 99
 100     [9]    ──►            1 2 4 5 10 20 25 50 100
 101     [2]    ──►  [prime]   1 101
 102     [8]    ──►            1 2 3 6 17 34 51 102
 103     [2]    ──►  [prime]   1 103
 104     [8]    ──►            1 2 4 8 13 26 52 104
 105     [8]    ──►            1 3 5 7 15 21 35 105
 106     [4]    ──►            1 2 53 106
 107     [2]    ──►  [prime]   1 107
 108    [12]    ──►            1 2 3 4 6 9 12 18 27 36 54 108
 109     [2]    ──►  [prime]   1 109
 110     [8]    ──►            1 2 5 10 11 22 55 110
 111     [4]    ──►            1 3 37 111
 112    [10]    ──►            1 2 4 7 8 14 16 28 56 112
 113     [2]    ──►  [prime]   1 113
 114     [8]    ──►            1 2 3 6 19 38 57 114
 115     [4]    ──►            1 5 23 115
 116     [6]    ──►            1 2 4 29 58 116
 117     [6]    ──►            1 3 9 13 39 117
 118     [4]    ──►            1 2 59 118
 119     [4]    ──►            1 7 17 119
 120    [16]    ──►            1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120
 121     [3]    ──►            1 11 121
 122     [4]    ──►            1 2 61 122
 123     [4]    ──►            1 3 41 123
 124     [6]    ──►            1 2 4 31 62 124
 125     [4]    ──►            1 5 25 125
 126    [12]    ──►            1 2 3 6 7 9 14 18 21 42 63 126
 127     [2]    ──►  [prime]   1 127
 128     [8]    ──►            1 2 4 8 16 32 64 128
 129     [4]    ──►            1 3 43 129
 130     [8]    ──►            1 2 5 10 13 26 65 130
 131     [2]    ──►  [prime]   1 131
 132    [12]    ──►            1 2 3 4 6 11 12 22 33 44 66 132
 133     [4]    ──►            1 7 19 133
 134     [4]    ──►            1 2 67 134
 135     [8]    ──►            1 3 5 9 15 27 45 135
 136     [8]    ──►            1 2 4 8 17 34 68 136
 137     [2]    ──►  [prime]   1 137
 138     [8]    ──►            1 2 3 6 23 46 69 138
 139     [2]    ──►  [prime]   1 139
 140    [12]    ──►            1 2 4 5 7 10 14 20 28 35 70 140
 141     [4]    ──►            1 3 47 141
 142     [4]    ──►            1 2 71 142
 143     [4]    ──►            1 11 13 143
 144    [15]    ──►            1 2 3 4 6 8 9 12 16 18 24 36 48 72 144
 145     [4]    ──►            1 5 29 145
 146     [4]    ──►            1 2 73 146
 147     [6]    ──►            1 3 7 21 49 147
 148     [6]    ──►            1 2 4 37 74 148
 149     [2]    ──►  [prime]   1 149
 150    [12]    ──►            1 2 3 5 6 10 15 25 30 50 75 150
 151     [2]    ──►  [prime]   1 151
 152     [8]    ──►            1 2 4 8 19 38 76 152
 153     [6]    ──►            1 3 9 17 51 153
 154     [8]    ──►            1 2 7 11 14 22 77 154
 155     [4]    ──►            1 5 31 155
 156    [12]    ──►            1 2 3 4 6 12 13 26 39 52 78 156
 157     [2]    ──►  [prime]   1 157
 158     [4]    ──►            1 2 79 158
 159     [4]    ──►            1 3 53 159
 160    [12]    ──►            1 2 4 5 8 10 16 20 32 40 80 160
 161     [4]    ──►            1 7 23 161
 162    [10]    ──►            1 2 3 6 9 18 27 54 81 162
 163     [2]    ──►  [prime]   1 163
 164     [6]    ──►            1 2 4 41 82 164
 165     [8]    ──►            1 3 5 11 15 33 55 165
 166     [4]    ──►            1 2 83 166
 167     [2]    ──►  [prime]   1 167
 168    [16]    ──►            1 2 3 4 6 7 8 12 14 21 24 28 42 56 84 168
 169     [3]    ──►            1 13 169
 170     [8]    ──►            1 2 5 10 17 34 85 170
 171     [6]    ──►            1 3 9 19 57 171
 172     [6]    ──►            1 2 4 43 86 172
 173     [2]    ──►  [prime]   1 173
 174     [8]    ──►            1 2 3 6 29 58 87 174
 175     [6]    ──►            1 5 7 25 35 175
 176    [10]    ──►            1 2 4 8 11 16 22 44 88 176
 177     [4]    ──►            1 3 59 177
 178     [4]    ──►            1 2 89 178
 179     [2]    ──►  [prime]   1 179
 180    [18]    ──►            1 2 3 4 5 6 9 10 12 15 18 20 30 36 45 60 90 180
 181     [2]    ──►  [prime]   1 181
 182     [8]    ──►            1 2 7 13 14 26 91 182
 183     [4]    ──►            1 3 61 183
 184     [8]    ──►            1 2 4 8 23 46 92 184
 185     [4]    ──►            1 5 37 185
 186     [8]    ──►            1 2 3 6 31 62 93 186
 187     [4]    ──►            1 11 17 187
 188     [6]    ──►            1 2 4 47 94 188
 189     [8]    ──►            1 3 7 9 21 27 63 189
 190     [8]    ──►            1 2 5 10 19 38 95 190
 191     [2]    ──►  [prime]   1 191
 192    [14]    ──►            1 2 3 4 6 8 12 16 24 32 48 64 96 192
 193     [2]    ──►  [prime]   1 193
 194     [4]    ──►            1 2 97 194
 195     [8]    ──►            1 3 5 13 15 39 65 195
 196     [9]    ──►            1 2 4 7 14 28 49 98 196
 197     [2]    ──►  [prime]   1 197
 198    [12]    ──►            1 2 3 6 9 11 18 22 33 66 99 198
 199     [2]    ──►  [prime]   1 199
 200    [12]    ──►            1 2 4 5 8 10 20 25 40 50 100 200

Alternate Version

<lang REXX>/* REXX ***************************************************************

  • Program to calculate and show divisors of positive integer(s).
  • 03.08.2012 Walter Pachl simplified the above somewhat
  • in particular I see no benefit from divAdd procedure
  • 04.08.2012 the reference to 'above' is no longer valid since that
  • was meanwhile changed for the better.
  • 04.08.2012 took over some improvements from new above
                                                                                                                                            • /

Parse arg low high . Select

 When low=  Then Parse Value '1 200' with low high
 When high= Then high=low
 Otherwise Nop
 End

do j=low to high

 say '   n = ' right(j,6) "   divisors = " divs(j)
 end

exit

divs: procedure; parse arg x

 if x==1 then return 1               /*handle special case of 1     */
 Parse Value '1' x With lo hi        /*initialize lists: lo=1 hi=x  */
 odd=x//2                            /* 1 if x is odd               */
 Do j=2+odd By 1+odd While j*j<x     /*divide by numbers<sqrt(x)    */
   if x//j==0 then Do                /*Divisible?  Add two divisors:*/
     lo=lo j                         /* list low divisors           */
     hi=x%j hi                       /* list high divisors          */
     End
   End
 If j*j=x Then                       /*for a square number as input */
   lo=lo j                           /* add its square root         */
 return lo hi                        /* return both lists           */</lang>

Ruby

<lang ruby>class Integer

 def factors() (1..self).select { |n| (self % n).zero? } end

end p 45.factors</lang>

[1, 3, 5, 9, 15, 45]

As we only have to loop up to , we can write <lang ruby>class Integer

 def factors()
   1.upto(Math.sqrt(self)).select {|i| (self % i).zero?}.inject([]) do |f, i| 
     f << i
     f << self/i unless i == self/i
     f
   end.sort
 end

end [45, 53, 64].each {|n| p n.factors}</lang> output

[1, 3, 5, 9, 15, 45]
[1, 53]
[1, 2, 4, 8, 16, 32, 64]


Run BASIC

<lang runbasic>PRINT "Factors of 45 are ";factorlist$(45) PRINT "Factors of 12345 are "; factorlist$(12345) END

function factorlist$(f) DIM L(100) FOR i = 1 TO SQR(f)

 IF (f MOD i) = 0 THEN
   L(c) = i
   c = c + 1
   IF (f <> i^2) THEN
     L(c) = (f / i)
     c = c + 1
   END IF
 END IF

NEXT i s = 1 while s = 1 s = 0 for i = 0 to c-1

if L(i) > L(i+1) and L(i+1) <> 0 then
 t = L(i)
 L(i) = L(i+1)
 L(i+1) = t
 s      = 1
end if

next i wend FOR i = 0 TO c-1

 factorlist$ = factorlist$ + STR$(L(i)) + ", "

NEXT end function</lang> output

Factors of 45 are 1, 3, 5, 9, 15, 45, 
Factors of 12345 are 1, 3, 5, 15, 823, 2469, 4115, 12345, 

Sather

Translation of: C++

<lang sather>class MAIN is

 factors(n :INT):ARRAY{INT} is
   f:ARRAY{INT};
   f := #;
   f := f.append(|1|); 
   f := f.append(|n|);
   loop i ::= 2.upto!( n.flt.sqrt.int );
     if n%i = 0 then
       f := f.append(|i|);

if (i*i) /= n then f := f.append(|n / i|); end;

     end;
   end;
   f.sort;
   return f;
 end;
 main is
   a :ARRAY{INT} := |3135, 45, 64, 53, 45, 81|;
   loop l ::= a.elt!;
     #OUT + "factors of " + l + ": ";
     r ::= factors(l);
     loop ri ::= r.elt!;
       #OUT + ri + " ";
     end;
     #OUT + "\n";
   end;
 end;

end;</lang>

Scala

<lang Scala>

 def factors(num: Int) = {
   (1 to num).filter { divisor =>
     num % divisor == 0
   }
 }</lang>

Scheme

This implementation uses a naive trial division algorithm. <lang scheme>(define (factors n)

 (define (*factors d)
   (cond ((> d n) (list))
         ((= (modulo n d) 0) (cons d (*factors (+ d 1))))
         (else (*factors (+ d 1)))))
 (*factors 1))

(display (factors 1111111)) (newline)</lang> Output:

(1 239 4649 1111111)

Seed7

<lang seed7>$ include "seed7_05.s7i";

const proc: writeFactors (in integer: number) is func

 local
   var integer: testNum is 0;
 begin
   write("Factors of " <& number <& ": ");
   for testNum range 1 to sqrt(number) do
     if number rem testNum = 0 then
       if testNum <> 1 then
         write(", ");
       end if;
       write(testNum);
       if testNum <> number div testNum then
         write(", " <& number div testNum);
       end if;
     end if;
   end for;
   writeln;
 end func;

const proc: main is func

 local
   const array integer: numsToFactor is [] (45, 53, 64);
   var integer: number is 0;
 begin
   for number range numsToFactor do
     writeFactors(number);
   end for;
 end func;</lang>

Output:

Factors of 45: 1, 45, 3, 15, 5, 9
Factors of 53: 1, 53
Factors of 64: 1, 64, 2, 32, 4, 16, 8

Slate

<lang slate>n@(Integer traits) primeFactors [

 [| :result |
  result nextPut: 1.
  n primesDo: [| :prime | result nextPut: prime]] writingAs: {}

].</lang> where primesDo: is a part of the standard numerics library: <lang slate>n@(Integer traits) primesDo: block "Decomposes the Integer into primes, applying the block to each (in increasing order)." [| div next remaining |

 div: 2.
 next: 3.
 remaining: n.
 [[(remaining \\ div) isZero]
    whileTrue:
      [block applyTo: {div}.

remaining: remaining // div].

  remaining = 1] whileFalse:
    [div: next.
     next: next + 2] "Just looks at the next odd integer."

].</lang>

Smalltalk

Copied from the Python example, but code added to the Integer built in class:

<lang smalltalk>Integer>>factors | a | a := OrderedCollection new. 1 to: (self / 2) do: [ :i | ((self \\ i) = 0) ifTrue: [ a add: i ] ]. a add: self. ^a</lang>

Then use as follows:

<lang smalltalk> 59 factors -> an OrderedCollection(1 59) 120 factors -> an OrderedCollection(1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120) </lang>

Swift

Simple implementation: <lang Swift>func factors(n: Int) -> [Int] {

   return filter(1...n) { n % $0 == 0 }

}</lang> More efficient implementation: <lang Swift>import func Darwin.sqrt

func sqrt(x:Int) -> Int { return Int(sqrt(Double(x))) }

func factors(n: Int) -> [Int] {

   var result = [Int]()
   
   filter (1 ... sqrt(n)) { n % $0 == 0 } .reduce(()) {
       
       result.append($1)
       result.append(n/$1)
   }
   
   return sorted(result)
   

}</lang> Shorter version of above: <lang Swift>func factors(n: Int) -> [Int] {

   return sorted( filter(1...sqrt(n)) { n % $0 == 0 }
                 .reduce( [Int]() )   { $0 + [$1, n/$1] } )
   

} </lang> Call: <lang Swift>println(factors(4)) println(factors(1)) println(factors(25)) println(factors(63)) println(factors(19)) println(factors(768))</lang>

Output:
[1, 2, 4]
[1]
[1, 5, 25]
[1, 3, 7, 9, 21, 63]
[1, 19]
[1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 256, 384, 768]

Tcl

<lang tcl>proc factors {n} {

   set factors {}
   for {set i 1} {$i <= sqrt($n)} {incr i} {
       if {$n % $i == 0} {
           lappend factors $i [expr {$n / $i}]
       }
   }
   return [lsort -unique -integer $factors]

} puts [factors 64] puts [factors 45] puts [factors 53]</lang> output

1 2 4 8 16 32 64
1 3 5 9 15 45
1 53

UNIX Shell

This should work in all Bourne-compatible shells, assuming the system has both sort and at least one of bc or dc.

<lang>factor() {

 r=`echo "sqrt($1)" | bc` # or `echo $1 v p | dc`
 i=1 
 while [ $i -lt $r ]; do
   if [ `expr $1 % $i` -eq 0 ]; then
     echo $i  
     expr $1 / $i
   fi
   i=`expr $i + 1`
 done | sort -nu

} </lang>

Ursala

The simple way: <lang Ursala>#import std

  1. import nat

factors "n" = (filter not remainder/"n") nrange(1,"n")</lang> The complicated way: <lang Ursala>factors "n" = nleq-<&@s <.~&r,quotient>*= "n"-* (not remainder/"n")*~ nrange(1,root("n",2))</lang> Another idea would be to approximate an upper bound for the square root of "n" with some bit twiddling such as &!*K31 "n", which evaluates to a binary number of all 1's half the width of "n" rounded up, and another would be to use the division function to get the quotient and remainder at the same time. Combining these ideas, losing the dummy variable, and cleaning up some other cruft, we have <lang Ursala>factors = nleq-<&@rrZPFLs+ ^(~&r,division)^*D/~& nrange/1+ &!*K31</lang> where nleq-<& isn't strictly necessary unless an ordered list is required. <lang Ursala>#cast %nL

example = factors 100</lang> output:

<1,2,4,5,10,20,25,50,100>

Wortel

<lang wortel>@let {

 factors1      &n !-\%%n @to n
 factors_tacit @(\\%% !- @to)
 [[
   !factors1 10 
   !factors_tacit 100 
   !factors1 720 
 ]]

}</lang>

Returns:
[
  [1 2 5 10]
  [1 2 4 5 10 20 25 50 100]
  [1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 30 36 40 45 48 60 72 80 90 120 144 180 240 360 720]
]

XPL0

<lang XPL0>include c:\cxpl\codes; int N0, N, F; [N0:= 1; repeat IntOut(0, N0); Text(0, " = ");

       F:= 2;  N:= N0;
       repeat  if rem(N/F) = 0 then
                       [if N # N0 then Text(0, " * ");
                       IntOut(0, F);
                       N:= N/F;
                       ]
               else F:= F+1;
       until   F>N;
       if N0=1 then IntOut(0, 1);      \1 = 1
       CrLf(0);
       N0:= N0+1;

until KeyHit; ]</lang>

Example output:

1 = 1
2 = 2
3 = 3
4 = 2 * 2
5 = 5
6 = 2 * 3
7 = 7
8 = 2 * 2 * 2
9 = 3 * 3
10 = 2 * 5
11 = 11
12 = 2 * 2 * 3
13 = 13
14 = 2 * 7
15 = 3 * 5
16 = 2 * 2 * 2 * 2
17 = 17
18 = 2 * 3 * 3
. . .
57086 = 2 * 17 * 23 * 73
57087 = 3 * 3 * 6343
57088 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 223
57089 = 57089
57090 = 2 * 3 * 5 * 11 * 173
57091 = 37 * 1543
57092 = 2 * 2 * 7 * 2039
57093 = 3 * 19031
57094 = 2 * 28547
57095 = 5 * 19 * 601
57096 = 2 * 2 * 2 * 3 * 3 * 13 * 61
57097 = 57097

zkl

Translation of: Chapel

<lang zkl>fcn f(n){ (1).pump(n.toFloat().sqrt(), List,

  'wrap(m){((n % m)==0) and T(m,n/m) or Void.Skip}) }

fcn g(n){ [[(m); [1..n.toFloat().sqrt()],'{n%m==0}; '{T(m,n/m)} ]] } // list comprehension</lang>

Output:
zkl: f(45)
L(L(1,45),L(3,15),L(5,9))

zkl: g(45)
L(L(1,45),L(3,15),L(5,9))