# Factors of an integer

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

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 every prime number has two factors; ‘1’ and itself.

 <:1:~>|~#:end:>~x}:str:/={^:wei:~%x<:a:x=$~=}:wei:x<:1:+{>~>x=-#:fin:^:str:}:fin:{{~%  ## 360 Assembly Very compact version. * Factors of an integer - 07/10/2015FACTOR CSECT USING FACTOR,R15 set base register LA R7,PG [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ LA R6,1 i L R3,N loop countLOOP L R5,N n LA R4,0 DR R4,R6 n/i LTR R4,R4 if mod(n,i)=0 BNZ NEXT XDECO R6,PG+120 edit i MVC 0(6,R7),PG+126 output i LA R7,6(R7) pgi=pgi+6NEXT LA R6,1(R6) i=i+1 BCT R3,LOOP loop XPRNT PG,120 print buffer XR R15,R15 set return code BR R14 return to callerN DC F'12345' <== input valuePG DC CL132' ' buffer YREGS END FACTOR Output:  1 3 5 15 823 2469 4115 12345  ## ACL2 (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)) ## 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;} ## 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; ## 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)) ## 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. 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);# FOR INT j IN # gen factors(num, # ) DO ( ### (INT j)VOID:( print((sep,whole(j,0))); sep:=", "# OD # )); print(new line)OD Output:  +45: 1, 45, 3, 15, 5, 9 +53: 1, 53 +64: 1, 64, 2, 32, 4, 16, 8  ## ALGOL W begin % return the factors of n ( n should be >= 1 ) in the array factor % % the bounds of factor should be 0 :: len (len must be at least 1) % % the number of factors will be returned in factor( 0 ) % procedure getFactorsOf ( integer value n ; integer array factor( * ) ; integer value len ) ; begin for i := 0 until len do factor( i ) := 0; if n >= 1 and len >= 1 then begin integer pos, lastFactor; factor( 0 ) := factor( 1 ) := pos := 1; % find the factors up to sqrt( n ) % for f := 2 until truncate( sqrt( n ) ) + 1 do begin if ( n rem f ) = 0 and pos <= len then begin % found another factor and there's room to store it % pos := pos + 1; factor( 0 ) := pos; factor( pos ) := f end if_found_factor end for_f; % find the factors above sqrt( n ) % lastFactor := factor( factor( 0 ) ); for f := factor( 0 ) step -1 until 1 do begin integer newFactor; newFactor := n div factor( f ); if newFactor > lastFactor and pos <= len then begin % found another factor and there's room to store it % pos := pos + 1; factor( 0 ) := pos; factor( pos ) := newFactor end if_found_factor end for_f; end if_params_ok end getFactorsOf ; % prpocedure to test getFactorsOf % procedure testFactorsOf( integer value n ) ; begin integer array factor( 0 :: 100 ); getFactorsOf( n, factor, 100 ); i_w := 1; s_w := 0; % set output format % write( n, " has ", factor( 0 ), " factors:" ); for f := 1 until factor( 0 ) do writeon( " ", factor( f ) ) end testFactorsOf ; % test the factorising % for i := 1 until 100 do testFactorsOf( i ) end. Output: 1 has 1 factors: 1 2 has 2 factors: 1 2 3 has 2 factors: 1 3 4 has 3 factors: 1 2 4 ... 96 has 12 factors: 1 2 3 4 6 8 12 16 24 32 48 96 97 has 2 factors: 1 97 98 has 6 factors: 1 2 7 14 49 98 99 has 6 factors: 1 3 9 11 33 99 100 has 9 factors: 1 2 4 5 10 20 25 50 100  ## APL  factors←{(0=(⍳⍵)|⍵)/⍳⍵} factors 123451 3 5 15 823 2469 4115 12345 factors 7201 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 ## 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} Output: 1,3,5,9,15,45 1,53 1,2,4,8,16,32,64 ## AutoIt ;AutoIt Version: 3.2.10.0$num = 45MsgBox (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&$intgEndFunc Output: Factors of 45 are: 1, 3, 5, 9, 15, 45  ## AWK  # syntax: GAWK -f FACTORS_OF_AN_INTEGER.AWKBEGIN { 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")}
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).

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);NEXTPRINT 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    NEXTEND SUB
Output:
 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:

@echo offset res=Factors of %1:for /L %%i in (1,1,%1) do call :fac %1 %%iecho %res%goto :eof :facset /a test = %1 %% %2if %test% equ 0 set res=%res% %2
Output:
>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:

@echo offset /p limit=Gimme a number:set res=Factors of %limit%:for /L %%i in (1,1,%limit%) do call :fac %limit% %%iecho %res%goto :eof :facset /a test = %1 %% %2if %test% equ 0 set res=%res% %2
Output:
>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

## C

#include <stdio.h>#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;}

### Prime factoring

#include <stdio.h>#include <stdlib.h>#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;}
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++

#include <iostream>#include <vector>#include <algorithm>#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;    }}

## C#

C# 3.0

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()));            }}

C# 1.0

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);}
Output:
Number:
32434243
For 1 through 5695
Found: 1. 32434243 / 1 = 32434243
Found: 307. 32434243 / 307 = 105649
Done.

## Ceylon

shared void run() {	{Integer*} getFactors(Integer n) =>		(1..n).filter((Integer element) => element.divides(n)); 	for(Integer i in 1..100) {		print("the factors of i are getFactors(i)");	}}

## Chapel

Inspired by the Clojure solution:

iter factors(n) {	for i in 1..floor(sqrt(n)):int {		if n % i == 0 then {			yield i;			yield n / i;		}	}}

## Clojure

(defn factors [n] 	(filter #(zero? (rem n %)) (range 1 (inc n)))) (print (factors 45))
(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.

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

Same idea, using for comprehensions.

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

## CoffeeScript

# Reference implementation for finding factors is slow, but hopefully# robust--we'll use it to verify the more complicated (but hopefully faster)# algorithm.slow_factors = (n) ->  (i for i in [1..n] when n % i == 0) # The rest of this code does two optimizations:#   1) When you find a prime factor, divide it out of n (smallest_prime_factor).#   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
Output:
> 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 ]

## 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.

(defun factors (n &aux (lows '()) (highs '()))  (do ((limit (1+ (isqrt n))) (factor 1 (1+ factor)))      ((= factor limit)       (when (= n (* limit limit))         (push limit highs))       (remove-duplicates (nreconc lows highs)))    (multiple-value-bind (quotient remainder) (floor n factor)      (when (zerop remainder)        (push factor lows)        (push quotient highs)))))

## D

### Procedural Style

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

### Functional Style

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

## EchoLisp

prime-factors gives the list of n's prime-factors. We mix them to get all the factors.

 ;; ppows;; input : a list g of grouped prime factors ( 3 3 3 ..);; returns (1 3 9 27 ...) (define (ppows g (mult 1))	(for/fold (ppows '(1)) ((a g))	    (set! mult (* mult a))	    (cons mult ppows))) ;; factors;; decomp n into ((2 2 ..) ( 3 3 ..)  ) prime factors groups;; combines (1 2 4 8 ..) (1 3 9 ..) lists (define (factors n)   (list-sort <   (if (<= n 1) '(1)         (for/fold (divs'(1)) ((g (map  ppows (group (prime-factors n)))))		    (for*/list ((a divs) (b g)) (* a b)))))) 
Output:
 (lib 'bigint)(factors 666)   → (1 2 3 6 9 18 37 74 111 222 333 666) (length (factors 108233175859200))   → 666 ;; 💀 (define huge 1200034005600070000008900000000000000000)(time ( length (factors huge)))    → (394ms 7776) 

## Ela

### Using higher-order function

open list factors m = filter (\x -> m % x == 0) [1..m]

### Using comprehension

factors m = [x \\ x <- [1..m] | m % x == 0]

## Elixir

defmodule RC do  def factor(1), do: [1]  def factor(n) do    (for i <- 1..div(n,2), rem(n,i)==0, do: i) ++ [n]  end   # Recursive (faster version);  def divisor(n), do: divisor(n, 1, []) |> Enum.sort   defp divisor(n, i, factors) when n < i*i    , do: factors  defp divisor(n, i, factors) when n == i*i   , do: [i | factors]  defp divisor(n, i, factors) when rem(n,i)==0, do: divisor(n, i+1, [i, div(n,i) | factors])  defp divisor(n, i, factors)                 , do: divisor(n, i+1, factors)end Enum.each([45, 53, 60, 64], fn n ->  IO.puts "#{n}: #{inspect RC.factor(n)}"end) IO.puts "\nRange: #{inspect range = 1..10000}"funs = [ factor:  &RC.factor/1,         divisor: &RC.divisor/1 ]Enum.each(funs, fn {name, fun} ->  {time, value} = :timer.tc(fn -> Enum.count(range, &length(fun.(&1))==2) end)  IO.puts "#{name}\t prime count : #{value},\t#{time/1000000} sec"end) 
Output:
45: [1, 3, 5, 9, 15, 45]
53: [1, 53]
60: [1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60]
64: [1, 2, 4, 8, 16, 32, 64]

Range: 1..10000
factor   prime count : 1229,    7.316 sec
divisor  prime count : 1229,    0.265 sec


## Erlang

### with Built in fuctions

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

### Recursive

Another, less concise, but faster version

  -module(divs).-export([divs/1]). divs(0) -> [];divs(1) -> [];divs(N) -> lists:sort(divisors(1,N))++[N]. divisors(1,N) ->      [1] ++ divisors(2,N,math:sqrt(N)). divisors(K,_N,Q) when K > Q -> [];divisors(K,N,_Q) when N rem K =/= 0 ->     [] ++ divisors(K+1,N,math:sqrt(N));divisors(K,N,_Q) when K * K  == N ->     [K] ++ divisors(K+1,N,math:sqrt(N));divisors(K,N,_Q) ->    [K, N div K] ++ divisors(K+1,N,math:sqrt(N)). 
Output:
58> timer:tc(divs, factors, [20000]).
{2237,
[1,2,4,5,8,10,16,20,25,32,40,50,80,100,125,160,200,250,400,
500,625,800,1000,1250,2000,2500,4000|...]}
59> timer:tc(divs, divs, [20000]).
{106,
[1,2,4,5,8,10,16,20,25,32,40,50,80,100,125,160,200,250,400,
500,625,800,1000,1250,2000,2500,4000|...]}


The first number is milliseconds. I'v ommitted repeating the first fuction.

## 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.

: 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 .factors53 .factors64 .factors100 .factors

## Fortran

Works with: Fortran version 90 and later
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

## 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:

allFactors[n]

## FunL

Function to compute set of factors:

def factors( n ) = {d | d <- 1..n if d|n}

Test:

for x <- [103, 316, 519, 639, 760]  println( 'The set of factors of ' + x + ' is ' + factors(x) )
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

# Built-in functionDivisorsInt(Factorial(5));# [ 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 ] # A possible implementation, not suitable to large ndiv := n -> Filtered([1 .. n], k -> n mod k = 0); div(Factorial(5));# [ 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 ] # 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, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 ]

## 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.

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))}
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:

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]}

Test:

((1..30) + [333333]).each { println ([number:it, factors:factorize(it)]) }
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]]

Using D. Amos module Primes [1] for finding prime factors

import HFM.Primes(primePowerFactors)import Data.List factors = map product.          mapM (uncurry((. enumFromTo 0) . map .(^) )) . primePowerFactors

### List comprehension

Naive, functional, no import

factors_naive n = [i | i <-[1..n], (mod n i) == 0]
factors_naive 6[1,2,3,6] 

Factor, cofactor. Rearrange a list of tuples to a sorted list

import Data.Listtuple_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]] )) 
factors_co 6[1,2,3,6] 

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)

import Data.Listfactors n = lows ++ (reverse $map (div n) lows) where lows = filter ((== 0) . mod n) [1..truncate . sqrt$ fromIntegral n] 
*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)

## 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

## Icon and Unicon

procedure main(arglist)numbers := arglist ||| [ 32767, 45, 53, 64, 100]    # combine command line provided and default set of valuesevery writes(lf,"factors of ",i := !numbers,"=") & writes(divisors(i)," ") do lf := "\n"end link factors
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 argument's prime factors.

q: 40 2 2 2 5

Alternatively, q: can produce provide a table of the exponents of the unique relevant prime factors

   __ q: 4202 3 5 72 1 1 1

With this, we can form lists of each of the potential relevant powers of each of these prime factors

   (^ i.@>:)&.>/ __ q: 420┌─────┬───┬───┬───┐│1 2 4│1 3│1 5│1 7│└─────┴───┴───┴───┘

From here, it's a simple matter (*/&>@{) to compute all possible factors of the original number

factrs=: */&>@{@((^ i.@>:)&.>/)@q:~&__   factrs 40 1  5 2 10 4 20 8 40

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:

   factors=: [: /:~@, */&>@{@((^ i.@>:)&.>/)@q:~&__   factors 4201 2 3 4 5 6 7 10 12 14 15 20 21 28 30 35 42 60 70 84 105 140 210 420

A less efficient, but concise variation on this theme:

    ~.,*/&> { 1 ,&.> q: 401 5 2 10 4 20 8 40

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:

factorsOfNumber=: monad define  Y=. y"_  /:~ ~. ( , Y%]) ( #~ 0=]|Y) 1+i.>.%:y)    factorsOfNumber 401 2 4 5 8 10 20 40

Another approach:

odometer =: #: i.@(*/)factors=: (*/@:^"1 odometer@:>:)[email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */:~&__

## Java

Works with: Java version 5+
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;}

## JavaScript

### Imperative

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]

### Functional (ES 5)

Translating the naive list comprehension example from Haskell, using a list monad for the comprehension

// Monadic bind (chain) for listsfunction chain(xs, f) {  return [].concat.apply([], xs.map(f));} // [m..n]function range(m, n) {  return Array.apply(null, Array(n - m + 1)).map(function (x, i) {    return m + i;  });} function factors_naive(n) {  return chain( range(1, n), function (x) {       // monadic chain/bind    return n % x ? [] : [x];                      // monadic fail or inject/return  });} factors_naive(6)

Output:

[1, 2, 3, 6]

Translating the Haskell (lows and highs) example

console.log(  (function (lstTest) {     // INTEGER FACTORS    function integerFactors(n) {      var rRoot = Math.sqrt(n),        intRoot = Math.floor(rRoot),         lows = range(1, intRoot).filter(function (x) {          return (n % x) === 0;        });       // for perfect squares, we can drop the head of the 'highs' list      return lows.concat(lows.map(function (x) {        return n / x;      }).reverse().slice((rRoot === intRoot) | 0));    }     // [m .. n]    function range(m, n) {      return Array.apply(null, Array(n - m + 1)).map(function (x, i) {        return m + i;      });    }     /*************************** TESTING *****************************/     // TABULATION OF RESULTS IN SPACED AND ALIGNED COLUMNS    function alignedTable(lstRows, lngPad, fnAligned) {      var lstColWidths = range(0, lstRows.reduce(function (a, x) {        return x.length > a ? x.length : a;      }, 0) - 1).map(function (iCol) {        return lstRows.reduce(function (a, lst) {          var w = lst[iCol] ? lst[iCol].toString().length : 0;          return (w > a) ? w : a;        }, 0);      });       return lstRows.map(function (lstRow) {        return lstRow.map(function (v, i) {          return fnAligned(v, lstColWidths[i] + lngPad);        }).join('')      }).join('\n');    }     function alignRight(n, lngWidth) {      var s = n.toString();      return Array(lngWidth - s.length + 1).join(' ') + s;    }     // TEST    return '\nintegerFactors(n)\n\n' + alignedTable(      lstTest.map(integerFactors).map(function (x, i) {        return [lstTest[i], '-->'].concat(x);      }), 2, alignRight    ) + '\n';   })([25, 45, 53, 64, 100, 102, 120, 12345, 32766, 32767]));

Output:

integerFactors(n)      25  -->  1   5  25     45  -->  1   3   5    9   15    45     53  -->  1  53     64  -->  1   2   4    8   16    32    64    100  -->  1   2   4    5   10    20    25     50  100    102  -->  1   2   3    6   17    34    51    102    120  -->  1   2   3    4    5     6     8     10   12   15   20   24    30     40     60    120  12345  -->  1   3   5   15  823  2469  4115  12345  32766  -->  1   2   3    6   43    86   127    129  254  258  381  762  5461  10922  16383  32766  32767  -->  1   7  31  151  217  1057  4681  32767 

## jq

Works with: jq version 1.4
# This implementation uses "sort" for tidinessdef 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
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 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 Output: julia> factors(45) 6-element Array{Int64,1}: 1 3 5 9 15 45  ## K  f:{d:&~x!'!1+_sqrt x;?d,_ x%|d} f 11 f 31 3 f 1201 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120 f 10241 2 4 8 16 32 64 128 256 512 1024 f 6008514751431 71 839 1471 6857 59569 104441 486847 1234169 5753023 10086647 87625999 408464633 716151937 8462696833 600851475143 #f 3491888400 / has 1920 factors1920 / Number of factors for 3491888400 .. 3491888409 #:'f' 3491888400+!101920 16 4 4 12 16 32 16 8 24 ## 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):  (defun factors (n) (list-comp ((<- i (when (== 0 (rem n i))) (lists:seq 1 (trunc (/ n 2))))) i))  ### 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):  (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)))  Output: > (factors 10677106534462215678539721403561279) (104729 104729 104729 98731 98731 32579 29269 1)  ## Liberty BASIC num = 10677106534462215678539721403561279maxnFactors = 1000dim primeFactors(maxnFactors), nPrimeFactors(maxnFactors)global nDifferentPrimeNumbersFound, nFactors, iFactor print "Start finding all factors of ";num; ":" nDifferentPrimeNumbersFound=0dummy = factorize(num,2)nFactors = showPrimeFactors(num)dim factors(nFactors)dummy = generateFactors(1,1)sort factors(), 0, nFactors-1for 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 ifend 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 Output: Start finding all factors of 10677106534462215678539721403561279:10677106534462215678539721403561279 = 29269^1 * 32579^1 * 98731^2 * 104729^31 12 292693 325794 987315 1047296 9535547517 28897576398 30653131019 321655724910 341196609111 974781036112 1033999889913 1096816344114 9414541412098115 9986483551747916 28530866145610917 30264142777483118 31757391375101919 32102717575462920 33686682413052121 35733179674433922 102087843129716923 108289774469337124 114868478901248925 929507088157857511126 985975507547621914927 1045874435891005819128 2988009080563683946129 3169533408943027579930 3325919841323046885131 3362085508960654054132 3527972562436533380933 3742300174123787913134 10691557723132121220135 11341079790399205145936 97346347835684259279991937 103260228929954895525562138 109533383796429148428523939 312931202998354055991106940 331942064385194335415347141 348320259061921377229637942 369481038491415704448276143 1119716148785903923259852944 10194985662483376790134271695145 10814340515605246253496593170946 32772971958814621929892634530147 36479232411295963915882747629148 10677106534462215678539721403561279done ## Logo to factors :n output filter [equal? 0 modulo :n ?] iseq 1 :nend show factors 28 ; [1 2 4 7 14 28] ## 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 fend ## Maple  numtheory:-divisors(n);  ## Mathematica / Wolfram Language Factorize[n_Integer] := Divisors[n] ## MATLAB / Octave  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;  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. (%i96) divisors(100);(%o96) {1,2,4,5,10,20,25,50,100} Such a function could be implemented like so: divisors2(n) := map( lambda([l], lreduce("*", l)), apply( cartesian_product, map( lambda([fac], setify(makelist(fac[1]^i, i, 0, fac[2]))), ifactors(n)))); ## MAXScript  fn factors n =( return (for i = 1 to n+1 where mod n i == 0 collect i))  Output:  factors 3#(1, 3)factors 7#(1, 7)factors 14#(1, 2, 7, 14)factors 60#(1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60)factors 54#(1, 2, 3, 6, 9, 18, 27, 54)  ## 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 :- 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. ### 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

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] ## NetRexx Translation of: REXX /* NetRexx ************************************************************ 21.04.2013 Walter Pachl* 21.04.2013 add method main to accept argument(s)*********************************************************************/options replace format comments java crossref symbols nobinaryclass 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 */ 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 ## Nim 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) ## 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 ) newline53 factors .s .clr ( => 1 53 ) newline64 factors .s .clr ( => 1 2 4 8 16 32 64 ) newline12 factors .s .clr ( => 1 2 3 4 6 12 )  ## Oberon-2 Oxford Oberon-2  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 vEND 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.LnEND Factors.  Output:  1 3 41 123 4 factors  ## 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(); }; } }} ## OCaml let rec range = function 0 -> [] | n -> range(n-1) @ [n] let factors n = List.filter (fun v -> (n mod v) = 0) (range n) ## Oforth Integer method: factors self seq filter(#[ self isMultiple ]) ; 120 factors println Output: [1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120]  ## 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.'<'} endin {Show {Factors 53}} ## PARI/GP divisors(n) ## Panda Panda has a factor function already, it's defined as: fun factor(n) type integer->integer f where n.mod(1..n=>f)==0 45.factor ## Pascal Translation of: Fortran Works with: Free Pascal version 2.6.2 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. 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 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. 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 sub factors{ my($n) = @_;        return grep { $n %$_ == 0 }(1 .. $n);}print join ' ',factors(64), "\n"; Or more intelligently: sub factors { my$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";

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

Library: ntheory
use ntheory qw/divisors/;print join " ", divisors(12345678), "\n";# Alternately something like:  fordivisors { say } 12345678; 

## Perl 6

Works with: Rakudo version 2015.12
sub factors (Int $n) { squish sort ($_, $n div$_ if $n %%$_ for 1 .. sqrt $n) } ## Phix There is a builtin factors(n), which takes an optional second parameter to include 1 and n, so eg ?factors(12345,1) displays Output: {1,3,5,15,823,2469,4115,12345}  You can find the implementation of factors() and prime_factors() in builtins\pfactors.e ## PHP function GetFactors($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 $factors;} ## PicoLisp (de factors (N) (filter '((D) (=0 (% N D))) (range 1 N) ) ) ## PL/I do i = 1 to n; if mod(n, i) = 0 then put skip list (i);end; ## PowerShell ### Straightforward but slow function Get-Factor ($a) {    1..$a | Where-Object {$a % $_ -eq 0 }} This one uses a range of integers up to the target number and just filters it using the Where-Object cmdlet. It's very slow though, so it is not very usable for larger numbers. ### A little more clever function Get-Factor ($a) {    1..[Math]::Sqrt($a)  | Where-Object {$a % $_ -eq 0 }  | ForEach-Object {$_; $a /$_ }         | Sort-Object -Unique}

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:

editvar /newvar /value=a /userinput=1 /title=Enter an integer:do /delimspaces %% -a- >bprintline Factors of -a-: -b-

## Prolog

Simple Brute Force Implementation

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

A Slightly Smarter Implementation

 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 )  .

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

  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)  .
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

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

>>> def factors(n):      return [i for i in range(1, n + 1) if not n%i]

Slightly better (realize that there are no factors between n/2 and n):

>>> 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]

Much better (realize that factors come in pairs, the smaller of which is no bigger than sqrt(n)):

>>> 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]

More efficient when factoring many numbers:

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

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

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


## Racket

 #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

## REALbasic

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 resultEnd Function

## REXX

### optimized version

This REXX version has no effective limits on the number of decimal 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.

/*REXX program displays divisors of any  [negative/zero/positive]  integer(s).*/parse arg bot top inc .                                       /*optional args.*/top=word(top bot 20,1); bot=word(bot 1,1); inc=word(inc 1,1)  /*range options.*/w=length(high)+1;       numeric digits max(9,w);      $='∞' /*digits for // */@.=left('',7); @.1='{unity}'; @.2='[prime]'; @.$='  {'$"} " /*some literals.*/say center('n',1+w) '#divisors' center('divisors',60) /*show a header.*/say copies('═',1+w) '═════════' copies('═' ,60) /* " " sep. */ do n=bot to top by inc; divs=divisors(n); #=words(divs) if divs==$  then do;  #=$; divs=' (infinite)'; end /*handle infinity*/ p=@.#; if n<0 then p=@.. /*handle negative*/ say center(n,w+1) center('['#"]",9) "──► " p ' ' divs end /*n*/ /* [↑] process a range of integers. */exit /*stick a fork in it, we're all done. *//*────────────────────────────────────────────────────────────────────────────*/divisors: procedure; parse arg x; x=abs(x); if x==1 then return 1odd=x//2; b=x; if x==0 then return '∞'a=1 /* [↓] process 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 factors to α&ß lists.*/ end /*j*/ /* [↑] % is REXX's integer division.*/ /* [↓] adjust for a square. ___ */if j*j==x then return a j b /*Was X a square? If so, insert √ x */ return a b /*return the divisors of both lists. */ 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 /* 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 Enddo j=low to high say ' n = ' right(j,6) " divisors = " divs(j) endexit 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 */ ## Ring  nArray = list(100)n = 45j = 0for i = 1 to n if n % i = 0 j = j + 1 nArray[j] = i oknext see "Factors of " + n + " = "for i = 1 to j see "" + nArray[i] + " "next  ## Ruby class Integer def factors() (1..self).select { |n| (self % n).zero? } endendp 45.factors [1, 3, 5, 9, 15, 45]  As we only have to loop up to ${\displaystyle {\sqrt {n}}}$, we can write class Integer def factors 1.upto(Math.sqrt(self)).select {|i| (self % i).zero?}.inject([]) do |f, i| f << self/i unless i == self/i f << i end.sort endend[45, 53, 64].each {|n| puts "#{n} : #{n.factors}"} Output: 45 : [1, 3, 5, 9, 15, 45] 53 : [1, 53] 64 : [1, 2, 4, 8, 16, 32, 64] ## Run BASIC PRINT "Factors of 45 are ";factorlist$(45)PRINT "Factors of 12345 are "; factorlist$(12345)END function factorlist$(f)DIM L(100)FOR i = 1 TO SQR(f)  IF (f MOD i) = 0 THEN    L(c) = i    c = c + 1    IF (f <> i^2) THEN      L(c) = (f / i)      c = c + 1    END IF  END IFNEXT is = 1while s = 1s = 0for i = 0 to c-1 if L(i) > L(i+1) and L(i+1) <> 0 then  t = L(i)  L(i) = L(i+1)  L(i+1) = t  s      = 1 end ifnext iwendFOR i = 0 TO c-1  factorlist$= factorlist$ + STR$(L(i)) + ", "NEXTend function 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++ 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; ## Scala  Brute force approach: def factors(num: Int) = { (1 to num).filter { divisor => num % divisor == 0 } Since factors can't be higher than sqrt(num), the code above can be edited as follows def factors(num: Int) = { (1 to sqrt(num)).filter { divisor => num % divisor == 0 }  ## Scheme This implementation uses a naive trial division algorithm. (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) Output:  (1 239 4649 1111111)  ## 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;
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


## SequenceL

Brute Force Method

A simple brute force method using an indexed partial function as a filter.

Factors(num(0))[i] := i when num mod i = 0 foreach i within 1 ... num;

Slightly More Efficient Method

A slightly more efficient method, only going up to the sqrt(n).

Factors(num(0)) :=	let		factorPairs[i] :=				[i] when i = sqrt(num)			else 				[i, num/i] when num mod i = 0 			foreach i within 1 ... floor(sqrt(num));	in		join(factorPairs);

## Sidef

func factors(n) {  var divs = []  range(1, n.sqrt.int).each { |d|    divs << d if n%%d  }  divs + [divs[-1]**2 == n ? divs.pop : ()] + divs.reverse.map{|d| n/d }} [53, 64, 32766].each { |n|    say "factors(#{n}): #{factors(n)}"}
Output:
factors(53): 1 53
factors(64): 1 2 4 8 16 32 64
factors(32766): 1 2 3 6 43 86 127 129 254 258 381 762 5461 10922 16383 32766


## Slate

n@(Integer traits) primeFactors[  [| :result |   result nextPut: 1.   n primesDo: [| :prime | result nextPut: prime]] writingAs: {}].

where primesDo: is a part of the standard numerics library:

n@(Integer traits) primesDo: block"Decomposes the Integer into primes, applying the block to each (in increasingorder)."[| 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."].

## Smalltalk

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

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

Then use as follows:

 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)

## Swift

Simple implementation:

func factors(n: Int) -> [Int] {     return filter(1...n) { n % $0 == 0 }} More efficient implementation: import func Darwin.sqrt func sqrt(x:Int) -> Int { return Int(sqrt(Double(x))) } func factors(n: Int) -> [Int] { var result = [Int]() for factor in filter (1...sqrt(n), { n %$0 == 0 }) {         result.append(factor)         if n/factor != factor { result.append(n/factor) }    }     return sorted(result) }

Call:

println(factors(4))println(factors(1))println(factors(25))println(factors(63))println(factors(19))println(factors(768))
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

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]
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.

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}

## Ursa

This program takes an integer from the command line and outputs its factors.

decl int nset n (int args<1>) decl int ifor (set i 1) (< i (+ (/ n 2) 1)) (inc i)        if (= (mod n i) 0)                out i "  " console        end ifend forout n endl console

## Ursala

The simple way:

#import std#import nat factors "n" = (filter not remainder/"n") nrange(1,"n")

The complicated way:

factors "n" = nleq-<&@s <.~&r,quotient>*= "n"-* (not remainder/"n")*~ nrange(1,root("n",2))

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

factors = nleq-<&@rrZPFLs+ ^(~&r,division)^*D/~& nrange/1+ &!*K31

where nleq-<& isn't strictly necessary unless an ordered list is required.

#cast %nL example = factors 100
Output:
<1,2,4,5,10,20,25,50,100>

## VBA

Function Factors(x As Integer) As String Application.Volatile Dim i As Integer Dim cooresponding_factors As String Factors = 1 corresponding_factors = x For i = 2 To Sqr(x)  If x Mod i = 0 Then   Factors = Factors & ", " & i   If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors  End If Next i If x <> 1 Then Factors = Factors & ", " & corresponding_factorsEnd Function
Output:
cell formula is "=Factors(840)"
resultant value is "1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840"

## Wortel

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

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;]
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
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
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))