# 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 ## AppleScript Translation of: JavaScript -- integerFactors :: Int -> [Int]on integerFactors(n) if n = 1 then {1} else set realRoot to n ^ (1 / 2) set intRoot to realRoot as integer set blnPerfectSquare to intRoot = realRoot -- isFactor :: Int -> Bool script isFactor on lambda(x) (n mod x) = 0 end lambda end script -- Factors up to square root of n, set lows to filter(isFactor, range(1, intRoot)) -- integerQuotient :: Int -> Int script integerQuotient on lambda(x) (n / x) as integer end lambda end script -- and quotients of these factors beyond the square root. lows & map(integerQuotient, ¬ items (1 + (blnPerfectSquare as integer)) thru -1 of reverse of lows) end ifend integerFactors -- TESTon run integerFactors(120) --> {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120}end run -- GENERIC LIBRARY FUNCTIONS -- filter :: (a -> Bool) -> [a] -> [a]on filter(f, xs) tell mReturn(f) set lst to {} set lng to length of xs repeat with i from 1 to lng set v to item i of xs if lambda(v, i, xs) then set end of lst to v end repeat return lst end tellend filter -- map :: (a -> b) -> [a] -> [b]on map(f, xs) tell mReturn(f) set lng to length of xs set lst to {} repeat with i from 1 to lng set end of lst to lambda(item i of xs, i, xs) end repeat return lst end tellend map -- range :: Int -> Int -> [Int]on range(m, n) if n < m then set d to -1 else set d to 1 end if set lst to {} repeat with i from m to n by d set end of lst to i end repeat return lstend range -- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Scripton mReturn(f) if class of f is script then f else script property lambda : f end script end ifend mReturn Output: {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120} ## 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)]))))

## COBOL

        IDENTIFICATION DIVISION.       PROGRAM-ID. FACTORS.       DATA DIVISION.       WORKING-STORAGE SECTION.       01  CALCULATING.           03  NUM  USAGE BINARY-LONG VALUE ZERO.           03  LIM  USAGE BINARY-LONG VALUE ZERO.           03  CNT  USAGE BINARY-LONG VALUE ZERO.           03  DIV  USAGE BINARY-LONG VALUE ZERO.           03  REM  USAGE BINARY-LONG VALUE ZERO.           03  ZRS  USAGE BINARY-SHORT VALUE ZERO.        01  DISPLAYING.           03  DIS  PIC 9(10) USAGE DISPLAY.        PROCEDURE DIVISION.       MAIN-PROCEDURE.           DISPLAY "Factors of? " WITH NO ADVANCING           ACCEPT NUM           DIVIDE NUM BY 2 GIVING LIM.            PERFORM VARYING CNT FROM 1 BY 1 UNTIL CNT > LIM               DIVIDE NUM BY CNT GIVING DIV REMAINDER REM               IF REM = 0                   MOVE CNT TO DIS                   PERFORM SHODIS               END-IF           END-PERFORM.            MOVE NUM TO DIS.           PERFORM SHODIS.           STOP RUN.        SHODIS.           MOVE ZERO TO ZRS.           INSPECT DIS TALLYING ZRS FOR LEADING ZERO.           DISPLAY DIS(ZRS + 1:)           EXIT PARAGRAPH.        END PROGRAM FACTORS. 

## 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.

## Fish

0v >i:0(?v'0'%+a*       >~a,:1:>r{%        ?vr:nr','ov              ^:&:;?(&:+1r:<        <  

Must be called with pre-polulated value (Positive Integer) in the input stack. Try at Fish Playground[1].

For Input Number :
 120

The following output was generated:

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

## 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

## FreeBASIC

' FB 1.05.0 Win64 Sub printFactors(n As Integer)  If n < 1 Then Return  Print n; " =>";  For i As Integer = 1 To n / 2    If n Mod i = 0 Then Print i; " ";  Next i  Print nEnd Sub  printFactors(11)printFactors(21)printFactors(32)printFactors(45)printFactors(67)printFactors(96)PrintPrint "Press any key to quit"Sleep
Output:
 11 => 1  11
21 => 1  3  7  21
32 => 1  2  4  8  16  32
45 => 1  3  5  9  15  45
67 => 1  67
96 => 1  2  3  4  6  8  12  16  24  32  48  96


## 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}


## FutureBasic

 include "ConsoleWindow" clear local modelocal fn IntegerFactors( f as long ) as Str255dim as long i, s, l(100), c : c = 0dim as Str255 factorStr for i = 1 to sqr(f)  if ( f mod i == 0 )    l(c) = i    c++      if ( f <> i ^ 2 )        l(c) = ( f / i )        c++      end if  end ifnext is = 1while ( s = 1 )s = 0  for i = 0 to c-1    if l(i) > l(i+1) and l(i+1) <> 0      swap l(i), l(i+1)      s = 1    end if  next iwendfor i = 0 to c-1  if ( i < c -1 )    factorStr = factorStr + str$(l(i)) + "," else factorStr = factorStr + str$(l(i))  end ifnextend fn = factorStr print "Factors of 25 are:"; fn IntegerFactors( 25 )print "Factors of 45 are:"; fn IntegerFactors( 45 )print "Factors of 103 are:"; fn IntegerFactors( 103 )print "Factors of 760 are:"; fn IntegerFactors( 760 )print "Factors of 12345 are:"; fn IntegerFactors( 12345 )print "Factors of 32766 are:"; fn IntegerFactors( 32766 )print "Factors of 32767 are:"; fn IntegerFactors( 32767 )print "Factors of 57097 are:"; fn IntegerFactors( 57097 )print "Factors of 12345678 are:"; fn IntegerFactors( 12345678 )print "Factors of 32434243 are:"; fn IntegerFactors( 32434243 ) 

Output:

Factors of 25 are: 1, 5, 25
Factors of 45 are: 1, 3, 5, 9, 15, 45
Factors of 103 are: 1, 103
Factors of 760 are: 1, 2, 4, 5, 8, 10, 19, 20, 38, 40, 76, 95, 152, 190, 380, 760
Factors of 12345 are: 1, 3, 5, 15, 823, 2469, 4115, 12345
Factors of 32766 are: 1, 2, 3, 6, 43, 86, 127, 129, 254, 258, 381, 762, 5461, 10922, 16383, 32766
Factors of 32767 are: 1, 7, 31, 151, 217, 1057, 4681, 32767
Factors of 57097 are: 1, 57097
Factors of 12345678 are: 1, 2, 3, 6, 9, 18, 47, 94, 141, 282, 423, 846, 14593, 29186, 43779, 87558, 131337, 262674, 685871, 1371742, 2057613, 4115226, 6172839, 12345678
Factors of 32434243 are: 1, 307, 105649, 32434243


## 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

## Gosu

var numbers = {11, 21, 32, 45, 67, 96}numbers.each(\ number -> printFactors(number)) function printFactors(n: int) {  if (n < 1) return  var result ="${n} => " (1 .. n/2).each(\ i -> {result += n % i == 0 ? "${i} " : ""})  print("${result}${n}")}
Output:
11 => 1 11
21 => 1 3 7 21
32 => 1 2 4 8 16 32
45 => 1 3 5 9 15 45
67 => 1 67
96 => 1 2 3 4 6 8 12 16 24 32 48 96


## 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'es Primes module for finding prime factors

import HFM.Primes (primePowerFactors)import Control.Monad (mapM)import Data.List (product) -- primePowerFactors :: Integer -> [(Integer,Int)] factors = map product .          mapM (\(p,m)-> [p^i | i<-[0..m]]) . primePowerFactors

Returns list of factors out of order, e.g.:

~> factors 42[1,7,3,21,2,14,6,42]

Or, prime decomposition task can be used (although, a trial division-only version will become very slow for large primes),

import Data.List (group)primePowerFactors = map (\x-> (head x, length x)) . group . factorize

The above function can also be found in the package arithmoi, as Math.NumberTheory.Primes.factorise :: Integer -> [(Integer, Int)], which performs "factorisation of Integers by the elliptic curve algorithm after Montgomery" and "is best suited for numbers of up to 50-60 digits".

Or, making do without further imports beyond the standard Prelude:

integerFactors  :: Integral a  => a -> [a]integerFactors n  | n < 1 = []  | otherwise = lows ++ if_ perfect tail id (reverse (quot n <$> lows)) where intRoot = floor (sqrt$ fromIntegral n)    perfect = intRoot * intRoot == n    lows = filter ((== 0) . rem n) [1 .. intRoot]  if_ :: Bool -> a -> a -> aif_ True x _ = xif_ False _ y = y  main :: IO ()main = print integerFactors 600 Output: [1,2,3,4,5,6,8,10,12,15,20,24,25,30,40,50,60,75,100,120,150,200,300,600] ### List comprehension Naive, functional, no import, in increasing order: factors_naive n = [i | i <-[1..n], mod n i == 0] ~> factors_naive 25[1,5,25] Factor, cofactor. Get the list of factor–cofactor pairs sorted, for a quadratic speedup: import Data.Listfactors_co n = sort [ i | i <- [1..floor (sqrt (fromIntegral n))] , (d,0) <- [divMod n i], i <- [i]++[d|d>i] ] A version of the above without the need for sorting, making it to be online (i.e. productive immediately, which can be seen in GHCi); factors in increasing order: import Data.Listfactors_o n = ds ++ [r | (d,0) <- [divMod n r], r <- [r]++[d|d>r]] ++ reverse (map (n div) ds) where r = floor (sqrt (fromIntegral n)) ds = [i | i <- [1..r-1], mod n i == 0] Testing: *Main> :set +s~> factors_o 120[1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120](0.00 secs, 0 bytes) ~> factors_o 12041111117[1,7,41,287,541,3787,22181,77551,155267,542857,3179591,22257137,41955091,293685637,1720158731,12041111117](0.09 secs, 50758224 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 #### ES5 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  #### ES6 (function (lstTest) { 'use strict'; // INTEGER FACTORS // integerFactors :: Int -> [Int] let integerFactors = (n) => { let rRoot = Math.sqrt(n), intRoot = Math.floor(rRoot), lows = range(1, intRoot) .filter(x => (n % x) === 0); // for perfect squares, we can drop // the head of the 'highs' list return lows.concat(lows .map(x => n / x) .reverse() .slice((rRoot === intRoot) | 0) ); }, // range :: Int -> Int -> [Int] range = (m, n) => Array.from({ length: (n - m) + 1 }, (_, i) => m + i); /*************************** TESTING *****************************/ // TABULATION OF RESULTS IN SPACED AND ALIGNED COLUMNS let alignedTable = (lstRows, lngPad, fnAligned) => { var lstColWidths = range( 0, lstRows .reduce( (a, x) => (x.length > a ? x.length : a), 0 ) - 1 ) .map((iCol) => lstRows .reduce((a, lst) => { let w = lst[iCol] ? lst[iCol].toString() .length : 0; return (w > a) ? w : a; }, 0)); return lstRows.map((lstRow) => lstRow.map((v, i) => fnAligned( v, lstColWidths[i] + lngPad )) .join('') ) .join('\n'); }, alignRight = (n, lngWidth) => { let s = n.toString(); return Array(lngWidth - s.length + 1) .join(' ') + s; }; // TEST return '\nintegerFactors(n)\n\n' + alignedTable(lstTest .map(integerFactors) .map( (x, i) => [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: . asnum  | 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:{i:{y[&x=y*x div y]}[x;1+!_sqrt x];?i,x div|i}equivalent to:q)f:{i:{y where x=y*x div y}[x ; 1+ til floor sqrt x]; distinct i,x div reverse i}    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

## Kotlin

// version 1.0.5-2 fun printFactors(n: Int) {    if (n < 1) return    print("$n => ") for (i in 1 .. n/2) if (n % i == 0) print("$i ")    println(n)      } fun main(args: Array<String>) {    val numbers = intArrayOf(11, 21, 32, 45, 67, 96)    for (number in numbers) printFactors(number)}
Output:
11 => 1 11
21 => 1 3 7 21
32 => 1 2 4 8 16 32
45 => 1 3 5 9 15 45
67 => 1 67
96 => 1 2 3 4 6 8 12 16 24 32 48 96


## 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

### A Simpler Approach

This is a somewhat simpler approach for finding the factors of smaller numbers (less than one million).

 print "ROSETTA CODE - Factors of an integer"'A simpler approach for smaller numbers[Start]printinput "Enter an integer (< 1,000,000): "; nn=abs(int(n)): if n=0 then goto [Quit]if n>999999 then goto [Start]FactorCount=FactorCount(n)select case FactorCount    case 1: print "The factor of 1 is: 1"    case else        print "The "; FactorCount; " factors of "; n; " are: ";        for x=1 to FactorCount            print " "; Factor(x);        next x        if FactorCount=2 then print " (Prime)" else printend selectgoto [Start] [Quit]print "Program complete."end function FactorCount(n)    dim Factor(100)    for y=1 to n        if y>sqr(n) and FactorCount=1 then'If no second factor is found by the square root of n, then n is prime.            FactorCount=2: Factor(FactorCount)=n: exit function        end if        if (n mod y)=0 then            FactorCount=FactorCount+1            Factor(FactorCount)=y        end if    next yend function 
Output:
ROSETTA CODE - Factors of an integer

Enter an integer (< 1,000,000): 1
The factor of 1 is: 1

Enter an integer (< 1,000,000): 2
The 2 factors of 2 are:  1 2 (Prime)

Enter an integer (< 1,000,000): 4
The 3 factors of 4 are:  1 2 4

Enter an integer (< 1,000,000): 6
The 4 factors of 6 are:  1 2 3 6

Enter an integer (< 1,000,000): 999999
The 64 factors of 999999 are:  1 3 7 9 11 13 21 27 33 37 39 63 77 91 99 111 117 143 189 231 259 273 297 333 351 407 429 481 693 777 819 999 1001 1221 1287 1443 2079 2331 2457 2849 3003 3367 3663 3861 4329 5291 6993 8547 9009 10101 10989 129
87 15873 25641 27027 30303 37037 47619 76923 90909 111111 142857 333333 999999

Enter an integer (< 1,000,000):
Program complete.


## Lingo

on factors(n)   res = [1]  repeat with i = 2 to n/2    if n mod i = 0 then res.add(i)  end repeat  res.add(n)  return resend
put factors(45)-- [1, 3, 5, 9, 15, 45]put factors(53)-- [1, 53]put factors(64)-- [1, 2, 4, 8, 16, 32, 64]

## 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:

## 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:

## 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.

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 listing as well as the number of factors. It also displays a final count of the number of primes found. /*REXX program displays divisors of any [negative/zero/positive] integer or a range.*/parse arg LO HI inc . /*obtain the optional args*/HI=word(HI LO 20, 1); LO=word(LO 1, 1); inc=word(inc 1, 1) /*define the range options*/w=length(high)+2; numeric digits max(9, w-2);$='∞'   /*decimal digits for  //  */@.=left('',7);  @.1="{unity}"; @.2='[prime]'; @.$=" {"$'}  ' /*define some literals.   */say center('n', w)    "#divisors"    center('divisors', 60)   /*display the  header.    */say copies('═', w)    "═════════"    copies('═'       , 60)   /*   "     "   separator. */p#=0                                                          /*count of prime numbers. */     do n=LO  to HI  by inc; divs=divisors(n); #=words(divs)  /*get list of divs; # divs*/     if divs==$then do; #=$ ; divs= '  (infinite)';  end   /*handle case for infinity*/     p=@.#;      if n<0  then if n\==-1  then p=@..           /*   "     "   "  negative*/     if p==@.2  then p#=p#+1                                  /*Prime? Then bump counter*/     say center(n, w)      center('['#"]", 9)       "──► "        p      ' '       divs     end   /*n*/                                 /* [↑]   process a range of integers.  */saysay left('', 17)     p#    ' primes were found.' /*display the number of primes found.  */exit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/divisors: procedure; parse arg x 1 b;     a=1    /*set  X  and  B  to the 1st argument. */if x<2  then do; x=abs(x);  if x==1  then return 1;  if x==0  then return '∞';  b=x;  endodd=x//2                                         /* [↓]  process EVEN or ODD ints.   ___*/        do j=2+odd  by 1+odd  while j*j<x        /*divide by all the integers up to √ x */        if x//j==0  then do; a=a j; b=x%j b; end /*÷?  Add factors to  α  and  ß  lists.*/        end   /*j*/                              /* [↑]  %  ≡  integer division.     ___*/if j*j==x  then  return  a j b                   /*Was  X  a square?   Then 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]    ──►  {unity}   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

Primes that were found:  46


### 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, 

## Rust

fn main() {    assert_eq!(vec![1, 2, 4, 5, 10, 10, 20, 25, 50, 100], factor(100)); // asserts that two expressions are equal to each other    assert_eq!(vec![1, 101], factor(101)); } fn factor(num: i32) -> Vec<i32> {    let mut factors: Vec<i32> = Vec::new(); // creates a new vector for the factors of the number     for i in 1..((num as f32).sqrt() as i32 + 1) {         if num % i == 0 {            factors.push(i); // pushes smallest factor to factors            factors.push(num/i); // pushes largest factor to factors        }    }    factors.sort(); // sorts the factors into numerical order for viewing purposes    factors // returns the factors}

## 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)


$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:

## 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))


## ZX Spectrum Basic

Translation of: AWK
10 INPUT "Enter a number or 0 to exit: ";n20 IF n=0 THEN STOP 30 PRINT "Factors of ";n;": ";40 FOR i=1 TO n50 IF FN m(n,i)=0 THEN PRINT i;" ";60 NEXT i70 DEF FN m(a,b)=a-INT (a/b)*b