Factors of an integer

Compute the factors of a number. The factors of a positive integer are those positive integers by which it 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 is has countably infinite members, and the factors of negative integers can be obtained from the factors of related positive numbers without difficulty; this task does not require handling of either of these cases). Note that even prime numbers will have at least two factors; ‘1’ and themselves.

AutoHotkey

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

Factors(n) { Loop, % floor(sqrt(n))

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

}</lang>

Output:

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

Clojure

This example is incorrect. Please fix the code and remove this message.

Details: It needs to include the number itself in the list of factors; right now it is only producing the proper divisors of the number, which is not quite right.

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

(print (factors 45))</lang>

(1 3 5 9 15)

Common Lisp

We iterate in the range 1..sqrt(n) collecting ‘low’ factors and corresponding ‘high’ factors, and combine at the end to produce an ordered list of factors.

<lang lisp>(defun factors (n &aux (lows '()) (highs '()))

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

D

<lang d> int[]factors(int num) {

       int[]ret;
       for(n=1;n<=num;n++) if (num%n==0) ret ~= n;

} </lang>

J

J has a primitive, q: which returns its prime factors.

<lang J>q: 40

2 2 2 5</lang>

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

<lang J>__ q: 40

2 5
3 1</lang>

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

┌───────┬───┐
│1 2 4 8│1 5│
└───────┴───┘

From here, it's a simple matter to compute all possible factors of the original number

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

<lang J>factors 40

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

Java

Works with: Java version 1.5+

<lang java5>public static TreeSet<Long> factors(long n){ TreeSet<Long> factors = new TreeSet<Long>(); factors.add(n); factors.add(1L); for(long test = n - 1; test >= Math.sqrt(n); test--){ if(n % test == 0){ factors.add(test); factors.add(n / test); } } return factors; }</lang>

PowerShell

Straightforward but slow

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

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

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

A little more clever

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

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

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

Python

<lang python>>>> def factors(n): return [i for i in range(1,n//2) if not n%i] + [n]

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

Alternative (after Ruby): <lang python>>>> from math import ceil, sqrt >>> def factor(n): return sorted(set(sum( ([x,n/x] for x in range(1, ceil(sqrt(n)) + 1) if not n%x), [] )))

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

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

R

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

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

} factors(60) </lang>

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

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

Ruby

<lang ruby>class Integer

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

end p 45.factors</lang>

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

As we only have to loop up to ${\sqrt {n}}$ , we can write <lang ruby>class Integer

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

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

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

Tcl

<lang tcl>proc factors {n} {

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

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

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

Ursala

Filter the sequence of numbers from 1 to n according to divisibility by n. <lang Ursala>

import std
import nat

factors "n" = (not remainder/"n")*~@t iota successor "n" </lang> test program: <lang Ursala>

cast %nL

example = factors 100 </lang> output:

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