Factors of an integer
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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.
See also:
Clojure
<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>
Java
<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>
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 , 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