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Count in factors

From Rosetta Code
Task
Count in factors
You are encouraged to solve this task according to the task description, using any language you may know.
Task

Write a program which counts up from   1,   displaying each number as the multiplication of its prime factors.

For the purpose of this task,   1   (unity)   may be shown as itself.


Example

      2   is prime,   so it would be shown as itself.
      6   is not prime;   it would be shown as   .
2144   is not prime;   it would be shown as   .


Related tasks



Ada[edit]

The solution uses the generic package Prime_Numbers from Prime decomposition#Ada

count.adb
with Ada.Command_Line, Ada.Text_IO, Prime_Numbers;
 
procedure Count is
package Prime_Nums is new Prime_Numbers
(Number => Natural, Zero => 0, One => 1, Two => 2); use Prime_Nums;
 
procedure Put (List : Number_List) is
begin
for Index in List'Range loop
Ada.Text_IO.Put (Integer'Image (List (Index)));
if Index /= List'Last then
Ada.Text_IO.Put (" x");
end if;
end loop;
end Put;
 
N  : Natural := 1;
Max_N : Natural := 15; -- the default for Max_N
begin
if Ada.Command_Line.Argument_Count = 1 then -- read Max_N from command line
Max_N := Integer'Value (Ada.Command_Line.Argument (1));
end if; -- else use the default
loop
Ada.Text_IO.Put (Integer'Image (N) & ": ");
Put (Decompose (N));
Ada.Text_IO.New_Line;
N := N + 1;
exit when N > Max_N;
end loop;
end Count;
Output:
 1:  1
 2:  2
 3:  3
 4:  2 x 2
 5:  5
 6:  2 x 3
 7:  7
 8:  2 x 2 x 2
 9:  3 x 3
 10:  2 x 5
 11:  11
 12:  2 x 2 x 3
 13:  13
 14:  2 x 7
 15:  3 x 5

ALGOL 68[edit]

Translation of: Euphoria
OP +:= = (REF FLEX []INT a, INT b) VOID:
BEGIN
[a + 1] INT c;
c[:⌈a] := a;
c[a+1:] := b;
a := c
END;
 
 
PROC factorize = (INT nn) []INT:
BEGIN
IF nn = 1 THEN (1)
ELSE
INT k := 2, n := nn;
FLEX[0]INT result;
WHILE n > 1 DO
WHILE n MOD k = 0 DO
result +:= k;
n := n % k
OD;
k +:= 1
OD;
result
FI
END;
 
FLEX[0]INT factors;
FOR i TO 22 DO
factors := factorize (i);
print ((whole (i, 0), " = "));
FOR j TO UPB factors DO
(j /= 1 | print (" × "));
print ((whole (factors[j], 0)))
OD;
print ((new line))
OD
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
19 = 19
20 = 2 × 2 × 5
21 = 3 × 7
22 = 2 × 11

AutoHotkey[edit]

Translation of: D
factorize(n){
if n = 1
return 1
if n < 1
return false
result := 0, m := n, k := 2
While n >= k{
while !Mod(m, k){
result .= " * " . k, m /= k
}
k++
}
return SubStr(result, 5)
}
Loop 22
out .= A_Index ": " factorize(A_index) "`n"
MsgBox % out
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
19: 19
20: 2 * 2 * 5
21: 3 * 7
22: 2 * 11

AWK[edit]

 
# syntax: GAWK -f COUNT_IN_FACTORS.AWK
BEGIN {
fmt = "%d=%s\n"
for (i=1; i<=16; i++) {
printf(fmt,i,factors(i))
}
i = 2144; printf(fmt,i,factors(i))
i = 6358; printf(fmt,i,factors(i))
exit(0)
}
function factors(n, f,p) {
if (n == 1) {
return(1)
}
p = 2
while (p <= n) {
if (n % p == 0) {
f = sprintf("%s%s*",f,p)
n /= p
}
else {
p++
}
}
return(substr(f,1,length(f)-1))
}
 

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
2144=2*2*2*2*2*67
6358=2*11*17*17

BBC BASIC[edit]

      FOR i% = 1 TO 20
PRINT i% " = " FNfactors(i%)
NEXT
END
 
DEF FNfactors(N%)
LOCAL P%, f$
IF N% = 1 THEN = "1"
P% = 2
WHILE P% <= N%
IF (N% MOD P%) = 0 THEN
f$ += STR$(P%) + " x "
N% DIV= P%
ELSE
P% += 1
ENDIF
ENDWHILE
= LEFT$(f$, LEN(f$) - 3)
 

Output:

         1 = 1
         2 = 2
         3 = 3
         4 = 2 x 2
         5 = 5
         6 = 2 x 3
         7 = 7
         8 = 2 x 2 x 2
         9 = 3 x 3
        10 = 2 x 5
        11 = 11
        12 = 2 x 2 x 3
        13 = 13
        14 = 2 x 7
        15 = 3 x 5
        16 = 2 x 2 x 2 x 2
        17 = 17
        18 = 2 x 3 x 3
        19 = 19
        20 = 2 x 2 x 5

Befunge[edit]

Lists the first 100 entries in the sequence. If you wish to extend that, the upper limit is implementation dependent, but may be as low as 130 for an interpreter with signed 8 bit data cells (131 is the first prime outside that range).

1>>>>:.48*"=",,::1-#v_.v
$<<<^_@#-"e":+1,+55$2<<<
v4_^#-1:/.:g00_00g1+>>0v
>8*"x",,:00g%!^!%g00:p0<
Output:
1 = 1 
2 = 2 
3 = 3 
4 = 2 x 2 
5 = 5 
6 = 2 x 3 
7 = 7 
8 = 2 x 2 x 2 
9 = 3 x 3 
10 = 2 x 5 
11 = 11 
12 = 2 x 2 x 3 
13 = 13 
14 = 2 x 7 
.
.
.

C[edit]

Code includes a dynamically extending prime number list. The program doesn't stop until you kill it, or it runs out of memory, or it overflows.

#include <stdio.h>
#include <stdlib.h>
 
typedef unsigned long long ULONG;
 
ULONG get_prime(int idx)
{
static long n_primes = 0, alloc = 0;
static ULONG *primes = 0;
ULONG last, p;
int i;
 
if (idx >= n_primes) {
if (n_primes >= alloc) {
alloc += 16; /* be conservative */
primes = realloc(primes, sizeof(ULONG) * alloc);
}
if (!n_primes) {
primes[0] = 2;
primes[1] = 3;
n_primes = 2;
}
 
last = primes[n_primes-1];
while (idx >= n_primes) {
last += 2;
for (i = 0; i < n_primes; i++) {
p = primes[i];
if (p * p > last) {
primes[n_primes++] = last;
break;
}
if (last % p == 0) break;
}
}
}
return primes[idx];
}
 
int main()
{
ULONG n, x, p;
int i, first;
 
for (x = 1; ; x++) {
printf("%lld = ", n = x);
 
for (i = 0, first = 1; ; i++) {
p = get_prime(i);
while (n % p == 0) {
n /= p;
if (!first) printf(" x ");
first = 0;
printf("%lld", p);
}
if (n <= p * p) break;
}
 
if (first) printf("%lld\n", n);
else if (n > 1) printf(" x %lld\n", n);
else printf("\n");
}
return 0;
}
Output:
1 = 1
2 = 2
3 = 3
4 = 2 x 2
5 = 5
6 = 2 x 3
7 = 7
8 = 2 x 2 x 2
9 = 3 x 3
10 = 2 x 5
11 = 11
12 = 2 x 2 x 3
13 = 13
14 = 2 x 7
.
.
.

C++[edit]

 
#include <iostream>
#include <sstream>
#include <iomanip>
using namespace std;
 
void getPrimeFactors( int li )
{
int f = 2; string res;
if( li == 1 ) res = "1";
else
{
while( true )
{
if( !( li % f ) )
{
stringstream ss; ss << f;
res += ss.str();
li /= f; if( li == 1 ) break;
res += " x ";
}
else f++;
}
}
cout << res << "\n";
}
 
int main( int argc, char* argv[] )
{
for( int x = 1; x < 101; x++ )
{
cout << right << setw( 4 ) << x << ": ";
getPrimeFactors( x );
}
cout << 2144 << ": "; getPrimeFactors( 2144 );
cout << "\n\n";
return system( "pause" );
}
 
Output:
   1: 1
   2: 2
   3: 3
   4: 2 x 2
   5: 5
   6: 2 x 3
   7: 7
   8: 2 x 2 x 2
   9: 3 x 3
  10: 2 x 5
  11: 11
  12: 2 x 2 x 3
  13: 13
  14: 2 x 7
  15: 3 x 5
  16: 2 x 2 x 2 x 2
  17: 17
  18: 2 x 3 x 3
  19: 19
  20: 2 x 2 x 5
  21: 3 x 7
  22: 2 x 11
  23: 23
  24: 2 x 2 x 2 x 3
  .
  .
  .

C#[edit]

using System;
using System.Collections.Generic;
 
namespace prog
{
class MainClass
{
public static void Main (string[] args)
{
for( int i=1; i<=22; i++ )
{
List<int> f = Factorize(i);
Console.Write( i + ": " + f[0] );
for( int j=1; j<f.Count; j++ )
{
Console.Write( " * " + f[j] );
}
Console.WriteLine();
}
}
 
public static List<int> Factorize( int n )
{
List<int> l = new List<int>();
 
if ( n == 1 )
{
l.Add(1);
}
else
{
int k = 2;
while( n > 1 )
{
while( n % k == 0 )
{
l.Add( k );
n /= k;
}
k++;
}
}
return l;
}
}
}

Clojure[edit]

(ns listfactors
(:gen-class))
 
(defn factors
"Return a list of factors of N."
([n]
(factors n 2 ()))
([n k acc]
(cond
(= n 1) (if (empty? acc)
[n]
(sort acc))
(>= k n) (if (empty? acc)
[n]
(sort (cons n acc)))
(= 0 (rem n k)) (recur (quot n k) k (cons k acc))
:else (recur n (inc k) acc))))
 
(doseq [q (range 1 26)]
(println q " = " (clojure.string/join " x "(factors q))))
 
Output:
1  =  1
2  =  2
3  =  3
4  =  2 x 2
5  =  5
6  =  2 x 3
7  =  7
8  =  2 x 2 x 2
9  =  3 x 3
10  =  2 x 5
11  =  11
12  =  2 x 2 x 3
13  =  13
14  =  2 x 7
15  =  3 x 5
16  =  2 x 2 x 2 x 2
17  =  17
18  =  2 x 3 x 3
19  =  19
20  =  2 x 2 x 5
21  =  3 x 7
22  =  2 x 11
23  =  23
24  =  2 x 2 x 2 x 3
25  =  5 x 5

CoffeeScript[edit]

count_primes = (max) ->
# Count through the natural numbers and give their prime
# factorization. This algorithm uses no division.
# Instead, each prime number starts a rolling odometer
# to help subsequent factorizations. The algorithm works similar
# to the Sieve of Eratosthenes, as we note when each prime number's
# odometer rolls a digit. (As it turns out, as long as your computer
# is not horribly slow at division, you're better off just doing simple
# prime factorizations on each new n vs. using this algorithm.)
console.log "1 = 1"
primes = []
n = 2
while n <= max
factors = []
for prime_odometer in primes
# digits are an array w/least significant digit in
# position 0; for example, [3, [0]] will roll as
# follows:
# [0] -> [1] -> [2] -> [0, 1]
[base, digits] = prime_odometer
i = 0
while true
digits[i] += 1
break if digits[i] < base
digits[i] = 0
factors.push base
i += 1
if i >= digits.length
digits.push 0
 
if factors.length == 0
primes.push [n, [0, 1]]
factors.push n
console.log "#{n} = #{factors.join('*')}"
n += 1
 
primes.length
 
num_primes = count_primes 10000
console.log num_primes

Common Lisp[edit]

Auto extending prime list:

(defparameter *primes*
(make-array 10 :adjustable t :fill-pointer 0 :element-type 'integer))
 
(mapc #'(lambda (x) (vector-push x *primes*)) '(2 3 5 7))
 
(defun extend-primes (n)
(let ((p (+ 2 (elt *primes* (1- (length *primes*))))))
(loop for i = p then (+ 2 i)
while (<= (* i i) n) do
(if (primep i t) (vector-push-extend i *primes*)))))
 
(defun primep (n &optional skip)
(if (not skip) (extend-primes n))
(if (= n 1) nil
(loop for p across *primes* while (<= (* p p) n)
never (zerop (mod n p)))))
 
(defun factors (n)
(extend-primes n)
(loop with res for x across *primes* while (> n (* x x)) do
(loop while (zerop (rem n x)) do
(setf n (/ n x))
(push x res))
finally (return (if (> n 1) (cons n res) res))))
 
(loop for n from 1 do
(format t "~a: ~{~a~^ × ~}~%" n (reverse (factors n))))
Output:
1: 
2: 2
3: 3
4: 4
5: 5
6: 2 × 3
7: 7
8: 2 × 2 × 2
9: 9
10: 2 × 5
11: 11
12: 2 × 2 × 3
13: 13
14: 2 × 7
...

Without saving the primes, and not all that much slower (probably because above code was not well-written):

(defun factors (n)
(loop with res for x from 2 to (isqrt n) do
(loop while (zerop (rem n x)) do
(setf n (/ n x))
(push x res))
finally (return (if (> n 1) (cons n res) res))))
 
(loop for n from 1 do
(format t "~a: ~{~a~^ × ~}~%" n (reverse (factors n))))

D[edit]

int[] factorize(in int n) pure nothrow
in {
assert(n > 0);
} body {
if (n == 1) return [1];
int[] result;
int m = n, k = 2;
while (n >= k) {
while (m % k == 0) {
result ~= k;
m /= k;
}
k++;
}
return result;
}
 
void main() {
import std.stdio;
foreach (i; 1 .. 22)
writefln("%d: %(%d × %)", i, i.factorize());
}
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
19: 19
20: 2 × 2 × 5
21: 3 × 7

Alternative Version[edit]

Library: uiprimes
Library uiprimes is a homebrew library to generate prime numbers upto the maximum 32bit unsigned integer range 2^32-1, by using a pre-generated bit array of Sieve of Eratosthenes (a dll in size of ~256M bytes :p ).
import std.stdio, std.math, std.conv, std.algorithm,
std.array, std.string, import xt.uiprimes;
 
pragma(lib, "uiprimes.lib");
 
// function _factorize_ included in uiprimes.lib
ulong[] factorize(ulong n) {
if (n == 0) return [];
if (n == 1) return [1];
ulong[] res;
uint limit = cast(uint)(1 + sqrt(n));
foreach (p; Primes(limit)) {
if (n == 1) break;
if (0UL == (n % p))
while((n > 1) && (0UL == (n % p ))) {
res ~= p;
n /= p;
}
}
if (n > 1)
res ~= [n];
return res;
}
 
string productStr(T)(in T[] nums) {
return nums.map!text().join(" x ");
}
 
void main() {
foreach (i; 1 .. 21)
writefln("%2d = %s", i, productStr(factorize(i)));
}

DCL[edit]

Assumes file primes.txt is a list of prime numbers;

$ close /nolog primes
$ on control_y then $ goto clean
$
$ n = 1
$ outer_loop:
$ x = n
$ open primes primes.txt
$
$ loop1:
$ read /end_of_file = prime primes prime
$ prime = f$integer( prime )
$ loop2:
$ t = x / prime
$ if t * prime .eq. x
$ then
$ if f$type( factorization ) .eqs. ""
$ then
$ factorization = f$string( prime )
$ else
$ factorization = factorization + "*" + f$string( prime )
$ endif
$ if t .eq. 1 then $ goto done
$ x = t
$ goto loop2
$ else
$ goto loop1
$ endif
$ prime:
$ if f$type( factorization ) .eqs. ""
$ then
$ factorization = f$string( x )
$ else
$ factorization = factorization + "*" + f$string( x )
$ endif
$ done:
$ write sys$output f$fao( "!4SL = ", n ), factorization
$ delete /symbol factorization
$ close primes
$ n = n + 1
$ if n .le. 2144 then $ goto outer_loop
$ exit
$
$ clean:
$ close /nolog primes
Output:
$ @count_in_factors
   1 = 1
   2 = 2
   3 = 3
   4 = 2*2
   5 = 5
   6 = 2*3
...
2144 = 2*2*2*2*2*67

DWScript[edit]

function Factorize(n : Integer) : String;
begin
if n <= 1 then
Exit('1');
var k := 2;
while n >= k do begin
while (n mod k) = 0 do begin
Result += ' * '+IntToStr(k);
n := n div k;
end;
Inc(k);
end;
Result:=SubStr(Result, 4);
end;
 
var i : Integer;
for i := 1 to 22 do
PrintLn(IntToStr(i) + ': ' + Factorize(i));
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
19: 19
20: 2 * 2 * 5
21: 3 * 7
22: 2 * 11

EchoLisp[edit]

 
(define (task (nfrom 2) (range 20))
(for ((i (in-range nfrom (+ nfrom range))))
(writeln i "=" (string-join (prime-factors i) " x "))))
 
 
Output:
(task 1_000_000_000)

1000000000     =     2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5    
1000000001     =     7 x 11 x 13 x 19 x 52579    
1000000002     =     2 x 3 x 43 x 983 x 3943    
1000000003     =     23 x 307 x 141623    
1000000004     =     2 x 2 x 41 x 41 x 148721    
1000000005     =     3 x 5 x 66666667    
1000000006     =     2 x 500000003    
1000000007     =     1000000007    
1000000008     =     2 x 2 x 2 x 3 x 3 x 7 x 109 x 109 x 167    
1000000009     =     1000000009    
1000000010     =     2 x 5 x 17 x 5882353    
1000000011     =     3 x 29 x 11494253    
1000000012     =     2 x 2 x 11 x 47 x 79 x 6121    
1000000013     =     7699 x 129887    
1000000014     =     2 x 3 x 13 x 103 x 124471    
1000000015     =     5 x 7 x 31 x 223 x 4133    
1000000016     =     2 x 2 x 2 x 2 x 62500001    
1000000017     =     3 x 3 x 111111113    
1000000018     =     2 x 500000009    
1000000019     =     83 x 12048193    

Eiffel[edit]

 
 
class
COUNT_IN_FACTORS
 
feature
 
display_factor (p: INTEGER)
-- Factors of all integers up to 'p'.
require
p_positive: p > 0
local
factors: ARRAY [INTEGER]
do
across
1 |..| p as c
loop
io.new_line
io.put_string (c.item.out + "%T")
factors := factor (c.item)
across
factors as f
loop
io.put_integer (f.item)
if f.is_last = False then
io.put_string (" x ")
end
end
end
end
 
 
factor (p: INTEGER): ARRAY [INTEGER]
-- Prime decomposition of 'p'.
require
p_positive: p > 0
local
div, i, next, rest: INTEGER
do
create Result.make_empty
if p = 1 then
Result.force (1, 1)
end
div := 2
next := 3
rest := p
from
i := 1
until
rest = 1
loop
from
until
rest \\ div /= 0
loop
Result.force (div, i)
rest := (rest / div).floor
i := i + 1
end
div := next
next := next + 2
end
ensure
is_divisor: across Result as r all p \\ r.item = 0 end
end
end
 
 

Test Output:

   1       1
   2       2
   3       3
   4       2 x 2
   5       5
   6       2 x 3
   7       7
   8       2 x 2 x 2
   9       3 x 3
  10       2 x 5
...
4990       2 x 5 x 499
4991       7 x 23 x 31
4992       2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 13
4993       4993
4994       2 x 11 x 227
4995       3 x 3 x 3 x 5 x 37
4996       2 x 2 x 1249
4997       19 x 263
4998       2 x 3 x 7 x 7 x 17
4999       4999
5000       2 x 2 x 2 x 5 x 5 x 5 x 5

Elixir[edit]

defmodule RC do
def factor(n), do: factor(n, 2, [])
 
def factor(n, i, fact) when n < i*i, do: Enum.reverse([n|fact])
def factor(n, i, fact) do
if rem(n,i)==0, do: factor(div(n,i), i, [i|fact]),
else: factor(n, i+1, fact)
end
end
 
Enum.each(1..20, fn n ->
IO.puts "#{n}: #{Enum.join(RC.factor(n)," x ")}" end)
Output:
1: 1
2: 2
3: 3
4: 2 x 2
5: 5
6: 2 x 3
7: 7
8: 2 x 2 x 2
9: 3 x 3
10: 2 x 5
11: 11
12: 2 x 2 x 3
13: 13
14: 2 x 7
15: 3 x 5
16: 2 x 2 x 2 x 2
17: 17
18: 2 x 3 x 3
19: 19
20: 2 x 2 x 5

Euphoria[edit]

function factorize(integer n)
sequence result
integer k
if n = 1 then
return {1}
else
k = 2
result = {}
while n > 1 do
while remainder(n, k) = 0 do
result &= k
n /= k
end while
k += 1
end while
return result
end if
end function
 
sequence factors
for i = 1 to 22 do
printf(1, "%d: ", i)
factors = factorize(i)
for j = 1 to length(factors)-1 do
printf(1, "%d * ", factors[j])
end for
printf(1, "%d\n", factors[$])
end for
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
19: 19
20: 2 * 2 * 5
21: 3 * 7
22: 2 * 11

F#[edit]

let factorsOf (num) =
Seq.unfold (fun (f, n) ->
let rec genFactor (f, n) =
if f > n then None
elif n % f = 0 then Some (f, (f, n/f))
else genFactor (f+1, n)
genFactor (f, n)) (2, num)
 
let showLines = Seq.concat (seq { yield seq{ yield(Seq.singleton 1)}; yield (Seq.skip 2 (Seq.initInfinite factorsOf))})
 
showLines |> Seq.iteri (fun i f -> printfn "%d = %s" (i+1) (String.Join(" * ", Seq.toArray f)))
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
:
2140 = 2 * 2 * 5 * 107
2141 = 2141
2142 = 2 * 3 * 3 * 7 * 17
2143 = 2143
2144 = 2 * 2 * 2 * 2 * 2 * 67
2145 = 3 * 5 * 11 * 13
2146 = 2 * 29 * 37
2147 = 19 * 113
:

Forth[edit]

: .factors ( n -- )
2
begin 2dup dup * >=
while 2dup /mod swap
if drop 1+ 1 or \ next odd number
else -rot nip dup . ." x "
then
repeat
drop . ;
 
: main ( n -- )
." 1 : 1" cr
1+ 2 ?do i . ." : " i .factors cr loop ;
 
15 main bye


Fortran[edit]

Please find the example output along with the build instructions in the comments at the start of the FORTRAN 2008 source. Compiler: gfortran from the GNU compiler collection. Command interpreter: bash. The code writes j assertions which don't prove primality of the factors but does prove they are the factors.

This algorithm creates a sieve of Eratosthenes, storing the largest prime factor to mark composites. It then finds prime factors by repeatedly looking up the value in the sieve, then dividing by the factor found until the value is itself prime. Using the sieve table to store factors rather than as a plain bitmap was to me a novel idea.

 
!-*- mode: compilation; default-directory: "/tmp/" -*-
!Compilation started at Thu Jun 6 23:29:06
!
!a=./f && make $a && echo -2 | OMP_NUM_THREADS=2 $a
!gfortran -std=f2008 -Wall -fopenmp -ffree-form -fall-intrinsics -fimplicit-none f.f08 -o f
! assert 1 = */ 1
! assert 2 = */ 2
! assert 3 = */ 3
! assert 4 = */ 2 2
! assert 5 = */ 5
! assert 6 = */ 2 3
! assert 7 = */ 7
! assert 8 = */ 2 2 2
! assert 9 = */ 3 3
! assert 10 = */ 2 5
! assert 11 = */ 11
! assert 12 = */ 3 2 2
! assert 13 = */ 13
! assert 14 = */ 2 7
! assert 15 = */ 3 5
! assert 16 = */ 2 2 2 2
! assert 17 = */ 17
! assert 18 = */ 3 2 3
! assert 19 = */ 19
! assert 20 = */ 2 2 5
! assert 21 = */ 3 7
! assert 22 = */ 2 11
! assert 23 = */ 23
! assert 24 = */ 3 2 2 2
! assert 25 = */ 5 5
! assert 26 = */ 2 13
! assert 27 = */ 3 3 3
! assert 28 = */ 2 2 7
! assert 29 = */ 29
! assert 30 = */ 5 2 3
! assert 31 = */ 31
! assert 32 = */ 2 2 2 2 2
! assert 33 = */ 3 11
! assert 34 = */ 2 17
! assert 35 = */ 5 7
! assert 36 = */ 3 3 2 2
! assert 37 = */ 37
! assert 38 = */ 2 19
! assert 39 = */ 3 13
! assert 40 = */ 5 2 2 2
 
module prime_mod
 
! sieve_table stores 0 in prime numbers, and a prime factor in composites.
integer, dimension(:), allocatable :: sieve_table
private :: PrimeQ
 
contains
 
! setup routine must be called first!
subroutine sieve(n) ! populate sieve_table. If n is 0 it deallocates storage, invalidating sieve_table.
integer, intent(in) :: n
integer :: status, i, j
if ((n .lt. 1) .or. allocated(sieve_table)) deallocate(sieve_table)
if (n .lt. 1) return
allocate(sieve_table(n), stat=status)
if (status .ne. 0) stop 'cannot allocate space'
sieve_table(1) = 1
do i=2,int(sqrt(real(n)))+1
if (sieve_table(i) .eq. 0) then
do j = i*i, n, i
sieve_table(j) = i
end do
end if
end do
end subroutine sieve
 
subroutine check_sieve(n)
integer, intent(in) :: n
if (.not. (allocated(sieve_table) .and. ((1 .le. n) .and. (n .le. size(sieve_table))))) stop 'Call sieve first'
end subroutine check_sieve
 
logical function isPrime(p)
integer, intent(in) :: p
call check_sieve(p)
isPrime = PrimeQ(p)
end function isPrime
 
logical function isComposite(p)
integer, intent(in) :: p
isComposite = .not. isPrime(p)
end function isComposite
 
logical function PrimeQ(p)
integer, intent(in) :: p
PrimeQ = sieve_table(p) .eq. 0
end function PrimeQ
 
subroutine prime_factors(p, rv, n)
integer, intent(in) :: p ! number to factor
integer, dimension(:), intent(out) :: rv ! the prime factors
integer, intent(out) :: n ! number of factors returned
integer :: i, m
call check_sieve(p)
m = p
i = 1
if (p .ne. 1) then
do while ((.not. PrimeQ(m)) .and. (i .lt. size(rv)))
rv(i) = sieve_table(m)
m = m/rv(i)
i = i+1
end do
end if
if (i .le. size(rv)) rv(i) = m
n = i
end subroutine prime_factors
 
end module prime_mod
 
program count_in_factors
use prime_mod
integer :: i, n
integer, dimension(8) :: factors
call sieve(40) ! setup
do i=1,40
factors = 0
call prime_factors(i, factors, n)
write(6,*)'assert',i,'= */',factors(:n)
end do
call sieve(0) ! release memory
end program count_in_factors
 

FreeBASIC[edit]

' FB 1.05.0 Win64
 
Sub getPrimeFactors(factors() As UInteger, n As UInteger)
If n < 2 Then Return
Dim factor As UInteger = 2
Do
If n Mod factor = 0 Then
Redim Preserve factors(0 To UBound(factors) + 1)
factors(UBound(factors)) = factor
n \= factor
If n = 1 Then Return
Else
factor += 1
End If
Loop
End Sub
 
Dim factors() As UInteger
 
For i As UInteger = 1 To 20
Print Using "##"; i;
Print " = ";
If i > 1 Then
Erase factors
getPrimeFactors factors(), i
For j As Integer = LBound(factors) To UBound(factors)
Print factors(j);
If j < UBound(factors) Then Print " x ";
Next j
Print
Else
Print i
End If
Next i
 
Print
Print "Press any key to quit"
Sleep
Output:
 1 = 1
 2 = 2
 3 = 3
 4 = 2 x 2
 5 = 5
 6 = 2 x 3
 7 = 7
 8 = 2 x 2 x 2
 9 = 3 x 3
10 = 2 x 5
11 = 11
12 = 2 x 2 x 3
13 = 13
14 = 2 x 7
15 = 3 x 5
16 = 2 x 2 x 2 x 2
17 = 17
18 = 2 x 3 x 3
19 = 19
20 = 2 x 2 x 5

Go[edit]

package main
 
import "fmt"
 
func main() {
fmt.Println("1: 1")
for i := 2; ; i++ {
fmt.Printf("%d: ", i)
var x string
for n, f := i, 2; n != 1; f++ {
for m := n % f; m == 0; m = n % f {
fmt.Print(x, f)
x = "×"
n /= f
}
}
fmt.Println()
}
}
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
...

Groovy[edit]

def factors(number) {
if (number == 1) {
return [1]
}
def factors = []
BigInteger value = number
BigInteger possibleFactor = 2
while (possibleFactor <= value) {
if (value % possibleFactor == 0) {
factors << possibleFactor
value /= possibleFactor
} else {
possibleFactor++
}
}
factors
}
Number.metaClass.factors = { factors(delegate) }
 
((1..10) + (6351..6359)).each { number ->
println "$number = ${number.factors().join(' x ')}"
}
Output:
1 = 1
2 = 2
3 = 3
4 = 2 x 2
5 = 5
6 = 2 x 3
7 = 7
8 = 2 x 2 x 2
9 = 3 x 3
10 = 2 x 5
6351 = 3 x 29 x 73
6352 = 2 x 2 x 2 x 2 x 397
6353 = 6353
6354 = 2 x 3 x 3 x 353
6355 = 5 x 31 x 41
6356 = 2 x 2 x 7 x 227
6357 = 3 x 13 x 163
6358 = 2 x 11 x 17 x 17
6359 = 6359

Haskell[edit]

Using factorize function from the prime decomposition task,

import Data.List (intercalate)
 
showFactors n = show n ++ " = " ++ (intercalate " * " . map show . factorize) n
-- Pointfree form
showFactors = ((++) . show) <*> ((" = " ++) . intercalate " * " . map show . factorize)

isPrime n = n > 1 && noDivsBy primeNums n

Output:
Main> print 1 >> mapM_ (putStrLn . showFactors) [2..]
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
. . .
 
Main> mapM_ (putStrLn . showFactors) [2144..]
2144 = 2 * 2 * 2 * 2 * 2 * 67
2145 = 3 * 5 * 11 * 13
2146 = 2 * 29 * 37
2147 = 19 * 113
2148 = 2 * 2 * 3 * 179
2149 = 7 * 307
2150 = 2 * 5 * 5 * 43
2151 = 3 * 3 * 239
2152 = 2 * 2 * 2 * 269
2153 = 2153
2154 = 2 * 3 * 359
. . .
 
Main> mapM_ (putStrLn . showFactors) [121231231232155..]
121231231232155 = 5 * 11 * 419 * 5260630559
121231231232156 = 2 * 2 * 97 * 1061 * 294487867
121231231232157 = 3 * 3 * 3 * 131 * 34275157261
121231231232158 = 2 * 19 * 67 * 1231 * 38681033
121231231232159 = 121231231232159
121231231232160 = 2 * 2 * 2 * 2 * 2 * 3 * 5 * 7 * 7 * 5154389083
121231231232161 = 121231231232161
121231231232162 = 2 * 60615615616081
121231231232163 = 3 * 13 * 83 * 191089 * 195991
121231231232164 = 2 * 2 * 253811 * 119410931
121231231232165 = 5 * 137 * 176979899609
. . .

The real solution seems to have to be some sort of a segmented offset sieve of Eratosthenes, storing factors in array's cells instead of just marks. That way the speed of production might not be diminishing as much.

Icon and Unicon[edit]

procedure main()
write("Press ^C to terminate")
every f := [i:= 1] | factors(i := seq(2)) do {
writes(i," : [")
every writes(" ",!f|"]\n")
}
end
 
link factors

factors.icn provides factors

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 ]
...

J[edit]

Solution:Use J's factoring primitive,
q:
Example (including formatting):
   ('1 : 1',":&> ,"1 ': ',"1 ":@q:) 2+i.10
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

Java[edit]

Translation of: Visual Basic .NET
public class CountingInFactors{ 
public static void main(String[] args){
for(int i = 1; i<= 10; i++){
System.out.println(i + " = "+ countInFactors(i));
}
 
for(int i = 9991; i <= 10000; i++){
System.out.println(i + " = "+ countInFactors(i));
}
}
 
private static String countInFactors(int n){
if(n == 1) return "1";
 
StringBuilder sb = new StringBuilder();
 
n = checkFactor(2, n, sb);
if(n == 1) return sb.toString();
 
n = checkFactor(3, n, sb);
if(n == 1) return sb.toString();
 
for(int i = 5; i <= n; i+= 2){
if(i % 3 == 0)continue;
 
n = checkFactor(i, n, sb);
if(n == 1)break;
}
 
return sb.toString();
}
 
private static int checkFactor(int mult, int n, StringBuilder sb){
while(n % mult == 0 ){
if(sb.length() > 0) sb.append(" x ");
sb.append(mult);
n /= mult;
}
return n;
}
}
Output:
1 = 1
2 = 2
3 = 3
4 = 2 x 2
5 = 5
6 = 2 x 3
7 = 7
8 = 2 x 2 x 2
9 = 3 x 3
10 = 2 x 5
9991 = 97 x 103
9992 = 2 x 2 x 2 x 1249
9993 = 3 x 3331
9994 = 2 x 19 x 263
9995 = 5 x 1999
9996 = 2 x 2 x 3 x 7 x 7 x 17
9997 = 13 x 769
9998 = 2 x 4999
9999 = 3 x 3 x 11 x 101
10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5

JavaScript[edit]

for(i = 1; i <= 10; i++)
console.log(i + " : " + factor(i).join(" x "));
 
function factor(n) {
var factors = [];
if (n == 1) return [1];
for(p = 2; p <= n; ) {
if((n % p) == 0) {
factors[factors.length] = p;
n /= p;
}
else p++;
}
return factors;
}
Output:
1 : 1
2 : 2
3 : 3
4 : 2 x 2
5 : 5
6 : 2 x 3
7 : 7
8 : 2 x 2 x 2
9 : 3 x 3
10 : 2 x 5

Julia[edit]

 
function factor_print{T<:Integer}(n::T)
const SEP = " \u00d7 "
-2 < n || return "-1"*SEP*factor_print(-n)
if isprime(n) || n < 2
return string(n)
end
a = T[]
for (k, v) in factor(n)
append!(a, k*ones(T, v))
end
sort!(a)
join(a, SEP)
end
 
lo = -4
hi = 40
println("Factor print ", lo, " to ", hi)
for i in lo:hi
println(@sprintf("%5d = ", i), factor_print(i))
end
 

I wrote this solution's factor_print function with ease rather than efficiency in mind. It may be more efficient to first sort the keys of the factor dictionary and to build the string in place, but I find the logic of the presented solution to be clearer. The factor built-in is relevant to this solution only for integers greater than 1, but I've constructed factor_print to return meaningful results for any representable proper integer.

Output:
Factor print -4 to 40
   -4 = -1 × 2 × 2
   -3 = -1 × 3
   -2 = -1 × 2
   -1 = -1
    0 = 0
    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
   19 = 19
   20 = 2 × 2 × 5
   21 = 3 × 7
   22 = 2 × 11
   23 = 23
   24 = 2 × 2 × 2 × 3
   25 = 5 × 5
   26 = 2 × 13
   27 = 3 × 3 × 3
   28 = 2 × 2 × 7
   29 = 29
   30 = 2 × 3 × 5
   31 = 31
   32 = 2 × 2 × 2 × 2 × 2
   33 = 3 × 11
   34 = 2 × 17
   35 = 5 × 7
   36 = 2 × 2 × 3 × 3
   37 = 37
   38 = 2 × 19
   39 = 3 × 13
   40 = 2 × 2 × 2 × 5

Liberty BASIC[edit]

 
'see Run BASIC solution
for i = 1000 to 1016
print i;" = "; factorial$(i)
next
wait
function factorial$(num)
if num = 1 then factorial$ = "1"
fct = 2
while fct <= num
if (num mod fct) = 0 then
factorial$ = factorial$ ; x$ ; fct
x$ = " x "
num = num / fct
else
fct = fct + 1
end if
wend
end function
Output:
1000 = 2 x 2 x 2 x 5 x 5 x 5
1001 = 7 x 11 x 13
1002 = 2 x 3 x 167
1003 = 17 x 59
1004 = 2 x 2 x 251
1005 = 3 x 5 x 67
1006 = 2 x 503
1007 = 19 x 53
1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7
1009 = 1009
1010 = 2 x 5 x 101
1011 = 3 x 337
1012 = 2 x 2 x 11 x 23
1013 = 1013
1014 = 2 x 3 x 13 x 13
1015 = 5 x 7 x 29
1016 = 2 x 2 x 2 x 127

Lua[edit]

function factorize( n )
if n == 1 then return {1} end
 
local k = 2
res = {}
while n > 1 do
while n % k == 0 do
res[#res+1] = k
n = n / k
end
k = k + 1
end
return res
end
 
for i = 1, 22 do
io.write( i, ": " )
fac = factorize( i )
io.write( fac[1] )
for j = 2, #fac do
io.write( " * ", fac[j] )
end
print ""
end

Maple[edit]

factorNum := proc(n)
local i, j, firstNum;
if n = 1 then
printf("%a", 1);
end if;
firstNum := true:
for i in ifactors(n)[2] do
for j to i[2] do
if firstNum then
printf ("%a", i[1]);
firstNum := false:
else
printf(" x %a", i[1]);
end if;
end do;
end do;
printf("\n");
return NULL;
end proc:
 
for i from 1 to 10 do
printf("%2a: ", i);
factorNum(i);
end do;
Output:
 1: 1
 2: 2
 3: 3
 4: 2 x 2
 5: 5
 6: 2 x 3
 7: 7
 8: 2 x 2 x 2
 9: 3 x 3
10: 2 x 5

Mathematica / Wolfram Language[edit]

n = 2; 
While[n < 100,
Print[Row[Riffle[Flatten[Map[Apply[ConstantArray, #] &, FactorInteger[n]]],"*"]]];
n++]

NetRexx[edit]

Translation of: Java
/* NetRexx */
options replace format comments java crossref symbols nobinary
 
runSample(arg)
return
 
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method factor(val) public static
rv = 1
if val > 1 then do
rv = ''
loop n_ = val until n_ = 1
parse checkFactor(2, n_, rv) n_ rv
if n_ = 1 then leave n_
parse checkFactor(3, n_, rv) n_ rv
if n_ = 1 then leave n_
loop m_ = 5 to n_ by 2 until n_ = 1
if m_ // 3 = 0 then iterate m_
parse checkFactor(m_, n_, rv) n_ rv
end m_
end n_
end
return rv
 
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method checkFactor(mult = long, n_ = long, fac) private static binary
msym = 'x'
loop while n_ // mult = 0
fac = fac msym mult
n_ = n_ % mult
end
fac = (fac.strip).strip('l', msym).space
return n_ fac
 
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method runSample(arg) private static
-- input is a list of pairs of numbers - no checking is done
if arg = '' then arg = '1 11 89 101 1000 1020 10000 10010'
loop while arg \= ''
parse arg lv rv arg
say
say '-'.copies(60)
say lv.right(8) 'to' rv
say '-'.copies(60)
loop fv = lv to rv
fac = factor(fv)
pv = ''
if fac.words = 1 & fac \= 1 then pv = '<prime>'
say fv.right(8) '=' fac pv
end fv
end
return
 
Output:
------------------------------------------------------------
       1 to 11
------------------------------------------------------------
       1 = 1
       2 = 2 <prime>
       3 = 3 <prime>
       4 = 2 x 2 
       5 = 5 <prime>
       6 = 2 x 3 
       7 = 7 <prime>
       8 = 2 x 2 x 2 
       9 = 3 x 3 
      10 = 2 x 5 
      11 = 11 <prime>

------------------------------------------------------------
      89 to 101
------------------------------------------------------------
      89 = 89 <prime>
      90 = 2 x 3 x 3 x 5 
      91 = 7 x 13 
      92 = 2 x 2 x 23 
      93 = 3 x 31 
      94 = 2 x 47 
      95 = 5 x 19 
      96 = 2 x 2 x 2 x 2 x 2 x 3 
      97 = 97 <prime>
      98 = 2 x 7 x 7 
      99 = 3 x 3 x 11 
     100 = 2 x 2 x 5 x 5 
     101 = 101 <prime>

------------------------------------------------------------
    1000 to 1020
------------------------------------------------------------
    1000 = 2 x 2 x 2 x 5 x 5 x 5 
    1001 = 7 x 11 x 13 
    1002 = 2 x 3 x 167 
    1003 = 17 x 59 
    1004 = 2 x 2 x 251 
    1005 = 3 x 5 x 67 
    1006 = 2 x 503 
    1007 = 19 x 53 
    1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7 
    1009 = 1009 <prime>
    1010 = 2 x 5 x 101 
    1011 = 3 x 337 
    1012 = 2 x 2 x 11 x 23 
    1013 = 1013 <prime>
    1014 = 2 x 3 x 13 x 13 
    1015 = 5 x 7 x 29 
    1016 = 2 x 2 x 2 x 127 
    1017 = 3 x 3 x 113 
    1018 = 2 x 509 
    1019 = 1019 <prime>
    1020 = 2 x 2 x 3 x 5 x 17 

------------------------------------------------------------
   10000 to 10010
------------------------------------------------------------
   10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5 
   10001 = 73 x 137 
   10002 = 2 x 3 x 1667 
   10003 = 7 x 1429 
   10004 = 2 x 2 x 41 x 61 
   10005 = 3 x 5 x 23 x 29 
   10006 = 2 x 5003 
   10007 = 10007 <prime>
   10008 = 2 x 2 x 2 x 3 x 3 x 139 
   10009 = 10009 <prime>
   10010 = 2 x 5 x 7 x 11 x 13 

Nim[edit]

Translation of: C
var primes = newSeq[int]()
 
proc getPrime(idx: int): int =
if idx >= primes.len:
if primes.len == 0:
primes.add 2
primes.add 3
 
var last = primes[primes.high]
while idx >= primes.len:
last += 2
for i, p in primes:
if p * p > last:
primes.add last
break
if last mod p == 0:
break
 
return primes[idx]
 
for x in 1 .. < int32.high.int:
stdout.write x, " = "
var n = x
var first = 1
 
for i in 0 .. < int32.high:
let p = getPrime(i)
while n mod p == 0:
n = n div p
if first == 0: stdout.write " x "
first = 0
stdout.write p
 
if n <= p * p:
break
 
if first > 0: echo n
elif n > 1: echo " x ", n
else: echo ""
1 = 1
2 = 2
3 = 3
4 = 2 x 2
5 = 5
6 = 2 x 3
7 = 7
8 = 2 x 2 x 2
9 = 3 x 3
10 = 2 x 5
11 = 11
12 = 2 x 2 x 3
13 = 13
14 = 2 x 7
...

Objeck[edit]

 
class CountingInFactors {
function : Main(args : String[]) ~ Nil {
for(i := 1; i <= 10; i += 1;){
count := CountInFactors(i);
("{$i} = {$count}")->PrintLine();
};
 
for(i := 9991; i <= 10000; i += 1;){
count := CountInFactors(i);
("{$i} = {$count}")->PrintLine();
};
}
 
function : CountInFactors(n : Int) ~ String {
if(n = 1) {
return "1";
};
 
sb := "";
n := CheckFactor(2, n, sb);
if(n = 1) {
return sb;
};
 
n := CheckFactor(3, n, sb);
if(n = 1) {
return sb;
};
 
for(i := 5; i <= n; i += 2;) {
if(i % 3 <> 0) {
n := CheckFactor(i, n, sb);
if(n = 1) {
break;
};
};
};
 
return sb;
}
 
function : CheckFactor(mult : Int, n : Int, sb : String) ~ Int {
while(n % mult = 0 ) {
if(sb->Size() > 0) {
sb->Append(" x ");
};
sb->Append(mult);
n /= mult;
};
 
return n;
}
}
 

Output:

1 = 1
2 = 2
3 = 3
4 = 2 x 2
5 = 5
6 = 2 x 3
7 = 7
8 = 2 x 2 x 2
9 = 3 x 3
10 = 2 x 5
9991 = 97 x 103
9992 = 2 x 2 x 2 x 1249
9993 = 3 x 3331
9994 = 2 x 19 x 263
9995 = 5 x 1999
9996 = 2 x 2 x 3 x 7 x 7 x 17
9997 = 13 x 769
9998 = 2 x 4999
9999 = 3 x 3 x 11 x 101
10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5

OCaml[edit]

open Big_int
 
let prime_decomposition x =
let rec inner c p =
if lt_big_int p (square_big_int c) then
[p]
else if eq_big_int (mod_big_int p c) zero_big_int then
c :: inner c (div_big_int p c)
else
inner (succ_big_int c) p
in
inner (succ_big_int (succ_big_int zero_big_int)) x
 
let () =
let rec aux v =
let ps = prime_decomposition v in
print_string (string_of_big_int v);
print_string " = ";
print_endline (String.concat " x " (List.map string_of_big_int ps));
aux (succ_big_int v)
in
aux unit_big_int
Execution:
$ ocamlopt -o count.opt nums.cmxa count.ml
$ ./count.opt
1 = 1
2 = 2
3 = 3
4 = 2 x 2
5 = 5
6 = 2 x 3
7 = 7
8 = 2 x 2 x 2
...
6351 = 3 x 29 x 73
6352 = 2 x 2 x 2 x 2 x 397
6353 = 6353
6354 = 2 x 3 x 3 x 353
6355 = 5 x 31 x 41
6356 = 2 x 2 x 7 x 227
6357 = 3 x 13 x 163
6358 = 2 x 11 x 17 x 17
6359 = 6359
^C

Octave[edit]

Octave's factor function returns an array:

for (n = 1:20)
printf ("%i: ", n)
printf ("%i ", factor (n))
printf ("\n")
endfor
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
19: 19
20: 2 2 5

PARI/GP[edit]

fnice(n)={
my(f,s="",s1);
if (n < 2, return(n));
f = factor(n);
s = Str(s, f[1,1]);
if (f[1, 2] != 1, s=Str(s, "^", f[1,2]));
for(i=2,#f[,1],
s1 = Str(" * ", f[i, 1]);
if (f[i, 2] != 1, s1 = Str(s1, "^", f[i, 2]));
s = Str(s, s1)
);
s
};
n=0;while(n++, print(fnice(n)))

Pascal[edit]

Works with: Free_Pascal
program CountInFactors(output);
 
type
TdynArray = array of integer;
 
function factorize(number: integer): TdynArray;
var
k: integer;
begin
if number = 1 then
begin
setlength(factorize, 1);
factorize[0] := 1
end
else
begin
k := 2;
while number > 1 do
begin
while number mod k = 0 do
begin
setlength(factorize, length(factorize) + 1);
factorize[high(factorize)] := k;
number := number div k;
end;
inc(k);
end;
end
end;
 
var
i, j: integer;
fac: TdynArray;
 
begin
for i := 1 to 22 do
begin
write(i, ': ' );
fac := factorize(i);
write(fac[0]);
for j := 1 to high(fac) do
write(' * ', fac[j]);
writeln;
end;
end.
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
19:  19
20:  2 * 2 * 5
21:  3 * 7
22:  2 * 11

Perl[edit]

Typically one would use a module for this. Note that these modules all return an empty list for '1'. This should be efficient to 50+ digits:

use ntheory qw/factor/;
print "$_ = ", join(" x ", factor($_)), "\n" for 1000000000000000000 .. 1000000000000000010;
Output:
1000000000000000000 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5
1000000000000000001 = 101 x 9901 x 999999000001
1000000000000000002 = 2 x 3 x 17 x 131 x 1427 x 52445056723
1000000000000000003 = 1000000000000000003
1000000000000000004 = 2 x 2 x 1801 x 246809 x 562425889
1000000000000000005 = 3 x 5 x 44087 x 691381 x 2187161
1000000000000000006 = 2 x 7 x 919 x 77724234416291
1000000000000000007 = 1370531 x 729644203597
1000000000000000008 = 2 x 2 x 2 x 3 x 3 x 97 x 26209 x 32779 x 166667
1000000000000000009 = 1000000000000000009
1000000000000000010 = 2 x 5 x 11 x 103 x 4013 x 21993833369

Giving similar output and also good for large inputs:

use Math::Pari qw/factorint/;
sub factor {
my ($pn,$pc) = @{Math::Pari::factorint(shift)};
return map { ($pn->[$_]) x $pc->[$_] } 0 .. $#$pn;
}
print "$_ = ", join(" x ", factor($_)), "\n" for 1000000000000000000 .. 1000000000000000010;

or, somewhat slower and limited to native 32-bit or 64-bit integers only:

use Math::Factor::XS qw/prime_factors/;
print "$_ = ", join(" x ", prime_factors($_)), "\n" for 1000000000000000000 .. 1000000000000000010;


If we want to implement it self-contained, we could use the prime decomposition routine from the Prime_decomposition task. This is reasonably fast and small, though much slower than the modules and certainly could have more optimization.

sub factors {
my($n, $p, @out) = (shift, 3);
return if $n < 1;
while (!($n&1)) { $n >>= 1; push @out, 2; }
while ($n > 1 && $p*$p <= $n) {
while ( ($n % $p) == 0) {
$n /= $p;
push @out, $p;
}
$p += 2;
}
push @out, $n if $n > 1;
@out;
}
 
print "$_ = ", join(" x ", factors($_)), "\n" for 100000000000 .. 100000000100;

We could use the second extensible sieve from Sieve_of_Eratosthenes#Extensible_sieves to only divide by primes.

tie my @primes, 'Tie::SieveOfEratosthenes';
 
sub factors {
my($n, $i, $p, @out) = (shift, 0, 2);
while ($n >= $p * $p) {
while ($n % $p == 0) {
push @out, $p;
$n /= $p;
}
$p = $primes[++$i];
}
push @out, $n if $n > 1 || !@out;
@out;
}
 
print "$_ = ", join(" x ", factors($_)), "\n" for 100000000000 .. 100000000010;
Output:
100000000000 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5
100000000001 = 11 x 11 x 23 x 4093 x 8779
100000000002 = 2 x 3 x 7 x 1543 x 1543067
100000000003 = 100000000003
100000000004 = 2 x 2 x 17573 x 1422637
100000000005 = 3 x 5 x 19 x 1627 x 215659
100000000006 = 2 x 3947 x 12667849
100000000007 = 353 x 283286119
100000000008 = 2 x 2 x 2 x 3 x 3 x 3 x 462962963
100000000009 = 7 x 13 x 53 x 1979 x 10477
100000000010 = 2 x 5 x 101 x 3541 x 27961

This next example isn't quite as fast and uses much more memory, but it is self-contained and shows a different approach. As written it must start at 1, but a range can be handled by using a map to prefill the p_and_sq array.

#!perl -C
use utf8;
use strict;
use warnings;
 
my $limit = 1000;
 
print "$_ = $_\n" for 1..3;
 
my @p_and_sq = ( [2, 4], [3, 9] );
 
N: for my $n ( 4 .. 1000 ) {
print $n, " = ";
for( my $i = 0; $i <= $#p_and_sq; ++$i ) {
my ($p, $sq) = @{ $p_and_sq[$i] };
if( $sq > $n ) {
print $n, "\n";
push @p_and_sq, [ $n, $n*$n ];
next N;
}
while( 0 == ($n % $p) ) {
print $p;
$n /= $p;
if( $n == 1 ) {
print "\n";
next N;
}
print " × ";
}
}
die "Ran out of primes?!";
}

Perl 6[edit]

Works with: rakudo version 2015-10-01
constant @primes = 2, |(3, 5, 7 ... *).grep: *.is-prime;
 
multi factors(1) { 1 }
multi factors(Int $remainder is copy) {
gather for @primes -> $factor {
 
# if remainder < factor², we're done
if $factor * $factor > $remainder {
take $remainder if $remainder > 1;
last;
}
 
# How many times can we divide by this prime?
while $remainder %% $factor {
take $factor;
last if ($remainder div= $factor) === 1;
}
}
}
 
say "$_: ", factors($_).join(" × ") for 1..*;

The first twenty numbers:

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
19: 19
20: 2 × 2 × 5

Here we use a multi declaration with a constant parameter to match the degenerate case. We use copy parameters when we wish to reuse the formal parameter as a mutable variable within the function. (Parameters default to readonly in Perl 6.) Note the use of gather/take as the final statement in the function, which is a common Perl 6 idiom to set up a coroutine within a function to return a lazy list on demand.

Note also the '×' above is not ASCII 'x', but U+00D7 MULTIPLICATION SIGN. Perl 6 does Unicode natively.

Here is a solution inspired from Almost_prime#C. It doesn't use &is-prime.

sub factor($n is copy) {
$n == 1 ?? 1 !!
gather {
$n /= take 2 while $n %% 2;
$n /= take 3 while $n %% 3;
loop (my $p = 5; $p*$p <= $n; $p+=2) {
$n /= take $p while $n %% $p;
}
take $n unless $n == 1;
}
}
 
say "$_ == ", join " \x00d7 ", factor $_ for 1 .. 20;
 

Phix[edit]

function factorise(atom n)
-- returns a list of all integer factors of n, that when multiplied together equal n
-- (adapted from the standard builtin factors(), which does not return duplicates)
sequence res = {}
integer p = 2,
step = 1,
lim = floor(sqrt(n))
 
while p<=lim do
while remainder(n,p)=0 do
res = append(res,sprintf("%d",p))
n = n/p
if n=p then exit end if
lim = floor(sqrt(n))
end while
p += step
step = 2
end while
return join(append(res,sprintf("%d",n))," x ")
end function
 
for i=1 to 10 do
printf(1,"%2d: %s\n",{i,factorise(i)})
end for
Output:
 1: 1
 2: 2
 3: 3
 4: 2 x 2
 5: 5
 6: 2 x 3
 7: 7
 8: 2 x 2 x 2
 9: 3 x 3
10: 2 x 5

PicoLisp[edit]

This is the 'factor' function from Prime decomposition#PicoLisp.

(de factor (N)
(make
(let (D 2 L (1 2 2 . (4 2 4 2 4 6 2 6 .)) M (sqrt N))
(while (>= M D)
(if (=0 (% N D))
(setq M (sqrt (setq N (/ N (link D)))))
(inc 'D (pop 'L)) ) )
(link N) ) ) )
 
(for N 20
(prinl N ": " (glue " * " (factor N))) )
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
19: 19
20: 2 * 2 * 5

PL/I[edit]

 
cnt: procedure options (main);
declare (i, k, n) fixed binary;
declare first bit (1) aligned;
 
do n = 1 to 40;
put skip list (n || ' =');
k = n; first = '1'b;
repeat:
do i = 2 to k-1;
if mod(k, i) = 0 then
do;
k = k/i;
if ^first then put edit (' x ')(A);
first = '0'b;
put edit (trim(i)) (A);
go to repeat;
end;
 
end;
if ^first then put edit (' x ')(A);
if n = 1 then i = 1;
put edit (trim(i)) (A);
end;
end cnt;
 

Results:

        1 = 1
        2 = 2
        3 = 3
        4 = 2 x 2
        5 = 5
        6 = 2 x 3
        7 = 7
        8 = 2 x 2 x 2
        9 = 3 x 3
       10 = 2 x 5
       11 = 11
       12 = 2 x 2 x 3
       13 = 13
       14 = 2 x 7
       15 = 3 x 5
       16 = 2 x 2 x 2 x 2
       17 = 17
       18 = 2 x 3 x 3
       19 = 19
       20 = 2 x 2 x 5
       21 = 3 x 7
       22 = 2 x 11
       23 = 23
       24 = 2 x 2 x 2 x 3
       25 = 5 x 5
       26 = 2 x 13
       27 = 3 x 3 x 3
       28 = 2 x 2 x 7
       29 = 29
       30 = 2 x 3 x 5
       31 = 31
       32 = 2 x 2 x 2 x 2 x 2
       33 = 3 x 11
       34 = 2 x 17
       35 = 5 x 7
       36 = 2 x 2 x 3 x 3
       37 = 37
       38 = 2 x 19
       39 = 3 x 13
       40 = 2 x 2 x 2 x 5

PowerShell[edit]

 
function eratosthenes ($n) {
if($n -ge 1){
$prime = @(1..($n+1) | foreach{$true})
$prime[1] = $false
$m = [Math]::Floor([Math]::Sqrt($n))
for($i = 2; $i -le $m; $i++) {
if($prime[$i]) {
for($j = $i*$i; $j -le $n; $j += $i) {
$prime[$j] = $false
}
}
}
1..$n | where{$prime[$_]}
} else {
"$n must be equal or greater than 1"
}
}
function prime-decomposition ($n) {
$array = eratosthenes $n
$prime = @()
foreach($p in $array) {
while($n%$p -eq 0) {
$n /= $p
$prime += @($p)
}
}
$prime
}
$OFS = " x "
"$(prime-decomposition 2144)"
"$(prime-decomposition 100)"
"$(prime-decomposition 12)"
 

Output:

2 x 2 x 2 x 2 x 2 x 67
2 x 2 x 5 x 5
2 x 2 x 3

PureBasic[edit]

Procedure Factorize(Number, List Factors())
Protected I = 3, Max
ClearList(Factors())
While Number % 2 = 0
AddElement(Factors())
Factors() = 2
Number / 2
Wend
Max = Number
While I <= Max And Number > 1
While Number % I = 0
AddElement(Factors())
Factors() = I
Number / I
Wend
I + 2
Wend
EndProcedure
 
If OpenConsole()
NewList n()
For a=1 To 20
text$=RSet(Str(a),2)+"= "
Factorize(a,n())
If ListSize(n())
ResetList(n())
While NextElement(n())
text$ + Str(n())
If ListSize(n())-ListIndex(n())>1
text$ + "*"
EndIf
Wend
Else
text$+Str(a) ; To handle the '1', which is not really a prime...
EndIf
PrintN(text$)
Next a
EndIf
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
19= 19
20= 2*2*5

Python[edit]

This uses the functools.lru_cache standard library module to cache intermediate results.

from functools import lru_cache
 
primes = [2, 3, 5, 7, 11, 13, 17] # Will be extended
 
@lru_cache(maxsize=2000)
def pfactor(n):
if n == 1:
return [1]
n2 = n // 2 + 1
for p in primes:
if p <= n2:
d, m = divmod(n, p)
if m == 0:
if d > 1:
return [p] + pfactor(d)
else:
return [p]
else:
if n > primes[-1]:
primes.append(n)
return [n]
 
if __name__ == '__main__':
mx = 5000
for n in range(1, mx + 1):
factors = pfactor(n)
if n <= 10 or n >= mx - 20:
print( '%4i %5s %s' % (n,
'' if factors != [n] or n == 1 else 'prime',
'x'.join(str(i) for i in factors)) )
if n == 11:
print('...')
 
print('\nNumber of primes gathered up to', n, 'is', len(primes))
print(pfactor.cache_info())
Output:
   1       1
   2 prime 2
   3 prime 3
   4       2x2
   5 prime 5
   6       2x3
   7 prime 7
   8       2x2x2
   9       3x3
  10       2x5
...
4980       2x2x3x5x83
4981       17x293
4982       2x47x53
4983       3x11x151
4984       2x2x2x7x89
4985       5x997
4986       2x3x3x277
4987 prime 4987
4988       2x2x29x43
4989       3x1663
4990       2x5x499
4991       7x23x31
4992       2x2x2x2x2x2x2x3x13
4993 prime 4993
4994       2x11x227
4995       3x3x3x5x37
4996       2x2x1249
4997       19x263
4998       2x3x7x7x17
4999 prime 4999
5000       2x2x2x5x5x5x5

Number of primes gathered up to 5000 is 669
CacheInfo(hits=3935, misses=7930, maxsize=2000, currsize=2000)

R[edit]

 
#initially I created a function which returns prime factors then I have created another function counts in the factors and #prints the values.
 
findfactors <- function(num) {
x <- c()
p1<- 2
p2 <- 3
everyprime <- num
while( everyprime != 1 ) {
while( everyprime%%p1 == 0 ) {
x <- c(x, p1)
everyprime <- floor(everyprime/ p1)
}
p1 <- p2
p2 <- p2 + 2
}
x
}
count_in_factors=function(x){
primes=findfactors(x)
x=c(1)
for (i in 1:length(primes)) {
x=paste(primes[i],"x",x)
}
return(x)
}
count_in_factors(72)
 
Output:
[1] "3 x 3 x 2 x 2 x 2 x 1"

Racket[edit]

See also #Scheme. This uses Racket’s math/number-theory package

#lang typed/racket
 
(require math/number-theory)
 
(define (factorise-as-primes [n : Natural])
(if
(= n 1)
'(1)
(let ((F (factorize n)))
(append*
(for/list : (Listof (Listof Natural))
((f (in-list F)))
(make-list (second f) (first f)))))))
 
(define (factor-count [start-inc : Natural] [end-inc : Natural])
(for ((i : Natural (in-range start-inc (add1 end-inc))))
(define f (string-join (map number->string (factorise-as-primes i)) " × "))
(printf "~a:\t~a~%" i f)))
 
(factor-count 1 22)
(factor-count 2140 2150)
; tb
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
19:	19
20:	2 × 2 × 5
21:	3 × 7
22:	2 × 11
2140:	2 × 2 × 5 × 107
2141:	2141
2142:	2 × 3 × 3 × 7 × 17
2143:	2143
2144:	2 × 2 × 2 × 2 × 2 × 67
2145:	3 × 5 × 11 × 13
2146:	2 × 29 × 37
2147:	19 × 113
2148:	2 × 2 × 3 × 179
2149:	7 × 307
2150:	2 × 5 × 5 × 43

REXX[edit]

simple approach[edit]

As per the task's requirements, the prime factors of   1   (unity) will be listed as   1,
even though, strictly speaking, it should be   null.         The same applies to   0.

Programming note:   if the HIGH argument is negative, its positive value is used and no displaying of the
prime factors are listed, but the number of primes found is always shown.   The showing of the count of
primes was included to help verify the factoring (of composites).

/*REXX program lists the prime factors of a specified integer  (or a range of integers).*/
@.=left('', 8); @.0="{unity} "; @.1='[prime] ' /*some tags and handy-dandy literals.*/
parse arg low high . /*get optional arguments from the C.L. */
if low=='' then do; low=1; high=40; end /*No LOW & HIGH? Then use the default.*/
if high=='' then high=low; tell= (high>0) /*No HIGH? " " " " */
w=length(high); high=abs(high) /*get maximum width for pretty output. */
numeric digits max(9, w+1) /*maybe bump the precision of numbers. */
#=0 /*the number of primes found (so far). */
do n=low to high; f=factr(n) /*process a single number or a range.*/
p=words(translate(f,,'x')) - (n==1) /*P: is the number of prime factors. */
if p==1 then #=#+1 /*bump the primes counter (exclude N=1)*/
if tell then say right(n, w) '=' @.p space(f, 0) /*show if prime and factors.*/
end /*n*/
say
say right(#, w) ' primes found.' /*display the number of primes found. */
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
factr: procedure; parse arg z 1 n,$; if z<2 then return z /*if Z too small, return Z*/
do while z// 2==0; $=$ 'x 2' ; z=z% 2; end /*maybe add factor of 2 */
do while z// 3==0; $=$ 'x 3' ; z=z% 3; end /* " " " " 3 */
do while z// 5==0; $=$ 'x 5' ; z=z% 5; end /* " " " " 5 */
do while z// 7==0; $=$ 'x 7' ; z=z% 7; end /* " " " " 7 */
 
do j=11 by 6 while j<=z /*insure that J isn't divisible by 3.*/
parse var j '' -1 _ /*get the last decimal digit of J. */
if _\==5 then do while z//j==0; $=$ 'x' j; z=z%j; end /*maybe reduce Z.*/
if _ ==3 then iterate /*if next number will be ÷ by 5, skip.*/
if j*j>n then leave /*are we higher than the √ N  ? */
y=j+2
do while z//y==0; $=$ 'x' y; z=z%y; end /*maybe reduce Z.*/
end /*j*/
 
if z==1 then z= /*if residual is unity, then nullify it*/
return strip( strip($ 'x' z), , "x") /*elide a possible leading (extra) "x".*/

output   when using the default inputs:

 1 = {unity}  1
 2 = [prime]  2
 3 = [prime]  3
 4 =          2x2
 5 = [prime]  5
 6 =          2x3
 7 = [prime]  7
 8 =          2x2x2
 9 =          3x3
10 =          2x5
11 = [prime]  11
12 =          2x2x3
13 = [prime]  13
14 =          2x7
15 =          3x5
16 =          2x2x2x2
17 = [prime]  17
18 =          2x3x3
19 = [prime]  19
20 =          2x2x5
21 =          3x7
22 =          2x11
23 = [prime]  23
24 =          2x2x2x3
25 =          5x5
26 =          2x13
27 =          3x3x3
28 =          2x2x7
29 = [prime]  29
30 =          2x3x5
31 = [prime]  31
32 =          2x2x2x2x2
33 =          3x11
34 =          2x17
35 =          5x7
36 =          2x2x3x3
37 = [prime]  37
38 =          2x19
39 =          3x13
40 =          2x2x2x5

12  primes found.

output   when the following input was used:   1   -10000

  1229  primes found.

output   when the following input was used:   1   -100000

  9592  primes found.

using integer SQRT[edit]

This REXX version computes the   integer square root   of the integer being factor   (to limit the range of factors),
this makes this version about 50% faster than the 1st REXX version.

Also, the number of early testing of prime factors was expanded.

Note that the   integer square root   section of code doesn't use any floating point numbers, just integers.

/*REXX program lists the prime factors of a specified integer  (or a range of integers).*/
@.=left('', 8); @.0="{unity} "; @.1='[prime] ' /*some tags and handy-dandy literals.*/
parse arg low high . /*get optional arguments from the C.L. */
if low=='' then do; low=1; high=40; end /*No LOW & HIGH? Then use the default.*/
if high=='' then high=low; tell= (high>0) /*No HIGH? " " " " */
w=length(high); high=abs(high) /*get maximum width for pretty output. */
numeric digits max(9, w+1) /*maybe bump the precision of numbers. */
#=0 /*the number of primes found (so far). */
do n=low to high; f=factr(n) /*process a single number or a range.*/
p=words(translate(f,,'x')) - (n==1) /*P: is the number of prime factors. */
if p==1 then #=#+1 /*bump the primes counter (exclude N=1)*/
if tell then say right(n, w) '=' @.p space(f, 0) /*show if prime & the factors.*/
end /*n*/
say
say right(#, w) ' primes found.' /*display the number of primes found. */
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
factr: procedure; parse arg z 1 n,$; if z<2 then return z /*if Z too small, return Z*/
do while z// 2==0; $=$ 'x 2'  ; z=z% 2; end /*maybe add factor of 2 */
do while z// 3==0; $=$ 'x 3'  ; z=z% 3; end /* " " " " 3 */
do while z// 5==0; $=$ 'x 5'  ; z=z% 5; end /* " " " " 5 */
do while z// 7==0; $=$ 'x 7'  ; z=z% 7; end /* " " " " 7 */
do while z//11==0; $=$ 'x 11' ; z=z%11; end /* " " " " 11 */
do while z//13==0; $=$ 'x 13' ; z=z%13; end /* " " " " 13 */
do while z//17==0; $=$ 'x 17' ; z=z%17; end /* " " " " 17 */
do while z//19==0; $=$ 'x 19' ; z=z%19; end /* " " " " 19 */
do while z//23==0; $=$ 'x 23' ; z=z%23; end /* " " " " 23 */
do while z//29==0; $=$ 'x 29' ; z=z%29; end /* " " " " 29 */
do while z//31==0; $=$ 'x 31' ; z=z%31; end /* " " " " 31 */
do while z//37==0; $=$ 'x 37' ; z=z%37; end /* " " " " 37 */
if z>40 then do
t=z; q=1; r=0; do while q<=t; q=q*4; end /*R: will be integer SQRT of Z.*/
 
do while q>1; q=q%4; _=t-r-q; r=r%2; if _>=0 then do; t=_; r=r+q; end
end /*while*/ /* [↑] find integer SQRT(z). */
 
do j=41 by 6 to r while j<=z /*insure J isn't divisible by 3*/
parse var j '' -1 _ /*get last decimal digit of J.*/
if _\==5 then do while z//j==0; $=$ 'x' j; z=z%j; end /*reduce Z?*/
if _ ==3 then iterate /*Next number ÷ by 5 ? Skip.*/
y=j+2
do while z//y==0; $=$ 'x' y; z=z%y; end /*reduce Z?*/
end /*j*/
end /*if z>40*/
 
if z==1 then z= /*if residual is unity, then nullify it*/
return strip(strip( $ 'x' z), , "x") /*elide a possible leading (extra) "x".*/

output   is identical to the 1st REXX version.

Ring[edit]

 
for i = 1 to 20
see "" + i + " = " + factors(i) + nl
next
 
func factors n
f = ""
if n = 1 return "1" ok
p = 2
while p <= n
if (n % p) = 0
f += string(p) + " x "
n = n/p
else p += 1 ok
end
return left(f, len(f) - 3)
 

Output:

1 = 1
2 = 2
3 = 3
4 = 2 x 2
5 = 5
6 = 2 x 3
7 = 7
8 = 2 x 2 x 2
9 = 3 x 3
10 = 2 x 5
11 = 11
12 = 2 x 2 x 3
13 = 13
14 = 2 x 7
15 = 3 x 5
16 = 2 x 2 x 2 x 2
17 = 17
18 = 2 x 3 x 3
19 = 19
20 = 2 x 2 x 5

Ruby[edit]

Starting with Ruby 1.9, 'prime' is part of the standard library and provides Integer#prime_division.

require 'optparse'
require 'prime'
 
maximum = 10
OptionParser.new do |o|
o.banner = "Usage: #{File.basename $0} [-m MAXIMUM]"
o.on("-m MAXIMUM", Integer,
"Count up to MAXIMUM [#{maximum}]") { |m| maximum = m }
o.parse! rescue ($stderr.puts $!, o; exit 1)
($stderr.puts o; exit 1) unless ARGV.size == 0
end
 
# 1 has no prime factors
puts "1 is 1" unless maximum < 1
 
2.upto(maximum) do |i|
# i is 504 => i.prime_division is [[2, 3], [3, 2], [7, 1]]
f = i.prime_division.map! do |factor, exponent|
# convert [2, 3] to "2 x 2 x 2"
([factor] * exponent).join " x "
end.join " x "
puts "#{i} is #{f}"
end
Example:
$ ruby prime-count.rb -h
Usage: prime-count.rb [-m MAXIMUM]
    -m MAXIMUM                       Count up to MAXIMUM [10]
$ ruby prime-count.rb -m 10000 | sed -e '11,9990d' 
1 is 1
2 is 2
3 is 3
4 is 2 x 2
5 is 5
6 is 2 x 3
7 is 7
8 is 2 x 2 x 2
9 is 3 x 3
10 is 2 x 5
9991 is 97 x 103
9992 is 2 x 2 x 2 x 1249
9993 is 3 x 3331
9994 is 2 x 19 x 263
9995 is 5 x 1999
9996 is 2 x 2 x 3 x 7 x 7 x 17
9997 is 13 x 769
9998 is 2 x 4999
9999 is 3 x 3 x 11 x 101
10000 is 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5

Run BASIC[edit]

for i = 1000 to 1016
print i;" = "; factorial$(i)
next
wait
function factorial$(num)
if num = 1 then factorial$ = "1"
fct = 2
while fct <= num
if (num mod fct) = 0 then
factorial$ = factorial$ ; x$ ; fct
x$ = " x "
num = num / fct
else
fct = fct + 1
end if
wend
end function
Output:
1000 = 2 x 2 x 2 x 5 x 5 x 5
1001 = 7 x 11 x 13
1002 = 2 x 3 x 167
1003 = 17 x 59
1004 = 2 x 2 x 251
1005 = 3 x 5 x 67
1006 = 2 x 503
1007 = 19 x 53
1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7
1009 = 1009
1010 = 2 x 5 x 101
1011 = 3 x 337
1012 = 2 x 2 x 11 x 23
1013 = 1013
1014 = 2 x 3 x 13 x 13
1015 = 5 x 7 x 29
1016 = 2 x 2 x 2 x 127

Scala[edit]

def primeFactors( n:Int ) = {
 
def primeStream(s: Stream[Int]): Stream[Int] = {
s.head #:: primeStream(s.tail filter { _ % s.head != 0 })
}
 
val primes = primeStream(Stream.from(2))
 
def factors( n:Int ) : List[Int] = primes.takeWhile( _ <= n ).find( n % _ == 0 ) match {
case None => Nil
case Some(p) => p :: factors( n/p )
}
 
if( n == 1 ) List(1) else factors(n)
}
 
// A little test...
{
val nums = (1 to 12).toList :+ 2144 :+ 6358
nums.foreach( n => println( "%6d : %s".format( n, primeFactors(n).mkString(" * ") ) ) )
}
 
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
  2144 : 2 * 2 * 2 * 2 * 2 * 67
  6358 : 2 * 11 * 17 * 17

Scheme[edit]

(define (factors n)
(let facs ((l '()) (d 2) (x n))
(cond ((= x 1) (if (null? l) '(1) l))
((< x (* d d)) (cons x l))
(else (if (= 0 (modulo x d))
(facs (cons d l) d (/ x d))
(facs l (+ 1 d) x))))))
 
(define (show l)
(display (car l))
(if (not (null? (cdr l)))
(begin
(display " × ")
(show (cdr l)))
(display "\n")))
 
(do ((i 1 (+ i 1))) (#f)
(display i)
(display " = ")
(show (reverse (factors i))))
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
...

Seed7[edit]

$ include "seed7_05.s7i";
 
const proc: writePrimeFactors (in var integer: number) is func
local
var boolean: laterElement is FALSE;
var integer: checker is 2;
begin
while checker * checker <= number do
if number rem checker = 0 then
if laterElement then
write(" * ");
end if;
laterElement := TRUE;
write(checker);
number := number div checker;
else
incr(checker);
end if;
end while;
if number <> 1 then
if laterElement then
write(" * ");
end if;
laterElement := TRUE;
write(number);
end if;
end func;
 
const proc: main is func
local
var integer: number is 0;
begin
writeln("1: 1");
for number range 2 to 2147483647 do
write(number <& ": ");
writePrimeFactors(number);
writeln;
end for;
end func;
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
. . .

Sidef[edit]

class Counter {
method factors(n, p=2) {
var out = [];
while (n >= p*p) {
while (n % p == 0) {
out.append(p);
n /= p;
}
p = self.next_prime(p);
}
(n > 1 || out.len.is_zero) && out.append(n);
return out;
}
 
method is_prime(n) {
self.factors(n).len == 1
}
 
method next_prime(p) {
do {
p == 2 ? (p = 3) : (p+=2)
} while (!self.is_prime(p));
return p;
}
}
 
100.times { |i|
say "#{i} = #{Counter().factors(i).join(' × ')}";
}

Tcl[edit]

This factorization code is based on the same engine that is used in the parallel computation task.

package require Tcl 8.5
 
namespace eval prime {
variable primes [list 2 3 5 7 11]
proc restart {} {
variable index -1
variable primes
variable current [lindex $primes end]
}
 
proc get_next_prime {} {
variable primes
variable index
if {$index < [llength $primes]-1} {
return [lindex $primes [incr index]]
}
variable current
while 1 {
incr current 2
set p 1
foreach prime $primes {
if {$current % $prime} {} else {
set p 0
break
}
}
if {$p} {
return [lindex [lappend primes $current] [incr index]]
}
}
}
 
proc factors {num} {
restart
set factors [dict create]
for {set i [get_next_prime]} {$i <= $num} {} {
if {$num % $i == 0} {
dict incr factors $i
set num [expr {$num / $i}]
continue
} elseif {$i*$i > $num} {
dict incr factors $num
break
} else {
set i [get_next_prime]
}
}
return $factors
}
 
# Produce the factors in rendered form
proc factors.rendered {num} {
set factorDict [factors $num]
if {[dict size $factorDict] == 0} {
return 1
}
dict for {factor times} $factorDict {
lappend v {*}[lrepeat $times $factor]
}
return [join $v "*"]
}
}

Demonstration code:

set max 20
for {set i 1} {$i <= $max} {incr i} {
puts [format "%*d = %s" [string length $max] $i [prime::factors.rendered $i]]
}

VBScript[edit]

Made minor modifications on the code I posted under Prime Decomposition.

Function CountFactors(n)
If n = 1 Then
CountFactors = 1
Else
arrP = Split(ListPrimes(n)," ")
Set arrList = CreateObject("System.Collections.ArrayList")
divnum = n
Do Until divnum = 1
'The -1 is to account for the null element of arrP
For i = 0 To UBound(arrP)-1
If divnum = 1 Then
Exit For
ElseIf divnum Mod arrP(i) = 0 Then
divnum = divnum/arrP(i)
arrList.Add arrP(i)
End If
Next
Loop
arrList.Sort
For i = 0 To arrList.Count - 1
If i = arrList.Count - 1 Then
CountFactors = CountFactors & arrList(i)
Else
CountFactors = CountFactors & arrList(i) & " * "
End If
Next
End If
End Function
 
Function IsPrime(n)
If n = 2 Then
IsPrime = True
ElseIf n <= 1 Or n Mod 2 = 0 Then
IsPrime = False
Else
IsPrime = True
For i = 3 To Int(Sqr(n)) Step 2
If n Mod i = 0 Then
IsPrime = False
Exit For
End If
Next
End If
End Function
 
Function ListPrimes(n)
ListPrimes = ""
For i = 1 To n
If IsPrime(i) Then
ListPrimes = ListPrimes & i & " "
End If
Next
End Function
 
'Testing the fucntions.
WScript.StdOut.Write "2 = " & CountFactors(2)
WScript.StdOut.WriteLine
WScript.StdOut.Write "2144 = " & CountFactors(2144)
WScript.StdOut.WriteLine
Output:
2 = 2
2144 = 2 * 2 * 2 * 2 * 2 * 67


Visual Basic .NET[edit]

Module CountingInFactors
 
Sub Main()
For i As Integer = 1 To 10
Console.WriteLine("{0} = {1}", i, CountingInFactors(i))
Next
 
For i As Integer = 9991 To 10000
Console.WriteLine("{0} = {1}", i, CountingInFactors(i))
Next
End Sub
 
Private Function CountingInFactors(ByVal n As Integer) As String
If n = 1 Then Return "1"
 
Dim sb As New Text.StringBuilder()
 
CheckFactor(2, n, sb)
If n = 1 Then Return sb.ToString()
 
CheckFactor(3, n, sb)
If n = 1 Then Return sb.ToString()
 
For i As Integer = 5 To n Step 2
If i Mod 3 = 0 Then Continue For
 
CheckFactor(i, n, sb)
If n = 1 Then Exit For
Next
 
Return sb.ToString()
End Function
 
Private Sub CheckFactor(ByVal mult As Integer, ByRef n As Integer, ByRef sb As Text.StringBuilder)
Do While n Mod mult = 0
If sb.Length > 0 Then sb.Append(" x ")
sb.Append(mult)
n = n / mult
Loop
End Sub
 
End Module
Output:
1 = 1
2 = 2
3 = 3
4 = 2 x 2
5 = 5
6 = 2 x 3
7 = 7
8 = 2 x 2 x 2
9 = 3 x 3
10 = 2 x 5
9991 = 97 x 103
9992 = 2 x 2 x 2 x 1249
9993 = 3 x 3331
9994 = 2 x 19 x 263
9995 = 5 x 1999
9996 = 2 x 2 x 3 x 7 x 7 x 17
9997 = 13 x 769
9998 = 2 x 4999
9999 = 3 x 3 x 11 x 101
10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5

XPL0[edit]

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;
]

Example output:

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

zkl[edit]

foreach n in ([1..*]){ println(n,": ",primeFactors(n).concat("\U2715;")) }

Using the fixed size integer (64 bit) solution from Prime decomposition#zkl

fcn primeFactors(n){  // Return a list of factors of n
acc:=fcn(n,k,acc,maxD){ // k is 2,3,5,7,9,... not optimum
if(n==1 or k>maxD) acc.close();
else{
q,r:=n.divr(k); // divr-->(quotient,remainder)
if(r==0) return(self.fcn(q,k,acc.write(k),q.toFloat().sqrt()));
return(self.fcn(n,k+1+k.isOdd,acc,maxD))
}
}(n,2,Sink(List),n.toFloat().sqrt());
m:=acc.reduce('*,1); // mulitply factors
if(n!=m) acc.append(n/m); // opps, missed last factor
else acc;
}
Output:
1: 
2: 2
3: 3
4: 2✕2
5: 5
6: 2✕3
...
591885: 3✕3✕5✕7✕1879
591886: 2✕295943
591887: 591887
591888: 2✕2✕2✕2✕3✕11✕19✕59
...

ZX Spectrum Basic[edit]

Translation of: BBC_BASIC
10 FOR i=1 TO 20
20 PRINT i;" = ";
30 IF i=1 THEN PRINT 1: GO TO 90
40 LET p=2: LET n=i: LET f$=""
50 IF p>n THEN GO TO 80
60 IF NOT FN m(n,p) THEN LET f$=f$+STR$ p+" x ": LET n=INT (n/p): GO TO 50
70 LET p=p+1: GO TO 50
80 PRINT f$( TO LEN f$-3)
90 NEXT i
100 STOP
110 DEF FN m(a,b)=a-INT (a/b)*b