Product of divisors
From Rosetta Code
Product of divisors is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Given a positive integer, return the product of its positive divisors.
- Task
Show the result for the first 50 positive integers.
Contents
11l[edit]
F product_of_divisors(n)
V ans = 1
V i = 1
V j = 1
L i * i <= n
I 0 == n % i
ans *= i
j = n I/ i
I j != i
ans *= j
i++
R ans
print((1..50).map(n -> product_of_divisors(n)))
- Output:
[1, 2, 3, 8, 5, 36, 7, 64, 27, 100, 11, 1728, 13, 196, 225, 1024, 17, 5832, 19, 8000, 441, 484, 23, 331776, 125, 676, 729, 21952, 29, 810000, 31, 32768, 1089, 1156, 1225, 10077696, 37, 1444, 1521, 2560000, 41, 3111696, 43, 85184, 91125, 2116, 47, 254803968, 343, 125000]
ALGOL 68[edit]
BEGIN # find the product of the divisors of the first 100 positive integers #
# calculates the number of divisors of v #
PROC divisor count = ( INT v )INT:
BEGIN
INT total := 1, n := v;
# Deal with powers of 2 first #
WHILE NOT ODD n DO
total +:= 1;
n OVERAB 2
OD;
# Odd prime factors up to the square root #
INT p := 3;
WHILE ( p * p ) <= n DO
INT count := 1;
WHILE n MOD p = 0 DO
count +:= 1;
n OVERAB p
OD;
p +:= 2;
total *:= count
OD;
# If n > 1 then it's prime #
IF n > 1 THEN total *:= 2 FI;
total
END # divisor count #;
# calculates the product of the divisors of v #
PROC divisor product = ( INT v )LONG INT:
BEGIN
INT count = divisor count( v );
LONG INT product := v ^ ( count OVER 2 );
IF ODD count THEN product *:= ENTIER sqrt( v ) FI;
product
END # divisor product # ;
BEGIN
INT limit = 50;
print( ( "Product of divisors for the first ", whole( limit, 0 ), " positive integers:", newline ) );
FOR n TO limit DO
print( ( whole( divisor product( n ), -10 ) ) );
IF n MOD 5 = 0 THEN print( ( newline ) ) FI
OD
END
END
- Output:
Product of divisors for the first 50 positive integers:
1 2 3 8 5
36 7 64 27 100
11 1728 13 196 225
1024 17 5832 19 8000
441 484 23 331776 125
676 729 21952 29 810000
31 32768 1089 1156 1225
10077696 37 1444 1521 2560000
41 3111696 43 85184 91125
2116 47 254803968 343 125000
ALGOL W[edit]
begin % find the product of the divisors of the first 100 positive integers %
% calculates the number of divisors of v %
integer procedure divisor_count( integer value v ) ; begin
integer total, n, p;
total := 1; n := v;
% Deal with powers of 2 first %
while not odd( n ) do begin
total := total + 1;
n := n div 2
end while_not_odd_n ;
% Odd prime factors up to the square root %
p := 3;
while ( p * p ) <= n do begin
integer count;
count := 1;
while n rem p = 0 do begin
count := count + 1;
n := n div p
end while_n_rem_p_eq_0 ;
p := p + 2;
total := total * count
end while_p_x_p_le_n ;
% If n > 1 then it's prime %
if n > 1 then total := total * 2;
total
end divisor_count ;
% calculates the product of the divisors of v %
integer procedure divisor_product( integer value v ) ; begin
integer count, product;
count := divisor_count( v );
product := 1;
for i := 1 until count div 2 do product := product * v;
if odd( count ) then product := product * entier( sqrt( v ) );
product
end divisor_product ;
begin
integer limit;
limit := 50;
write( i_w := 1, s_w := 0, "Product of divisors for the first ", limit, " positive integers:" );
for n := 1 until limit do begin
if n rem 5 = 1 then write();
writeon( i_w := 10, s_w := 1, divisor_product( n ) )
end for_n
end
end.
- Output:
Product of divisors for the first 50 positive integers:
1 2 3 8 5
36 7 64 27 100
11 1728 13 196 225
1024 17 5832 19 8000
441 484 23 331776 125
676 729 21952 29 810000
31 32768 1089 1156 1225
10077696 37 1444 1521 2560000
41 3111696 43 85184 91125
2116 47 254803968 343 125000
APL[edit]
divprod ← ×/(⍸0=⍳|⊢)
10 5 ⍴ divprod¨ ⍳50
- Output:
1 2 3 8 5
36 7 64 27 100
11 1728 13 196 225
1024 17 5832 19 8000
441 484 23 331776 125
676 729 21952 29 810000
31 32768 1089 1156 1225
10077696 37 1444 1521 2560000
41 3111696 43 85184 91125
2116 47 254803968 343 125000
AWK[edit]
# syntax: GAWK -f PRODUCT_OF_DIVISORS.AWK
# converted from Go
BEGIN {
limit = 50
printf("The products of positive divisors for the first %d positive integers are:\n",limit)
for (i=1; i<=limit; i++) {
printf("%12d ",product_divisors(i))
if (i % 10 == 0) {
printf("\n")
}
}
exit(0)
}
function product_divisors(n, ans,i,j,k) {
ans = 1
i = 1
k = (n % 2 == 0) ? 1 : 2
while (i*i <= n) {
if (n % i == 0) {
ans *= i
j = n / i
if (j != i) {
ans *= j
}
}
i += k
}
return(ans)
}
- Output:
The products of positive divisors for the first 50 positive integers are:
1 2 3 8 5 36 7 64 27 100
11 1728 13 196 225 1024 17 5832 19 8000
441 484 23 331776 125 676 729 21952 29 810000
31 32768 1089 1156 1225 10077696 37 1444 1521 2560000
41 3111696 43 85184 91125 2116 47 254803968 343 125000
C[edit]
#include <math.h>
#include <stdio.h>
// See https://en.wikipedia.org/wiki/Divisor_function
unsigned int divisor_count(unsigned int n) {
unsigned int total = 1;
unsigned int p;
// Deal with powers of 2 first
for (; (n & 1) == 0; n >>= 1) {
++total;
}
// Odd prime factors up to the square root
for (p = 3; p * p <= n; p += 2) {
unsigned int count = 1;
for (; n % p == 0; n /= p) {
++count;
}
total *= count;
}
// If n > 1 then it's prime
if (n > 1) {
total *= 2;
}
return total;
}
// See https://mathworld.wolfram.com/DivisorProduct.html
unsigned int divisor_product(unsigned int n) {
return pow(n, divisor_count(n) / 2.0);
}
int main() {
const unsigned int limit = 50;
unsigned int n;
printf("Product of divisors for the first %d positive integers:\n", limit);
for (n = 1; n <= limit; ++n) {
printf("%11d", divisor_product(n));
if (n % 5 == 0) {
printf("\n");
}
}
return 0;
}
- Output:
Product of divisors for the first 50 positive integers:
1 2 3 8 5
36 7 64 27 100
11 1728 13 196 225
1024 17 5832 19 8000
441 484 23 331776 125
676 729 21952 29 810000
31 32768 1089 1156 1225
10077696 37 1444 1521 2560000
41 3111696 43 85184 91125
2116 47 254803968 343 125000
C++[edit]
#include <cmath>
#include <iomanip>
#include <iostream>
// See https://en.wikipedia.org/wiki/Divisor_function
unsigned int divisor_count(unsigned int n) {
unsigned int total = 1;
// Deal with powers of 2 first
for (; (n & 1) == 0; n >>= 1)
++total;
// Odd prime factors up to the square root
for (unsigned int p = 3; p * p <= n; p += 2) {
unsigned int count = 1;
for (; n % p == 0; n /= p)
++count;
total *= count;
}
// If n > 1 then it's prime
if (n > 1)
total *= 2;
return total;
}
// See https://mathworld.wolfram.com/DivisorProduct.html
unsigned int divisor_product(unsigned int n) {
return static_cast<unsigned int>(std::pow(n, divisor_count(n)/2.0));
}
int main() {
const unsigned int limit = 50;
std::cout << "Product of divisors for the first " << limit << " positive integers:\n";
for (unsigned int n = 1; n <= limit; ++n) {
std::cout << std::setw(11) << divisor_product(n);
if (n % 5 == 0)
std::cout << '\n';
}
}
- Output:
Product of divisors for the first 50 positive integers:
1 2 3 8 5
36 7 64 27 100
11 1728 13 196 225
1024 17 5832 19 8000
441 484 23 331776 125
676 729 21952 29 810000
31 32768 1089 1156 1225
10077696 37 1444 1521 2560000
41 3111696 43 85184 91125
2116 47 254803968 343 125000
Cowgol[edit]
include "cowgol.coh";
sub divprod(n: uint32): (prod: uint32) is
prod := 1;
var d := n;
while d > 1 loop
if n % d == 0 then
prod := prod * d;
end if;
d := d - 1;
end loop;
end sub;
var n: uint32 := 1;
while n <= 50 loop
var dp := divprod(n);
print_i32(dp);
print_char('\t');
if dp < 10000000 then
print_char('\t');
end if;
if n % 5 == 0 then
print_nl();
end if;
n := n + 1;
end loop;
- Output:
1 2 3 8 5 36 7 64 27 100 11 1728 13 196 225 1024 17 5832 19 8000 441 484 23 331776 125 676 729 21952 29 810000 31 32768 1089 1156 1225 10077696 37 1444 1521 2560000 41 3111696 43 85184 91125 2116 47 254803968 343 125000
Factor[edit]
USING: grouping io math.primes.factors math.ranges prettyprint
sequences ;
"Product of divisors for the first 50 positive integers:" print
50 [1,b] [ divisors product ] map 5 group simple-table.
- Output:
Product of divisors for the first 50 positive integers: 1 2 3 8 5 36 7 64 27 100 11 1728 13 196 225 1024 17 5832 19 8000 441 484 23 331776 125 676 729 21952 29 810000 31 32768 1089 1156 1225 10077696 37 1444 1521 2560000 41 3111696 43 85184 91125 2116 47 254803968 343 125000
FreeBASIC[edit]
dim p as ulongint
for n as uinteger = 1 to 50
p = n
for i as uinteger = 2 to n/2
if n mod i = 0 then p *= i
next i
print p,
next n
- Output:
1 2 3 8 5 36 7 64 27 100 11 1728 13 196 225 1024 17 5832 19 8000 441 484 23 331776 125 676 729 21952 29 810000 31 32768 1089 1156 1225 10077696 37 1444 1521 2560000 41 3111696 43 85184 91125 2116 47 254803968
Go[edit]
package main
import "fmt"
func prodDivisors(n int) int {
prod := 1
i := 1
k := 2
if n%2 == 0 {
k = 1
}
for i*i <= n {
if n%i == 0 {
prod *= i
j := n / i
if j != i {
prod *= j
}
}
i += k
}
return prod
}
func main() {
fmt.Println("The products of positive divisors for the first 50 positive integers are:")
for i := 1; i <= 50; i++ {
fmt.Printf("%9d ", prodDivisors(i))
if i%5 == 0 {
fmt.Println()
}
}
}
- Output:
The products of positive divisors for the first 50 positive integers are:
1 2 3 8 5
36 7 64 27 100
11 1728 13 196 225
1024 17 5832 19 8000
441 484 23 331776 125
676 729 21952 29 810000
31 32768 1089 1156 1225
10077696 37 1444 1521 2560000
41 3111696 43 85184 91125
2116 47 254803968 343 125000
GW-BASIC[edit]
10 FOR N = 1 TO 50
20 P# = N
30 FOR I = 2 TO INT(N/2)
40 IF N MOD I = 0 THEN P# = P# * I
50 NEXT I
60 PRINT P#,
70 NEXT N
- Output:
1 2 3 8 5 36 7 64 27 100 11 1728 13 196 225 1024 17 5832 19 8000 441 484 23 331776 125 676 729 21952 29 810000 31 32768 1089 1156 1225 10077696 37 1444 1521 2560000 41 3111696 43 85184 91125 2116 47 254803968 343 125000
Haskell[edit]
import Data.List.Split (chunksOf)
------------------------- DIVISORS -----------------------
divisors
:: Integral a
=> a -> [a]
divisors n =
((<>) <*> (rest . reverse . fmap (quot n))) $
filter ((0 ==) . rem n) [1 .. root]
where
root = (floor . sqrt . fromIntegral) n
rest
| n == root * root = tail
| otherwise = id
-------------- SUMS AND PRODUCTS OF DIVISORS -------------
main :: IO ()
main =
mapM_
putStrLn
[ "Sums of divisors of [1..100]:"
, test sum
, "Products of divisors of [1..100]:"
, test product
]
test
:: (Show a, Integral a)
=> ([a] -> a) -> String
test f =
let xs = show . f . divisors <$> [1 .. 100]
w = maximum $ length <$> xs
in unlines $ unwords <$> fmap (fmap (justifyRight w ' ')) (chunksOf 5 xs)
justifyRight :: Int -> Char -> String -> String
justifyRight n c = (drop . length) <*> (replicate n c <>)
- Output:
Sums of divisors of [1..100]:
1 3 4 7 6
12 8 15 13 18
12 28 14 24 24
31 18 39 20 42
32 36 24 60 31
42 40 56 30 72
32 63 48 54 48
91 38 60 56 90
42 96 44 84 78
72 48 124 57 93
72 98 54 120 72
120 80 90 60 168
62 96 104 127 84
144 68 126 96 144
72 195 74 114 124
140 96 168 80 186
121 126 84 224 108
132 120 180 90 234
112 168 128 144 120
252 98 171 156 217
Products of divisors of [1..100]:
1 2 3 8 5
36 7 64 27 100
11 1728 13 196 225
1024 17 5832 19 8000
441 484 23 331776 125
676 729 21952 29 810000
31 32768 1089 1156 1225
10077696 37 1444 1521 2560000
41 3111696 43 85184 91125
2116 47 254803968 343 125000
2601 140608 53 8503056 3025
9834496 3249 3364 59 46656000000
61 3844 250047 2097152 4225
18974736 67 314432 4761 24010000
71 139314069504 73 5476 421875
438976 5929 37015056 79 3276800000
59049 6724 83 351298031616 7225
7396 7569 59969536 89 531441000000
8281 778688 8649 8836 9025
782757789696 97 941192 970299 1000000000
Julia[edit]
using Primes
function proddivisors(n)
f = [one(n)]
for (p, e) in factor(n)
f = reduce(vcat, [f * p^j for j in 1:e], init = f)
end
return prod(f)
end
for i in 1:50
print(lpad(proddivisors(i), 10), i % 10 == 0 ? " \n" : "")
end
- Output:
1 2 3 8 5 36 7 64 27 100
11 1728 13 196 225 1024 17 5832 19 8000
441 484 23 331776 125 676 729 21952 29 810000
31 32768 1089 1156 1225 10077696 37 1444 1521 2560000
41 3111696 43 85184 91125 2116 47 254803968 343 125000
Perl[edit]
#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Product_of_divisors
use warnings;
my @products = ( 1 ) x 51;
for my $n ( 1 .. 50 )
{
$n % $_ or $products[$n] *= $_ for 1 .. $n;
}
printf '' . (('%11d' x 5) . "\n") x 10, @products[1 .. 50];
- Output:
1 2 3 8 5
36 7 64 27 100
11 1728 13 196 225
1024 17 5832 19 8000
441 484 23 331776 125
676 729 21952 29 810000
31 32768 1089 1156 1225
10077696 37 1444 1521 2560000
41 3111696 43 85184 91125
2116 47 254803968 343 125000
Phix[edit]
imperative[edit]
for i=1 to 50 do printf(1,"%,12d",{product(factors(i,1))}) if remainder(i,5)=0 then puts(1,"\n") end if end for
- Output:
1 2 3 8 5
36 7 64 27 100
11 1,728 13 196 225
1,024 17 5,832 19 8,000
441 484 23 331,776 125
676 729 21,952 29 810,000
31 32,768 1,089 1,156 1,225
10,077,696 37 1,444 1,521 2,560,000
41 3,111,696 43 85,184 91,125
2,116 47 254,803,968 343 125,000
functional[edit]
same output
sequence r = apply(apply(true,factors,{tagset(50),{1}}),product) puts(1,join_by(apply(true,sprintf,{{"%,12d"},r}),1,5,""))
Python[edit]
Finding divisors efficiently[edit]
def product_of_divisors(n):
assert(isinstance(n, int) and 0 < n)
ans = i = j = 1
while i*i <= n:
if 0 == n%i:
ans *= i
j = n//i
if j != i:
ans *= j
i += 1
return ans
if __name__ == "__main__":
print([product_of_divisors(n) for n in range(1,51)])
- Output:
[1, 2, 3, 8, 5, 36, 7, 64, 27, 100, 11, 1728, 13, 196, 225, 1024, 17, 5832, 19, 8000, 441, 484, 23, 331776, 125, 676, 729, 21952, 29, 810000, 31, 32768, 1089, 1156, 1225, 10077696, 37, 1444, 1521, 2560000, 41, 3111696, 43, 85184, 91125, 2116, 47, 254803968, 343, 125000]
Choosing the right abstraction[edit]
This is really an exercise in defining a divisors function, and the only difference between the suggested Sum and Product draft tasks boils down to two trivial morphemes:
reduce(add, divisors(n), 0) vs reduce(mul, divisors(n), 1)
The goal of Rosetta code (see the landing page) is to provide contrastive insight (rather than comprehensive coverage of homework questions :-). Perhaps the scope for contrastive insight in the matter of divisors is already exhausted by the trivially different Proper divisors task.
'''Sums and products of divisors'''
from math import floor, sqrt
from functools import reduce
from operator import add, mul
# divisors :: Int -> [Int]
def divisors(n):
'''List of all divisors of n including n itself.
'''
root = floor(sqrt(n))
lows = [x for x in range(1, 1 + root) if 0 == n % x]
return lows + [n // x for x in reversed(lows)][
(1 if n == (root * root) else 0):
]
# ------------------------- TEST -------------------------
# main :: IO ()
def main():
'''Product and sums of divisors for each integer in range [1..50]
'''
print('Products of divisors:')
for n in range(1, 1 + 50):
print(n, '->', reduce(mul, divisors(n), 1))
print('Sums of divisors:')
for n in range(1, 1 + 100):
print(n, '->', reduce(add, divisors(n), 0))
# MAIN ---
if __name__ == '__main__':
main()
Raku[edit]
Yet more tasks that are tiny variations of each other. Tau function, Tau number, Sum of divisors and Product of divisors all use code with minimal changes. What the heck, post 'em all.
use Prime::Factor:ver<0.3.0+>;
use Lingua::EN::Numbers;
say "\nTau function - first 100:\n", # ID
(1..*).map({ +.&divisors })[^100]\ # the task
.batch(20)».fmt("%3d").join("\n"); # display formatting
say "\nTau numbers - first 100:\n", # ID
(1..*).grep({ $_ %% +.&divisors })[^100]\ # the task
.batch(10)».&comma».fmt("%5s").join("\n"); # display formatting
say "\nDivisor sums - first 100:\n", # ID
(1..*).map({ [+] .&divisors })[^100]\ # the task
.batch(20)».fmt("%4d").join("\n"); # display formatting
say "\nDivisor products - first 100:\n", # ID
(1..*).map({ [×] .&divisors })[^100]\ # the task
.batch(5)».&comma».fmt("%16s").join("\n"); # display formatting
- Output:
Tau function - first 100:
1 2 2 3 2 4 2 4 3 4 2 6 2 4 4 5 2 6 2 6
4 4 2 8 3 4 4 6 2 8 2 6 4 4 4 9 2 4 4 8
2 8 2 6 6 4 2 10 3 6 4 6 2 8 4 8 4 4 2 12
2 4 6 7 4 8 2 6 4 8 2 12 2 4 6 6 4 8 2 10
5 4 2 12 4 4 4 8 2 12 4 6 4 4 4 12 2 6 6 9
Tau numbers - first 100:
1 2 8 9 12 18 24 36 40 56
60 72 80 84 88 96 104 108 128 132
136 152 156 180 184 204 225 228 232 240
248 252 276 288 296 328 344 348 360 372
376 384 396 424 441 444 448 450 468 472
480 488 492 504 516 536 560 564 568 584
600 612 625 632 636 640 664 672 684 708
712 720 732 776 792 804 808 824 828 852
856 864 872 876 880 882 896 904 936 948
972 996 1,016 1,040 1,044 1,048 1,056 1,068 1,089 1,096
Divisor sums - first 100:
1 3 4 7 6 12 8 15 13 18 12 28 14 24 24 31 18 39 20 42
32 36 24 60 31 42 40 56 30 72 32 63 48 54 48 91 38 60 56 90
42 96 44 84 78 72 48 124 57 93 72 98 54 120 72 120 80 90 60 168
62 96 104 127 84 144 68 126 96 144 72 195 74 114 124 140 96 168 80 186
121 126 84 224 108 132 120 180 90 234 112 168 128 144 120 252 98 171 156 217
Divisor products - first 100:
1 2 3 8 5
36 7 64 27 100
11 1,728 13 196 225
1,024 17 5,832 19 8,000
441 484 23 331,776 125
676 729 21,952 29 810,000
31 32,768 1,089 1,156 1,225
10,077,696 37 1,444 1,521 2,560,000
41 3,111,696 43 85,184 91,125
2,116 47 254,803,968 343 125,000
2,601 140,608 53 8,503,056 3,025
9,834,496 3,249 3,364 59 46,656,000,000
61 3,844 250,047 2,097,152 4,225
18,974,736 67 314,432 4,761 24,010,000
71 139,314,069,504 73 5,476 421,875
438,976 5,929 37,015,056 79 3,276,800,000
59,049 6,724 83 351,298,031,616 7,225
7,396 7,569 59,969,536 89 531,441,000,000
8,281 778,688 8,649 8,836 9,025
782,757,789,696 97 941,192 970,299 1,000,000,000
REXX[edit]
/*REXX program displays the first N product of divisors (shown in a columnar format).*/
numeric digits 20 /*ensure enough decimal digit precision*/
parse arg n cols . /*obtain optional argument from the CL.*/
if n=='' | n=="," then n= 50 /*Not specified? Then use the default.*/
if cols=='' | cols=="," then cols= 5 /* " " " " " " */
say ' index │'center("product of divisors", 102) /*display title for the column #s*/
say '───────┼'center("" , 102,'─') /* " " separator (above)*/
w= max(20, length(n) ) /*W: used to align 1st output column. */
$=; idx= 1 /*$: the output list, shown in columns*/
do j=1 for N /*process N positive integers. */
$= $ || right( commas( sigma(j) ), 20) /*add a sigma (sum) to the output list.*/
if j//cols\==0 then iterate /*Not a multiple of cols? Don't display*/
say center(idx, 7)'│' $ /*display partial list to the terminal.*/
idx= idx + cols /*bump the index number for the output.*/
$= /*start with a blank line for next time*/
end /*j*/
if $\=='' then say center(idx, 7)' ' $ /*any residuals sums left to display? */
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?
/*──────────────────────────────────────────────────────────────────────────────────────*/
sigma: procedure; parse arg x; if x==1 then return 1; odd=x // 2 /* // ◄──remainder.*/
p= x /* [↓] only use EVEN or ODD integers.*/
do k=2+odd by 1+odd while k*k<x /*divide by all integers up to √x. */
if x//k==0 then p= p * k * (x % k) /*multiple the two divisors to product.*/
end /*k*/ /* [↑] % is the REXX integer division*/
if k*k==x then return p * k /*Was X a square? If so, add √ x */
return p /*return (sigma) sum of the divisors. */
- output when using the default input:
index │ product of divisors ───────┼────────────────────────────────────────────────────────────────────────────────────────────────────── 1 │ 1 2 3 8 5 6 │ 36 7 64 27 100 11 │ 11 1,728 13 196 225 16 │ 1,024 17 5,832 19 8,000 21 │ 441 484 23 331,776 125 26 │ 676 729 21,952 29 810,000 31 │ 31 32,768 1,089 1,156 1,225 36 │ 10,077,696 37 1,444 1,521 2,560,000 41 │ 41 3,111,696 43 85,184 91,125 46 │ 2,116 47 254,803,968 343 125,000
Ring[edit]
limit = 50
row = 0
see "working..." + nl
for n = 1 to limit
pro = 1
for m = 1 to n
if n%m = 0
pro = pro*m
ok
next
see "" + pro + " "
row = row + 1
if row % 5 = 0
see nl
ok
next
see "done..." + nl
- Output:
working... 1 2 3 8 5 36 7 64 27 100 11 1728 13 196 225 1024 17 5832 19 8000 441 484 23 331776 125 676 729 21952 29 810000 31 32768 1089 1156 1225 10077696 37 1444 1521 2560000 41 3111696 43 85184 91125 2116 47 254803968 343 125000 done...
Ruby[edit]
def divisor_count(n)
total = 1
# Deal with powers of 2 first
while n % 2 == 0 do
total = total + 1
n = n >> 1
end
# Odd prime factors up to the square root
p = 3
while p * p <= n do
count = 1
while n % p == 0 do
count = count + 1
n = n / p
end
total = total * count
p = p + 2
end
# If n > 1 then it's prime
if n > 1 then
total = total * 2
end
return total
end
def divisor_product(n)
return (n ** (divisor_count(n) / 2.0)).floor
end
LIMIT = 50
print "Product of divisors for the first ", LIMIT, " positive integers:\n"
for n in 1 .. LIMIT
print "%11d" % [divisor_product(n)]
if n % 5 == 0 then
print "\n"
end
end
- Output:
Product of divisors for the first 50 positive integers:
1 2 3 8 5
36 7 64 27 100
11 1728 13 196 225
1024 17 5832 19 8000
441 484 23 331776 125
676 729 21952 29 810000
31 32768 1089 1156 1225
10077696 37 1444 1521 2560000
41 3111696 43 85184 91125
2116 47 254803968 343 125000
Wren[edit]
import "/math" for Int, Nums
import "/fmt" for Fmt
System.print("The products of positive divisors for the first 50 positive integers are:")
for (i in 1..50) {
Fmt.write("$9d ", Nums.prod(Int.divisors(i)))
if (i % 5 == 0) System.print()
}
- Output:
The products of positive divisors for the first 50 positive integers are:
1 2 3 8 5
36 7 64 27 100
11 1728 13 196 225
1024 17 5832 19 8000
441 484 23 331776 125
676 729 21952 29 810000
31 32768 1089 1156 1225
10077696 37 1444 1521 2560000
41 3111696 43 85184 91125
2116 47 254803968 343 125000
