Product of divisors
Given a positive integer, return the product of its positive divisors.
- Task
Show the result for the first 50 positive integers.
ALGOL 68
<lang algol68>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 := 1;
FOR i TO count OVER 2 DO product *:= v OD;
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</lang>
- 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
<lang algolw>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.</lang>
- 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++
<lang cpp>#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';
}
}</lang>
- 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
Factor
<lang factor>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.</lang>
- 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
Go
<lang go>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()
}
}
}</lang>
- 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
Python
Finding divisors efficiently
<lang Python>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)])</lang>
- 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
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.
<lang python>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()</lang>
Raku
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.
<lang perl6>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</lang>
- 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
<lang rexx>/*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 . /*obtain optional argument from the CL.*/ if n== | n=="," then n= 50 /*Not specified? Then use the default.*/ say 'the sums of divisors for ' n " integers:"; say /*show what output is being shown*/ say '─index─' center("sum of divisors", 101,'─') /*display a title for the tau numbers. */ w= max(7, length(n) ) /*W: used to align 1st output column. */ $= /*$: the output list, shown ten/line. */
do j=1 for N /*process N positive integers. */
$= $ || right( commas(sigma(j)), 20) /*add a sigma (sum) to the output list.*/
if j//5\==0 then iterate /*Not a multiple of 10? Don't display.*/
say center(commas(j-4), 7)' ' $ /*display partial list to the terminal.*/
$= /*start with a blank line for next line*/
end /*j*/
if j<=5 then j= 2 /handle case if this is the 1st display*/ if $\== then say center((j-1), 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 ? /*──────────────────────────────────────────────────────────────────────────────────────*/ iSqrt: procedure; parse arg x; q= 1; r= 0; do while q<=x; q= q*4; end
do while q>1; q=q%4; _=x-r-q; r=r%2; if _>=0 then do;x=_;r=r+q; end; end; return r
/*──────────────────────────────────────────────────────────────────────────────────────*/ 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. */</lang>
- output when using the default input:
the sums of divisors for 50 integers: ─index─ ───────────────────────────────────────────sum 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
Wren
<lang ecmascript>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()
}</lang>
- 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