Product of divisors

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.

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

1. 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):
   ]

1. ------------------------- TEST -------------------------
2. 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))

1. MAIN ---

if __name__ == '__main__':

   main()</lang>

Wren

Library: Wren-math

Library: Wren-fmt

<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