Numbers which are the cube roots of the product of their proper divisors

Numbers which are the cube roots of the product of their proper 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.

Example

Consider the number 24. Its proper divisors are: 1, 2, 3, 4, 6, 8 and 12. Their product is 13,824 and the cube root of this is 24. So 24 satisfies the definition in the task title.

Task

Compute and show here the first 50 positive integers which are the cube roots of the product of their proper divisors.

Also show the 500th and 5,000th such numbers.

Stretch

Compute and show the 50,000th such number.

Reference

• OEIS:A111398 - Numbers which are the cube roots of the product of their proper divisors.

Note

OEIS considers 1 to be the first number in this sequence even though, strictly speaking, it has no proper divisors. Please therefore do likewise.

J

Note that the cube root of the product of the proper divisors is the fourth root of the product of all divisors of a positive integer. That said, we do not need to find roots here -- we only need to inspect the powers of the prime factors of the number:

F=: ('';1 1 1;1 3;,7) e.~ __ <@/:~@{:@q:"0 ]

Task examples:

   N=: 1+I.F 1+i.2^18
   5 10$N
  1  24  30  40  42  54  56  66  70  78
 88 102 104 105 110 114 128 130 135 136
138 152 154 165 170 174 182 184 186 189
190 195 222 230 231 232 238 246 248 250
255 258 266 273 282 285 286 290 296 297
   499{N
2526
   4999{N
23118
   49999{N
223735

Phix

with javascript_semantics
sequence n50 = {}
integer count = 0, n = 1, n5 = 500
printf(1,"First 50 numbers which are the cube roots\n")
printf(1," of the products of their proper divisors:\n")
while count<50000 do
    if product(factors(n))=n*n*n then
        count += 1
        if count<=50 then
            n50 &= n
            if count=50 then
                printf(1,"%s\n",join_by(n50,1,10,"",fmt:="%4d"))
            end if
        elsif count=n5 then
            printf(1,"%,6dth: %,d\n",{n5,n})
            n5 *= 10
        end if
    end if
    n += 1
end while

Output:

First 50 numbers which are the cube roots
 of the products of their proper divisors:
   1  24  30  40  42  54  56  66  70  78
  88 102 104 105 110 114 128 130 135 136
 138 152 154 165 170 174 182 184 186 189
 190 195 222 230 231 232 238 246 248 250
 255 258 266 273 282 285 286 290 296 297

   500th: 2,526
 5,000th: 23,118
50,000th: 223,735

Python

''' Rosetta code rosettacode.org/wiki/Numbers_which_are_the_cube_roots_of_the_product_of_their_proper_divisors '''

from functools import reduce
from sympy import divisors


FOUND = 0
for num in range(1, 1_000_000):
    divprod = reduce(lambda x, y: x * y, divisors(num)[:-1])if num > 1 else 1
    if num * num * num == divprod:
        FOUND += 1
        if FOUND <= 50:
            print(f'{num:5}', end='\n' if FOUND % 10 == 0 else '')
        if FOUND == 500:
            print(f'\nFive hundreth: {num:,}')
        if FOUND == 5000:
            print(f'\nFive thousandth: {num:,}')
        if FOUND == 50000:
            print(f'\nFifty thousandth: {num:,}')
            break

Output:

    1   24   30   40   42   54   56   66   70   78
   88  102  104  105  110  114  128  130  135  136
  138  152  154  165  170  174  182  184  186  189
  190  195  222  230  231  232  238  246  248  250
  255  258  266  273  282  285  286  290  296  297

Five hundreth: 2,526

Five thousandth: 23,118

Fifty thousandth: 223,735

Raku

use Prime::Factor;
use Lingua::EN::Numbers;
my @cube-div = lazy 1, |(2..∞).hyper.grep: { .³ == [×] .&proper-divisors }

put "First 50 numbers which are the cube roots of the products of their proper divisors:\n" ~
  @cube-div[^50]».fmt("%3d").batch(10).join: "\n";

printf "\n%16s: %s\n", .Int.&ordinal.tc, comma @cube-div[$_ - 1] for 5e2, 5e3, 5e4;

Output:

First 50 numbers which are the cube roots of the products of their proper divisors:
  1  24  30  40  42  54  56  66  70  78
 88 102 104 105 110 114 128 130 135 136
138 152 154 165 170 174 182 184 186 189
190 195 222 230 231 232 238 246 248 250
255 258 266 273 282 285 286 290 296 297

  Five hundredth: 2,526

 Five thousandth: 23,118

Fifty thousandth: 223,735

Wren

Library: Wren-math

Library: Wren-fmt

import "./math" for Int, Nums
import "./fmt" for Fmt

var numbers50 = []
var count = 0
var n = 1
System.print("First 50 numbers which are the cube roots of the products of their proper divisors:")
while (true) {
    var pd = Int.properDivisors(n)
    if (Nums.prod(pd) == n * n * n) {
        count = count + 1
        if (count <= 50) {
            numbers50.add(n)
            if (count == 50) Fmt.tprint("$3d", numbers50, 10)
        } else if (count == 500) {
            Fmt.print("\n500th   : $,d", n)
        } else if (count == 5000) {
            Fmt.print("5,000th : $,d", n)
        } else if (count == 50000) {
            Fmt.print("50,000th: $,d", n)  
            break
        }
    }
    n = n + 1
}

Output:

First 50 numbers which are the cube roots of the products of their proper divisors:
  1  24  30  40  42  54  56  66  70  78 
 88 102 104 105 110 114 128 130 135 136 
138 152 154 165 170 174 182 184 186 189 
190 195 222 230 231 232 238 246 248 250 
255 258 266 273 282 285 286 290 296 297 

500th   : 2,526
5,000th : 23,118
50,000th: 223,735