Talk:Jordan-Pólya numbers

Isn't that like N-smooth_numbers using n! instead of primes?

In N-smooth numbers one can change p1..pn by 2! to n!
I think , only using 2!,3!,5!,..pn! is sufficient. 4! = (2!)^2*3!, 6! = 5!*3!... so no extra numbers will be created.

7213895789838336 =(4!)^8 * (2!)^16 == ((2!)^2 *(3!))^8 * (2!)^16 = (2!)^32*(3!)^8

Horst (talk) 08:48, 9 June 2023 (UTC)

There may, of course, be more than one way to decompose a J-P number into a product of factorials but the idea is to choose the way which uses the largest factorials and present these in highest to lowest order. I've added a sentence to the task description to try and clarify this. --PureFox (talk) 09:57, 9 June 2023 (UTC)

Factoring the 1050th number

The 1050th jp number is 139,345,920,000. One and I suspect the only possible factorisation is 7! * 5!^3 * 2!^4, but the Phix/Julia/Wren entries are not getting that. I have thought of a better strategy, based on the prime powers, but it is not straightforward and would naturally produce the lowest factorials - I think I might have just thought of a way to convert that to the highest factorials... In the meantime, some examples:

 92,160 = 6! * 2!^7, or
 92,160 = 5! * 3! * 2!^7
 18,345,885,696 = 4!^7 * 2!^2, or
 18,345,885,696 = 3!^7 * 2!^16
 139,345,920,000 = 7! * 5!^3 * 2!^4 (only one?)
 18,345,885,696 = 4!^7 * 2!^2, or
 18,345,885,696 = 3!^7 * 2!^16
 724,775,731,200 = 6! * 5! * 2!^23, or
 724,775,731,200 = 5!^2 * 3! * 2!^23
 9,784,472,371,200 = 6!^2 * 4!^2 * 2!^15, or
 9,784,472,371,200 = 5!^2 * 3!^4 * 2!^19
 439,378,587,648,000 = 14! * 4!^2 * 2!^15 (only one?)
 7,213,895,789,838,336 = 4!^8 * 2!^16, or
 7,213,895,789,838,336 = 3!^8 * 2!^32

There are probably only a few ways to do each, maybe show them all? --Petelomax (talk) 23:52, 9 June 2023 (UTC)