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)
- I believe (bicbw) there is no way to make 439,378,587,648,000 = 14! * 7! with only prime factorials. --Petelomax (talk) 02:53, 10 June 2023 (UTC)
- Yes, your right.14! = 13!*2*7 2= 2! but no way to get a lonely 7. Of course there 16!= 2!^4 * 15! . To much work for to less advantage --Horst (talk) 16:28, 11 June 2023 (UTC)
Factoring the 1050th number
The 1050th jp number is 139,345,920,000. It factors as 7! * 5!^3 * 2!^4 or 8! * 5!^3 * 2! but the Phix/Julia/Wren entries are not getting that. Update: new algorithm posted, however my faffing about with prime powers is probably not the point, one-at-a-time/less greedy probably is. --Petelomax (talk) 05:31, 10 June 2023 (UTC)