Talk:Jordan-Pólya numbers: Difference between revisions
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(Responded to Horst.) |
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<pre>7213895789838336 =(4!)^8 * (2!)^16 == ((2!)^2 *(3!))^8 * (2!)^16 = (2!)^32*(3!)^8</pre> |
<pre>7213895789838336 =(4!)^8 * (2!)^16 == ((2!)^2 *(3!))^8 * (2!)^16 = (2!)^32*(3!)^8</pre> |
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[[User:Horst|Horst]] ([[User talk:Horst|talk]]) 08:48, 9 June 2023 (UTC) |
[[User:Horst|Horst]] ([[User talk:Horst|talk]]) 08:48, 9 June 2023 (UTC) |
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: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. --[[User:PureFox|PureFox]] ([[User talk:PureFox|talk]]) 09:57, 9 June 2023 (UTC) |
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Revision as of 09:57, 9 June 2023
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)