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        riddles >> hard >> Pythagorean factorials
        
(Message started by: JocK on May 20^th, 2006, 3:54am)

Title: Pythagorean factorials
Post by JocK on May 20^th, 2006, 3:54am

0!, 1!, 2! and 6! can all be written as the sum of two perfect squares. How many such 'Pythagorean factorials' exist?

Title: Re: Pythagorean factorials
Post by Icarus on May 20^th, 2006, 8:42am

Any number is expressible as the sum of two squares iff its prime factorization does not contain any entries of the form p^e where p = 3 mod 4 and e is odd.

This makes it very hard for n! to be the sum of two squares, particularly for larger values of n. The 4k+3 primes up to 71 alone show that no other numbers less than 1000 have factorials that are the sum of perfect squares. I suspect that the answer to the question is "4".

Title: Re: Pythagorean factorials
Post by Eigenray on May 25^th, 2006, 12:31pm

There are only finitely many Pythagorean factorials. Since

pi₃(x) / (x/log x) -> 1/2,

where pi₃(x) is the number of primes <x which are 3 mod 4, we have

[ pi₃(x) - pi₃(x/2) ] / (x/log x) -> 1/4.

Hence for sufficiently large n, there is always a prime p=3 mod 4 with n/2 < p <n, which therefore divides n! exactly once. A constructive bound could be obtained from a careful proof of Dirichlet's theorem, but I'm pretty sure "sufficiently large" here is n>6.

Title: Re: Pythagorean factorials
Post by x2862 on Feb 12^th, 2011, 10:25pm

I have listed all the fasctorial from 1 to 2000 on my website
http://www.x2862.com/Factorial/

This may or may not be of some use as these are very large numbers and hard to work with.