Hyperfactorial Problem
We're used to dealing with n!
How about n?
Define n? = 1! 2! ... (n-1)!
Do not include an n! factor.
So, 1? = 1, 2? = 1, 3? = 2, 4? = 12, and 5? = 288.
It is known that for any positive integers a, b, and c, that f(a,b,c)=
(a+b+c)? a? b? c?
-------------------
(a+b)?(b+c)?(c+a)?
counts something. Prove that f(a,b,c) is an integer without using that it counts something. (Note that for a=1, f(a,b,c) specializes to b+c choose b,c.)
Please post solutions in white.
|