Re: Hyperfactorial Problem
Isn't this "superfactorical" instead?
Answer:
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(m+n)?/m? = (m+n-1)!...m!
f(a,b,c) = [(a+b+c)?/(a+b)?]c?/{[(a+c)?/a?][(b+c)?/b?]}
= [(a+b+c-1)!...(a+b)!]c?/{[(a+c-1)!...a!][(b+c-1)!...b!]}
Now, there is a "natural" grouping into c factors.
For 0<=k<=c-1,
(a+b+k)!k!/[(a+k)!.(b+k)!] = C(a+b+k,a+k)/C(b+k,b), so
f(a,b,c) is the product of these terms with 0<=k<=c-1.
Similarly, it is a product of terms C(a+b+k,b+k)/C(a+k,a)
[ by symmetry ].
The square of f(a,b,c) is then the product of terms
C(a+b+k,a+k)C(a+b+k,b+k)/[C(a+k,a)C(b+k,b)]
but C(a+b+k,a+k)/C(a+k,a) and
C(a+b+k,b+k)/C(b+k,b) are multinomial coefficients and
are nonnegative integers. Thus, the square of f(a,b,c) is
a product of these and hence a nonnegative integer. Thus,
f(a,b,c) is the nonnegative square root of this product and
must be a nonnegative integer.
Is there something simpler you have in mind?
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