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Re: Hyperfactorial Problem
Isn't this "superfactorical" instead?
Answer: <font color="white"> (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? </font> |
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