Re: Hyperfactorial Problem
Okay, the problem is equivalent to showing:
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C(x+b,b)C(x+b+1,b)...C(x+b+m,b) is divisible by
C(b,b)C(b+1,b)...C(b+m,b) for any integer x>=0.
This can be shown, but is almost certainly not the simplest solution.
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