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#1
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Re: Hyperfactorial Problem
[ QUOTE ]
Isn't this "superfactorical" instead? [/ QUOTE ] Not exactly. The index is off by one, which actually does make the expression more complicated if you use n$ instead of n?. I hadn't seen superfactorial before you mentioned it. Thanks for pointing it out. I think I'll try to prove that Bell number determinant identity, equation 17 on the second linked page. By the way, the formula f(a,b,c) is usually written quite differently (not necessarily in a way that helps here), and I think most people familiar with it would not recognize the expression I used. |
#2
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Re: Hyperfactorial Problem
Okay, the problem is equivalent to showing:
<font color="white"> 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. </font> |
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