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.

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&lt;=k&lt;=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&lt;=k&lt;=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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Re: Hyperfactorial Problem
 

Ignore my last post!

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.

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&gt;=0.

This can be shown, but is almost certainly not the simplest solution.

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