Two Plus Two Newer Archives

Two Plus Two Newer Archives (http://archives1.twoplustwo.com/index.php)
-   Science, Math, and Philosophy (http://archives1.twoplustwo.com/forumdisplay.php?f=49)
-   -   Hyperfactorial Problem (http://archives1.twoplustwo.com/showthread.php?t=556738)

pzhon 11-28-2007 08:09 PM

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.

bigpooch 12-01-2007 12:25 AM

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&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?
</font>

bigpooch 12-01-2007 02:55 AM

Re: Hyperfactorial Problem
 
Ignore my last post!

pzhon 12-01-2007 05:05 AM

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.

bigpooch 12-01-2007 11:04 AM

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&gt;=0.

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

</font>


All times are GMT -4. The time now is 06:12 AM.

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, vBulletin Solutions Inc.