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Topic: Sum of squares of binomial coefficients
Replies: 5   Last Post: Oct 18, 2012 4:28 AM

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Jérôme Collet

Posts: 1
Registered: 10/18/12
Re: Sum of squares of binomial coefficients
Posted: Oct 18, 2012 4:28 AM
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I stated, by trial and errors that:
\sum_{r,s}{ (\binom{r+s}{r} \binom{2m-r-s}{m-r})^2 } = \binom{4m+1}{2m}
It is true for m from 1 to 250, with infinite precision calculations, so
I think it is true for all m.

But it is not a proof.
I will try do do it using manually Sister Celine's method, but if anyone
has a combinatorial proof, I am interested.

Le 13/10/2012 00:13, Gavin Wraith a écrit :
> In message <121020120727017676%edgar@math.ohio-state.edu.invalid>
> Jérôme Collet <Jerome.Collet@laposte.net> wrote:
>

>> So my question is
>> \sum_{r,s}{ (\binom{r+s}{r} \binom{2m-r-s}{m-r})^2 }
>> seems to be equivalent to
>> \sqrt{2\pi m} \binom{2m}{m}^2
>> How can I prove it ?

>
> A bit more attention; maybe not very useful. Suppose you have
> a 2m-element set. Call a quadruple of subsets (A,B,U,V) such that
> 1) #U = #V = m
> 2) #A = #B
> 3) #(A\cap U) = #(B\cap V)
> "jolly". Then I believe that
> \sum_{r,s}{ (\binom{r+s}{r} \binom{2m-r-s}{m-r})^2 }
> is the number of jolly quadruples. Now \binom{2m}{m}^2 is
> the number of ways of choosing pairs of subsets (U,V) such that
> #U = #V = m. I am sceptical that all the rest of the jollity
> counts asymptotically for \sqrt{2\pi m}. I would have expected something
> a lot bigger, but I cannot give you any good reasons right now.
>





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