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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: 3 Registered: 1/16/07
Re: Sum of squares of binomial coefficients
Posted: Oct 12, 2012 9:27 AM

Le 11/10/2012 23:55, Gavin Wraith a écrit :

> In message <111020121212107442%edgar@math.ohio-state.edu.invalid>
>         Jérôme Collet <Jerome.Collet@laposte.net> wrote:

> >
> > I need to compute the sum : \sum_{r,s}{ (\binom{r+s}{r}
> > \binom{2m-r-s}{m-r})^2  } I know, because I used Stirling formula,
> > Taylor-polynomials, and ignored some problems on the borders, that
> > this sum should be close to \sqrt{2\pi m}. The convergence is very
> > fast, error is less than .5% if m>7. Nevertheless, I do not know
> > how to prove it correctly.

>
> This statement worries me. The expression you give   \sum_{r,s}{
> (\binom{r+s}{r} \binom{2m-r-s}{m-r})^2  } is certainly larger than
>   \sum_{r,s}{ (\binom{r+s}{r} \binom{2m-r-s}{m-r})  } which evaluates
> to \binom{2m}{m}, unless I am mistaken. And that grows much faster
> than \sqrt{2\pi m}.

I made a stupid mistake, I forgot a normalizing term, which is
\binom{2m}{m}^2

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 ?

Date Subject Author
10/11/12 Jérôme Collet
10/11/12 Gavin Wraith
10/12/12 Gavin Wraith
10/12/12 Jérôme Collet
10/12/12 Gavin Wraith
10/18/12 Jérôme Collet