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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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 Gavin Wraith Posts: 20 Registered: 9/29/06
Re: Sum of squares of binomial coefficients
Posted: Oct 12, 2012 9:25 AM

In message <111020121555505579%edgar@math.ohio-state.edu.invalid>
Gavin Wraith <gavin@wra1th.plus.com> wrote:

> 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}.

Erratum: the penultimate line should read

"which evaluates to 4^m\binom{2m}{m}, unless I am mistaken."

This is the number of ways of choosing from a 2m-element set
an m-element subset and, independently a subset, call it A,
of any size. If we suppose that A has r+s elements, r of which
are in the m-element set, you can see why.
In any case, the expression given by Jerome has to exceed this
number but be less than its square.

--
Gavin Wraith (gavin@wra1th.plus.com)