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Possible Duplicate:
Identity involving binomial coefficients

Simplify the sum: $\sum_{k=0}^n {2k\choose k}{2n-2k\choose n-k}$

So we can denote $a_n=\sum\limits_{k=0}^n {2k\choose k}{2n-2k\choose n-k}$ and see that $\sum\limits_{n=0}^{+\infty}\sum\limits_{k=0}^n a_n x^n$ is a convolution of $b_n$ and $b_n$, where $b_n={2n\choose n}$, hence all we need to finish this is generating function for Catalan numbers. One derivative and we are done, $a_n=4^n$.

But this sum looks very nice and it's bugging me if has it any interesting combinatorial interpretation. Any ideas?

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    See also http://math.stackexchange.com/q/80649/1522012-08-24

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