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Is there any formula for calculating $\sum_{k=0}^n {n\choose k} {2n\choose 2k}$ ?

One possible way is to use Stirling's approximation, but couldn't reach a reasonable answer.

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  • $\begingroup$ @labbhattacharjee Didn't get your meaning. Can you be more specific? $\endgroup$ – m0_as Feb 18 '16 at 5:00
  • $\begingroup$ $$(1+x^2)^n(x+1)^{2n}=\sum_{k=0}^n\binom nk(x^2)^k\cdot\sum_{k=0}^{2n}\binom{2n}k x^{2n-k}$$ Compare the coefficients of $x^{2n}$ $\endgroup$ – lab bhattacharjee Feb 18 '16 at 5:05
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This is probably not the most pleasant result. $$S_n=\sum_{k=0}^n {n\choose k} {2n\choose 2k}=\, _3F_2\left(\frac{1}{2}-n,-n,-n;\frac{1}{2},1;-1\right)$$ where appears the generalized hypergeometric function.

However, plotting $\log(S_n)$ as a function of $n$ reveals almost a straight line and a simple linear regression gives $$\log(S_n)\approx 2.06597 n-2.27715$$ ($R^2=0.999998$) which gives at least the order of magnitude.

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