It is well-known that \begin{align*} \sum_{i=0}^n \binom{2i}{i}\binom{2n-2i}{n-i} = 4^n, \end{align*} where one might use combinatorial arguments or generating function technique to prove this.

Now I am interested in finding a closed-form formula for \begin{align*} \sum_{i=0}^n \binom{2i}{i}\binom{2n-2i}{n-i}x^i. \end{align*}

I had a look around but there are not so many literatures that mention this. I came across a paper, and in equation $(5)$ the author did talk about this, but it just doesn't look simple enough to me.

Does anyone have any idea how to proceed? Any idea would be very much appreciated.


If we define $$R_n(x)=\sum_{i=0}^{n} {2i\choose i}{2n-2i\choose n-i}x^i$$ then the generating function for these polynomials can be calculated:

$$G(x,t)=\sum_{n=0}^{\infty}R_n(x)t^n\\=\sum_{i=0}^{\infty}{2i\choose i}(xt)^i\sum_{n=0}^{\infty}{2n\choose n}t^n\\=\frac{1}{\sqrt{(1-4t)(1-4xt)}}$$

(we can combine the square roots in their common region of convergence) However we note that


and using the expression for the generating function of the Legendre polynomial $P_n(x)$ and it's uniqueness we may deduce that:

$$R_n(x)=4^n x^{n/2} P_n\Big(\frac{x+1}{2\sqrt{x}}\Big)$$

The generating function approach is in general very useful for calculating quantities like $R_n^{(m)}(1)$ as well.

For example, a simple calculation shows that

$$R_n^{'}(1)=\frac{\partial}{\partial x}G(e^x,t)\Big|_{x=0}=\frac{1}{n!}\frac{d^{n}}{dt^n}\Big(\frac{2t}{(1-4t)^2}\Big)\Big|_{t=0}=n2^{2n-1}$$

  • 1
    $\begingroup$ This is beautiful, for sure. $\to +1$ $\endgroup$ – Claude Leibovici May 10 at 8:35

I think that you are facing a gaussian hypergeometric function (this hides an infinite summation) $$\sum_{i=0}^n \binom{2i}{i}\binom{2n-2i}{n-i}x^i=\binom{2 n}{n} \, _2F_1\left(\frac{1}{2},-n;-n+\frac{1}{2};x\right)$$


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