I came across the following combinatorial identity in a paper 'a remarkable sequence of integers'.

$\sum_{k=0}^m 2^{-2k} \binom{2k}{k} \binom{2m-k}{m} =4^{-m} \binom{4m+1}{2m}. $

I gave an elementary proof as follows, yet I a combinatorial intepretation seems difficult to a layman like me. So I post it here for discussion.

My elementary proof is through the method of coefficients.

Let $[t^n]f(t)$ be the coefficient of $t^n$ in $f(t)$.

**Lemma:** $[t^k]\frac{1}{\sqrt{1-t}}=4^{-k} \binom{2k}{k}$.

**Proof:**

$$ \begin{aligned} [t^k]\frac{1}{\sqrt{1-t}} &=\binom{-1/2}{k} (-1)^k \\\\ &=\binom{1/2+k-1}{k} \\\\ &=\frac{(k-1/2)(k-3/2)\cdots (1/2)}{k!}\\\\ &=\frac{(2k-1)(2k-3)\cdots 1}{k!}2^{-k} \\\\ &=\frac{(2k)(2k-1)(2k-2)(2k-3)\cdots 2\cdot1}{k!\cdot k!} 4^{-k} \\\\ &= 4^{-k} \binom{2k}{k} \end{aligned} $$

QED

Moreover, it is easy to see

$$ \begin{aligned} \binom{2m-k}{m}&=\binom{2m-k}{m-k} \\\\ &=\binom{-(2m-k)+m-k-1}{m-k}(-1)^{m-k}\\\\ &= \binom{-m-1}{m-k} (-1)^{m-k} \\\\ &=[t^{m-k}]\frac{1}{(1-t)^{m+1}} \end{aligned} $$

**Proposition:**

$$\sum_{k=0}^m 2^{-2k} \binom{2k}{k} \binom{2m-k}{m}= 4^{-m} \binom{4m+1}{2m}.$$

**Proof:**

$$ \begin{aligned} \sum_{k=0}^m 2^{-2k} \binom{2k}{k} \binom{2m-k}{m} &= [t^m]\left(\frac{1}{\sqrt{1-t}} \frac{1}{(1-t)^{m+1}}\right) \\\\ &= [t^m]\frac{1}{(1-t)^{m+(3/2)}} \\\\ &=\binom{-m-(3/2)}{m} (-1)^m \\\\ &=\binom{2m+(1/2)}{m} \\\\ &= 2^{-m} \frac{(4m+1)(2m-1)\cdots(2m+3)}{m!} \\\\ &=2^{-m} \frac{(4m+1)(2m-1)\cdots(2m+3)}{m!} \frac{4m(4m-2)\cdots(2m+2)}{4m(4m-2)\cdots(2m+2)} \\\\ &=2^{-2m}\frac{(4m+1)!}{(2m+1)!(2m)!} \\\\ &=2^{-2m} \binom{4m+1}{2m} \end{aligned} $$

QED