I am wondering if a bijective proof of the following identity involving Catalan generating functions has appeared anywhere in the literature. (It's not difficult to simply verify it for the functions involved, so a bijective proof is the interesting part here.)
Let $C_n=\displaystyle\frac{1}{n+1}\binom{2n}{n}$, $C(z)=\displaystyle\sum_{n=0}^{\infty}{C_nz^n}=\dfrac{1-\sqrt{1-4z}}{2z}$, and $C_{\text{odd}}(z)=\displaystyle\sum_{n=0}^{\infty}{C_{2n+1}z^n}=\dfrac{C(\sqrt{z})-C(-\sqrt{z})}{2\sqrt{z}}$. Then $$ (zC(z))\circ(zC(4z))=zC_{\text{odd}}(z), $$ where $\circ$ denotes composition of functions. Expressed in terms of the coefficients of these functions, this yields, after a few cancellations, $$ \sum_{k=0}^{n}{\binom{2n-2k}{n-k}\binom{n+k}{k}4^k}=\binom{4n+1}{2n}, $$ or, equivalently, $$ \sum_{k=0}^{n}{\binom{2k}{k}\binom{2n-k}{n-k}4^{n-k}}=\binom{4n+1}{2n}. $$
Update: The identity $(zC(z))\circ(zC(4z))=zC_{\text{odd}}(z)$ can be thought of as a companion to Shapiro's Catalan identity for $C_{\text{even}}(z)=\displaystyle\sum_{n=0}^{\infty}{C_{2n}z^n}=\dfrac{C(\sqrt{z})+C(-\sqrt{z})}{2}$, namely, that $$ C(4z)=C_{\text{even}}(z)^2. $$