Verifying Touchard's Identity $$C_{n+1} = \sum_{k=0}^{\lfloor n/2\rfloor}{n\choose 2k}\cdot C_k\cdot 2^{n-2k}$$
where $C_n$ are the Catalan numbers.
I think we start by diving both sides by $2^n$, but unsure of where to go from there
 A: Notice that $2^{-2k}\binom{n}{2k}\binom{2k}{k}=\binom{n/2}{k}\cdot \binom{n/2-1/2}{k}$. So
$$A_n=\sum_{k=0}^{\lfloor n/2 \rfloor}2^n \cdot 2^{-2k}\cdot \binom{n}{2k}\binom{2k}{k}\cdot\dfrac{1}{k+1}=2^n\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n/2}{k}\cdot \binom{n/2-1/2}{k}\cdot\dfrac{1}{k+1}$$
Without loss of generality, suppose $n/2 \in \mathbb{N}$. By Vandermonde's identity, we have
$$A_n=\dfrac{2^n}{n/2+1} \cdot \binom{n+1/2}{n/2}=\dfrac{1}{n+2}\binom{2(n+1)}{n+1}=C_{n+1}$$
A: Here is a proof using the method by Wilf from
Generatingfunctionology.

Suppose we are trying to show that
$$C_{n+1} = 2^n \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} C_k 2^{-2k}$$
where $C_n = \frac{1}{n+1}{2n\choose n}$ is the $n$-th Catalan number.

Recall the Catalan number recurrence
$$C_{n+1} = \sum_{k=0}^n C_k C_{n-k}$$
which implies for the generating function $C(z)$ of these numbers that
$$\frac{C(z)-1}{z} = C(z)^2$$
which has solutions $$C_{1,2}(z) = \frac{1\pm \sqrt{1-4z}}{2z}$$
of which the second one is analytic at zero so that
$$C(z) = \frac{1- \sqrt{1-4z}}{2z}.$$
Following Wilf we introduce the generating function $D(z)$ where
$$D(z) = \sum_{n\ge 0} z^n 2^n 
\sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} C_k 2^{-2k}
= \sum_{k\ge 0} C_k 2^{-2k}
\sum_{n\ge 2k} {n\choose 2k} z^n 2^n
\\ = \sum_{k\ge 0} C_k 2^{-2k}
\sum_{n\ge 0} {n+2k\choose 2k} z^{n+2k} 2^{n+2k}
= \sum_{k\ge 0} C_k z^{2k}
\sum_{n\ge 0} {n+2k\choose 2k} z^n 2^n
\\ = \sum_{k\ge 0} C_k z^{2k} \frac{1}{(1-2z)^{2k+1}}.$$
This gives that
$$D(z) = \frac{1}{1-2z} 
\frac{1-\sqrt{1-4z^2/(1-2z)^2}}{2z^2/(1-2z)^2}
\\ = \frac{(1-2z)^2}{1-2z} 
\frac{1-\sqrt{1-4z^2/(1-2z)^2}}{2z^2}
= (1-2z)
\frac{1-\sqrt{1-4z^2/(1-2z)^2}}{2z^2}
\\= \frac{1-2z-\sqrt{(1-2z)^2-4z^2}}{2z^2}
= \frac{1-2z-\sqrt{1-4z}}{2z^2}.$$
On the other hand
$$\sum_{n\ge 0} C_{n+1} z^n
= \frac{C(z)-1}{z}
= \frac{1}{z} \left(\frac{1- \sqrt{1-4z}}{2z} -1 \right)
\\ = \frac{1}{z} \left(\frac{1- 2z -\sqrt{1-4z}}{2z}\right)
= \frac{1- 2z -\sqrt{1-4z}}{2z^2}.$$
We have equality, QED.
