TITLE: Binomial identity arising from Catalan recurrence.
QUESTION [7 upvotes]: The following identity features right at the end of the answer to another question I asked: https://math.stackexchange.com/a/4019598/155881.
$$F(x) = \frac{1}{\sqrt{1-4x}} \left(\frac{1- \sqrt{1-4x}}{2x}\right)^n = \sum\limits_{k=0}^{\infty}{n+2k \choose k}x^k$$
Haven't been able to prove this. We know the expansions of both terms on the left side, making the identity:
$$\left(\sum\limits_{j=0}^{\infty} \left(\binom{n+2j-1}{j} - \binom{n+2j-1}{j-1}\right)x^j\right)
\left(\sum\limits_{l=0}^{\infty}{2l \choose l} x^l\right) = \sum\limits_{k=0}^{\infty}{2n+k \choose k}x^k$$

My attempt:
An obvious thing to do is to try a convolution of the left side and equate coefficients of $x^k$ (also using an alternate expression for the first summation in the product of the left side):
$$\sum\limits_{l=0}^k {n+2l-1\choose l}\frac{l}{n+l} {2k-2l\choose k-l} = {2n+k\choose k}$$
Not sure how to make progress on this. Expanding the binomial terms doesn't lead anywhere.

REPLY [2 votes]: Here is an approach based upon the Lagrange inversion theorem. We follow the paper Lagrange Inversion: when and how by R. Sprugnoli etal.
We consider the generating function
\begin{align*}
F(x)=\sum_{k=0}^{\infty}a_kx^k=\sum_{k=0}^\infty\binom{n+2k}{k}x^k
\end{align*}
and apply formula (G6) from the paper. The formula (G6) says that if there are functions $A(u)$ and $\phi(u)$, so that the coefficient $a_k$ admits a representation
\begin{align*}
a_k=[u^k]A(u)\phi(u)^k
\end{align*}
then the following is valid:
\begin{align*}
F(x)&=\sum_{k=0}^{\infty}[u^k]A(u)\phi(u)^kx^k\\
&=\left.\frac{A(u)}{1-x\phi^{\prime}(u)}\right|_{u=x\phi(u)}\tag{1}
\end{align*}

We obtain
\begin{align*}
\color{blue}{F(x)}&\color{blue}{=\sum_{k=0}^\infty\binom{n+2k}{k}x^k}\\
&=\sum_{k=0}^\infty[u^k](1+u)^{n+2k}x^k\tag{2}\\
&=\left.\frac{(1+u)^n}{1-x\,\frac{d}{du}(1+u)^2}\right|_{u=x(1+u)^2}\tag{3}\\
&=\frac{(1+u)^n}{1-\frac{u}{(1+u)^2}\cdot 2(1+u)}\tag{4}\\
&=\frac{(1+u)^{n+1}}{1-u}\tag{5}\\
&=\frac{\left(\frac{1}{2x}\left(1-\sqrt{1-4x}\right)\right)^{n+1}}{2-\frac{1}{2x}\left(1-\sqrt{1-4x}\right)}\tag{6}\\
&\,\,\color{blue}{=\frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^n}\tag{7}
\end{align*}
and the claim follows.

Comment:

In (2) we use the coefficient of operator $[x^k]$ to denote the coefficient of a series. We observe we can apply the Lagrange Inversion theorem with $A(u)=(1+u)^n$ and $\phi(u)=(1+u)^2$.

In (3) we use formula (1) with $A$ and $\phi$ as stated in the comment above.

In (4) we substitute $x=\frac{u}{1+u}$ and do the derivation.

In (5) we simplify the expression.

In (6) we recall $u=x(1+u)^2$ which is a quadratic equation in $u=u(x)$. We calculate
\begin{align*}
u(x)&=\frac{1}{2x}\left(1-2x\pm\sqrt{1-4x}\right)\\
&=\frac{1}{2x}\left(1\pm\sqrt{1-4x}\right)-1
\end{align*}
and take the solution with the minus sign, since this one can be expanded as generating function.

In (7) we make again some simplifications to obtain the wanted form.