<s>[INST] Binomial identity arising from Catalan recurrence. 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.
 [/INST] We seek to show that with
$$Q(z) = \frac{1}{\sqrt{1-4z}}
\left(\frac{1-\sqrt{1-4z}}{2z}\right)^n$$
we have
$$[z^k] Q(z) = {n+2k\choose k}.$$
Now with the branch cut on $[1/4, \infty)$ for $\sqrt{1-4z}$ we have
analyticity of $Q(z)$ in a neighborhood of the origin (note that the
exponentiated term does not in fact have a pole at $z=0$) and the Cauchy
Coefficient Formula applies. We obtain
$$[z^k] Q(z) = \frac{1}{2\pi i}
\int_{|z|=\varepsilon} \frac{1}{z^{k+1}}
\frac{1}{\sqrt{1-4z}}
\left(\frac{1-\sqrt{1-4z}}{2z}\right)^n \; dz.$$
We put $\sqrt{1-4z} = w$ so that  $\frac{1}{\sqrt{1-4z}} \; dz =
-\frac{1}{2} \; dw$  and $z=(1-w^2)/4.$ With $w = 1 - 2z - \cdots$ we get
as the image of $|z|=\varepsilon$ a contour that winds around $w=1$
counterclockwise once and may be  deformed to a circle, so that we obtain
$$[z^k] Q(z) =
-\frac{1}{2}
\frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{4^{k+1}}{(1-w^2)^{k+1}}
(1-w)^n
\frac{1}{2^n} \frac{4^n}{(1-w^2)^n}
\; dw
\\ = \frac{(-1)^k \times 2^{n+2k+1}}{2\pi i}
\int_{|w-1|=\gamma}
\frac{(w-1)^n}{(w^2-1)^{n+k+1}} \; dw
\\ = \frac{(-1)^k \times 2^{n+2k+1}}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(w-1)^{k+1}} \frac{1}{(w+1)^{n+k+1}} \; dw
\\ = \frac{(-1)^k \times 2^{k}}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(w-1)^{k+1}} \frac{1}{(1+(w-1)/2)^{n+k+1}} \; dw$$
Apply the Cauchy Residue Theorem to get
$$(-1)^k \times 2^k \times (-1)^k \frac{1}{2^k}
{n+k+k\choose n+k} = \bbox[5px,border:2px solid #00A000]{
{n+2k\choose k}}$$
as claimed.
</s>