Catalan triangle created by $[t^{n+1}](\frac{1-\sqrt{1-4t}}{2})^{k+1}=\frac{k+1}{n+1}{2n-k \choose n-k}$? I tried to get the catalan triangle from its Riordan array. I failed at $d_{n,k}=[t^{n+1}](\frac{1-\sqrt{1-4t}}{2})^{k+1}=\frac{k+1}{n+1}{2n-k \choose n-k}$. How do I get there?

Working with the generating function $C(z)$ of the Catalan numbers
the claim is equivalent to
$$[z^{n-k}] C(z)^{k+1} = \frac{k+1}{n+1}
{2n-k\choose n-k}.$$
The generating function
$$C(z) = \frac{1-\sqrt{1-4z}}{2z}$$
has the functional equation
$$C(z) = 1 + z C(z)^2.$$
Note that $[z^0] C(z)^{k+1} = 1$ which matches the RHS of the claim
when $n=k,$ so we may suppose that $n\gt k.$ We then find
$$[z^{n-k}] C(z)^{k+1} = 
\frac{1}{n-k} [z^{n-k-1}] (C(z)^{k+1})'
= \frac{k+1}{n-k} [z^{n-k-1}] C(z)^k C'(z).$$
We get from the Cauchy Coefficient Formula
$$\frac{k+1}{n-k} [z^{n-k-1}] C(z)^k C'(z)
= \frac{k+1}{n-k} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k}}
C(z)^k C'(z) \; dz.$$
Now we put $C(z) = w$ and get from the functional equation that
$(w-1)/w^2 = z.$ This yields the integral
$$\frac{k+1}{n-k} \frac{1}{2\pi i}
\int_{|w-1|=\gamma} \frac{w^{2n-2k}}{(w-1)^{n-k}} w^k \; dw
\\ = \frac{k+1}{n-k} \frac{1}{2\pi i}
\int_{|w-1|=\gamma} \frac{w^{2n-k}}{(w-1)^{n-k}} \; dw
\\ = \frac{k+1}{n-k} \frac{1}{2\pi i}
\int_{|w-1|=\gamma} \frac{1}{(w-1)^{n-k}}
\sum_{q=0}^{2n-k} {2n-k\choose q} (w-1)^q\; dw
\\ = \frac{k+1}{n-k} {2n-k\choose n-k-1}
= \frac{k+1}{n+1} {2n-k\choose n-k}.$$
This is the claim. Note that $C(z)$ is analytic in a neighborhood of
the origin. The above can also be obtained using Lagrange
Inversion,
in fact this entry treats the very same generating function.
