We have a recurrence relation in two variables $n, k \in\{0,1\dots \}$
$$c(n, k) = \sum_{j=1} a(j-1, k-1)a(n-j, k), \space n\geq 1, k\geq 1 \tag{1}$$ where the generating function $a(x,y)=\sum_{n=0}^\infty \sum_{k=0}^\infty a(n,k)x^ny^k = \frac{2}{\sqrt{1-4x^2}-2xy+1}.$
We also know that
$$\sum_{n=0}^\infty c(n,0)x^n = \frac{2}{1+\sqrt{1-4x^2}} =: c(x)$$ and $$c(0,k) = 0, \text{ when } k>0.$$
How to find the generating function $c(x,y)=\sum_{n=0}^\infty \sum_{k=0}^\infty c(n,k)x^ny^k$?
My try:
Denote $f(x,y) = xy\times a(x,y)$. Then $f(n, k)=a(n-1, k-1), n,k>0$ and $f(0,k)=f(n,0)=0$.
Multiply term-wise by $x^ny^k$ and sum $(1)$ over $n\geq 1, k\geq 1$ to get
$$\sum_ {n=1}^\infty \sum_{k=1}^\infty a(n,k)x^ny^k = \sum_ {n=1}^\infty \sum_{k=1}^\infty \sum_{j=1}^n a(j-1, k-1) a(n-j, k) x^n y^k.$$
The left hand side is missing the terms of $c(x)$ and since $c(0, k)=0,$ when $k>0$ we can let the index $k$ run from $0$ (we already add the missing $1$=($c(0,0)$) when adding $c(x)$), so the left hand side is
$$c(x,y) - c(x).$$
On the right hand side we replace $a(j-1, k-1)$ by $f(j, k)$ and since $f(0,k)=f(n,0)=0$ we can let the indices $j$ and $k$ run from zero. Also $n$ can be let run from zero. So the left hand side is
$$\sum_ {n=0}^\infty \sum_{k=0}^\infty \sum_{j=0}^n f(j, k) a(n-j, k) x^n y^k$$
It seems like this is some sort of product of generating functions (I'm trying to make it be $fa$ i.e $xya^2$), but shouldn't there be double sum for the coefficient of the product since this is bivariate power series?
Could someone please tell me what is the correct solution and how to get it?
PS.
By inspection I'm guessing
$$a(n, k) = \frac{k+1}{\frac{n+k}{2}+1} {{n}\choose{\frac{n-k}{2}}}, \text{ when }n+k \text{ is even and } 0 \text{ otherwise}$$
and
$$c(n,k) = \frac{2k+1}{\frac{n}{2}+k+1} {{n}\choose{\frac{n}{2}-k}}\text{ when }n \text{ is even and } 0 \text{ otherwise}$$