# Borel Bivariate Generating Function

I want to prove the following statement:

$$\beta(t,x)=C(1+t,x)= \frac {C((1+t)x)} {1-xC((1+t)x)}$$

Where $$C(x)$$ is the generating function for the Catalan Numbers and $$\beta(x)$$ is the Borel generating function.

I know I have to use the fact that $$C(t,x)= \frac {C(tx)} {1-xC(tx)}$$ But unsure what other statements i need to use. Any help would be great thanks.

We assume \begin{align*} C(t,x)&=\sum_{n=0}^\infty\sum_{k=0}^n C_{n,k}t^kx^n=\frac{C(tx)}{1-xC(tx)}\tag{1}\\ B_{n,k}&=\sum_{s=k}^n\binom{s}{k}C_{n,s}\tag{2} \end{align*}
We obtain from (2) \begin{align*} \color{blue}{\sum_{k=0}^nB_{n,k}t^k}&=\sum_{k=0}^n\sum_{s=k}^n\binom{s}{k}C_{n,s}t^k\\ &=\sum_{s=0}^nC_{n,s}\sum_{k=0}^s\binom{s}{k}t^k\tag{3}\\ &\,\,\color{blue}{=\sum_{s=0}^nC_{n,s}(1+t)^s}\tag{4} \end{align*}
• In (3) we use $$\sum_{k=0}^n\sum_{s=k}^n a_{k,s}=\sum_{\color{blue}{0\leq k\leq s\leq n}} a_{k,s}=\sum_{s=0}^n\sum_{k=0}^s a_{k,s}$$.
We finally obtain from (4) \begin{align*} \color{blue}{\mathcal{B}(t,x)}&=\sum_{n=0}^\infty\sum_{k=0}^n B_{n,k}t^kx^n\\ &=\sum_{n=0}^\infty\sum_{s=0}^nC_{n,s}(1+t)^s x^n\tag{4 \leftarrow}\\ &\,\,\color{blue}{=\frac{C((1+t)x)}{1-xC((1+t)x)}}\tag{1 \leftarrow } \end{align*} and the claim follows.