Generating function of a parametrized binomial coefficient

I still think the way you go is as good as one gets. Precisely, if $$\color{darkblue}{F_m(z)}:=\sum_{p\geqslant 0}\binom{mp}{p}\frac{z^p}{(m-1)p+1}\color{darkblue}{=1+z\big(F_m(z)\big)^m}\tag{1}$$ (the equality is from here, where $F_m(z)=B_{m,1}(z)$ in that notation), then $$\color{darkblue}{B_m(z)}=F_m(z)+(m-1)zF_m'(z)\color{darkblue}{=\frac{F_m(z)}{m-(m-1)F_m(z)}}\tag{2}$$ (the first equality is clear; taking derivative of $(1)$ helps to get the second one).

Thus computing $B_m(z)$ amounts to solving $(1)$ and plugging the result into $(2)$.

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Combinatorics

Generating Functions

Catalan Numbers

Power Series

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