Let be $m$ an integer and $A_p(m) = \binom{mp}{p}$.
I'd like to know more about $B_m(z) = \sum_{p \geq 0} A_p(m) z^p$.
At least, I'd love to be able to compute $B_m\left(\dfrac{1}{q}\right)$ for some $q$ integers.
What I tried:
- Look at Fuss-Catalan numbers and its generating function to derive a relation with $B_m$
- But as there is no closed form of the generating function, I cannot derive an interesting enough relation here.
Intuitively, I could try to interpret $A_p(m)$ as something combinatorial and look for a recurrence relation to explicit $B_m$, but I have no idea.
Best Answer
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 $(2)$.