# Generating function of a parametrized binomial coefficient

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.

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 $$(2)$$.

• Can you solve $(1)$ for all values of $m$? (even if $m \geq 5$)? Jun 24, 2019 at 8:17
• Speaking of computation, yes, I can solve it numerically. No, we can't express the solution in radicals, but analytically $(1)$ and $(2)$ are the simplest form of the solution to expect (at least I think so). Jun 24, 2019 at 8:39
• @metamorphy: ... and (+1) also for the interesting reference ... Jul 1, 2019 at 15:42
• Here is my answer handling non-integer $m$ (following this idea). Aug 1, 2019 at 17:03
• More general $G_{m,k}(z)=\sum_{p\geqslant 0}\binom{mp+k}{p}z^p$ obtained here. Mar 22, 2021 at 9:00

For $$m=1,2,3$$ there are closed forms for $$B_m(z)$$.
For $$m\geq 4$$ come again hypergeometric functions with interesting patterns $$B_m(z)=\, _{m-1}F_{m-2}\left(\frac{1}{m},\frac{2}{m},\cdots,\frac{m-1}m; \frac{1}{m-1},\frac{2}{m-1},\cdots,\frac{m-2}{m-1};\frac{m^m}{(m-1)^{m-1}}z\right)$$