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 $(2)$.

  • $\begingroup$ Can you solve $(1)$ for all values of $m$? (even if $m \geq 5$)? $\endgroup$ – Raito Jun 24 '19 at 8:17
  • $\begingroup$ 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). $\endgroup$ – metamorphy Jun 24 '19 at 8:39
  • $\begingroup$ @metamorphy: ... and (+1) also for the interesting reference ... $\endgroup$ – Markus Scheuer Jul 1 '19 at 15:42
  • $\begingroup$ Here is my answer handling non-integer $m$ (following this idea). $\endgroup$ – metamorphy Aug 1 '19 at 17:03

This is not an answer.

For $m=1,2,3$ there are closed forms for $B_m(z)$.

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


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