# Generalized Catalan generating series

Let $$\mathscr{B}_k(z) = \sum_{n\geq 0}{kn+1\choose n}\frac{1}{kn+1}z^n\,,$$ then it is well known that $$\tag{1}\label{1} \text{log}\mathscr{B}_k(z)= \sum_{n\geq 1}{kn\choose n}\frac{1}{kn}z^n\,.$$ Let $$\tag{2}\label{2} \mathscr{B}_{k,a}(z) = \sum_{n\geq 0}{kan+1\choose an}\frac{1}{kan+1}z^{an}\,,$$ then we can write $$\mathscr{B}_{k,a}(z) =\frac{1}{a}\sum_{l=1}^a\mathscr{B}_{k}(\omega_a^lz)$$, where $$\omega_a=e^{2\pi i/a}$$. Using this, one should be able to show that $$\text{log}\mathscr{B}_{k,a}(z)$$ is not equal to $$\mathscr{L}_{k,a}(z) =\sum_{n\geq 0}{kan\choose an}\frac{1}{kan}z^{an} = \frac{1}{a}\sum_{l=1}^a\text{log}\mathscr{B}_{k}(\omega_a^lz)\,.$$ So is there some other formula (similar to \ref{1}) expresing $$\mathcal{L}_{k,a}(z)$$ in terms of logarithms of $$\mathscr{B}_{k,a}$$?

Edit: I obtain a generating series of certain geometric invariants given by $$\text{exp}\Big[\sum_{n>0}n^2\sum_{m>0}\frac{1}{amn}{anm\choose nm}z^{nm}\Big]\,.$$ My motivation is that I expected something of the form $$\prod_{n>0}F_n(z)$$. So I was hoping that instead of $$F_n(z) = \prod_{k>1}^n\mathscr{B}_a(-\omega^k_n)^n$$ I would be able to obtain something in terms of \eqref{2}.

• Why do you think other formula exists? Dec 6, 2020 at 1:55
• I have edited the question to explain why I was hopping for something nicer. Do you think this is a pointless search? Dec 6, 2020 at 9:36