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}.

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


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