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I hope it is OK to ask the following reference request. If my question is not suitable, please let me know and I will do my best to modify it!

Let $N\in\mathbb{N}$, let $q$ be a point in the open unit disc in the complex plane, and let $0\leq m \leq N$ be an integer. Some personal research I have been doing on iterated integrals of Eisenstein series has led me to consider certain series of the following form: $$S_m(N;q):=\sum_{k\geq 1}\frac{\sigma_N(k)}{k^N} k^m q^k,$$ where $\sigma_N(k):=\sum_{d\vert k} d^N$ is the $N$th power sum-of-divisors function.

There are two "extreme" cases that connect to well-known series: $m=0$ and $m=N$. In the $m=0$ case we recover a Lambert series $$S_0(N;q) = \sum_{k\geq 1}\frac{q^k}{k^N(1-q^k)}$$ and in the $m=N$ case (for $N\geq 3$ odd) we recover the nonconstant part of the $\mathbb{Q}$-normalised Eisenstein series $\mathbb{G}_{N+1}(\tau)$, with $q=\exp(2\pi i \tau)$ and $\tau$ in the upper half plane. Another connection to well-known objects comes from taking the limit $$\lim_{q\to 1}S_m(N;q) = \zeta(-m)\zeta(N-m).$$


In the cases $1\leq m\leq N-1$ I do not have a good name for these series, and I am not aware of them in the literature on modular forms/$L$-functions. I hope it is OK to ask: have these series been studied before, and are there any references for them if so? Is anything known about relations between the $S_m(N;q)$ (for a fixed $N$), or the dimension of the $\mathbb{Q}$-vector space generated by $\pi^m S_m(N;q)$ for $0\leq m\leq N$?

In fact, I am interested primarily not in the "functional" case, but with the specific case $q = e^{-2\pi}$. As an example of what I mean about relations in this case, Ramanujan's formula for $\zeta(N)$ in terms of a Lambert series gives a $\mathbb{Q}$-relation between $\zeta(N)$, $\pi^N$ and $S_0(N;e^{-2\pi})$ (at least for $N\equiv 3\pmod{4}$). I have been studying this case in my research and have found more relations between these numbers as well as $\pi^m S_m(N;e^{-2\pi})$ for larger $m$. If anything can be said about this case, I would be very grateful!

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The sum which you consider could be rewritten as a double sum of the following:
\begin{align} S_{m}(N,q)=\sum_{i,j\ge 1}i^{m-N}j^me^{-2ij\alpha}:=f_{N,m}(\alpha) \end{align} by setting $q=e^{-2\alpha}$. It is easy to check that $$f_{N,m}(\alpha)=(-1)^m\frac{1}{2^m}\frac{\,d^m}{\,d \alpha^m}\sum_{i,j\ge 1}i^{-N}e^{-2ij\alpha}=\frac{(-1)^m}{2^m}\frac{\,d^m}{\,d \alpha^m}\sum_{k\ge 1}\frac{1}{k^N(e^{2k\alpha}-1)}.$$ For $N=2n+1\ge 3$ be an odd positive integer, we have the Ramanujan’s formula for $\zeta(2n+1)$, (which could be find in [Bruce C Berndt: Ramanujan's Notebooks: Part IV]) \begin{align} \frac{1}{\alpha^{n}}\left(\frac{\zeta(2n+1)}{2} + \sum_{m=1}^\infty \frac{1}{m^{2n+1}(e^{2m\alpha}-1)}\right)\quad- \frac{1}{(-\beta)^{n}}\left(\frac{\zeta(2n+1)}{2} + \sum_{m=1}^\infty \frac{1}{m^{2n+1}(e^{2m\beta}-1)}\right)\\ =2^{2n}\sum_{k=0}^{n+1} (-1)^{k-1}\frac{B_{2k}}{(2k)!}\frac{B_{2n+2-2k}}{(2n+2-2k)!}\alpha^{n+1-k}\beta^k, \end{align} where $\alpha,\beta>0$ such that $\alpha\beta=\pi^2$ and $B_r , r\ge 0$, denote the $r$-th Bernoulli number. Which means $f_{N,m}(\alpha)$ is a quasi-modular function. We also have the modularity for the cases $N=2n$ is even follows the modularity of $$\sum_{k\ge 1}\frac{1}{k^{2n}(e^{2k\alpha}-1)}.$$ However, the above function is not modular or quasi-modular.

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  • $\begingroup$ Thanks for your answer! Unfortunately I am a little unclear about your point regarding $f_{N,m}(\alpha)$ being quasi-modular (for $N=2n$ even). You mention that quasi-modularity of $f_{N,m}(\alpha)$ follows from modularity of $\sum k^{-2n}(e^{2k\alpha}-1)^{-1}$, but you then say that this function is not (quasi-)modular. So is $f_{N,m}(\alpha)$ only modular when $N$ is odd? $\endgroup$ – Alex Saad Aug 23 '18 at 12:40
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    $\begingroup$ Yes, $f_{N,m}(\alpha)$ is modular only $N$ is odd. $\endgroup$ – Zhou Aug 23 '18 at 12:56
  • $\begingroup$ Great - thank you! $\endgroup$ – Alex Saad Aug 23 '18 at 12:57
  • $\begingroup$ In this case, is it possible to say what the weight of $f_{N,m}(\alpha)$ is? $\endgroup$ – Alex Saad Aug 23 '18 at 13:16
  • $\begingroup$ I am not sure, but I think this article [Ramanujan’s Formula for $\zeta(2n+1)$, arxiv.org/pdf/1701.02964.pdf ] gives the answer which you need, possibly. $\endgroup$ – Zhou Aug 23 '18 at 14:34

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