arXiv2025
Let $A(q)=\sum_{n=0}^{\infty}a_n q^n$ and $B(q)=\sum_{n=0}^{\infty}b_n q^n$ be two eta quotients. Previously, we considered the problem of when \[ a_n=0 <=> b_n=0. \] Here we consider the ``mod $m$'' version of this problem, i.e. eta quotients $A(q)$ and $B(q)$ and integers $m>1$ such that \[ a_n \equiv 0 \pmod m <=> b_n \equiv 0 \pmod m? \] We found results for $m=p^2$, $p=2, 3$ and $5$. For $m=4,9$, we found results which apply to infinite families of eta quotients. For example: Let $A(q)$ have the form \begin{equation} A(q) = f_1^{3j_1+1}\prod_{3\nmid i}f_i^{3j_i}\prod_{3|i}f_i^{j_i} =: \sum_{n=0}^{\infty}a_nq^n,\,\,B(q) = \frac{f_3}{f_1^3}A(q) =: \sum_{n=0}^{\infty}b_nq^n \end{equation} with $f_{k}=\prod_{n=1}^{\infty}(1-q^{kn})$. Then \begin{align*} a_{3n}-b_{3n}&\equiv 0\pmod 9,\\ 2a_{3n+1}+b_{3n+1}&\equiv0\pmod 9,\\ a_{3n+2}+2b_{3n+2}&\equiv0\pmod 9. \end{align*} Some of these theorems also had some combinatorial implications, such as the following: Let $p_2^{(3)}(n)$ denote the number of bipartitions $(\pi_1, \pi_2)$ of $n$ where $\pi_1$ is 3-regular. Then \begin{equation*} p_2^{(3)}(n)\equiv0\pmod 9 <=> n\text{ is not a generalized pentagonal number}. \end{equation*} In the case of $m=25$, we do not have any general theorems that apply to an infinite family of eta quotients. Instead we give two tables of results that appear to hold experimentally. We do prove some individual results (using theory of modular forms), such as the following: Let the sequences $\{c_n\}$ and $\{d_n\}$ be defined by \begin{equation*} f_1^{10}=:\sum_{n=0}^{\infty}c_nq^n, \hspace{25pt} f_1^{5}f_5=:\sum_{n=0}^{\infty}d_nq^n. \end{equation*} Then \begin{equation*} c_n \equiv 0 \pmod{25} <=> d_n \equiv 0 \pmod{25}. \end{equation*}