The power series expansion of certain infinite products $q^{r}\prod_{n=1}^{\infty}(1-q^{n})^{a_{1}}(1-q^{2n})^{a_{2}}\cdots(1-q^{mn})^{a_{m}}$

Abstract

Let ℕ, $\mathbb{N}_{0}$, ℤ, ℚ, and ℂ denote the sets of positive integers, nonnegative integers, integers, rational numbers, and complex numbers, respectively. If $f(q)$ is a complex-valued function with

$$f(q)= \sum_{n=0}^{\infty} f_{n} q^{n} \quad\bigl(q \in\mathbb{C}, \vert q \vert <1 \bigr) $$

we define

$$\bigl[f(q)\bigr]_{n}:=f_{n} \quad(n \in\mathbb{N}_{0}). $$

For $k \in\mathbb{N}$ we define

$$E_{k}:= \prod_{n=1}^{\infty} \bigl(1-q^{kn}\bigr) \quad\bigl(q \in\mathbb{C}, \vert q \vert <1 \bigr). $$

We show how modular equations of a special form can be used in conjunction with the representation numbers of certain quadratic forms to determine

$$\bigl[q^{r}E_{1}^{a_{1}}\cdots E_{m}^{a_{m}} \bigr]_{n} \quad(r \in\mathbb{N}_{0},m \in \mathbb{N},a_{1}, \ldots,a_{m} \in\mathbb{Z}) $$

for certain products $q^{r}E_{1}^{a_{1}}\cdots E_{m}^{a_{m}}$. For example, we show that

$$\biggl[q^{2}\frac{E_{1}^{4}E_{16}^{4}}{E_{2}^{2}E_{8}^{2}} \biggr]_{n} = \begin{cases} 0 & \mbox{if}\ n \equiv1\ (\mbox{mod}\ 4),\\ - \sigma(N) & \mbox{if}\ n \equiv3\ (\mbox{mod}\ 4),\\ \sigma(N) & \mbox{if}\ n \equiv2\ (\mbox{mod}\ 4),\\ 4\sigma(N) & \mbox{if}\ n \equiv4\ (\mbox{mod}\ 8),\\ 0 & \mbox{if}\ n \equiv0\ (\mbox{mod}\ 8), \end{cases} $$

where $N$ denotes the odd part of the positive integer $n$ and

$$\sigma(n): =\sum_{ \begin{array}{c} \scriptstyle d \in\mathbb{N}\\[-3pt] \scriptstyle d \mid n \end{array}} d. $$