Finding Exponents of Generating Function $A(x)=\prod\limits_{k=1}^{\infty}\frac{1-x^{6k}}{(1-x^{2k})(1-x^{3k})} $

Question

I have the following generating function:

$$A(x)=\prod\limits_{k=1}^{\infty}\frac{1-x^{6k}}{(1-x^{2k})(1-x^{3k})} $$

Among multiple of $2$, we will get a multiple of 6 whenever we take a multiple that is a multiple of 3: possible multiples: $3m, 3m+1, 3m+2$, where $3m$ cancels with the numerator.

So, we will have $1$ left in the numerator. What is left are factors in which the exponent is either $2\times(3m+1) $, $2\times(3m+2)$ or a multiple of $3$ ($3\times(2m+1), 3\times(3m+2)$).

Questions:

1. Why $3m$ cancels with the numerator?
2. Why what is left are factors in which the exponent is either $2\times(3m+1) $, $2\times(3m+2)$ or a multiple of $3$: $3\times(2m+1), 3\times(3m+2)$.

Answer 1

We obtain \begin{align*} \color{blue}{A(x)}&=\prod_{k=1}^\infty \frac{1-x^{6k}}{\left(1-x^{2k}\right)\left(1-x^{3k}\right)}\\ &=\prod_{k=1}^\infty \frac{1+x^{3k}}{1-x^{2k}}\tag{1}\\ &=\prod_{k=1}^\infty \frac{1-x^k+x^{2k}}{1-x^k}\tag{2}\\ &\,\,\color{blue}{=\prod_{k=1}^\infty \left(1+\frac{x^{2k}}{1-x^k}\right)}\tag{3} \end{align*}

Comment:

• In (1) we cancel $1-x^{3k}$.

• In (2) we cancel $1+x^k$.

Using the coefficient of operator $[x^N]$ to denote the coefficient of $x^N$ of a series, we derive from (3) \begin{align*} \color{blue}{[x^N]A(x)}&\color{blue}{=[x^N]\prod_{k=1}^\infty \left(1+\frac{x^{2k}}{1-x^k}\right)}\\ &=[x^N]\left(1+x^2+x^3+x^4+\cdots\right)\\ &\qquad\qquad\cdot\left(1+x^4+x^6+x^8+\cdots\right)\\ &\qquad\qquad\cdot\left(1+x^6+x^9+x^{12}+\cdots\right)\\ &\qquad\qquad\cdot\ \cdots \end{align*} which is the number of partitions of $N$ where each part of $N$ occurs at least twice.