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

*

*Why $3m$ cancels with the numerator?

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


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.

