Let $\mathcal{A}$ be the set of partitions in which each part may occur 0, 1, 4, or 5 times and let $\mathcal{B}$ be the set of partitions which have no parts congruent to 2mod4, and in which parts divisible by four occur at most once each. Then for any $n \in \mathbb{N}$, I need to show that $|\mathcal{A}_n| = |\mathcal{B}_n|$. In other words, I need to show that the number of partitions in $\mathcal{A}$ of size n equals the number of partitions $\mathcal{B}$ of size n.
What I've done so far was the following. I've managed to derive both generating functions, namely:
$$\begin{equation*}\Phi_\mathcal{A} = \prod_{j=1}^{\infty}(1 + x^j + x^{4j} + x ^{5j})\end{equation*}$$
$$\begin{equation*}\Phi_\mathcal{B} = \prod_{j=1}^{\infty}(1 + x^{4j})\left(\frac{1}{1-x^{4j-3}}\right)\left(\frac{1}{1-x^{4j-1}}\right)\end{equation*}$$
Furthermore, in order to show that $|\mathcal{A}_n| = |\mathcal{B}_n|$ for any $n \in \mathbb{N}$, it suffices to show that the two generating functions are equal, $\Phi_\mathcal{A} = \Phi_\mathcal{B}$. However, I am unsure how to go about to manupulate them in order to show this final desired result. Any help would be appreciated.
Thanks!
Yes, your own observation nicely works. Start with ${\cal B}$.
First use the trick also mentioned in wikipedia (for partitions in odd parts) to show that
$\frac{1}{(1-x)(1-x^3)(1-x^5)(1-x^7)\dots} =$ $ (1+x)(1+x^2)(1+x^3) \dots$; this is the product of the fractions of odd powers $\prod(\frac{1}{1-x^{4j-3}})(\frac{1}{1-x^{4j-1}})$.
Now you have to multiply that result with the multiples of $4$, i.e, $(1+x^4)(1+x^8)(1+x^{12}) \dots$, combining $(1+x^j)(1+x^{4j})$ -- here $(1+x^j)$ is a term from the previous computation and $(1+x^{4j})$ one of the multiples of $4$. Voila.