I want to prove that:
The number of partitions of $n$ into parts congruent to $1$ or $5$ ($\text{mod}$ 6) equals number of partitions of $n$ into distinct parts all congruent to $1$ or $2$ ($\text{mod}$ 3).
Is there a way to prove this using the well known Mobius inversion formula? I have been trying to establish a bijection between the set of partitions of $n$ of the first kind and the set of partitions of $n$ of the latter kind, but it is very difficult to establish uniqueness. Any help is appreciated. Thank you.
Once approach is to use the generating functions. Let $S_1, S_2$ denote the sets $\{m \in \mathbb{N} : m \equiv 1,2 \pmod 3\}$ and $\{m \in \mathbb{N} : m \equiv 1,5 \pmod 6\}$. If $p_{S}(n)$ denotes the number of partitions of $n$ into elements of $S$ and $\overline{p_S}(n)$ denotes the number of partitions of $n$ into distinct elements of $S$, then as formal power series $$\sum_{n=0}^{\infty}\overline{p_{S_1}}(n)q^n = \prod_{j=0}^{\infty}(1 + q^{3j + 1})(1 + q^{3j + 2})$$ and $$\sum_{n=0}^{\infty}p_{S_2}(n)q^n = \prod_{j=0}^{\infty}\frac{1}{(1 - q^{6j + 1})(1 - q^{6j + 5})}.$$ We have \begin{equation*} \begin{aligned} &\mathrel{\phantom{=}}\prod_{j=0}^{\infty}(1 + q^{3j + 1})(1 + q^{3j + 2}) \\ &= \prod_{j=0}^{\infty}\frac{(1 - q^{6j + 2})(1 - q^{6j + 4})}{(1 - q^{3j + 1})(1 - q^{3j + 2})} \\ &= \prod_{j=0}^{\infty}\frac{(1 - q^{6j + 2})(1 -q^{6j+ 4})}{(1 - q^{6j + 1})(1 - q^{6j + 4})(1 - q^{6j + 2})(q - q^{6j + 5})} \\ &= \prod_{j=0}^{\infty}\frac{1}{(1 - q^{6j + 1})(1 - q^{6j + 5})}. \end{aligned} \end{equation*} thus $$\overline{p_{S_1}}(n) = p_{S_2}(n).$$ Here the second line follows from the difference of two squares identity and the third line follows by noting that $\{m \in \mathbb{N} : m \equiv 1,2 \pmod 3\} = \{m \in \mathbb{N} : m \equiv 1,2,4,6 \pmod 6\}$.