Prove that number of partitions of $n$ into parts $2,5,11$ modulo $12$ is the same as the number of partitions into distinct parts $2,4,5$ modulo 6

Question

Let $A(n)$ denote the number of partitions of the positive integer $n$ into parts congruent to $2$, $5$, or $11$ modulo $12$. Let $B(n)$ denote the number of partitions of $n$ into distinct parts congruent to either $2$, $4$, or $5$ modulo $6$. Prove that $A(n) = B(n)$.

I tried to do this with generating functions. Now \begin{align}\sum_{n=0}^{\infty}A(n)x^n &=\prod_{k=0}^{\infty} \left(\sum_{i=0}^{\infty}x^{i(2+12k)} \right)\left(\sum_{i=0}^{\infty}x^{i(5+12k)} \right)\left(\sum_{i=0}^{\infty}x^{i(11+12k)} \right) \\ &=\prod_{k=0}^{\infty}\frac1{1-x^{2+12k}}\frac1{1-x^{5+12k}}\frac1{1-x^{11+12k}} \end{align} And $$\sum_{n=0}^{\infty} B(n)x^n=\prod_{k=0}^{\infty} (1+x^{2+6k})(1+x^{4+6k})(1+x^{5+6k})$$ How can these two be same?

Answer 1

\begin{align}\sum_{n=0}^{\infty}A(n)x^n &=\prod_{k=0}^\infty\frac{1}{(1-x^{2+12k})(1-x^{5+12k})(1-x^{11+12k})} \\ &=\prod_{k=0}^\infty\frac{1}{(1-x^{2+12k})(1-x^{5+6k})} \\ &=\prod_{k=0}^\infty\frac{1}{(1+x^{1+6k})(1-x^{1+6k})(1-x^{5+6k})} \\ &=\prod_{k=0}^\infty\frac{1-x^{3+6k}}{(1+x^{1+6k})(1-x^{1+6k})(1-x^{3+6k})(1-x^{5+6k})} \\ &=\prod_{k=0}^\infty\frac{1-x^{3+6k}}{(1+x^{1+6k})(1-x^{1+2k})} \\ &=\prod_{k=0}^\infty\frac{(1-x^{3+6k})(1-x^{2k})}{(1+x^{1+6k})(1-x^{1+2k})(1-x^{2k})} \\ &=\prod_{k=0}^\infty\frac{(1-x^{3+6k})(1-x^{2k})}{(1+x^{1+6k})(1-x^k)} \\ &=\prod_{k=0}^\infty\frac{(1+x^k)(1-x^{3+6k})}{1+x^{1+6k}} \\ &=\prod_{k=0}^\infty\frac{(1+x^{6k})(1+x^{1+6k})(1+x^{2+6k})(1+x^{3+6k})(1+x^{4+6k})(1+x^{5+6k})(1-x^{3+6k})}{1+x^{1+6k}} \\ &=\prod_{k=0}^\infty(1+x^{6k})(1+x^{2+6k})(1+x^{3+6k})(1+x^{4+6k})(1+x^{5+6k})(1-x^{3+6k}) \\ &=\prod_{k=0}^\infty(1+x^{6k})(1+x^{2+6k})(1+x^{4+6k})(1+x^{5+6k})(1-x^{6+12k}) \\ &=\prod_{k=0}^\infty(1+x^{6k})(1+x^{2+6k})(1+x^{4+6k})(1+x^{5+6k})(1-x^{6+12k})\frac{1-x^{12k}}{1-x^{12k}} \\ &=\prod_{k=0}^\infty\frac{(1+x^{6k})(1+x^{2+6k})(1+x^{4+6k})(1+x^{5+6k})(1-x^{6k})}{1-x^{12k}} \\ &=\prod_{k=0}^\infty (1+x^{2+6k})(1+x^{4+6k})(1+x^{5+6k}) \\ &= \sum_{n=0}^\infty B(n)x^n \end{align}