Parity of Partition Function

Question

Let $T(n)$ denote the number of partitions of $n$ into parts not congruent to $3$ mod $6$. Deduce that $T(n)$ is also the number of partitions of $n $ in which odd parts appear at most twice (even parts appear without restriction). Can you give a combinatorial proof that $T(n)$ is even whenever $n$ is congruent to $2$ or $3$ mod $4$?

Answer 1

For the first part, note that the generating function for partitions with parts not congruent to three mod 6 is $$\prod_{k\ge 1} \frac{1}{1-z^k} \prod_{k \ge 1} (1-z^{6k-3}) = \prod_{k\ge 1} \frac{1}{1-z^{2k}} \prod_{k\ge 1} \frac{1}{1-z^{2k-1}} \prod_{k \ge 1} (1-z^{6k-3}) \\ = \prod_{k\ge 1} \frac{1}{1-z^{2k}} \prod_{k\ge 1} \frac{1-z^{6k-3}}{1-z^{2k-1}} = \prod_{k\ge 1} \frac{1}{1-z^{2k}} \prod_{k\ge 1} \left(1+z^{2k-1}+z^{4k-2}\right).$$ This concludes the first part of the exercise.