Number of Partitions of $n$ No Part Appears Exactly Once Equals to $n$ Partitioned into 0,2 ,3 or 4 (Mod 6 ) I want to prove that the number of partitions of $n$ in which no part appears exactly once is equal to the numbers of partitions of $n$ into parts that are congruent to one of $0,2,3,$ or $4$ mod $6$.

My approach
Intuitively, no part appearing not just once dually requires each part appears $3,5,7... <n$ number of times which means odd times just without 1.
Now I want to develop bijection between the first combinatorics and the second combinatorics. However, how could I make a function that relates each $3,5,7...<n $, which the number of cases depends on the size of $n$ into only 4 basic cases which rendered by $6$?
I think I am pretty bit lost. Any guidance to proceed further?
 A: The generating function for partitions with no part appearing once is
$$\prod_{n=1}^\infty(1+q^{2n}+q^{3n}+q^{4n}+\cdots)
=\prod_{n=1}^\infty\frac{1+q^{3n}}{1-q^{2n}}.$$
The generating function for partitions into parts
congruent to $0$, $2$, $3$ or $4$ modulo $6$ is
$$\prod_{k=1}^\infty\frac1{(1-q^{6k-4})(1-q^{6k-3})(1-q^{6k-2})(1-q^{6k})}
=\frac{\prod_{k=1}^\infty(1-q^{6k-5})
(1-q^{6k-1})}{\prod_{m=1}^\infty(1-q^m)}.$$
Can you convert one of these to the other?
A: Let
\begin{align}
&A(n):= \text{no. of partitions of $n$ in which no part appears exactly once,}
\\&B(n):= \text{no. of partitions of $n$ into parts that are congruent to one of 0, 2, 3, or 4 mod 6.}
\end{align}
Now, for some indeterminate $q$, we have
\begin{align}
\sum_{n=0}^\infty A(n)q^n&=(1+q^2+q^3+\cdots)(1+q^4+q^6+\cdots)\cdots(1+q^{2k}+q^{3k}+\cdots)\cdots
\\&=\frac{(1-q^2)(1+q^2+q^3+\cdots)(1-q^4)(1+q^4+q^6+\cdots)\cdots(1-q^{2k})(1+q^{2k}+q^{3k}+\cdots)\cdots}{(1-q^2)(1-q^4)\cdots(1-q^{2k})\cdots}
\\&=\frac{(1+q^3)(1+q^6)\cdots(1+q^{3k})\cdots}{(1-q^2)(1-q^4)\cdots(1-q^{2k})\cdots}
\\&=\frac{(1-q^3)(1+q^3)(1-q^6)(1+q^6)\cdots(1-q^{3k})(1+q^{3k})\cdots}{((1-q^2)(1-q^4)\cdots(1-q^{2k})\cdots)((1-q^3)(1-q^6)\cdots(1-q^{3k})\cdots)}
\\&=\frac{1}{(1-q^2)(1-q^3)(1-q^4)(1-q^6)\cdots(1-q^{6k-4})(1-q^{6k-3})(1-q^{6k-2})(1-q^{6k})\cdots}
\\&=\sum_{n=0}^\infty B(n)q^n
\end{align}
Comparing coefficients on both sides, we obtain $$A(n)=B(n).$$
