Congruence modulo 2 for q-series

Question

This quest arose from certain calculations with integer partitions (having distinct parts) and the corresponding values of their Dyson ranks.

I would like to ask:

QUESTION. Is this congruence true modulo $2$? $$\sum_{n\geq0}\frac{q^{\binom{n+1}2}}{\prod_{j=1}^n(1-q^j)}\sum_{m=1}^n\frac{q^m}{1-q^m}\equiv \sum_{n\geq0}q^{3n+2}\prod_{j=1}^n(1-q^{4j-1}).$$

Clarification. As Richard Stanley commented on this, I meant to say that after expanding the two $q$-series, the coefficients agree term-by-term modulo $2$.

Answer 1

Just for the record, this problem is now solved due to the answer given by Gjergji Zaimi to my follow up question here. Proceed as follows:

The LHS is computed modulo $2$ by \begin{align}\sum_{n\geq1}\frac{q^{\binom{n+1}2}}{(q;q)_n}\sum_{m=1}^n\frac{q^m}{1-q^m} &=\sum_{n\geq1}\frac{q^{\frac{n(3n+1)}2}(1+q^{2n+1})(-q;q)_n}{(q;q)_n} \sum_{j=1}^n\frac{1+q^{2j}}{1-q^{2j}} \\ &=\sum_{n\geq1}q^{\frac{n(3n+1)}2}(1+q^{2n+1}). \tag2 \end{align} The RHS is computed with the help of well-known identity (replace $x\rightarrow q^{-1}, q\rightarrow q^4$) $$\sum_{k\geq0}x^{k+2}q^{k+1}\prod_{j=1}^k(1-xq^j)=\sum_{n\geq0}(-1)^n\left(x^{3n+2}q^{\frac{(n+1)(3n+2)}2}+x^{3n+3}q^{\frac{(n+1)(3n+4)}2}\right)$$ so that $$\sum_{n\geq0}q^{3n+2}\prod_{j=1}^n(1-q^{4k-1}) \equiv \sum_{n\geq0} \left(q^{-3n-2}q^{2(n+1)(3n+2)}+q^{-3n-3}q^{2(n+1)(3n+4)}\right). \tag3$$ Now, just check (2) and (3) agree.