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

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 $2$.