Given an integer partition $\lambda=(\lambda_1,\dots,\lambda_{\ell(\lambda)})$ of $n$ where $\ell(\lambda)$ is the length of $\lambda$, associate its conjugate partition $\lambda'$. Denote by $\lambda''=\lambda',0$ found by appending one extra zero at the right end of $\lambda'$. Further, define the following two numerics $a(\lambda'')_j=\lambda_j''-\lambda_{j+1}''$ for $j=1,2,\dots,\ell(\lambda')$ and also that $b(\lambda'')=\#\{j: a(\lambda'')_j>0\}$.

For example, if $\lambda=(4,2,1)$ then $\lambda'=(3,2,1,1)$ and $\lambda''=(3,2,1,1,0)$ and $a(\lambda'')=(1,1,0,1)$ and $b(\lambda'')=3$.

QUESTION.If $n=2^m$ then are these two polynomials equal? $$\sum_{\lambda\vdash n}(q-1)^{2b(\lambda'')}q^{n-\ell(\lambda)} \prod_{a(\lambda'')_j\geq1}\frac{q^{2a(\lambda'')_j}-1}{q^2-1}=(q-1)(q^{2n-1}-1).\tag1$$

**Remark 1.** To get some motivation, consider dividing the left-hand side of (1) by $(q-1)^2$, for any $n\in\mathbb{N}$. Taking the limit $q\rightarrow1$ in the resulting expression forces $b(\lambda'')=1$ which means the corresponding Young diagram of the partition $\lambda'$ (hence $\lambda$ itself) must be rectangular. Therefore, the final expression equals the sum of divisors (arithmetic) function
$$\sum_{d\,\vert\, n}d.$$

**Remark 2.** I also observe that if $q\rightarrow-1$ in (1), then the left-hand side counts the number of ways of writing $n\in\mathbb{N}$ as a sum of two squares, which is this sequence $r_2(n)$.