Let $\nu_p(n)$ denote the $p$-adic valuation of $n$, i.e. the highest power of $p$ dividing $n$.

Consider the following two $q$-series formed by infinite products
$$\prod_{n\geq1}\left(\frac{1+q^n}{1-q^n}\right)^2=\sum_{k\geq0}a_k\,q^k \qquad \text{and} \qquad
\prod_{n\geq1}\left(\frac{1+q^n}{1-q^n}\right)^n=\sum_{k\geq0}b_k\,q^k.$$
Both $a_k$ and $b_k$ have combinatorial interpretations in the context of partitions. One reference for $b_k$ would be: Corteel, S., Savelief, C., Vuletić, M.: Plane overpartitions and cylindric partitions. J. Combin. Theory Ser. A 118(4), 1239–1269 (2011). Some references for $a_k$ include: Jeremy Lovejoy, Overpartition pairs, Annales de l'institut Fourier, vol.56, no.3, p.781-794, 2006.

I would like to ask:

>**QUESTION.** Is this true? If $k=j^2\geq1$ is a perfect square, then we have $\nu_2(a_k)=2=\nu_2(2b_k)$.

**ADDED.** I thought it might be proper to record the following extension: if 
$$\prod_{n\geq1}\left(\frac{1+q^n}{1-q^n}\right)^r=\sum_{k\geq0}a_k^{(r)}q^k$$
and (once again) $k=j^2$, then $\nu_2(a_k^{(r)})=\nu_2(2r)$ independent of $j$.