Background:
While trying to answer this question, I came up with a question of my own.
Let $|a|,|b|,|q|<1$ and $$A(a,b;q)=\sum_{n\ge1}\frac{(aq)^n}{1-bq^n}.$$ One can show that $$A(a,b;q)=\sum_{n\ge1}\phi_n(a,b)q^n,$$ where $$\phi_n(a,b)=\sum_{d|n}a^db^{n/d-1}.$$ Preforming the index shift $j=n/d$, we can write $$\phi_n(a,b)=\sum_{j|n}a^{n/j}b^{j-1}=\frac{a}{b}\phi_n(b,a).$$ Thus $$A(a,b;q)=\frac{a}{b}A(b,a;q).\tag1$$ Then let $$B(a,b;q)=\sum_{n\ge1}\frac{(aq)^n}{(1-bq^n)^2},$$ and we can see that $$B(a,b;q)=\frac{\partial}{\partial b}A(a/q,b;q).$$ Then let $\sigma_n(a,b)=[q^n]B(a,b;q)$.
Do these coefficients satisfy a similar symmetry under $(a,b)\mapsto (b,a)$ that $\phi_n$ does? In other words, is there a 'nice' relationship between $B(a,b;q)$ and $B(b,a;q)$?
My attempts are below.
I tried setting $M(a,b;q)=A(a/q,b;q)$, and I noticed that $$\begin{align} M(a,b;q)&=\frac{a}{bq}A(b,a/q;q)\\ &=a\sum_{n\ge1}\frac{(bq)^{n-1}}{1-aq^{n-1}}\\ &=\frac{a}{1-a}+a\sum_{n\ge1}\frac{(bq)^{n}}{1-aq^{n}}\\ &=\frac{a}{1-a}+bA(a,b;q). \end{align}$$ Thus $$\begin{align} B(a,b;q)&=\partial_b bA(a,b;q)\\ &=A(a,b;q)+b\partial_b A(a,b;q)\\ &=\sum_{n\ge1}q^n\phi_n(a,b)+\sum_{n\ge1}q^nb\partial_b \phi_n(a,b)\\ &=\sum_{n\ge1}q^n\left\{\sum_{d|n}a^db^{n/d-1}+b\sum_{d|n}\left(\tfrac{n}{d}-1\right)a^db^{n/d-2}\right\}\\ &=\sum_{n\ge1}q^n\sum_{d|n}\tfrac{n}{d}a^db^{n/d-1}. \end{align}$$ Thus $$\sigma_n(a,b)=\sum_{d|n}\tfrac{n}{d}a^db^{n/d-1},$$ which we can write as $$\sigma_n(a,b)=\frac{a}{b}\sum_{d|n}db^da^{n/d-1},$$ which looks a lot like $\frac{a}{b}\sigma_n(b,a)$, but this is not so.
At this point I am at a loss for a formula that relates $\sigma_n(a,b)$ and $\sigma_n(b,a)$, so could I have some help? Thanks!
Edit: I took the liberty of posting this question on Math Overflow here.