TITLE: Partitions, $q$-polynomials and generating functions
QUESTION [6 upvotes]: Recall the integer partition function $P(n)$ with generating function
$$\sum_{n\geq0}P(n)x^n=\prod_{k=1}^{\infty}\frac1{1-x^k}.$$
Let $[n]_q=\frac{1-q^n}{1-q}$ denote the $q$-analogue of the integer $n$ and let $\lambda=(\lambda_1,\lambda_2,\dots)\vdash n$ be a partition of $n$. Now, define the function
$$\Psi_q(n)=\sum_{\lambda\vdash n}\,\,\sum_{j\geq1}\,\, [\lambda_j]_q.$$
For example, $\Psi_q(3)=[3]_q+([2]_q+[1]_q)+([1]_q+[1]_q+[1]_q)=q^2+2q+6$. In particular, when $q=1$, we obtain $\Psi_q(n)=nP(n)$ with generating function
$$\sum_{n\geq1}nP(n)x^n=x\frac{d}{dx}\prod_{k=1}^{\infty}\frac1{1-x^k}.$$
I would like to ask

Question 1. Is there a generating function for the sequence of $q$-polynomials $\Psi_q(n)$?
Question 2. What is the combinatorial interpretation of the coefficients of $\Psi_q(n)$?

For example, the constant term of $\Psi_q(n)$ enumerates the number of $1$'s in all partitions of $n$.

REPLY [3 votes]: A generating function of sorts is given by
  $$ \sum_{n\geq 1}\Psi_q(n)x^n = P(x)\sum_{m\geq 0}q^m\sum_{k\geq m+1}
       \frac{x^k}{1-x^k}, $$
where $P(x)=\prod_{i\geq 1}(1-x^i)^{-1}$.