Generating function related to 2-residues of partitions Question
Express the following power series in two commuting variables $x,y$ as an infinite product, or a short sum of infinite products:
$$
\frac{P(xy)^2}{(1-x)}\sum_{k=-\infty}^\infty(2k+1)x^{k^2}y^{k^2+k}.
$$
Here $P(x)=\prod_{i=1}^\infty(1-x^i)^{-1}$ is the partition generating function.
Does it have any special properties e.g. automorphic form?
Motivation
The $2$-residue of an integer node $(n,m)$ in the plane is $m-n$ mod $2$. So the $2$-residues alternate as $0,1$ in a checkerboard pattern. The Young diagram $[\lambda]$ of a partition $\lambda$ is a set of nodes in the plane. An addable node of $\lambda$ does not belong to $[\lambda]$, but can be adjoined to give the Young diagram of a partition (of $|\lambda|+1$).
Now define, for $i=0,1$:
$c_i(\lambda)$ is the number of nodes in $[\lambda]$ with $2$-residue $i$.
$a_i(\lambda)$ is the number of addable nodes of $\lambda$ with $2$-residue $i$.
Then my power series is the generating function of
$$
\sum_\lambda a_0(\lambda)x^{c_0(\lambda)}y^{c_1(\lambda)}
$$
Here $\lambda$ ranges over all partitions. I'll leave it as an exercise to work out the corresponding identity for the other generating function $\sum_\lambda a_1(\lambda)x^{c_0(\lambda)}y^{c_1(\lambda)}$, using the first.
The coefficient of a given monomial $x^ay^b$ is the dimension of a certain algebra, naturally associated to the symmetric group $S_{a+b}$, defined in characteristic $2$.
Other Information
Using the Jacobi Triple Product identity I can factorize the generating function for the $2$-residues of partitions as
$$
\sum_{\lambda}x^{c_0(\lambda)}y^{c_1(\lambda)}
=P(xy)^2\sum\limits_{k=-\infty}^\infty x^{k^2}y^{k^2+k}
$$
$$
=\prod\limits_{i=1}^\infty\frac{(1+x^{2i-1}y^{2i})(1+x^{2i-1}y^{2i-2})(1+x^iy^i)}{(1-x^iy^i)}.
$$
Experts on the modular representation of the symmetric group will understand the significance of the left hand side and the first equality. The identity has a combinatorial proof that can be deduced from Cilanne E. Boulet, A four-parameter partition identity, arXiv:math/0308012v1
The generating function I'm interested in can be got (almost) using partial differentiation from this.
Also if we set $x=y$ in the original, standard results give:
$$
\sum_\lambda a_0(\lambda)x^{|\lambda|}=\frac{1}{2(1-x)} \frac{P(x)^4+P(x^2)^2}{P(x)^3}
$$
$$
\sum_\lambda a_1(\lambda)x^{|\lambda|}=\frac{1}{2(1-x)} \frac{P(x)^4-P(x^2)^2}{P(x)^3}
$$
 A: Let us do the following transform:
\begin{align}
\begin{cases}e^{\pi i \tau}=q=xy\\
e^{\pi i z}=u=\sqrt{y}.
\end{cases}
\end{align}
Then, we just need consider the function
\begin{align}
f(z;\tau)&=\frac{P(q)^2}{\left(1-q/u^2\right)^2}\frac{\partial}{\partial u}\left(u\sum_{k\in\mathbb{Z}}q^{k^2}u^{2k}\right)\\
&=\frac{P(q)^2}{\left(1-q/u^2\right)^2}\sum_{k\in\mathbb{Z}}q^{k^2}u^{2k}+\frac{uP(q)^2}{\left(1-q/u^2\right)^2}\frac{\partial}{\partial u}\left(\sum_{k\in\mathbb{Z}}q^{k^2}u^{2k}\right).
\end{align}
Let
\begin{align}
\vartheta(z;\tau):=\sum_{k\in\mathbb{Z}}q^{k^2}u^{2k}=\prod_{n\ge 1}(1-q^{2n})(1+u^{-2}q^{2n-1})(1+u^2q^{2n-1})
\end{align}
be the Jacobi Theta function, then it is easy seen that
\begin{align}
f(z;\tau)=\frac{\vartheta(z;\tau)P(q)^2}{\left(1-q/u^2\right)^2}\left(1+2\sum_{n\ge 1}\left(\frac{1}{1+u^{-2}q^{2n-1}}-\frac{1}{1+u^{2}q^{2n-1}}\right)\right).
\end{align}
Namely,
\begin{align}
f(z;\tau)=\frac{\vartheta(z;\tau)\eta(\tau)^2e^{\frac{\pi i\tau}{12}}}{\left(1-e^{\pi i \tau}e^{-4\pi i z}\right)^2}\left(1+2\sum_{n\ge 1}\left(\frac{1}{1+e^{-4\pi i z}e^{(2n-1)\pi i \tau}}-\frac{1}{1+e^{4\pi i z}e^{(2n-1)\pi i \tau}}\right)\right),
\end{align}
where $\eta(\tau)$ is the Dedekind eta function.
