Let $\mathcal{OD}$ be the set of all odd and distinct integer partitions. This has a generating function given by $$\sum_{\lambda\vdash\mathcal{OD}}q^{\vert\lambda\vert}=\prod_{j\geq1}(1+q^{2j-1})$$ where $\vert\lambda\vert$ stand for the sum of the parts $\lambda_1+\lambda_2+\cdots$ of $\lambda$.
Define the sequence of number $s(\lambda):=\#\{i: \lambda_i-\lambda_{i+1}>2\}$.
Is there a formula as a bivariate generating function for the below sum? $$\sum_{\lambda\in\mathcal{OD}} x^{s(\lambda)}q^{\vert\lambda\vert}=$$
Supposing that the partition has $m$ elements we have to select $m-1$ gaps between consecutive values (even ones). The first gap contributes $m-1$ times to the sum, the second one $m-2$ times and gap number $m-1$ contributes one time. The initial value contributes $m$ times. This yields
$$\sum_{m\ge 1} (q^m + q^{3m} + q^{5m}+ \cdots) \prod_{p=1}^{m-1} (q^{2(m-p)} + x q^{4(m-p)} + x q^{6(m-p)} + \cdots) \\ = \sum_{m\ge 1} q^m (1 + q^{2m} + q^{4m} + \cdots) \prod_{p=1}^{m-1} q^{2(m-p)} (1 + x q^{2(m-p)} + x q^{4(m-p)} + \cdots) \\ = \sum_{m\ge 1} q^m \frac{1}{1-q^{2m}} \prod_{p=1}^{m-1} q^{2(m-p)} \left(1 + x \frac{q^{2(m-p)}}{1-q^{2(m-p)}}\right) \\ = \sum_{m\ge 1} \frac{q^m}{1-q^{2m}} \prod_{p=1}^{m-1} q^{2p} \left(1 + x \frac{q^{2p}}{1-q^{2p}}\right).$$
This is
$$\bbox[5px,border:2px solid #00A000]{ \sum_{m\ge 1} \frac{q^{m^2}}{1-q^{2m}} \prod_{p=1}^{m-1} \left(1 + x \frac{q^{2p}}{1-q^{2p}}\right).}$$