Generalized Andrews mod 8 identity for indexed partitions

The conjecture holds in the stronger setting of an arbitrary prescribed set of positions, not merely a union of residue classes.

Let $T\subseteq\mathbb N_{>0}$, and let $\mathcal P_T$ consist of partitions $\lambda$ with distinct odd parts for which $\lambda_j$ is even whenever $j\in T$. For independent variables $x_1,x_2,\ldots$, put

$$x^\lambda=\prod_{j\ge1}x_j^{\lambda_j}, \qquad X_j=x_1\cdots x_j,\qquad X_0=1.$$

Then

$$\boxed{\displaystyle \sum_{\lambda\in\mathcal P_T}x^\lambda = \prod_{j\ge1}\frac1{1-X_j^2} \prod_{\substack{j\ge1\j\notin T}} (1+X_{j-1}X_j).}$$

To prove this, extend every partition by trailing zeros and write

$$\lambda_j=2\mu_j+\varepsilon_j, \qquad \varepsilon_j\in\{0,1\}.$$

Weak decrease together with distinctness of odd parts is equivalent to

$$\mu_j-\mu_{j+1}\ge\varepsilon_{j+1} \qquad(j\ge1). \tag{1}$$

Indeed, if $\varepsilon_{j+1}=0$, weak decrease is equivalent to $\mu_j\ge\mu_{j+1}$. If $\varepsilon_{j+1}=1$, either weak decrease against an even preceding part or strict decrease between distinct odd parts forces $\mu_j\ge\mu_{j+1}+1$. Conversely, (1) gives weak decrease and excludes repeated odd parts.

Define

$$e_j=\mu_j-\mu_{j+1}-\varepsilon_{j+1}.$$

Then the $e_j$ are arbitrary independent nonnegative integers, while the $\varepsilon_j$ are arbitrary independent bits subject only to

$$\varepsilon_j=0\qquad(j\in T).$$

The inverse bijection is explicit:

$$\mu_j=\sum_{r\ge j}(e_r+\varepsilon_{r+1}), \qquad \lambda_j=2\mu_j+\varepsilon_j.$$

Its weight factors as

$$x^\lambda = \prod_{r\ge1}X_r^{2e_r} \prod_{j\ge1}(X_{j-1}X_j)^{\varepsilon_j}.$$

Summing the independent nonnegative-integer and binary choices proves the boxed identity. One can first impose a finite length cutoff and then pass coefficientwise to the limit, making the formal-series argument exact.

Finally, choose

$$T=\{j\ge1:j\equiv t_1,\ldots,t_i\pmod k\}.$$

This gives precisely the conjectured product for every admissible modulus and every prescribed set of residue classes, including residue zero. The specialization $x_j=q$ also yields

$$\sum_{\lambda\in\mathcal P_T}q^{|\lambda|} = \frac{\prod_{j\notin T}(1+q^{2j-1})} {\prod_{j\ge1}(1-q^{2j})}.$$

Source: R. Li, arXiv:2510.23233, Conjecture 8.1, equation (8.1).