For integers k>i≥ 0, call a part lambda_m in a partition (k,i)-indexed if m equiv i pmodk. Let P_(t_1,t_2,…,t_i;k) be the set of partitions such that, for every 1≤ j≤ i, the (t_j,k)-indexed parts are even and the odd parts are distinct. Writing lambda=lambda_1+lambda_2+… and letting X_n=x^n, the proposed identity is Generalized Andrews identity. ∑_lambda∈P_(t_1,t_2,…,t_i;k)x^lambda=∏_n=1^∞frac11-X_n^2∏_n notequiv t_1,…,t_i pmodk(1+X_n-1X_n). This is presented as a proposed generalization of Andrews' mod 8 identity and its companions, motivated by the computation in the paper; the general identity is left as a problem for further study.

Generalized Andrews mod 8 identity for indexed partitions

About 1 year old · traced to 2025

For integers $k>i\geq 0$ , call a part $\lambda_m$ in a partition $(k,i)$ -indexed if $m\equiv i\pmod{k}$ . Let $\mathcal{P}_{(t_1,t_2,\ldots,t_i;k)}$ be the set of partitions such that, for every $1\leq j\leq i$ , the $(t_j,k)$ -indexed parts are even and the odd parts are distinct. Writing $\lambda=\lambda_1+\lambda_2+\cdots$ and letting $X_n=x^n$ , the proposed identity is

Generalized Andrews identity.

\sum_{\lambda\in\mathcal{P}_{(t_1,t_2,\ldots,t_i;k)}}x^{\lambda}=\prod_{n=1}^{\infty}\frac{1}{1-X_n^2}\prod_{n\not\equiv t_1,\ldots,t_i\pmod{k}}(1+X_{n-1}X_n).

This is presented as a proposed generalization of Andrews' mod $8$ identity and its companions, motivated by the computation in the paper; the general identity is left as a problem for further study.

References

Primary source

Runqiao Li, “Partition analysis and the little Göllnitz identites”, arXiv:2510.23233 (2025).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader-posted argument claims a complete stronger proof for arbitrary allowed positions, but no independent verification has appeared.

Runqiao Li's 2025 preprint formulates the identity as Conjecture $8.1$ , extending Andrews-type partition identities modulo $8$ . The preprint leaves the general case as a problem for further study.

Posted attempt

An unverified reader-posted argument claims a complete proof for every prescribed set $T$ of indexed positions, using a bijection with independent nonnegative variables and binary choices. Specializing $T$ to the relevant residue classes would imply the proposed identity, and even a multivariate strengthening; this attempt has not been independently verified.

Current status (as of August 2026): The identity remains unverified; a complete stronger proof has been claimed, but no independent confirmation is recorded.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

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).