Conjectural congruences for the $k$ -elongated plane partition function modulo powers of $5$

From papers

Let $d_k(n)$ denote the $k$ -elongated plane partition function, and let $c,n\geq 0$ . For a list of integers in an argument, interpret the corresponding congruence as holding for each listed integer. Conjectural congruences. The following congruences are conjectured:

\begin{aligned} d_{125c+58}(25n+16)&\equiv 0\pmod{125},\\ d_{125c+83}(125n+41)&\equiv 0\pmod{125},\\ d_{125c+83}(125n+91)&\equiv 0\pmod{125},\\ d_{125c+100}(125n+124)&\equiv 0\pmod{125},\\ d_{125c+5}(125n+69)&\equiv 0\pmod{125},\\ d_{125c+5}(125n+119)&\equiv 0\pmod{125},\\ d_{125c+30}(125n+69)&\equiv 0\pmod{125},\\ d_{125c+60}(125n+14)&\equiv 0\pmod{125},\\ d_{125c+60}(125n+64)&\equiv 0\pmod{125},\\ d_{125c+60}(125n+89)&\equiv 0\pmod{125},\\ d_{125c+60}(125n+114)&\equiv 0\pmod{125},\\ d_{125c+58}(125n+91)&\equiv 0\pmod{625},\\ d_{125c+58}(125n+66)&\equiv 0\pmod{3125},\\ d_{125c+58}(125n+116)&\equiv 0\pmod{3125}. \end{aligned}

These congruences extend the families proved earlier in the paper, but the source reports them only from numerical calculations and does not establish them theoretically.

Progress summary

Open

The conjecture remains unproved: a 2025 paper reports numerical evidence for several divisibility patterns but gives neither a proof nor a disproof.

The problem asks whether the listed congruence families for the $k$ -elongated plane partition function hold for all nonnegative $c$ and $n$ . They were presented as Conjecture $7.1$ in an April 2025 preprint, which says the list arose from numerical calculations and is not exhaustive.

Known results

The same preprint proves other congruence families modulo $5$ , $25$ , and $125$ , but explicitly separates them from Conjecture $7.1$ ; no cited result establishes the displayed families.

April 2025 numerical conjectures

The authors suggest that cases with $c=0$ might be approachable using Radu’s algorithm or RaduRK, followed by localization, but report no such proof. The available record contains no verified proof, disproof, correction, or claimed settlement of these specific congruences.

Current status (as of August 2026): The congruences remain conjectural, with numerical support but no publicly verified proof or disproof recorded.

Sources

Sources & referencesView supporting material

Primary source

Russelle Guadalupe, “The k-elongated plane partition function modulo small powers of 5”, arXiv:2504.08627 (2025).

1 week ago

The two claimed congruences modulo $3125$ are false for infinitely many values of $c$ . In fact,

\boxed{ \begin{aligned} d_{125c+58}(66)&\equiv625c(c-2)\pmod{3125},\\ d_{125c+58}(116)&\equiv-625c(c-2)\pmod{3125} \end{aligned}} \qquad(c\ge0).

Write

f_j(q)=\prod_{r\ge1}(1-q^{jr}), \qquad F_k(q)=\sum_{n\ge0}d_k(n)q^n =\frac{f_2(q)^k}{f_1(q)^{3k+1}},

and put

H(q)=\frac{f_2(q)}{f_1(q)^3}, \qquad U(q)=H(q)^{125}.

The freshman's-dream congruence gives

U(q)\equiv H(q^{125})\equiv1 \pmod{(5,q^{125})}.

Consequently, in

R=(\mathbb Z/3125\mathbb Z)[[q]]/(q^{125}),

we have

U-1\in5R, \qquad (U-1)^5=0.

Since

F_{125c+58}=F_{58}U^c,

the binomial theorem yields

F_{125c+58} \equiv F_{58}\sum_{j=0}^4\binom cj(U-1)^j \pmod{(3125,q^{125})}.

Thus every coefficient of degree less than $125$ , modulo $3125$ , is determined for all $c\ge0$ by its values at $c=0,1,2,3,4$ .

These values can be computed entirely with integers using

d_k(0)=1,

and the logarithmic-derivative recurrence

n\,d_k(n) = \sum_{j=1}^n \left( (3k+1)\sigma_1(j) - 2k\,\mathbf1_{2\mid j}\sigma_1(j/2) \right)d_k(n-j).

It gives

\begin{array}{c|rrrrr} c&0&1&2&3&4\\ \hline d_{125c+58}(66)&0&2500&0&1875&1875\\ d_{125c+58}(116)&0&625&0&1250&1250 \end{array} \qquad\pmod{3125}.

Newton interpolation in the binomial basis gives respectively

2500\binom c1+1250\binom c2 \equiv625c(c-2),

and

625\binom c1+1875\binom c2 \equiv-625c(c-2) \pmod{3125}.

In particular, taking $c=1$ and $n=0$ in the claimed $66$ -progression gives $k=183$ and

d_{183}(66) = 41811774646706534706935593169839977287497341782757382366770687360668596327065364582071240000 \equiv2500\not\equiv0\pmod{3125}.

Both claimed congruences fail for every

c\equiv1,3,4\pmod5.

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Shivam Patel · 23 hours ago