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

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.

References

Primary source

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

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A 2025 paper proposed these divisibility claims from computation, but a later posted calculation claims that two of them fail; that disproof has not been independently verified.

Guadalupe’s 2025 preprint states these families as Conjecture 7.1, based on numerical calculations rather than proof. The conjecture includes the two claimed congruences modulo $3125$ for $d_{125c+58}(125n+66)$ and $d_{125c+58}(125n+116)$ .

Known results

Guadalupe (2025) proves other infinite congruence families for $d_k(n)$ modulo $5$ , $25$ , and $125$ , but not the displayed conjectural families.

Posted attempt

An unverified calculation claims both $3125$ congruences are false for infinitely many $c$ , giving $d_{125c+58}(66)\equiv625c(c-2)\pmod{3125}$ and $d_{125c+58}(116)\equiv-625c(c-2)\pmod{3125}$ . In particular, it reports $d_{183}(66)\equiv2500\not\equiv0\pmod{3125}$ at $c=1,n=0$ . The attempt has not been independently verified.

Current status (as of August 2026): The original families have no verified proof, and a reader-written calculation claims a disproof of the two modulo $3125$ families; independent confirmation or refutation is absent.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

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.