Let a_m(n) denote the generalized cubic partition function indexed by m, as in the paper. For every non-negative integer n, the following congruences are conjectured, with r ranging over the indicated residue sets: Further generalized cubic partition congruence conjecture. a_25(31^2n+r)equiv 0 pmod31^2quad r∈41,81,585, a_37(43^2n+r)equiv 0 pmod43^2quad r∈716,987, a_53(59^2n+100)equiv 0 pmod59^2, a_65(71^2n+r)equiv 0 pmod71^2quad r∈41,47, a_77(83^2n+157)equiv 0 pmod83^2. The source states that these congruences cannot be proved by the same procedure used for the preceding theorem. It also notes that elementary proofs for these conjectures would be desirable.

Further generalized cubic partition congruences modulo squared primes

About 1 year old · traced to 2025

Let $a_m(n)$ denote the generalized cubic partition function indexed by $m$ , as in the paper. For every non-negative integer $n$ , the following congruences are conjectured, with $r$ ranging over the indicated residue sets:

Further generalized cubic partition congruence conjecture.

a_{25}(31^2n+r)\equiv 0 \pmod{31^2}\quad r\in\{41,81,585\},

a_{37}(43^2n+r)\equiv 0 \pmod{43^2}\quad r\in\{716,987\},

a_{53}(59^2n+100)\equiv 0 \pmod{59^2},

a_{65}(71^2n+r)\equiv 0 \pmod{71^2}\quad r\in\{41,47\},

a_{77}(83^2n+157)\equiv 0 \pmod{83^2}.

The source states that these congruences cannot be proved by the same procedure used for the preceding theorem. It also notes that elementary proofs for these conjectures would be desirable.

References

Primary source

Hirakjyoti Das, Saikat Maity and Manjil P. Saikia, “Arithmetic Properties of Generalized Cubic and Overcubic Partitions”, arXiv:2503.19399 (2026).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed progress

A reader claims a direct calculation disproves one of the five conjectured divisibility patterns, but the calculation has not been independently verified.

Das, Maity, and Saikia (2025) recorded five congruence families modulo squared primes as Conjecture 7.2. They state that their method for the preceding theorem does not apply and that elementary proofs would be desirable.

Posted attempt

A reader claims the first family fails already at $n=1$ : the displayed calculation gives $a_{25}(1002)\equiv87\pmod{961}$ , although $1002=31^2+41$ . This would disprove that residue class, hence constitute a partial disproof of the conjecture; the attempt has not been independently verified.

Current status (as of August 2026): the published source leaves all five families unproved, while one reader-written calculation claims a counterexample to the first family and awaits verification.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

Write

f_d(q)=\prod_{r\ge1}(1-q^{dr}).

The generalized cubic partition function is defined by

A_c(q)=\sum_{n\ge0}a_c(n)q^n =\frac1{f_1(q)f_2(q)^{c-1}}.

Logarithmic differentiation gives

\frac{qA_c'(q)}{A_c(q)} = \sum_{j\ge1} \left( \sigma(j)+2(c-1)\mathbf1_{2\mid j}\sigma(j/2) \right)q^j, \qquad \sigma(j)=\sum_{d\mid j}d.

Consequently, every coefficient can be obtained from the exact integer recurrence

a_c(0)=1,\qquad n\,a_c(n) = \sum_{j=1}^{n} \left( \sigma(j)+2(c-1)\mathbf1_{2\mid j}\sigma(j/2) \right)a_c(n-j).

For $c=25$ , this gives

a_{25}(41)=793193782332525\equiv0\pmod{961},

but

a_{25}(1002) = 407239135983429585593226172302712545925477100479625790566986356836123195908257686174012648188411513945732551 \equiv87\pmod{961}.

Since

1002=31^2\cdot1+41,

and $41$ is one of the explicitly conjectured residues, the asserted congruence

a_{25}(31^2n+41)\equiv0\pmod{31^2}

already fails at $n=1$ . In fact, the coefficient is $25\pmod{31}$ , so even divisibility by $31$ fails.