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

From papers

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.

Progress summary

Open

The five proposed divisibility patterns remain unproved, and no verified proof or disproof was found.

The problem asks whether five families of congruences for generalized cubic partition numbers hold for every nonnegative integer $n$ . Das, Maity, and Saikia recorded them as Conjecture 7.2 in 2025.

March 2025 statement

The authors state that the method used for the preceding theorem cannot establish these congruences and note that elementary proofs would be desirable. The paper gives no proof, disproof, or resolution of Conjecture 7.2; a separate proof announcement mentioned there concerns a different conjecture.

Current status (as of August 2026): all five squared-prime congruence families remain conjectural, with no verified proof or disproof recorded.

Sources

Sources & referencesView supporting material

Primary source

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

1 week ago

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.

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Shivam Patel · 22 hours ago