Thejitha–Fathima overcolored partition congruence conjecture

Less than 1 year old · traced to 2026

Let an overcolored partition of $n$ be a partition in which even parts may appear in one of $r$ colors and odd parts may appear in one of $s$ colors, with the first occurrence of each part optionally overlined. Let $\bar{a}_{r,s}(n)$ denote the number of such partitions, where $k\geq 1$ and $i,j\geq 0$ . Thejitha–Fathima congruence conjecture. For all $n\geq 0$ ,

\bar{a}_{2^{k+1}j+2^k-1,2i+1}(3n+2)\equiv 0 \pmod{2^{k+1}},

\bar{a}_{2^{k+1}j+2^k-1,2i+1}(9n+3)\equiv 0 \pmod{2^{k+2}},

\bar{a}_{2^{k+1}j+2^k-1,2i+1}(9n+6)\equiv 0 \pmod{2^{k+2}}.

These congruences are Ramanujan-type divisibility results for overcolored partitions and were proposed by Thejitha and Fathima as families of congruences modulo powers of $2$ . The present paper states that it provides an elementary proof using classical $q$ -series manipulations and properties of Ramanujan's theta function; the supplied text does not specify whether that proof establishes all three displayed congruences, so the database status remains open.

References

Primary source

Imdadul Hussain, Suparno Ghoshal and Arijit Jana, “Proof of a Conjecture on Overcolored Partition Restricted by Parity of the Parts”, arXiv:2603.12401 (2026).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A March 2026 preprint claims a proof, but a posted calculation alleges the unrestricted statement is false, so neither the proof nor the alleged counterexample is independently verified.

Thejitha and Fathima proposed three divisibility families for the overcolored partition function with odd-color parameter $s=2i+1$ . The March 2026 proof paper claims all three families hold, but its displayed argument uses the narrower specialization $s=2^k i+1$ .

March 2026 claimed proof

Hussain, Ghoshal, and Jana state that the conjecture is true and give an elementary $q$ -series and theta-function argument for the three congruences. The parameter mismatch means the proof does not visibly establish the stated family with every odd $s=2i+1$ ; no independent verification or correction was found.

Posted attempt

A reader derives $\bar a_{r,s}(2)=2(r+s^2)$ and claims a counterexample at $(k,j,i,n)=(4,0,1,0)$ , with $(r,s)=(15,3)$ , contradicting the first congruence modulo $32$ . This is an unverified posted attempt, not independent evidence.

Current status (as of August 2026): A preprint claims to prove all three congruence families, but the unrestricted case $s=2i+1$ remains unverified and is challenged by an unverified counterexample.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

The first claimed congruence fails for infinitely many admissible parameter choices, despite the paper's claim that all three congruences were proved.

Write

f_d(q)=\prod_{m\ge1}(1-q^{dm}).

The generating function is

\sum_{n\ge0}\overline a_{r,s}(n)q^n = \frac{f_2(q)^{3s-2r}} {f_1(q)^{2s}f_4(q)^{s-r}}.

Modulo $q^3$ ,

\begin{aligned} f_1(q)^{-2s} &=(1-q)^{-2s}(1-q^2)^{-2s}+O(q^3)\\ &=1+2sq+(2s^2+3s)q^2+O(q^3),\\ f_2(q)^{3s-2r} &=1+(2r-3s)q^2+O(q^4),\\ f_4(q)^{r-s}&=1+O(q^4). \end{aligned}

Therefore, for every $r,s\ge1$ ,

\boxed{\overline a_{r,s}(2)=2(r+s^2).}

Now take any $k\ge4$ and $j\ge0$ , and choose $i=1$ and $n=0$ in the conjecture. Its parameters become

r=2^{k+1}j+2^k-1, \qquad s=2i+1=3.

Thus

\overline a_{r,3}(3n+2) = \overline a_{r,3}(2) = 2^{k+2}j+2^{k+1}+16 \equiv16\not\equiv0\pmod{2^{k+1}}.

The smallest instance is

(k,j,i,n)=(4,0,1,0), \qquad (r,s)=(15,3),

for which

\overline a_{15,3}(2)=48\equiv16\not\equiv0\pmod{32}.

More generally, the initial instance of the first asserted congruence holds exactly when

2^k\mid s^2-1.

The discrepancy in the claimed proof is that the conjecture and theorem allow every odd $s=2i+1$ , whereas their proof immediately replaces this by the narrower assumption $s=2^ki+1$ . The narrower original conjecture is unaffected, but the stated unrestricted theorem and congruence are false.