Let p_0(n) denote the number of partitions counted by the paper's function p_0. Let ell>3 be a prime number with ell notequiv23 pmod24, and let delta_ell be the least positive integer satisfying 24 delta_ell equiv1 pmodell. Mod 4 congruence conjecture. For every n≥0 and every integer j with 0≤ j≤ell-2, p_0 big(2 ell^2n+2 ell j+2 delta_ell big)equiv0 pmod4. This conjecture predicts infinite arithmetic families of divisibility-by-four congruences for the partition function p_0 and motivates further investigation of its arithmetic properties.

A mod 4 congruence for partitions with parts separated by parity

At least 2 years old · documented by 2023

Let $p_0(n)$ denote the number of partitions counted by the paper's function $p_0$ . Let $\ell>3$ be a prime number with $\ell\not\equiv23\pmod{24}$ , and let $\delta_\ell$ be the least positive integer satisfying

24\delta_\ell\equiv1\pmod{\ell}.

Mod 4 congruence conjecture. For every $n\geq0$ and every integer $j$ with $0\leq j\leq\ell-2$ ,

p_0\big(2\ell^2n+2\ell j+2\delta_\ell\big)\equiv0\pmod{4}.

This conjecture predicts infinite arithmetic families of divisibility-by-four congruences for the partition function $p_0$ and motivates further investigation of its arithmetic properties.

References

Primary source

Shishuo Fu and Dazhao Tang, “Partitions with parts separated by parity: conjugation, congruences and the mock theta functions”, arXiv:2306.13309 (2023).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader-written attempt claims the conjecture is false, giving an explicit counterexample and an infinite family, but neither has been independently verified.

Fu and Tang (2023) state this divisibility assertion as Conjecture 5.1 for their partition function $p_0$ . The paper presents it as motivation for further investigation, not as a proved result.

Known results

The paper proves $p_0(2m)\equiv p_\psi(m)\pmod{2}$ .
It records that $p_\psi(m)$ is odd exactly when $24m-1=P^{4a+1}s^2$ for a prime $P$ with $P\nmid s$ .

Posted attempt

The attempt claims that for every admissible prime $\ell\ge29$ , choosing $j=(23\ell-r)/24$ with $r=(24\delta_\ell-1)/\ell$ yields infinitely many odd conjectured values, using primes $24n+23$ . It gives $p_0(1612)=132350955536194689031169297511\equiv3\pmod4$ for $\ell=29$ . This is a claimed complete disproof, but it has not been independently verified.

Current status (as of August 2026): The conjecture has an unverified claimed disproof, while no independently verified proof or counterexample is recorded.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

The conjecture fails for every admissible prime ℓ≥29, and for each such prime it fails for infinitely many n. In fact the relevant partition values are odd, so the proposed congruence fails even modulo 2.

Two results already proved in the source are decisive:

(1) Equation (4.2) states that p_ψ(m) is odd if and only if

24m−1=P^{4a+1}s²

for a prime P with P∤s.

(2) Remark 4.5 gives

p₀(2m)≡p_ψ(m) (mod 2).

Now fix any prime ℓ≥29 with ℓ≢23 (mod 24). Let δ be the least positive integer satisfying 24δ≡1 (mod ℓ), and define

r=(24δ−1)/ℓ, j=(23ℓ−r)/24.

Here 1≤r≤23. Since ℓ²≡1 (mod 24) and ℓr≡−1 (mod 24), we have r≡−ℓ (mod 24); hence j is an integer. Moreover j≥0. The same congruence shows that ℓ+r is a positive multiple of 24; because ℓ≥29, necessarily ℓ+r≥48. Therefore

j≤ℓ−2,

so j lies in precisely the range required by the conjecture.

For every n≥0 put

m=ℓ²n+ℓj+δ.

Direct substitution gives

24m−1 =24ℓ²n+24ℓj+24δ−1 =ℓ²(24n+23).

Whenever P=24n+23 is prime, we have P≠ℓ because ℓ≢23 (mod 24). Consequently

24m−1=Pℓ²

has the form in (1), with a=0 and s=ℓ. It follows that

p₀(2ℓ²n+2ℓj+2δ) ≡p_ψ(m) ≡1 (mod 2).

Dirichlet's theorem supplies infinitely many primes P≡23 (mod 24), hence infinitely many such n for every admissible ℓ≥29.

For a concrete exact instance, take

ℓ=29, δ=23, r=19, j=27, n=0.

Then the conjectured argument is

2·29·27+2·23=1612,

and the source's defining generating function

∑{v≥0}p₀(v)q^v =∑{h≥1}q^{2h²}∏_{a=1}^h(1−q^{2a−1})^{−2}

gives the exact coefficient

p₀(1612)=132350955536194689031169297511≡3 (mod 4).

Thus the published mod-4 conjecture is false both at this explicit first example and throughout the infinite all-prime family above.