Density conjecture for even values of the partition function p_0

From papers

Let p0(m)p_0(m) denote the partition function introduced in the paper, and for n1n\geq1 define the counting set

{mp0(m)0(mod2),  0m<n}.\{m\mid p_0(m)\equiv0\pmod{2},\;0\leq m<n\}.

Density conjecture. The natural density of indices mm for which p0(m)p_0(m) is even exists and equals

limn#{mp0(m)0(mod2),  0m<n}n=15.\lim_{n\to\infty}\frac{\#\{m\mid p_0(m)\equiv0\pmod{2},\;0\leq m<n\}}{n}=\frac15.

This predicts a precise parity distribution for the values of p0p_0, but the supplied text gives no evidence of a resolution.

Progress summary

Open

No publicly verified progress on this density conjecture was found.

No public discussion or published progress addressing this conjecture was found.

Current status (as of August 2026): The conjecture appears open, with no recorded public activity establishing or refuting it.

Sources & referencesView supporting material

Primary source

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

Additional references

2 papers in this index state this conjecture (2014–2023). The statement above is taken from the most recent of them; the others are arXiv:1409.2909.

Solutions 1

Counterexample

The conjectured density 1/5 is false. In fact the exact corrected asymptotic is

#{1≤n≤X : p₀(n) is odd} ∼(π²/6)X/log X,

so

#{1≤n≤X : p₀(n) is even} =X−(π²/6+o(1))X/log X.

In particular, the natural density of even values is 1.

The primary source's Corollary 4.4, equation (4.7), gives

{n≥0}p₀(n)qⁿ =∑{r≥1}q^{2r²}∏{j=1}^r(1−q^{2j−1})^{−2} ≡∑{r≥1}q^{2r²}∏_{j=1}^r(1−q^{4j−2})^{−1} (mod 2).

Every exponent on the right is even. Therefore p₀(2m+1) is even for every m≥0, already ruling out the proposed density 1/5.

For the sharp asymptotic, the same source's equation (4.2) and Remark 4.5 give

p₀(2m)≡p_ψ(m) (mod 2),

p_ψ(m) odd ⟺ 24m−1=P^{4a+1}s²,

where P is prime, a≥0, and P∤s. Writing t=P^{2a}s, this becomes the exact characterization

p₀(n) odd ⟺ 12n−1=Pt²,

with P prime, (t,6)=1, P≡23 (mod 24), and v_P(t) even.

Indeed, 12n−1 must be 23 modulo 24 because n is even; every integer coprime to 6 has square 1 modulo 24, forcing P≡23 modulo 24. Conversely, these conditions give even n and exactly the required exponent 4a+1. The representation Pt² is unique because P is the squarefree kernel.

Set Y=12X+O(1). If we temporarily drop the even-valuation condition on t, the number of excess pairs is at most

∑_{P≤Y}⌊√Y/P^{3/2}⌋ =O(√Y).

Hence

#{1≤n≤X : p₀(n) is odd} =∑_{t≤√Y,(t,6)=1}π(Y/t²;24,23)+O(√Y).

The prime number theorem in the fixed progression 23 modulo 24, together with Chebyshev's bound to control the summable tail, gives

{t≤√Y,(t,6)=1}π(Y/t²;24,23) ∼Y/(8log Y) ∑{(t,6)=1}1/t².

For completeness, first fix t≤T and apply the prime number theorem termwise; for T<t≤Y^{1/4}, use π(u;24,23)≪u/log u and then let T→∞; the remaining t>Y^{1/4} contribute O(Y^{3/4}).

Finally,

∑_{(t,6)=1}1/t² =ζ(2)(1−2^{−2})(1−3^{−2}) =π²/9.

Since Y∼12X and log Y∼log X, the exact leading constant is

(12/8)(π²/9)=π²/6.

Thus the published conjecture has the wrong density, and the complete corrected parity asymptotic above follows from its own cited mock-theta parity theorem. The relevant primary source is Fu–Tang, arXiv:2306.13309; the additionally attached arXiv:1409.2909 concerns a different function.

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Shivam Patel · · edited