Density conjecture for even values of the partition function p_0
Let denote the partition function introduced in the paper, and for define the counting set
Density conjecture. The natural density of indices for which is even exists and equals
This predicts a precise parity distribution for the values of , but the supplied text gives no evidence of a resolution.
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).
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.
Progress summary
A posted attempt claims the conjecture is false and that the even values have density one, but this proposed correction has not been independently verified.
The conjecture predicts that the indices where is even have natural density . The stored 2023 paper by Shishuo Fu and Dazhao Tang studies the relevant parity-related partition families and mock theta functions, but the retrieved record does not itself state a resolution.
Posted attempt
A posted attempt claims a complete refutation: it derives that odd values satisfy , so the even values have density . It also claims that every is even, using the paper’s parity generating-function identity. The attempt has not been independently verified.
Current status (as of August 2026): The proposed density has an unverified claimed refutation, while the corrected asymptotic and density-one conclusion remain mathematically unconfirmed.
Sources
Solutions 1
CounterexampleThis solution needs a summarySee full solution
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.