Let a(m) be defined by the paper's generating function, and suppose that m is a positive integer such that a(m)equiv 0 pmod8. Write the prime factorization of 4m+3 as 4m+3=∏_i=1^uh_i∏_j=1^vg_j^alpha_j, where each alpha_j≥ 2. Congruence conjecture. For every n≥ 1 satisfying (n,2∏_j=1^vg_j^alpha_j)=1, one has b(18mn^2+frac9n^2-12)equiv 0 pmod72. This conjecture proposes an infinite family of congruences modulo 72 for the coefficients b(n) of the second-order mock theta function B(q). The preceding discussion explains that related congruences can be obtained using Radu's algorithm, while a q-series proof and the general conjectured family remain to be established.

Congruence conjecture for the second-order mock theta function coefficients

From papers

Let $a(m)$ be defined by the paper's generating function, and suppose that $m$ is a positive integer such that

a(m)\equiv 0 \pmod{8}.

Write the prime factorization of $4m+3$ as

4m+3=\prod_{i=1}^{u}h_i\prod_{j=1}^{v}g_j^{\alpha_j},

where each $\alpha_j\geq 2$ . Congruence conjecture. For every $n\geq 1$ satisfying

\left(n,2\prod_{j=1}^{v}g_j^{\alpha_j}\right)=1,

one has

b\left(18mn^2+\frac{9n^2-1}{2}\right)\equiv 0\pmod{72}.

This conjecture proposes an infinite family of congruences modulo $72$ for the coefficients $b(n)$ of the second-order mock theta function $\mathcal{B}(q)$ . The preceding discussion explains that related congruences can be obtained using Radu's algorithm, while a $q$ -series proof and the general conjectured family remain to be established.

Progress summary

Open

The conjecture remains unproved, while a later paper settles related identities without addressing this infinite family of congruences.

H. Nath and H. Das proposed this conjecture in 2025 for coefficients of the second-order mock theta function $\mathcal{B}(q)$ . It predicts infinitely many congruences modulo $72$ from the condition $a(m)\equiv0\pmod{8}$ and a factorization condition on $4m+3$ .

Known results

Nath and Das, 2025: proved the analogous family using the squarefull part of $4m+1$ , not $4m+3$ .
Nath and Das, 2025: proved several related congruence families modulo $2$ , $4$ , $8$ , $36$ , $54$ , and $72$ .
Radu’s algorithm verifies various individual congruences for the auxiliary coefficients $a(n)$ , but does not prove this conjecture.

January 2026 related development

A later paper claims analytic proofs of three identities for the second-order mock theta functions $A(q)$ , $B(q)$ , and $\mu_2(q)$ requested by Nath and Das. The retrieved description does not claim that it proves or refutes this specific congruence conjecture.

Current status (as of August 2026): The related $4m+1$ theorem is proved, but the stated $4m+3$ congruence family remains neither proved nor independently refuted in the retrieved sources.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

Hemjyoti Nath and Hirakjyoti Das, “Infinite families of congruences for the second order mock theta function B(q)”, arXiv:2509.20708 (2025).

Additional references

12 papers in this index state this conjecture (2005–2025). The statement above is taken from the most recent of them; the others are arXiv:2503.08517, arXiv:2501.01178, arXiv:2402.08340, arXiv:2109.15243, arXiv:2105.10975, arXiv:2010.13256, arXiv:1703.01955, arXiv:1511.04005, arXiv:1303.0568, arXiv:1011.0975, arXiv:math/0504569.

1 week ago

Conjecture 9.1 is false. Its coprimality condition uses the squarefull part of $4m+3$ , whereas the adjacent proved theorem correctly uses the squarefull part of $4m+1$ .

Write

\sum_{t\ge0}a(t)q^t =(q;q)_\infty^4(q^2;q^2)_\infty.

Choose

m=31,\qquad n=5.

Exact Euler-product coefficient extraction gives

a(31)=8,\qquad a(781)=484.

Thus the required hypothesis $a(m)\equiv0\pmod8$ holds. Moreover,

4m+3=127

is prime, so its squarefull factor is the empty product $1$ . The conjectured coprimality hypothesis therefore reduces to

\gcd(5,2)=1,

which also holds.

The conjectured coefficient index is

18mn^2+\frac{9n^2-1}{2} =18\cdot31\cdot25+\frac{225-1}{2} =14062 =18\cdot781+4.

But equation (8.27), already proved in the primary source, states

b(18t+4)\equiv9a(t)\pmod{72}.

Consequently

\boxed{ b(14062)\equiv9a(781) =9\cdot484 =4356 \equiv36\not\equiv0\pmod{72}.}

Hence $(m,n)=(31,5)$ satisfies every stated hypothesis and contradicts the conclusion.

The structural error is transparent: the source's proved Theorem 1.13 uses the squarefull part of

4m+1=125=5^3,

and therefore requires $\gcd(n,250)=1$ , correctly excluding $n=5$ . Replacing $4m+1$ by $4m+3=127$ removes exactly the obstructing prime.

The value $a(781)\equiv4\pmod8$ also follows immediately from the source's Newman recurrence (8.9) with $p=5$ , since $a(31)\equiv0\pmod8$ and $a(1)=-4$ :

a(781)\equiv-5^3a(1) =500\equiv4\pmod8.

Source: H. Nath and H. Das, Infinite families of congruences for the second order mock theta function $\mathcal B(q)$ , arXiv:2509.20708, Theorem 1.13, equations (8.9), (8.27), and Conjecture 9.1.

Infinitely many counterexamples. The obstruction persists through the infinite admissible family

m=31,\qquad n_j=5\cdot3^{4j}\qquad(j\ge0).

Since $4m+3=127$ is prime, every $n_j$ satisfies the conjectured condition $\gcd(n_j,2)=1$ . Put

t_j=\frac{125n_j^2-1}{4} =781\cdot3^{8j}+\frac{3^{8j}-1}{4}.

The source's proved recurrence (8.28), together with its explicit values $\nu(3)=4$ and $g(3)=297\equiv1\pmod8$ , yields

a\left(3^{8j}t+\frac{3^{8j}-1}{4}\right) \equiv a(t)\pmod8.

Taking $t=781$ gives $a(t_j)\equiv4\pmod8$ for every $j$ . Equation (8.27) therefore proves

\boxed{ b\left(18\cdot31n_j^2+\frac{9n_j^2-1}{2}\right) =b(18t_j+4) \equiv36\pmod{72} \qquad(j\ge0).}

Thus the conjecture has infinitely many source-admissible counterexamples; all are correctly excluded by the squarefull- $(4m+1)$ hypothesis of the adjacent proved theorem.

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Shivam Patel · 20 hours ago · edited