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

About 21 years old · traced to 2005

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.

References

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.

Posted 1 month ago

Progress summary

Refreshed 1 month ago

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

CounterexampleThis solution needs a summarySee full solutionHide full solution

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.