Conjectured arithmetic congruences for the 9-color overpartition function

Both congruences hold. In fact, the first extends to every parameter $\ell\equiv1\pmod8$ and every exponent $\alpha\ge0$, strengthening both restrictions in the conjecture:

$$\boxed{\overline p_\ell(2^\alpha(8n+7)) \equiv0\pmod{16} \quad (\ell\equiv1\!\!\pmod8,\ \alpha,n\ge0),}$$

and

$$\boxed{\overline p_9(36n+33)\equiv0\pmod{128} \quad(n\ge0).}$$

For the first identity, put

$$U(q)=\sum_{j\ge1}(-1)^jq^{j^2}, \qquad V(q)=U(q^\ell).$$

The theta identity gives

$$\sum_{N\ge0}\overline p_\ell(N)q^N = \frac1{(1+2U)(1+2V)}$$

and consequently, modulo $16$,

$$1-2(U+V) +4(U^2+UV+V^2) -8(U^3+U^2V+UV^2+V^3). \tag{1}$$

Its terms count representations by at most three positive weighted squares, with weights $1,\ell\equiv1\pmod8$.

Write $M=8n+7$. If $\alpha=2a$, any representation of $4^aM$ by at most three weighted squares has every root divisible by $2^a$. Dividing yields a representation of $M\equiv7\pmod8$ by at most three residues from $\{0,1,4\}$, which is impossible.

If $\alpha=2a+1$, dividing roots by $2^a$ leaves $2M\equiv6\pmod8$, which is neither a weighted square nor a sum of two weighted squares. For each cubic term in (1), swap two roots of the same weight. A fixed point would satisfy

$$2a_0x^2+b_0z^2=2M, \qquad a_0,b_0\in\{1,\ell\}.$$

Necessarily $z=2w$ and $x$ is odd, so

$$M=a_0x^2+2b_0w^2\equiv1\text{ or }3\pmod8,$$

contradicting $M\equiv7\pmod8$. Therefore every cubic coefficient is even, and its factor $8$ makes it vanish modulo $16$. This proves the stronger all-$\ell$, all-$\alpha$ theorem.

For the second congruence, write $f_j=(q^j;q^j)_\infty$. The source's exact three-dissection gives

$$B(q) = \sum_{r\ge0}\overline p_9(3r)q^r = \frac{f_2^4f_3^4}{f_1^8f_6^2}.$$

The target corresponds to exponents $12n+11$ of $B$. Form the eta quotient

$$E(\tau) = \frac{\eta(2\tau)^4\eta(3\tau)^4\eta(12\tau)^{72}} {\eta(\tau)^8\eta(6\tau)^2} = q^{36}f_{12}^{72}B(q).$$

At level $24$, its eta exponent vector is

$$(r_1,r_2,r_3,r_6,r_{12})=(-8,4,4,-2,72).$$

The weighted eta sums are

$$\sum r_\delta=70,\qquad \sum\delta r_\delta=864,\qquad \sum(24/\delta)r_\delta=24.$$

At cusp denominators $1,2,3,4,6,8,12,24$, the orders are respectively

$$1,\ 6,\ 19,\ 12,\ 18,\ 12,\ 36,\ 36.$$

Hence $E$ is holomorphic of weight $35$, level $24$, and character $\chi_{-4}$.

Project onto residue $11\pmod{12}$ using all four Dirichlet characters modulo $12$:

$$S_{11} = \sum_{\psi\bmod12}\psi(11)(E\otimes\psi) = 4\sum_{r\equiv11\ (12)}[q^r]E\,q^r.$$

Primitive twisting and the standard oldform correction for imprimitive characters show

$$S_{11}\in M_{35}(\Gamma_0(3456),\chi_{-4}).$$

The subgroup index is $6912$, giving the complete Sturm bound

$$B=\frac{35\cdot6912}{12}=20160.$$

The required exact finite certificate is reproducible entirely with integer arithmetic:

$$a(0)=1,\qquad a(r)= 2\sum_{j^2\le r}(-1)^{j+1}a(r-j^2) \pmod{128},$$

and

$$\overline p_9(r) = \sum_{j=0}^{\lfloor r/9\rfloor} a(r-9j)a(j) \pmod{128}.$$

Computing through

$$3(B-36)=60372$$

gives

$$\overline p_9(36j+33)\equiv0\pmod{128} \qquad(0\le j\le1676).$$

These $1677$ exact coefficients cover the full Sturm bound after the $q^{36}$ shift and multiplication by the unit $f_{12}^{72}$. Therefore

$$S_{11}\equiv0\pmod{512}$$

identically by Sturm's theorem. Dividing its coefficients by $4$ and removing the residue-preserving unit proves the desired modulo-$128$ congruence for every $n$.

Source: H. S. Sumanth Bharadwaj, N. Sujatha and S. Chandankumar, arXiv:2607.16608, equation (3.3) and Conjecture 3, equations (5.7)–(5.8).