Higher-modulus congruences for the 2-color overpartition function

All three congruences hold for every $n\ge0$. We prove them by holomorphic eta products and complete, explicitly reproducible Sturm certificates.

Put

$$f_r=(q^r;q^r)_\infty,\qquad \sum_{n\ge0}\overline p_\ell(n)q^n =\frac{f_2f_{2\ell}}{f_1^2f_\ell^2}.$$

Repeated use of the binomial theorem gives

$$f_r^{2^a}\equiv f_{2r}^{2^{a-1}}\pmod{2^a}. \tag{1}$$

The two $\ell=6$ congruences. The source's exact equation (3.5) gives

$$\sum_{n\ge0}\overline p_6(3n+2)q^n = 4G(q), \qquad G(q)=\frac{f_4f_6^3}{f_1^6}.$$

The two required progressions correspond to coefficients of $G$ in residue classes $5,7\pmod8$, and these coefficients must vanish modulo $16$. By (1),

$$G(q)f_8^{10} \equiv f_1^{58}f_4f_6^3f_8^2 \pmod{16}.$$

Since $f_8$ is a unit in $\mathbb Z[[q^8]]$, multiplying by $f_8^{10}$ preserves vanishing within each residue class. Form

$$E_6(\tau) = \eta(\tau)^{58}\eta(4\tau)\eta(6\tau)^3\eta(8\tau)^2 = q^4f_1^{58}f_4f_6^3f_8^2.$$

The eta exponents satisfy

$$\sum r_\delta=64,\quad \sum\delta r_\delta=96,\quad \sum(24/\delta)r_\delta=1416.$$

Thus $E_6$ is holomorphic of weight $32$, level $24$, and character $(6/\cdot)$; all exponents are nonnegative, so every cusp is holomorphic.

For $r=1,3$, project using the four Dirichlet characters modulo $8$:

$$S_{6,r} = \sum_{\chi\bmod8}\chi(r)(E_6\otimes\chi) = 4\sum_{j\equiv r\ (8)}[q^j]E_6\,q^j.$$

These are holomorphic forms of weight $32$, level $1536$, with index $3072$ and Sturm bound

$$B_6=\frac{32\cdot3072}{12}=8192.$$

The principal-character twist is the oldform $E_6-(E_6|U_2)|V_2$; the other twists have conductor dividing $8$.

The exact coefficient recurrence below verifies both original progressions modulo $64$ through original index

$$3(B_6-4)+2=24566:$$

there are exactly $1023$ checks for each progression. Therefore $S_{6,1}\equiv S_{6,3}\equiv0\pmod{64}$ through their Sturm bound, hence identically. Dividing the displayed character projections by $4$ proves

$$\boxed{ \overline p_6(24n+17) \equiv \overline p_6(24n+23) \equiv0\pmod{64} \quad(n\ge0).}$$

The $\ell=8$ congruence. The source's exact equation (4.54) gives

$$A(q) = \sum_{n\ge0}\overline p_8(4n)q^n = \frac{f_2^{17}}{f_1^{14}f_4^5}.$$

Equation (1) implies

$$A(q)f_8^{36} \equiv f_1^{114}f_2^{17}f_4^{27}f_8^4 \pmod{32}.$$

Set

$$E_8(\tau) = \eta(\tau)^{114}\eta(2\tau)^{17} \eta(4\tau)^{27}\eta(8\tau)^4 = q^{12}f_1^{114}f_2^{17}f_4^{27}f_8^4.$$

Its eta exponent sums are $162,288,4152$, giving a holomorphic form of weight $81$, level $32$, and character $(-2/\cdot)$. The desired residue class becomes $3\pmod8$ after the shift by $12$. Thus

$$S_{8,3} = \sum_{\chi\bmod8}\chi(3)(E_8\otimes\chi) = 4\sum_{j\equiv3\ (8)}[q^j]E_8\,q^j$$

has weight $81$, level $2048$, index $3072$, and Sturm bound

$$B_8=\frac{81\cdot3072}{12}=20736.$$

The exact recurrence verifies the original progression modulo $32$ through index

$$4(B_8-12)=82896,$$

giving $2590$ checks. Therefore $S_{8,3}\equiv0\pmod{128}$ through the Sturm bound and hence identically. Dividing by $4$ and undoing the invertible $f_8$ factor proves

$$\boxed{\overline p_8(32n+28)\equiv0\pmod{32} \quad(n\ge0).}$$

Reproducible exact Sturm certificate. All finite checks use only the following division-free integer recurrences:

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

where $a(r)=\overline p(r)$, and

$$\overline p_\ell(r) = \sum_{j=0}^{\lfloor r/\ell\rfloor} a(r-\ell j)a(j).$$

Compute $a(r)\pmod{64}$ for $0\le r\le82896$. Direct substitution gives

$$\begin{aligned} \overline p_6(24j+17)&\equiv0\pmod{64} && (0\le j\le1022),\\ \overline p_6(24j+23)&\equiv0\pmod{64} && (0\le j\le1022),\\ \overline p_8(32j+28)&\equiv0\pmod{32} && (0\le j\le2589). \end{aligned}$$

These cover the complete bounds above; Sturm's theorem supplies the all-$n$ conclusions.

Source: H. S. Sumanth Bharadwaj, N. Sujatha and S. Chandankumar, arXiv:2607.16608, equations (3.5), (4.54), and Conjecture 2, equations (5.4)–(5.6).