Conjecture on congruences for overcubic partition tuples

Write

$$f_j=\prod_{m\ge1}(1-q^{jm}),\qquad B(q)=\frac{f_4}{f_1^2f_2},\qquad B(q)^t=\sum_{n\ge0}\overline b_t(n)q^n.$$

We prove, for every $i\ge3$, $k,n\ge0$, and $t=2^ik+2^{i-1}-1$, that

$$\overline b_t(8n+5)\equiv0\pmod{2^{i+2}},\qquad \overline b_t(8n+7)\equiv0\pmod{2^{i+3}}.$$

All power-series computations take place in $\mathbb Z[[q]]$.

First, Jacobi's identity gives

$$A(q):=B(q)^{-1}=\varphi(-q)\varphi(-q^2)=1+2C(q),$$

where

$$U=\sum_{j\ge1}(-1)^jq^{j^2},\quad V=\sum_{j\ge1}(-1)^jq^{2j^2},\quad W=U+V,\quad Z=UV,\quad C=W+2Z.$$

Since no integer congruent to $5$ or $7\pmod8$ is represented by $x^2+2y^2$,

$$[q^N]A=[q^N]C=0 \qquad(N\equiv5,7\pmod8). \tag{1}$$

We establish the two base congruences

$$[q^{8n+5}]B^3\equiv0\pmod{32},\qquad [q^{8n+7}]B^3\equiv0\pmod{64}. \tag{2}$$

Suppose first that $N\equiv5\pmod8$, and let

$$R=\#\{(x,y)\in\mathbb Z_{>0}^2:x,y\text{ odd},\ x^2+4y^2=N\}.$$

Because $C^2\equiv W^2\pmod4$, the $UV$ term contributes nothing by (1), and $V^2$ has only even exponents,

$$[q^N]C^2\equiv[q^N]U^2=-2R\pmod4. \tag{3}$$

Indeed, each representation by two positive squares has one odd entry and one entry $2y$ with $y$ odd, and its two orders both contribute sign $-1$. Modulo $2$, Frobenius gives

$$C^3\equiv W^3\equiv(U+V)(U(q^2)+V(q^2))\pmod2.$$

At an odd exponent, the first factor contributes an odd square from $U$; the term $U(q^2)$ cannot contribute because it would represent $N$ as $x^2+2y^2$, whereas $V(q^2)$ gives exactly $R$. Therefore

$$[q^N]C^3\equiv R\pmod2,\qquad [q^N]C^4\equiv0\pmod2.$$

Expanding $B^3=(1+2C)^{-3}$ modulo $32$ and using (1) yields

$$[q^N]B^3\equiv[q^N](24C^2-80C^3+240C^4) \equiv8\bigl(3(-2R)+2R\bigr)=-32R\equiv0\pmod{32}. \tag{4}$$

Next suppose $N\equiv7\pmod8$, and let

$$R=\#\{(x,y,z)\in\mathbb Z_{>0}^3: x,y,z\text{ odd},\ x^2+4y^2+2z^2=N\}.$$

The square residues modulo $8$ give

$$[q^N]W^2=0,\qquad[q^N]UV^2=0,\qquad[q^N]U^2V=2R. \tag{5}$$

In the last equality, the two $U$-indices are an odd integer and $2y$, with $y$ odd, while the $V$-index is odd; both orders contribute sign $+1$. Also $[q^N]Z^2$ is even because $Z^2\equiv Z(q^2)\pmod2$. Hence

$$[q^N]C^2=[q^N](W^2+4WZ+4Z^2)\equiv0\pmod8. \tag{6}$$

The only term in $W^3$ contributing at residue $7$ is $3U^2V$, so $[q^N]W^3=6R$. Moreover,

$$W^2Z\equiv\bigl(U(q^2)+V(q^2)\bigr)UV\pmod2.$$

The term $U(q^2)UV$ cannot contribute at residue $7$, since it would require two square residues to sum to $3\pmod4$; $V(q^2)UV$ counts exactly the same triples as $R$. Thus

$$[q^N]C^3\equiv[q^N](W^3+2W^2Z)\equiv6R+2R\equiv0\pmod4. \tag{7}$$

Writing $C^2=C(q^2)+2D$, one has $C^4\equiv C(q^2)^2\pmod4$, so all odd-index coefficients of $C^4$ are divisible by $4$. Finally,

$$C^5\equiv W(q)W(q^4)\pmod2.$$

Its odd exponents have the form $x^2+4y^2$ or $x^2+8y^2$, neither of which is $7\pmod8$. Hence

$$[q^N]C^4\equiv0\pmod4,\qquad[q^N]C^5\equiv0\pmod2. \tag{8}$$

The expansion modulo $64$ now gives

$$[q^N]B^3 \equiv8[q^N](3C^2-10C^3+30C^4-84C^5) \equiv0\pmod{64},$$

proving both assertions of (2).

For the uniform lifting step, put

$$M=2^ik+2^{i-1}=2^{i-1}(2k+1),\qquad L=M/4=2^{i-3}(2k+1),\qquad e=v_2(L)=i-3.$$

Since $A\equiv B\equiv1\pmod2$,

$$X:=B^4-1\in8\mathbb Z[[q]],$$

and

$$B^{M-1}=A(1+X)^L =A+LAX+\sum_{j=2}^{L}\binom Lj AX^j. \tag{9}$$

For every $j$,

$$v_2\binom Lj\ge e-v_2(j),$$

by $\binom Lj=(L/j)\binom{L-1}{j-1}$. Thus for odd $j\ge3$, every coefficient of $\binom Lj AX^j$ is divisible by $2^{e+9}$. For even $j\ge2$, write $X=8Y$. Since $A\equiv1\pmod2$ and $Y^j$ is a square modulo $2$, its odd-index coefficients vanish modulo $2$, and consequently

$$v_2\left([q^N]\binom Lj AX^j\right) \ge3j+1+e-v_2(j)\ge e+6 \qquad(N\text{ odd}). \tag{10}$$

Thus every term with $j\ge2$ vanishes modulo $2^{e+6}=2^{i+3}$ at odd indices.

For $N\equiv5,7\pmod8$, (1) and $AX=B^3-A$ reduce (9) to

$$\overline b_{M-1}(N) \equiv L\,\overline b_3(N)\pmod{2^{i+3}}. \tag{11}$$

Since $v_2(L)=i-3$, the first base congruence in (2) implies divisibility by $2^{i+2}$ at $N=8n+5$, and the second implies divisibility by $2^{i+3}$ at $N=8n+7$. Both conjectured families therefore hold for all admissible $i,k,n$.