Let n≥ 0, i≥ 3, and k≥ 0. The overcubic partition tuple function overlineb_m(r) is considered at the indicated arithmetic progressions. Congruence conjecture. For all n≥ 0, i≥ 3, and k≥ 0, overlineb_2^i k + 2^i-1-1(8n+5) equiv 0 pmod2^i+2, and overlineb_2^i k + 2^i-1-1(8n+7) equiv 0 pmod2^i+3. This conjecture proposes further infinite families of arithmetic congruences for overcubic partition tuples, extending the preceding divisibility results; the authors indicate that it may follow from techniques similar to those used in the paper.

Conjecture on congruences for overcubic partition tuples

The conjecture asserts two infinite families of divisibility congruences for overcubic partition tuples, at indices congruent to $5$ and $7$ modulo $8$ . It is presented as a conjectural extension of previously established results.

Known results

For $\alpha,l\geq0$ and $n\geq1$ , $\overline{b}_{2^\alpha l}(n)\equiv0\pmod{2^{\alpha+1}}$ .
For odd $k$ , $\overline{b}_k(n)$ has a modulo- $4$ characterization: it is $2$ when $n=l^2$ or $n=2l^2$ , and $0$ otherwise.
Further modulo- $4$ congruences are proved on progressions such as $64m+48$ , $128m+80$ , and $256\cdot2^\alpha m+192$ .

July 2026 statement

Saikat Maity and Manjil P. Saikia state the target assertions as Conjecture 1.5 and suggest that similar techniques may apply. No proof, counterexample, verification, or retraction of this specific conjecture was found.

Current status (as of August 2026): Earlier divisibility results are proved, but the two stated congruence families remain an open conjecture with no publicly recorded proof or disproof.

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 $i,k,n$ .

Conjecture on congruences for overcubic partition tuples

Progress summary

Primary source

Solutions 1