Conjectured variation of the three-colored partition identity for Q^_1(y)

Prior proof and attribution. This conjecture was already proved by Shane Chern, Linked partition ideals and Russell's three-colored partitions, https://arxiv.org/abs/2607.03977 , Theorem 1.2 and Section 6. Its Theorem 1.1 proves an even stronger refinement with all three colors marked independently. Therefore the original problem is not open; the text below is only an alternative derivation and does not claim the first proof or priority for its refinement.

Proof, with a stronger independent refinement by the number of red parts.

Let $A_1(x,z)$ count the source's admissible three-colored partitions with no $1_R$, marking each part by $x$, each red part additionally by $z$, and total size by $q$. Let $A_3(x,z)$ denote the analogous boundary class forbidding $1_R,2_R,1_B$. Then the conjectured weighted generating function is

$$Q_1^*(y)=A_1(y,y),$$

because a nonred part has weight $y$, while a red part has weight $y^2$.

The source's smallest-part decompositions, with the additional red marker retained, reduce to

$$\begin{aligned} A_1(x,z) &=(1+xq)A_1(xq,z) +xqA_1(xq^2,z) +xzq^2A_3(xq^2,z),\\ A_3(x,z) &=xqA_3(xq,z)+A_1(xq,z). \end{aligned} \tag{1}$$

Write $T_L=L(L+1)/2$. The stronger two-statistic formulas are

$$\boxed{ [x^Lz^r]A_1(x,z) = \frac{ q^{T_L+r^2}(-q;q)_{L-1}(1+q^r) }{ (q^2;q^2)_r(q;q)_{L-r} } } \quad(L\ge1,\ 0\le r\le L), \tag{2}$$

and

$$\boxed{ [x^Lz^r]A_3(x,z) = \frac{ q^{T_L+r(r+1)}(-q;q)_L }{ (q^2;q^2)_r(q;q)_{L-r} } } \quad(L\ge0,\ 0\le r\le L). \tag{3}$$

Both series have constant term $1$. Indeed, writing the coefficients as $a_{L,r}$, $c_{L,r}$, equations (1) become

$$(1-q^L)a_{L,r} =(q^L+q^{2L-1})a_{L-1,r} +q^{2L}c_{L-1,r-1},$$ $$c_{L,r}=q^L(a_{L,r}+c_{L-1,r}).$$

Substitution of (2)–(3) reduces these recurrences to

$$\frac{q^{L-r}(1+q^r)+(1-q^{L-r})}{1+q^L}=1$$

and

$$(1-q^L)(1+q^r) =(1+q^r)(1-q^{L-r}) +q^{L-r}(1-q^{2r}).$$

Hence the refinement holds identically as a formal power series.

Next, Euler's identity

$$\sum_{j+k=M} \frac{q^k}{(q^2;q^2)_j(q^2;q^2)_k} =\frac1{(q;q)_M}$$

reduces the source's triple sums to

$$S_{a,b,b+1}(y) = \sum_{L\ge0}\sum_{r=0}^L \frac{ y^{L+r} q^{T_L+r(r-1+a)+b(L-r)} (-q;q)_L }{ (q^2;q^2)_r(q;q)_{L-r} }. \tag{4}$$

Comparing (2) and (4) gives

$$Q_1^*(y) =S_{1,0,1}(y)+yqS_{3,1,2}(y). \tag{5}$$

The same reduction, by writing $M=j+k$, canceling $1-q^M$, and shifting $M\mapsto M+1$, gives

$$S_{3,0,1}-S_{3,1,2} =yqS_{4,1,2}+yq^2S_{5,2,3}. \tag{6}$$

The source's already proved atomic relation is

$$S_{1,0,1}-S_{3,0,1} =y^2q^2S_{4,1,2}+y^2q^3S_{5,2,3}. \tag{7}$$

By (6), the right side of (7) is $yq(S_{3,0,1}-S_{3,1,2})$. Combining with (5) yields

$$\boxed{ Q_1^*(y) =(1+yq)S_{3,0,1}(y), }$$

exactly as conjectured. The source's proved equation (3.1) additionally gives

$$\boxed{ Q_1^*(y) =(-yq;q)_\infty \sum_{r\ge0}\frac{y^rq^{r^2}}{(q;q)_r}. }$$

Formulas (2)–(3) are strictly stronger: they keep the number of parts and the number of red parts independent.

Source: Matthew C. Russell, On a pair of three-colored (mod 10) partition identities, The Ramanujan Journal 70, article 28 (2026), Conjecture 3, https://doi.org/10.1007/s11139-026-01375-9 ; https://arxiv.org/abs/2509.07169 .