Refined decomposition conjecture for alternating sign matrix enumerations

Let An(z,ρ):=An(z,ρ,1)A_n(z,\rho):=A_n(z,\rho,1), and for odd order define

B2m+1(z,ρ)=AHSASM(2m+1)zμ(A)m2ρf(A)2,B_{2m+1}(z,\rho)=\sum_{A\in\operatorname{HSASM}(2m+1)}z^{\frac{\mu(A)-m}{2}}\rho^{f(A)-2},

where HSASM(2m+1)\operatorname{HSASM}(2m+1) is the set of horizontally symmetric ASMs of order 2m+12m+1, μ(A)\mu(A) is the statistic used in the preceding enumeration, and f(A)f(A) is the position of the 11 in the first row. Refined decomposition conjecture. There exists a family of polynomials B2m(z,ρ)B_{2m}(z,\rho) in {z,ρ}\{z,\rho\} such that, for every m1m\geq 1,

A2m+1(z,ρ)=ρB2m+1(z,1)B2m+2(z,ρ),A_{2m+1}(z,\rho)=\rho B_{2m+1}(z,1)B_{2m+2}(z,\rho), A2m(z,ρ)=ρ(ρ+1)B2m(z,1)B2m+1(z,ρ).A_{2m}(z,\rho)=\rho(\rho+1)B_{2m}(z,1)B_{2m+1}(z,\rho).

This refines the previously conjectured decomposition by tracking the position of the 11 in the first row. The source presents it as a proposal; no proof or resolution is supplied in the given text.

Sources & referencesView supporting material

Primary source

Guo-Niu Han and Lihong Yang, “Yet another doubly refined enumeration of Alternating Sign Matrices”, arXiv:2601.10870 (2026).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.