RESEARCH / Focused target
Joint Symmetry of the Mesh Patterns 123 and 321 with Shading {0,1,2}² ∪ {(3,3)}
RELEASED FOUNDATIONS / PROPOSED CONNECTION
Research connections
All source anchors (1)
- D5/S3/Combinatorics/MeshPattern/MeshPatternS21.resultAbsent from this graph
Problem
The question
Shuzhen Lv and Philip B. Zhang, Joint Equidistributions of Mesh Patterns 123 and 321 with Symmetric and Minus-Antipodal Shadings, arXiv:2501.00357v3, Section 2.4, Conjecture 2 (printed page 9), pair S21: the mesh patterns (123, R) and (321, R) are jointly equidistributed on Sₙ, where R = {0,1,2}² ∪ {(3,3)} as rendered in the source's diagram.
An occurrence of (123, R) in π is a triple of positions with increasing values such that no other point of π lies in a shaded cell; (321, R) is the same with decreasing values. Joint equidistribution means that, for all n, k and l, the number of π ∈ Sₙ with k occurrences of (123, R) and l occurrences of (321, R) equals the number with l and k.
Motivation
Our foothold
The theorem D5/S3/Combinatorics/MeshPattern/MeshPatternS21.result establishes the statement for every n.
Gap
Missing bridges
Pre-registration issue 12358 records the literature screen: the source proves only Wilf-equivalence, and the arXiv, GitHub and repository searches located no proof of the joint statement. This is a bounded negative finding.
Route
Proposed approach
- An occurrence is determined by its largest value m: it exists exactly when the origin-anchored rectangle Cₘ (positions at most n − m + 3, values at most m) contains exactly three points, including the point of row m and the point of column n − m + 3, and the order type of these three points decides between 123 and 321.
- The rectangles Cₘ form a Ferrers shape. A seven-state involution acting inside the union of the rectangles that contain exactly three points preserves the number of points in every Cₘ and both incidences, and exchanges the increasing and decreasing order types in each rectangle with three points.
- The involution therefore exchanges the two occurrence counts at every largest value simultaneously.
Falsifier
What would falsify this route
The statement would fail if some occurrence were not captured by its rectangle, if the involution changed a rectangle count or an incidence, or if it failed to be an involution.
Evidence
Evidence to collect
The joint distributions are symmetric for every n ≤ 10; an independent referee implementation checked the occurrence criterion, the rectangle reformulation and the involution on all 4,037,914 permutations with n ≤ 10.
Triage
Scope assessment
theorem; the statement is Conjecture 2 (pair S21) of arXiv:2501.00357v3 and is quantified over every n.
- Proved (formalized): the stronger simultaneous exchange of the occurrence indicators at every largest value, and hence the joint symmetry.
- Proved (paper argument, not formalized): the same exchange follows from Fomin growth diagrams on the Ferrers shape formed by the rectangles, with the cells outside the shape held fixed. Pair S22 follows from S21 by reverse and complement, as the source notes.
- Computed: the numbers of permutations avoiding both patterns are 1, 2, 4, 14, 70, 440, 3262, 27578, 260596 for n = 1, …, 9.
- Open: a formula for these numbers, and for the full joint distribution; no such formula is proved here.
ASSUMED-UNVERIFIED
Unverified assumptions
The literature screen is limited to the arXiv, GitHub and repository searches recorded above; the shading was read from the rendered diagram of arXiv v3.