RESEARCH / Focused target
Cyclic Permutations Avoiding 4132 with 1324-Avoiding Cycle Forms
RELEASED FOUNDATIONS / PROPOSED CONNECTION
Research connections
All source anchors (1)
- D5/S3/Combinatorics/ArcherCyclicPadovan.resultAbsent from this graph
Problem
The question
Kassie Archer, Ethan Borsh, Jensen Bridges, Christina Graves and Millie Jeske, Pattern-restricted cyclic permutations with a pattern-restricted cycle form, arXiv:2408.15000v1, Section 4:
we conjecture that … |A°_n(4132;1324)| is the (3n)th Padovan number
Here A°_n(4132; 1324) is the set of cyclic permutations of 1, …, n whose one-line notation avoids
4132 and all of whose cycle forms, the rotations of the cycle word, avoid 1324.
Motivation
Our foothold
Let P_0 = 1, P_1 = P_2 = 0 and P_{k+3} = P_{k+1} + P_k (OEIS A000931). The theorem
D5/S3/Combinatorics/ArcherCyclicPadovan.result establishes |A°_n(4132; 1324)| = P_{3n} for every
n ≥ 1; the counts for n = 1, …, 9 are 1, 1, 2, 5, 12, 28, 65, 151, 351. Equivalently, their
generating function is x(1 − x)² / (1 − 3x + 2x² − x³).
Gap
Missing bridges
Pre-registration issue 11215 records the literature screen. Of the three papers citing
arXiv:2408.15000, Pan's arXiv:2409.17482 proves only the Pell case (2431; 1324) and Pan's
arXiv:2505.02045 settles conjectures of the earlier paper arXiv:2312.05145 on the standard cycle form.
The repository had no result for this class. These are bounded negative findings.
Route
Proposed approach
- Encode a cyclic permutation by its cycle word rooted at
1. A selected quadruple is a1324in some rotation exactly when the three entries read cyclically after its minimum form a213. - Write a valid cycle word as
1, L, 2, R. IfLis empty, deleting2gives a valid word of lengthn − 1. Otherwise every entry ofLexceeds every entry ofR,Ris increasing, and1, Lstandardizes to a valid word. - In that block form the one-line word avoids
4132exactly when its tail avoids4132and the tail entries below the first entry ofLincrease. This defines an auxiliary classD_kwithd_k = b_k + d_{k−1}, andb_n = b_{n−1} + Σ_{k=1}^{n−2} d_k. - Eliminating
dgivesb_n = 3b_{n−1} − 2b_{n−2} + b_{n−3}forn ≥ 4, the recurrence satisfied byP_{3n}, with matching initial values1, 1, 2.
Falsifier
What would falsify this route
The statement would fail if some n had a count different from P_{3n}. It depends on reading
"all cycle forms" as every rotation of the cycle word and on the Padovan indexing of OEIS A000931.
Evidence
Evidence to collect
Exhaustive enumeration of all cycle words through n = 10 agrees with the decomposition, both one-line
criteria, the auxiliary split and the counts 1, 1, 2, 5, 12, 28, 65, 151, 351, 816.
Triage
Scope assessment
theorem; the conjecture is stated in the Open Questions of arXiv:2408.15000v1 and is quantified over
every n.
ASSUMED-UNVERIFIED
Unverified assumptions
The literature screen is limited to the citation list, arXiv searches and repository checks recorded above.