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RESEARCH / Focused target

Cyclic Permutations Avoiding 4132 with 1324-Avoiding Cycle Forms

Literature status: not recheckedRepository record: proved

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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

  1. Encode a cyclic permutation by its cycle word rooted at 1. A selected quadruple is a 1324 in some rotation exactly when the three entries read cyclically after its minimum form a 213.
  2. Write a valid cycle word as 1, L, 2, R. If L is empty, deleting 2 gives a valid word of length n − 1. Otherwise every entry of L exceeds every entry of R, R is increasing, and 1, L standardizes to a valid word.
  3. In that block form the one-line word avoids 4132 exactly when its tail avoids 4132 and the tail entries below the first entry of L increase. This defines an auxiliary class D_k with d_k = b_k + d_{k−1}, and b_n = b_{n−1} + Σ_{k=1}^{n−2} d_k.
  4. Eliminating d gives b_n = 3b_{n−1} − 2b_{n−2} + b_{n−3} for n ≥ 4, the recurrence satisfied by P_{3n}, with matching initial values 1, 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.

Dossier & anchor history

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