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Pattern Avoidance for Fibonacci Sequences using $k$-Regular Words

Combinatorics 2026-03-11 v4 Discrete Mathematics

Abstract

Two kk-ary Fibonacci recurrences are ak(n)=ak(n1)+kak(n2)a_k(n) = a_k(n-1) + k \cdot a_k(n-2) and bk(n)=kbk(n1)+bk(n2)b_k(n) = k \cdot b_k(n-1) + b_k(n-2). We provide a simple proof that ak(n)a_k(n) is the number of kk-regular words over [n]={1,2,,n}[n] = \{1,2,\ldots,n\} that avoid patterns {121,123,132,213}\{121, 123, 132, 213\} when using base cases ak(0)=ak(1)=1a_k(0) = a_k(1) = 1 for any k1k \geq 1. This was previously proven by Kuba and Panholzer in the context of Wilf-equivalence for restricted Stirling permutations, and it creates Simion and Schmidt's classic result on the Fibonacci sequence when k=1k=1, and the Jacobsthal sequence when k=2k=2. We complement this theorem by proving that bk(n)b_k(n) is the number of kk-regular words over [n][n] that avoid {122,213}\{122, 213\} with bk(0)=bk(1)=1b_k(0) = b_k(1) = 1 for any~k2k \geq 2. Finally, we conjecture that Avn2(121,123,132,213)=a1(n)2|Av^{2}_{n}(\underline{121}, 123, 132, 213)| = a_1(n)^2 for n0n \geq 0. That is, vincularizing the Stirling pattern in Kuba and Panholzer's Jacobsthal result gives the Fibonacci-squared numbers.

Keywords

Cite

@article{arxiv.2312.16052,
  title  = {Pattern Avoidance for Fibonacci Sequences using $k$-Regular Words},
  author = {Emily Downing and Elizabeth Hartung and Cody Lucido and Aaron Williams},
  journal= {arXiv preprint arXiv:2312.16052},
  year   = {2026}
}

Comments

20 pages, submitted to special journal issue for Permutation Patterns 2023 (PP23) in DMTCS

R2 v1 2026-06-28T14:02:09.251Z