English
Related papers

Related papers: Pattern Avoidance for Fibonacci Sequences using $k…

200 papers

Let (F_n^{(k)})_{n\geq -(k-2)} be the k-generalized Fibonacci sequence, defined as the linear recurrence sequence whose first k terms are \(0, 0, \ldots, 0, 1\), and whose subsequent terms are determined by the sum of the preceding k terms.…

Number Theory · Mathematics 2025-01-08 Roberto Alvarenga , Ana Paula Chaves , Maria Eduarda Ramos , Matheus Silva , Marcos Sosa

We use combinatorial and generating function techniques to enumerate various sets of involutions which avoid 231 or contain 231 exactly once. Interestingly, many of these enumerations can be given in terms of $k$-generalized Fibonacci…

Combinatorics · Mathematics 2007-05-23 Eric S. Egge , Toufik Mansour

Let T_k^m={\sigma \in S_k | \sigma_1=m}. We prove that the number of permutations which avoid all patterns in T_k^m equals (k-2)!(k-1)^{n+1-k} for k <= n. We then prove that for any \tau in T_k^1 (or any \tau in T_k^k), the number of…

Combinatorics · Mathematics 2007-05-23 T. Mansour

The method we have applied in "A. Bernini, L. Ferrari, R. Pinzani, Enumerating permutations avoiding three Babson-Steingrimsson patterns, Ann. Comb. 9 (2005), 137--162" to count pattern avoiding permutations is adapted to words. As an…

Combinatorics · Mathematics 2007-11-22 Antonio Bernini , Luca Ferrari , Renzo Pinzani

Let a tribonacci sequence be a sequence of integers satisfying $a_k=a_{k-1}+a_{k-2}+a_{k-3}$ for all $k\ge 4$. For any positive integers $k$ and $n$, denote by $f_k(n)$ the number of tribonacci sequences with $a_1, a_2, a_3>0$ and with…

Number Theory · Mathematics 2023-01-31 Luke Pebody

Nonnesting permutations are permutations of the multiset $\{1,1,2,2,\dots,n,n\}$ that avoid subsequences of the form $abba$ for any $a\neq b$. These permutations have recently been studied in connection to noncrossing (also called…

Combinatorics · Mathematics 2026-01-21 Sergi Elizalde , Amya Luo

A word $w=w_1w_2\cdots w_n$ is alternating if either $w_1<w_2>w_3<w_4>\cdots$ (when the word is up-down) or $w_1>w_2<w_3>w_4<\cdots$ (when the word is down-up). In this paper, we initiate the study of (pattern-avoiding) alternating words.…

Combinatorics · Mathematics 2015-05-18 Emma L. L. Gao , Sergey Kitaev , Philip B. Zhang

A generalization of the well--known Fibonacci sequence is the $k$--Fibonacci sequence with some fixed integer $k\ge 2$. The first $k$ terms of this sequence are $0,\ldots,0,1$, and each term afterwards is the sum of the preceding $k$ terms.…

Number Theory · Mathematics 2020-08-25 Eric F. Bravo , Jhon J. Bravo , Carlos A. Gómez

A word $w=w_1w_2\cdots w_n$ is alternating if either $w_1<w_2>w_3<w_4>\cdots$ (when the word is up-down) or $w_1>w_2<w_3>w_4<\cdots$ (when the word is down-up). The study of alternating words avoiding classical permutation patterns was…

Combinatorics · Mathematics 2016-03-02 Alice L. L. Gao , Sergey Kitaev , Philip B. Zhang

We study the avoidability of long $k$-abelian-squares and $k$-abelian-cubes on binary and ternary alphabets. For $k=1$, these are M\"akel\"a's questions. We show that one cannot avoid abelian-cubes of abelian period at least $2$ in infinite…

Discrete Mathematics · Computer Science 2015-07-10 Michaël Rao , Matthieu Rosenfeld

A finite word $w$ is an abelian square if $w = xx^\prime$ with $x^\prime$ a permutation of $x$. In 1972, Entringer, Jackson, and Schatz proved that every binary word of length $k^2 + 6k$ contains an abelian square of length $\geq 2k$. We…

Combinatorics · Mathematics 2010-12-03 Elyot Grant

In [GM] Guibert and Mansour studied involutions on n letters avoiding (or containing exactly once) 132 and avoiding (or containing exactly once) an arbitrary pattern on k letters. They also established a bijection between 132-avoiding…

Combinatorics · Mathematics 2007-05-23 O. Guibert , T. Mansour

We consider the enumeration of ordered set partitions avoiding a permutation pattern, as introduced by Godbole, Goyt, Herdan and Pudwell. Let $\op_{n,k}(p)$ be the number of ordered set partitions of $\{1,2,\ldots,n\}$ into $k$ blocks that…

Combinatorics · Mathematics 2013-07-02 Anisse Kasraoui

The sequence of partial sums of Fibonacci numbers, beginning with $2$, $4$, $7$, $12$, $20$, $33,\dots$, has several combinatorial interpretations (OEIS A000071). For instance, the $n$-th term in this sequence is the number of length-$n$…

Combinatorics · Mathematics 2025-03-17 Erik Bates , Blan Morrison , Mason Rogers , Arianna Serafini , Anav Sood

We study the problem of counting alternating permutations avoiding collections of permutation patterns including 132. We construct a bijection between the set S_n(132) of 132-avoiding permutations and the set A_{2n + 1}(132) of alternating,…

Combinatorics · Mathematics 2021-03-30 Joel Brewster Lewis

We introduce new avoidability problems for words by considering equivalence relations, k-abelian equivalences, which lie properly in between equality and commutative equality, i.e. abelian equality. For two k-abelian equivalent words the…

Combinatorics · Mathematics 2015-03-19 Mari Huova , Juhani Karhumäki

It is well known that permutations avoiding any 3-length pattern are enumerated by the Catalan numbers. If the three patterns 123, 132 and 213 are avoided at the same time we obtain a class of permutations enumerated by the Fibonacci…

Combinatorics · Mathematics 2007-05-23 E. Barcucci , A. Bernini , M. Poneti

Recently, Babson and Steingrimsson (see \cite{BS}) introduced generalized permutations patterns that allow the requirement that two adjacent letters in a pattern must be adjacent in the permutation. Following \cite{BCS}, let $e_k\pi$…

Combinatorics · Mathematics 2007-05-23 T. Mansour

The extension of pattern avoidance from ordinary permutations to those on multisets gave birth to several interesting enumerative results. We study permutations on regular multisets, i.e., multisets in which each element occurs the same…

Combinatorics · Mathematics 2013-06-21 Marie-Louise Bruner

We prove that the number of permutations which avoid 132-patterns and have exactly one 123-pattern equals (n-2)2^(n-3). We then give a bijection onto the set of permutations which avoid 123-patterns and have exactly one 132-pattern.…

Combinatorics · Mathematics 2007-05-23 Aaron Robertson