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Related papers: Some bijections for restricted Motzkin paths

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We consider a certain abstract of RNA secondary structures, which is closely related to RNA shapes. The generating function counting the number of the abstract structures is obtained by means of Narayana numbers and 2-Motzkin paths, through…

Combinatorics · Mathematics 2019-07-18 Sang Kwan Choi

Chen and collaborators give a recursively defined bijection from 021-avoiding ascent sequences to 021-avoiding (aka 132-avoiding) permutations. Here we give an algorithmic bijection from 021-avoiding ascent sequences to Dyck paths. Our…

Combinatorics · Mathematics 2014-02-25 David Callan

A lattice-path description of $K$-restricted jagged partitions is presented. The corresponding lattice paths can have peaks only at even $x$ coordinate and the maximal value of the height cannot be larger than $K-1$. Its weight is twice…

Combinatorics · Mathematics 2007-05-23 P. Jacob , P. Mathieu

Counting the number of permutations of a given total displacement is equivalent to counting weighted Motzkin paths of a given area (Guay-Paquet and Petersen, 2014). The former combinatorial problem is still open. In this work, we show that…

Data Structures and Algorithms · Computer Science 2020-08-27 Andreas Bärtschi , Barbara Geissmann , Daniel Graf , Tomas Hruz , Paolo Penna , Thomas Tschager

We present a substantial generalization of the equinumeracy of grand Dyck paths and Dyck-path prefixes, constrained within a band. The number of constrained paths starting at level $i$ and ending in a window of size $2j+2$ is equal to the…

Combinatorics · Mathematics 2021-02-02 Nachum Dershowitz

A {\em Motzkin path} of length $n$ is a lattice path from $(0,0)$ to $(n,0)$ in the plane integer lattice $\mathbb{Z}\times\mathbb{Z}$ consisting of horizontal-steps $(1, 0)$, up-steps $(1,1)$, and down-steps $(1,-1)$, which never passes…

Combinatorics · Mathematics 2008-05-29 Yidong Sun

We give a combinatorial interpretation of a matrix identity on Catalan numbers and the sequence $(1, 4, 4^2, 4^3, ...)$ which has been derived by Shapiro, Woan and Getu by using Riordan arrays. By giving a bijection between weighted partial…

Combinatorics · Mathematics 2007-05-23 William Y. C. Chen , Nelson Y. Li , Louis W. Shapiro , Sherry H. F. Yan

We introduce a subfamily of skew Dyck paths called box paths and show that they are in bijection with pairs of ternary trees, confirming an observation stated previously on the On-Line Encyclopedia of Integer Sequences. More generally, we…

Combinatorics · Mathematics 2024-01-23 Yuxuan Zhang , Yan Zhuang

We present an extension of a theorem by Michael Drmota and Mich\`ele Soria [Images and Preimages in Random Mappings, 1997] that can be used to identify the limiting distribution for a class of combinatorial schemata. This is achieved by…

Combinatorics · Mathematics 2016-05-11 Michael Wallner

This paper develops a structural theory of unique shortest paths in real-weighted graphs. Our main goal is to characterize exactly which sets of node sequences, which we call path systems, can be realized as unique shortest paths in a graph…

Data Structures and Algorithms · Computer Science 2018-10-18 Greg Bodwin

A bijection is presented between (1): partitions with conditions $f_j+f_{j+1}\leq k-1$ and $ f_1\leq i-1$, where $f_j$ is the frequency of the part $j$ in the partition, and (2): sets of $k-1$ ordered partitions $(n^{(1)}, n^{(2)}, ...,…

Combinatorics · Mathematics 2008-01-15 P Jacob , P. Mathieu

The theme of this article is a "reciprocity" between bounded up-down paths and bounded alternating sequences. Roughly speaking, this ``reciprocity" manifests itself by the fact that the extension of the sequence of numbers of paths of…

Combinatorics · Mathematics 2024-07-30 Johann Cigler , Christian Krattenthaler

The number of inversion sequences avoiding two patterns $101$ and $102$ is known to be the same as the number of permutations avoiding three patterns $2341$, $2431$, and $3241$. This sequence also counts the number of Schr\"{o}der paths…

Combinatorics · Mathematics 2024-04-08 JiSun Huh , Sangwook Kim , Seunghyun Seo , Heesung Shin

A variation of Dyck paths allows for down-steps of arbitrary length, not just one. Credits for this invention are given to Emeric Deutsch. Surprisingly, the enumeration of them is somewhat akin to the analysis of Motzkin-paths; the last…

Combinatorics · Mathematics 2020-04-16 Helmut Prodinger

In this article, we study the enumeration by length of several walk models on the square lattice. We obtain bijections between walks in the upper half-plane returning to the $x$-axis and walks in the quarter plane. A recent work by Bostan,…

Combinatorics · Mathematics 2020-02-18 Frédéric Chyzak , Karen Yeats

We consider Motzkin paths of length $L$, not fixed at zero at both end points, with constant weights on the edges and general weights on the end points. We investigate, as the length $L$ tends to infinity, the limit behaviors of (a)…

Probability · Mathematics 2023-09-12 Wlodzimierz Bryc , Yizao Wang

A notion of cyclic descents on standard Young tableaux (SYT) of rectangular shape was introduced by Rhoades, and extended to certain skew shapes by Adin, Elizalde and Roichman. The cyclic descent set restricts to the usual descent set when…

Combinatorics · Mathematics 2023-01-04 Bin Han

A k-triangulation of a convex polygon is a maximal set of diagonals so that no k+1 of them mutually cross in their interiors. We present a bijection between 2-triangulations of a convex n-gon and pairs of non-crossing Dyck paths of length…

Combinatorics · Mathematics 2007-05-23 Sergi Elizalde

We consider Dyck paths having height at most two with some constraints on the number of consecutive valleys at height one which must be followed by a suitable number of valleys at height zero. We prove that they are enumerated by so-called…

Discrete Mathematics · Computer Science 2024-06-25 Elena Barcucci , Antonio Bernini , Stefano Bilotta , Renzo Pinzani

Suppose 2n voters vote sequentially for one of two candidates. For how many such sequences does one candidate have strictly more votes than the other at each stage of the voting? The answer is \binom{2n}{n} and, while easy enough to prove…

Combinatorics · Mathematics 2007-05-23 Glenn Hurlbert , Vikram Kamat
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