arXivDaily arXiv每日学术速递 周一至周五更新
arXiv周末暂无论文更新,休息一下吧,周末愉快~~
arXiv 1507.06154math.CO
收藏

避免在交替的单词上出现纽状模式

Avoiding vincular patterns on alternating words

Alice L. L. Gao, Sergey Kitaev, Philip B. Zhang

摘要:

如果是$w_1<w_2>w_3<w_4>\cdots$(当单词为上下时)或$w_1>w_2<w_3>w_4<\cdots$(当单词为上下时),则单词$w=w_1w_2\cdots w_n$是交替的。作者在~ \cite{GKZ}中开始了对避免经典排列模式的交替字的研究,其中特别指出,避免偶数长度的123个上下字由Narayana数计数。然而,对123个回避上下字的结构了解不多。在本文中,我们填补了这一空白,引入了一个切割对的概念,使我们能够细分成等价类的一组字的问题。我们提供了一个组合论证,证明了等价类的个数是由Catalan数给出的,从而导出了\cite{GKZ}中相应结果的另一个(组合)证明.此外,我们推广的枚举结果在~ \cite{GKZ}的情况下,交替的话,避免了一个vincular模式的长度为3。我们表明,它是足够的,以避免连续模式$\underline{132}$和避免连续模式$\underline{312}$,以回答我们所有的枚举问题的奇数长度的上下字的枚举长度的上下字。两种关键情形中的前者由第二类Stirling数枚举。

英文摘要:

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 initiated by the authors in~\cite{GKZ}, where, in particular, it was shown that 123-avoiding up-down words of even length are counted by the Narayana numbers. However, not much was understood on the structure of 123-avoiding up-down words. In this paper, we fill in this gap by introducing the notion of a cut-pair that allows us to subdivide the set of words in question into equivalence classes. We provide a combinatorial argument to show that the number of equivalence classes is given by the Catalan numbers, which induces an alternative (combinatorial) proof of the corresponding result in~\cite{GKZ}. Further, we extend the enumerative results in~\cite{GKZ} to the case of alternating words avoiding a vincular pattern of length 3. We show that it is sufficient to enumerate up-down words of even length avoiding the consecutive pattern $\underline{132}$ and up-down words of odd length avoiding the consecutive pattern $\underline{312}$ to answer all of our enumerative questions. The former of the two key cases is enumerated by the Stirling numbers of the second kind.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

↑