避免123的非嵌套排列
Nonnesting permutations avoiding 123
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中文总结 AI 辅助
本文解决了多重集{1,1,2,2,...,n,n}上避免子序列123的非嵌套排列的计数问题，通过构造生成树并两次应用核方法，给出了生成函数的代数表达式及渐近公式。
中文摘要 AI 辅助
多重集{1,1,2,2,...,n,n}的非嵌套排列是一个不包含子序列abba（其中a与b不同）的单词。Elizalde和Luo枚举了避免每个包含两个或更多长度为三的模式集合的非嵌套排列，并留下了避免单一模式123的枚举问题（他们论文中的问题1）。我们解决了这个问题。记$c_n$为{1,1,...,n,n}中避免123的非嵌套排列的数量，$C(z)=\sum_{n\ge 0}c_nz^n$，我们证明$C(z) = 1 + z(1+U)/(2 - z(1+U)(1+V))$，其中$V = 1+zV^2$和$U = 1+3z+zU^2$是常数项为1的形式幂级数分支。特别地，$C$在$\mathbb{Q}(z)$上是次数恰好为4的代数函数，且$c_n \sim \frac{9+6\sqrt{3}}{8\sqrt{\pi}}\\, 6^n n^{-3/2}$。证明为这些对象构造了一个生成树，其标签由两个非负整数组成，并两次应用核方法。
英文摘要
A nonnesting permutation of the multiset {1,1,2,2,...,n,n} is a word containing no subsequence abba with a distinct from b. Elizalde and Luo enumerated nonnesting permutations avoiding each set of two or more patterns of length three, and left open the enumeration of those avoiding the single pattern 123 (Problem 1 of their paper). We solve this problem. Writing $c_n$ for the number of nonnesting permutations of {1,1,...,n,n} avoiding 123 and $C(z)=\sum_{n\ge 0}c_nz^n$, we prove $C(z) = 1 + z(1+U)/(2 - z(1+U)(1+V))$, where $V = 1+zV^2$ and $U = 1+3z+zU^2$ are the formal power series branches with constant term 1. In particular $C$ is algebraic of degree exactly 4 over $\mathbb{Q}(z)$, and $c_n \sim \frac{9+6\sqrt{3}}{8\sqrtπ}\, 6^n n^{-3/2}$. The proof constructs a generating tree for these objects whose labels consist of two nonnegative integers, and applies the kernel method twice.
