A Combinatorial Proof for 132-Avoiding Permutations with a Unique Longest Increasing Subsequence
Combinatorics
2023-03-07 v1
Abstract
We provide a simple injective proof that the number of 132-avoiding permutations with a unique longest increasing subsequence is at least as large as the number of 132-avoiding permutations without a unique longest increasing subsequence.
Cite
@article{arxiv.2303.02808,
title = {A Combinatorial Proof for 132-Avoiding Permutations with a Unique Longest Increasing Subsequence},
author = {Nicholas Van Nimwegen},
journal= {arXiv preprint arXiv:2303.02808},
year = {2023}
}
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4 Pages