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# Journal of Combinatorics

## Volume 8 (2017)

### Number 4

### The number of permutations with the same peak set for signed permutations

Pages: 631 – 652

DOI: https://dx.doi.org/10.4310/JOC.2017.v8.n4.a4

#### Authors

#### Abstract

A signed permutation $\pi = \pi_1 \pi_2 \dotsc \pi_n$ in the hyperoctahedral group $B_n$ is a word such that each $\pi_i \in \lbrace \bar{n}, \dotsc , \bar{1}, 1, \dotsc , n \rbrace$ where $i = \bar{i}$, and $\lbrace \lvert \pi_1 \rvert , \lvert \pi_2 \rvert , \dotsc , \lvert \pi n \rvert \rbrace = \lbrace 1, 2, \dotsc , n \rbrace$. An index $i$ is a peak of $\pi$ if $\pi_{i-1} \lt \pi_i \gt \pi_{i+1}$ and $P_B(\pi)$ denotes the set of all peaks of $\pi$. Given any set $S$, we define $P_B(S, n)$ to be the set of signed permutations $\pi \in B_n$ with $P_B(\pi) = S$. In this paper we show that $\lvert P_B(S, n) \rvert = p(n)2^{2n - \lvert S \rvert - 1}$ where $p(n)$ is a polynomial that takes integral values when evaluated at integers. In addition, we have partially extended these results to the more complicated case where we add $\pi_0 = 0$ at the beginning of the permutations, which gives rise to the possibility of a peak at position $1$, for both the symmetric and the hyperoctahedral groups. In both cases we establish recursive formulae to compute the number of permutations (signed permutations in the case of $B_n$) with a given peak set.

#### Keywords

peak sets, permutations, signed permutations

Received 17 August 2015

Published 17 July 2017