Upper bound for the number of privileged words

TL;DR

This paper establishes a new upper bound on the number of privileged words of length n over an alphabet of size q, improving previous bounds by incorporating iterated logarithms.

Contribution

It provides a nontrivial upper bound for the number of privileged words, advancing the understanding of their combinatorial properties and addressing an open problem from prior research.

Findings

Derived an upper bound involving iterated logarithms for privileged words

Improved upon the previous upper bound by Rukavicka (2020)

Applicable for alphabet sizes greater than one and sufficiently large n

Abstract

A non-empty word $w$ is a \emph{border} of a word $u$ if $\vert w\vert<\vert u\vert$ and $w$ is both a prefix and a suffix of $u$. A word $u$ is \emph{privileged} if $\vert u\vert\leq 1$ or if $u$ has a privileged border $w$ that appears exactly twice in $u$. Peltom\"aki (2016) presented the following open problem: ``Give a nontrivial upper bound for $B(n)$'', where $B(n)$ denotes the number of privileged words of length $n$. Let $\ln^{[0]}{(n)}=n$ and let $\ln^{[j]}{(n)}=\ln{(\ln^{[j-1]}{(n)})}$, where $j,n$ are positive integers. We show that if $q>1$ is a size of the alphabet and $j\geq 3$ is an integer then there are constants $\alpha_j$ and $n_j$ such that \[B(n)\leq \alpha_j\frac{q^{n}\sqrt{\ln{n}}}{\sqrt{n}}\ln^{[j]}{(n)}\prod_{i=2}^{j-1}\sqrt{\ln^{[i]}(n)}\mbox{, where }n\geq n_j\mbox{.}\] This result improves the upper bound of Rukavicka (2020).