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Subexponential upper bound on the number of rich words

Published: November 14, 2025 | arXiv ID: 2511.11069v1

By: Josef Rukavicka

Potential Business Impact:

Finds a way to count special word patterns.

Business Areas:
A/B Testing Data and Analytics

Let $R(n)$ denote the number of rich words of length $n$ over a given finite alphabet. In 2017 it was proved that $\lim_{n\rightarrow\infty} \sqrt[n]{R(n)}=1$; it means the number of rich words has a subexponential growth. However, up to now, no subexponential upper bound on $R(n)$ has been presented. The current paper fills this gap. Let $\frac{1}{2}<λ<1$ and $γ>1$ be real constants, let $q$ be the size of the alphabet, and let $φ$ be a positive function with $\lim_{n\rightarrow\infty}φ(n)=\infty$ and $\lim_{n\rightarrow\infty}\frac{n}{φ(n)}=\infty$. Let $\ln^*(x)$ denote the iterated logarithm of $x>0$. We prove that there are $n_0$ and $c>0$ such that if $n>n_0$, \[f(n)=\sqrt[γ]{c\ln^*{(\frac{n}{φ(n)}}\ln{q})}\quad\mbox{ and }\quad B(n)=q^{\frac{n}{φ(n)}+\frac{n}{(2λ)^{f(n)-1}}}\mbox{}\] then $\lim_{n\rightarrow\infty}\sqrt[n]{B(n)}=1$ and $R(n)\leq B(n)$.

Page Count
12 pages

Category
Mathematics:
Combinatorics