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Subexponential upper bound on the number of rich words

Combinatorics 2025-11-17 v1 Discrete Mathematics

Abstract

Let R(n)R(n) denote the number of rich words of length nn over a given finite alphabet. In 2017 it was proved that limnR(n)n=1\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)R(n) has been presented. The current paper fills this gap. Let 12<λ<1\frac{1}{2}<\lambda<1 and γ>1\gamma>1 be real constants, let qq be the size of the alphabet, and let ϕ\phi be a positive function with limnϕ(n)=\lim_{n\rightarrow\infty}\phi(n)=\infty and limnnϕ(n)=\lim_{n\rightarrow\infty}\frac{n}{\phi(n)}=\infty. Let ln(x)\ln^*(x) denote the iterated logarithm of x>0x>0. We prove that there are n0n_0 and c>0c>0 such that if n>n0n>n_0, f(n)=cln(nϕ(n)lnq)γ\mboxandB(n)=qnϕ(n)+n(2λ)f(n)1\mboxf(n)=\sqrt[\gamma]{c\ln^*{(\frac{n}{\phi(n)}}\ln{q})}\quad\mbox{ and }\quad B(n)=q^{\frac{n}{\phi(n)}+\frac{n}{(2\lambda)^{f(n)-1}}}\mbox{} then limnB(n)n=1\lim_{n\rightarrow\infty}\sqrt[n]{B(n)}=1 and R(n)B(n)R(n)\leq B(n).

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Cite

@article{arxiv.2511.11069,
  title  = {Subexponential upper bound on the number of rich words},
  author = {Josef Rukavicka},
  journal= {arXiv preprint arXiv:2511.11069},
  year   = {2025}
}