Numbers with countable expansions in base of generalized golden ratios

TL;DR

This paper generalizes previous results on the number of expansions in base of generalized golden ratios, showing that certain algebraic numbers have countably many expansions while others have continuum many, thus solving an open question.

Contribution

It extends the understanding of expansions in bases of generalized golden ratios, characterizing numbers with countably many expansions and solving an open problem by Baker.

Findings

Numbers of the form (pβ+q)/ (k+1)^n have countably many expansions.

Other elements in the interval have continuum many expansions.

The results generalize known cases for the classical golden ratio.

Abstract

Sidorov and Vershik showed that in base $G=\frac{\sqrt{5}+1}{2}$ and with the digits $0,1$ the numbers $x=nG ~(\text {mod} 1)$ have $\aleph_{0}$ expansions for any $n\in\mathbb{Z}$, while the other elements of $(0, \frac{1}{G-1})$ have $2^{\aleph_{0}}$ expansions. In this paper, we generalize this result to the generalized golden ratio base $\beta=\mathcal{G}(m)$. With the digit-set $\{0,1,\cdots, m\}$, if $m=2k+1$, $\mathcal{G}(m)=\frac{k+1+\sqrt{k^{2}+6k+5}}{2}$, the numbers $x=\frac{p\beta+q}{(k+1)^{n}}\in(0, \frac{m}{\beta-1})$ (where $n, p, q\in\mathbb{Z}$) have $\aleph_{0}$ expansions, while the other elements of $(0, \frac{m}{\beta-1})$ have $2^{\aleph_{0}}$ expansions; if $m=2k$, $\mathcal{G}(m)=k+1$, the numbers with countably many expansions are $\frac{p}{(k+1)^{n}}\in(0, 2) ~(n, p\in\mathbb{N}\cup\{0\})$. This solves an open question by Baker.