广义黄金比例基上的可数展开数
Numbers with countable expansions in base of generalized golden ratios
摘要：
Sidorov和Vershik表明，在基$G=\frac{\sqrt{5}+1}{2}$和数字$0,1$中，数字$x=nG ~(\text {mod} 1)$对任何$n\in\mathbb{Z}$都有$\aleph_{0}$扩展，而$(0, \frac{1}{G-1})$的其他元素都有$2^{\aleph_{0}}$扩展。本文将这一结果推广到广义黄金比例基$β=\mathcal{G}(m)$上。对于数字集$\{0,1,\cdots, m\}$，如果$m=2k+1$，$\mathcal{G}(m)=\frac{k+1+\sqrt{k^{2}+6k+5}}{2}$，则数字$x=\frac{pβ+q}{(k+1)^{n}}\in(0, \frac{m}{β-1})$（其中$n, p, q\in\mathbb{Z}$）具有$\aleph_{0}$展开式，而$(0, \frac{m}{β-1})$的其他元素具有$2^{\aleph_{0}}$展开式;如果$m=2k$，$\mathcal{G}(m)=k+1$，则具有可数个展开式的数字是$\frac{p}{(k+1)^{n}}\in(0, 2) ~(n, p\in\mathbb{N}\cup\{0\})$。这解决了贝克提出的一个悬而未决的问题。
英文摘要：
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 $β=\mathcal{G}(m)$. 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 $x=\frac{pβ+q}{(k+1)^{n}}\in(0, \frac{m}{β-1})$ (where $n, p, q\in\mathbb{Z}$) have $\aleph_{0}$ expansions, while the other elements of $(0, \frac{m}{β-1})$ 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.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
