We use cookies to distinguish you from other users and to provide you with a better experience on our websites. Close this message to accept cookies or find out how to manage your cookie settings.
We use cookies to distinguish you from other users and to provide you with a better experience on our websites. Close this message to accept cookies or find out how to manage your cookie settings.
Let 
$q\in (1,2)$ . A 
$q$ -expansion of a number 
$x$ in 
$[0,1/(q-1)]$ is a sequence 
$({\it\delta}_{i})_{i=1}^{\infty }\in \{0,1\}^{\mathbb{N}}$ satisfying Let 
$$\begin{eqnarray}x=\mathop{\sum }_{i=1}^{\infty }\frac{{\it\delta}_{i}}{q^{i}}.\end{eqnarray}$$
${\mathcal{B}}_{\aleph _{0}}$ denote the set of 
$q$ for which there exists 
$x$ with a countable number of 
$q$ -expansions, and let 
${\mathcal{B}}_{1,\aleph _{0}}$ denote the set of 
$q$ for which 
$1$ has a countable number of 
$q$ -expansions. In Erdős et al [On the uniqueness of the expansions 
$1=\sum _{i=1}^{\infty }q^{-n_{i}}$ . Acta Math. Hungar.58 (1991), 333–342] it was shown that 
$\min {\mathcal{B}}_{\aleph _{0}}=\min {\mathcal{B}}_{1,\aleph _{0}}=(1+\sqrt{5})/2$ , and in S. Baker [On small bases which admit countably many expansions. J. Number Theory147 (2015), 515–532] it was shown that 
${\mathcal{B}}_{\aleph _{0}}\cap ((1+\sqrt{5})/2,q_{1}]=\{q_{1}\}$ , where 
$q_{1}\,({\approx}1.64541)$ is the positive root of 
$x^{6}-x^{4}-x^{3}-2x^{2}-x-1=0$ . In this paper we show that the second smallest point of 
${\mathcal{B}}_{1,\aleph _{0}}$ is 
$q_{3}\,({\approx}1.68042)$ , the positive root of 
$x^{5}-x^{4}-x^{3}-x+1=0$ . En route to proving this result, we show that 
${\mathcal{B}}_{\aleph _{0}}\cap (q_{1},q_{3}]=\{q_{2},q_{3}\}$ , where 
$q_{2}\,({\approx}1.65462)$ is the positive root of 
$x^{6}-2x^{4}-x^{3}-1=0$ .
