Erdős Problem #1096

declared status 'proved'. Formalized: yes. Let $1<q<1+\epsilon$ and consider the set of numbers of the shape $\sum_{i\in S}q^i$ (for all finite $S$), ordered by size as $0=x_1<x_2<\cdots$. Is it true that, provided $\epsilon>0$ is sufficiently small, $x_{k+1}-x_k \to 0$? Current best: In [EJK90] Erd\H{o}, Jo\'{o}, and Komornik prove that any Pisot-Vijayaraghavan number cannot have this property, and also prove that, for any $1<q\leq 2$, $x_{k+1}-x_k\leq 1$ for all $k$. Bugeaud [Bu96] proved that $1<q\leq 2$ is a Pisot-Vijayaraghavan number if and only if$$\liminf (x_{k+1}^m-x_k^m)>0$$for all $m\geq 1$, where $x_k^m$ is the set of those numbers which can be written as a finite sum $\sum_{n\geq 0}c_nq^n$ for some $c_n\in \{0,\ldots,m\}$ (so that the sequence in the question is $x_k^1$). Prize: no. Tags: number theory.

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