Problem Statement
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, $x_{k+1}-x_k \to 0$?
Is it true that, provided $\epsilon>0$ is sufficiently small, $x_{k+1}-x_k \to 0$?
Categories:
Number Theory
Progress
A problem of Erdős and Joó posed in the 1991 problem session of Great Western Number Theory.They speculate that the threshold may be $q_0$, where $q_0\approx 1.3247$ is the real root of $x^3=x+1$, and is the smallest Pisot-Vijayaraghavan number.
In [EJK90] Erdő, Joó, 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 $k$.
The sequence always begins $0,1,q$.
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 $x_k^1$). Erdős, Joó, and Schnitzer [EJS96] improved this to show that, if $1<q<(1+\sqrt{5})/2$, then $q$ is a Pisot-Vijayaraghavan number if and only if\[\liminf (x_{k+1}^2-x_k^2)>0.\]
Source: erdosproblems.com/1096 | Last verified: January 19, 2026