Problem Statement
Let $\alpha,\beta\in \mathbb{R}_{>0}$ such that $\alpha/\beta$ is irrational. Is the multiset\[\{ \lfloor \alpha\rfloor,\lfloor 2\alpha\rfloor,\lfloor 4\alpha\rfloor,\ldots\}\cup \{ \lfloor \beta\rfloor,\lfloor 2\beta\rfloor,\lfloor 4\beta\rfloor,\ldots\}\]complete? That is, can all sufficiently large natural numbers $n$ be written as\[n=\sum_{s\in S}\lfloor 2^s\alpha\rfloor+\sum_{t\in T}\lfloor 2^t\beta\rfloor\]for some finite $S,T\subset \mathbb{N}$?
What if $2$ is replaced by some $\gamma\in(1,2)$?
What if $2$ is replaced by some $\gamma\in(1,2)$?
Categories:
Number Theory Complete Sequences
Progress
This question was first mentioned by Graham [Gr71].Hegyvári [He89] proved that this holds if $\alpha=m/2^n$ is a dyadic rational and $\beta$ is not. He later [He91] proved that, for any fixed $\alpha>0$, the set of $\beta$ for which this holds either has measure $0$ or infinite measure. In [He94] he proved that the set of $(\alpha,\beta)$ for which the corresponding set of sums does not contain an infinite arithmetic progression has cardinality continuum.
Hegyvári [He89] proved that the sequence is not complete if $\alpha\geq 2$ and $\beta =2^k\alpha$ for some $k\geq 0$. Jiang and Ma [JiMa24] and Fang and He [FaHe25] prove that the sequence is not complete if $1<\alpha<2$ and $\beta=2^k\alpha$ for some sufficiently large $k$.
It is likely (and Hegyvári conjectures) that the assumption $\alpha/\beta$ irrational can be weakened to $\alpha/\beta \neq 2^k$ and either $\alpha$ or $\beta$ not a dyadic rational.
In the comments van Doorn proves the sequence is complete if $\alpha < 2<\beta<3$, and also proves that if either $\alpha$ or $\beta$ is not a dyadic rational then the corresponding sequence with ceiling functions replacing the floor functions is complete.
Source: erdosproblems.com/354 | Last verified: January 14, 2026