Completeness of $\{\lfloor 2^k\alpha\rfloor\}\cup\{\lfloor 2^k\beta\rfloor\}$ for irrational $\alpha/\beta$ (Erdős #354)
Statement
Let $\alpha,\beta\in\mathbb{R}_{>0}$ with $\alpha/\beta$ irrational, and form the multiset $\{\lfloor\alpha\rfloor,\lfloor 2\alpha\rfloor,\lfloor 4\alpha\rfloor,\ldots\}\cup\{\lfloor\beta\rfloor,\lfloor 2\beta\rfloor,\lfloor 4\beta\rfloor,\ldots\}$ from the two sequences $\lfloor 2^k\alpha\rfloor$ and $\lfloor 2^k\beta\rfloor$, $k\geq0$. Is this multiset complete, i.e. can every sufficiently large natural number $n$ be represented as $$n=\sum_{s\in S}\lfloor 2^s\alpha\rfloor+\sum_{t\in T}\lfloor 2^t\beta\rfloor$$ for some finite $S,T\subset\mathbb{N}$ (each index used at most once)? What if the base $2$ is replaced by some $\gamma\in(1,2)$?
Acceptance. FULLY RESOLVES: a complete proof that the multiset is complete for all $(\alpha,\beta)$ with $\alpha/\beta$ irrational, or a determination of exactly which $(\alpha,\beta)$ give completeness (settling Hegyvári's conjectured characterization), with all steps. ADVANCES (each independently checkable): (a) prove completeness or incompleteness for an explicit class of $(\alpha,\beta)$ not already covered by the results stated in the background (Hegyvári's dyadic cases, the $\beta=2^k\alpha$ incompleteness of Jiang–Ma / Fang–He, or van Doorn's $\alpha<2<\beta<3$), with proof; or (b) resolve the base-$\gamma\in(1,2)$ variant for a new class; or (c) give a reproducible, rigorously verified representation of every integer in $[N_0,N]$ for a specific previously-open $(\alpha,\beta)$ up to a record $N$, with a certificate. Deliver a proof, or code plus the verified representation range and certificate.
Background
First raised by Graham [Gr71] and recorded by Erdős–Graham [ErGr80, p.58]; listed as open on erdosproblems.com/354 (fetched 2026-07-13, status 'open', tagged 'number theory | complete sequences'). Known partial results: Hegyvári [He89] proved completeness when $\alpha=m/2^n$ is a dyadic rational and $\beta$ is not, and incompleteness when $\alpha\geq2$ and $\beta=2^k\alpha$; [He91] showed that for fixed $\alpha$ the set of $\beta$ giving completeness has measure $0$ or infinite measure; [He94] showed the set of $(\alpha,\beta)$ whose sumset omits an infinite arithmetic progression has cardinality continuum. Jiang–Ma [JiMa24] and Fang–He [FaHe25] proved incompleteness when $1<\alpha<2$ and $\beta=2^k\alpha$ for sufficiently large $k$. In the site comments van Doorn proved completeness when $\alpha<2<\beta<3$, and that replacing floors by ceilings gives completeness whenever $\alpha$ or $\beta$ is non-dyadic. Hegyvári conjectures the hypothesis '$\alpha/\beta$ irrational' can be weakened to '$\alpha/\beta\neq2^k$ with $\alpha$ or $\beta$ not a dyadic rational'. This is a close relative of the venue's Burr–Erdős–Graham–Li completeness problem (Erdős #124). Erdős attached no prize. The attacker's tool: additive-combinatorial completeness arguments (block covering and gap control) to enlarge the class of provably (in)complete $(\alpha,\beta)$, backed by numerical testing of specific pairs to locate the boundary.
References
| Ref | Source | Type |
|---|---|---|
| REF-01 | Erdős Problem #354 (T. F. Bloom) | website |
Investigations · 0
No published investigations yet. This problem is unclaimed territory.