Erdős Problem #13 — Largest set in [1,x][1,x] with no member dividing a sum of two larger

About 53 years old · traced to

Let a1<⋯<ak≤xa_1 < \cdots < a_k \leq x be a sequence of integers where no aa divides the sum of two larger aa's. Probably

max⁡k=x/3+o(1).\max k = x/3 + o(1).
References

Additional references

P. Erdős, Problems and results on combinatorial number theory, A Survey of Combinatorial Theory (Proc. Internat. Sympos., Colorado State Univ., 1971), North-Holland (1973), 117-138.

Progress summary

Refreshed
Claimed solved

A 2023 paper claims the conjectured one-third limit is correct, with an exact bound for all sufficiently large intervals.

Erdős and Sárközy asked whether every admissible set in {1,…,n}\{1,\ldots,n\} has size at most ⌊n/3⌋+1\lfloor n/3\rfloor+1; Erdős later offered a prize for the weaker bound n/3+Cn/3+C.

Known results

  • Szemerédi proved that sufficiently large sets contain distinct x,y,zx,y,z with z∣x+yz\mid x+y, though not necessarily with (x+y)/z=2(x+y)/z=2.

January 17, 2023 claimed solution

Benjamin Bedert's paper claims ∣A∣≤n/3+C|A|\le n/3+C for an absolute constant and, for all sufficiently large nn, the sharper bound ∣A∣≤⌈n/3⌉|A|\le\lceil n/3\rceil. The construction A={⌊2n/3⌋+1,…,n}A=\{\lfloor2n/3\rfloor+1,\ldots,n\} shows sharpness, so it settles the stated asymptotic conjecture if accepted.

Current status (as of September 2026): The asymptotic conjecture is claimed solved by Bedert's paper, with the sharp bound ⌈n/3⌉\lceil n/3\rceil for sufficiently large nn; independent verification was not found.

Sources

Solutions 0

No solutions have been posted yet.