Erdős Problem #12 — Reciprocal sums when no term divides a sum of two larger terms

About 53 years old · traced to

Let a1<⋯a_1 < \cdots be an infinite sequence of integers where no aia_i divides the sum of two greater aa's. Sárközi and I proved that the aa's then have density 00 and this result is best possible (Erdös and Sárközi [1970]). Probably ∑1/ai<∞\sum 1/a_i < \infty holds.

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
Open

A DeepMind-attributed construction claims that dense divisor-avoiding sets exist and defeats the proposed power-saving bound, but the proof is not yet independently checkable.

Erdős and Sárközy asked whether such a set can have size about the square root of NN, and whether every such set must be much smaller than NN infinitely often. The problem page records affirmative and negative answers claimed through a DeepMind construction, while still labeling the problem open.

Known results

  • Erdős and Sárközy, 1970: every such infinite set has density 00.
  • Erdős and Sárközy, 1970: examples have ∣A∩[1,N]∣>N/f(N)|A\cap[1,N]|>N/f(N) infinitely often for every f(N)→∞f(N)\to\infty.
  • Elsholtz and Planitzer, 2017: ∣A∩[1,N]∣≫N1/2/((log⁡N)1/2(log⁡log⁡N)2(log⁡log⁡log⁡N)2)|A\cap[1,N]|\gg N^{1/2}/((\log N)^{1/2}(\log\log N)^2(\log\log\log N)^2).
  • Schoen, 2001, and Baier, 2004: pairwise-coprime sets satisfy upper bounds ≪N2/3\ll N^{2/3} and ≪N2/3/log⁡N\ll N^{2/3}/\log N infinitely often.

April 2026 DeepMind construction

The construction is reported to give ∣A∩[1,N]∣≥N/(log⁡N)O(log⁡log⁡log⁡N)|A\cap[1,N]|\ge N/(\log N)^{O(\log\log\log N)} for all sufficiently large NN. Thus the first question is answered yes and the second no. The associated Lean declarations remain by sorry, so this is an unverified claim rather than a checkable resolution.

Current status (as of April 2026): The reported construction settles both growth questions if validated, but no independently checkable proof was found; the reciprocal-sum question remains open.

Sources

Solutions 0

No solutions have been posted yet.