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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. The answer to Problem 1196 is yes, with the constant of the secondary term determined: writing S(x)S(x) for the supremum of ∑a∈A1/(alog⁡a)\sum_{a\in A}1/(a\log a) over the primitive sets A⊂[x,∞)A\subset[x,\infty),

S(x)≤1+γlog⁡x+O ⁣(1log⁡2x),S(x)\le1+\frac{\gamma}{\log x}+O\!\left(\frac{1}{\log^{2}x}\right),

which gives the asked bound 1+o(1)1+o(1) as x→∞x\to\infty. Nat Sothanaphan posted three dated notes in the problem's discussion thread, each linked above with its post. The first, A short note on the secondary constant in the divisibility-chain bound (posted 16 April 2026, printed date 17 April 2026), optimizes the divisibility-chain argument recorded on Price's page: its Theorem 1 says that for a fixed threshold Y≥2Y\ge2, with cY=∑q<YΛ(q)/qc_Y=\sum_{q<Y}\Lambda(q)/q, one has S(x)≤1+(γ+cY)/log⁡x+O(1/log⁡2x)S(x)\le1+(\gamma+c_Y)/\log x+O(1/\log^{2}x), so the choice Y=2Y=2, where c2=0c_2=0, gives the displayed bound, and γ\gamma is the best constant this fixed-threshold variant of the method yields. The second, A divisor-coupling lemma and primitive sets above xx (posted 20 April 2026), recasts Terence Tao's flow-graph version of the thread's argument as a pure resummation: its Lemma 2 bounds ν(A)\nu(A), for a primitive A⊂(x,∞)A\subset(x,\infty) and a nonnegative measure ν\nu on N\mathbb N, by the mass that a measure on the pairs (n,m)(n,m) with m∣nm\mid n, m<nm<n gives to the pairs crossing xx, plus the positive parts of the differences between ν\nu and the measure's two marginals above xx; its Theorem 1 derives $\sum_{a\in A}1/(a\log a)\le 1+O(1/\log x)$ from it, without naming flow graphs or Markov chains. The third, A logarithmic refinement of the resummation bound for Erdős problem 1196 (posted 21 April 2026, printed date 22 April 2026), refines the second note's bound to the displayed one with the explicit measures ν(n)=1/(nlog⁡n)\nu(n)=1/(n\log n) and μ(n,m)=Λ(n/m)/(nlog⁡2n)\mu(n,m)=\Lambda(n/m)/(n\log^{2}n) and Mertens's theorem in the form $\sum_{q\le t}\Lambda(q)/q=\log t-\gamma+O(e^{-c\sqrt{\log t}})$, and concludes that the coefficient of 1/log⁡x1/\log x the resummation proof furnishes is γ\gamma. The thread post of 21 April 2026 notes that this is the same secondary term as the original argument's. No independent check of the proofs is recorded.

Submission note. Posted to the site's forum by Nat Sothanaphan on 16 April 2026:

Here are the notes proving, for A⊂[x,∞)A \subset [x, \infty),

>∑a∈A1alog⁡a≤1+γlog⁡x+O(1log⁡2x)>> \sum_{a \in A} \frac{1}{a \log a} \le 1 + \frac{\gamma}{\log x} + O\left(\frac{1}{\log^2 x}\right) >

by optimizing GPT-5.4 Pro's argument.

There's nothing particularly exciting here, but it seems worth written down.

Posted to the site's forum by Nat Sothanaphan on 20 April 2026:

Thanks for this reformulation!

I have managed to find a variant that does not explicitly mention either flow graphs or Markov chains. In fact, it is a pure resummation argument.

Here are the notes.

Posted to the site's forum by Nat Sothanaphan on 21 April 2026:

Optimizing the resummation variant gives the bound

>∑a∈A1alog⁡a≤1+γlog⁡x+O(1log⁡2x)>> \sum_{a \in A} \frac{1}{a \log a} \le 1 + \frac{\gamma}{\log x} + O\left(\frac{1}{\log^2 x}\right) >

which has the same secondary term as the original Price's proof. While not surprising, it is informative as different variants seem to give slightly different bounds.

Here are the notes.

AI system. Each note carries the disclosure "Produced with use of GPT-5.4 Thinking", with links to the sessions, and the author signs alone; the author is the claimant, and GPT-5.4 Thinking is the system they name.

Standing. The notes are dated manuscripts posted on Google Drive and linked from the thread, not filed on the proof-claims tab, not refereed and not formalized. Remark 4.1 of the paper of Alexeev, Barreto, Li, Lichtman, Price, Shah, Tang and Tao (arXiv:2605.00301v1), whose card is alexeev_2026_primitive_sets_von_mangoldt_chains_erdos, credits the sharper bound to Nat Sothanaphan, using GPT, as an analysis of a variant of the paper's argument, and its footnote links the note of 21 April 2026; the paper's Remark 4.2 gives Tao's alternate Markov chain proof with the weaker constant 2γ2\gamma. The site's curator labels the problem proved, credits the solution to GPT-5.4 Pro prompted by Price, and points to the comment section for further refinements without crediting them to anyone, so no reviewer is named for this claim and it stays claimed.

Depends on. Price's page: the first note optimizes the argument recorded there; the second and third notes give a self-contained resummation proof.