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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. There is a set AA of positive integers and a constant c>0c>0 with f(n)≥clog⁡nf(n)\ge c\log n for all large nn, which answers both questions of Problem 358 yes; it is the bound that Tao's later Theorem 1.1 states. Nat Sothanaphan announced on the site's discussion thread on 17 February 2026 (post 4343) that GPT-5.2 Thinking, working under Sothanaphan's supervision, appeared to have found a full solution, and posted the write-up on 18 February 2026, asking for errors to be reported. It follows the strategy that Terence Tao proposed on the thread on 7 February 2026.

Submission note. Posted to the site's forum by Nat Sothanaphan on 18 February 2026:

Alright, here goes. Link to PDF file. Link to Tex files.

I also check your concern with GPT just in case which produces this response; you may find it useful.

Please point out any error or concern - I (or GPT) will try to address them.

Hope it works 🙏.

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

After some amount of work, it does not appear that GPT is able to fix the gap. (At least with the current process I'm using.)

The upside is that GPT seems to understand the difficulty well and is not claiming false solutions or doing unproductive things, when operated in the current process.

It also makes a final wrap up material of the attempt, which may or may not contain workable ideas.

Withdrawal. On 18 February 2026 Tao reported in the thread a gap in Lemma 17, in the allocation of intervals to the exceptional integers. On 20 February 2026 the author reported that, after some work, GPT did not appear able to fix the gap, and posted no repaired version; this withdraws the claim. Tao's manuscript (claim page) cites the write-up as a previous claim of its Theorem 1.1 that fell short of a complete proof, and credits it with the observation that ∑n≤xf(n)≤xlog⁡x+O(x)\sum_{n\le x}f(n)\le x\log x+O(x) for every AA. The problem's standing takes nothing from this page.

Depends on. Nothing in this wiki.