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ProvedAI × Math·2026

Erdős Problem #1196

GPT-5.4 Pro produced the key proof idea for a 1968 primitive-sets conjecture on the Erdős sum.

Formula

Primitive set
AN{1},abAabA\subset\mathbb{N}\setminus\{1\},\quad a\ne b\in A\Longrightarrow a\nmid b

No element of the set divides another; the primes are the basic example.

Erdős Problem #1196
supA[x,)Aprimitive  aA1aloga=1+o(1)(x)\sup_{\substack{A\subset [x,\infty)\\ A\;\mathrm{primitive}}}\;\sum_{a\in A}\frac{1}{a\log a}=1+o(1)\qquad (x\to\infty)

The conjecture asks for the asymptotic upper bound of the Erdős sum once all elements of A are large.

Von Mangoldt chain
nnqwith probabilityΛ(q)logn(qn)n\longmapsto \frac{n}{q}\quad\text{with probability}\quad \frac{\Lambda(q)}{\log n}\qquad(q\mid n)

The identity sum_{q|n} Lambda(q) = log n makes this a natural downward divisibility chain on the integers.

Refined bound
aA1aloga1+2γlogx+O ⁣(1(logx)2)\sum_{a\in A}\frac{1}{a\log a}\le 1+\frac{2\gamma}{\log x}+O\!\left(\frac{1}{(\log x)^2}\right)

The zeta-weight reformulation gives an explicit sharpening of the asymptotic result.

Summary

Erdős Problem #1196 asks whether every primitive set A whose elements are all at least x has Erdős sum at most 1 + o(1) as x tends to infinity. In April 2026, Liam Price prompted GPT-5.4 Pro on the problem; Kevin Barreto recognized the result and specialists on the Erdős Problems forum quickly refined and checked the argument. The decisive idea is to keep the problem on the integers: a von Mangoldt-weighted divisibility Markov chain, together with an adjoint hitting-probability bound, controls how often a primitive set can be hit. Tao, Lichtman, Sawin, Barreto, and others then recast the proof using an invariant weight governed by 1/zeta, yielding a clean bound with an explicit error term.

Sources