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Let k≥3k\geq 3 and fk(N)f_k(N) be the maximum of ∑n∈A1n\sum_{n\in A}\frac{1}{n} over all A⊆{1,…,N}A\subseteq\{1,\ldots,N\} containing no kk subsets with the same pairwise least common multiple. Estimate fk(N)f_k(N). The claimed answer: fk(N)=(log⁡N)γk+o(1)f_k(N)=(\log N)^{\gamma_k+o(1)}, where γk\gamma_k is a weighted generalization of the Tang-Zhang sunflower capacity.

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No person has examined this. Nothing here has been checked at all. say whether it holds →

  1. proof attempt · #1

    Przemysław Chojecki, using GPT-5.4 Pro

    That credit came with the record as it was imported. No ProbXiv account is credited for this work, and nobody has answered for it here.

    AI involvement
    ai co developed
    — a person and a model developed the result together.

    A weighted version of the Tang-Zhang sunflower-capacity argument giving the exact logarithmic exponent was developed with GPT-5.4 Pro, using a mass-transport idea from the forum's discussion of problem #1196.

    Identifies the exponent as a variational sunflower-capacity constant, sharpening the Tang-Zhang bounds; the value of that constant itself remains open, as does site acceptance

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