Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Claim. Let with and ; the sequence need not be monotone and its ratios need not converge. Theorem 1.1 (p. 1) of Kan Jiang and Lifeng Xi, The Erdős similarity problem for null sequences with adjacent ratios bounded below, arXiv:2610.11449v1, posted 2026-10-08 (10 pages, the only version), states that for every there is a compact set of Lebesgue measure greater than such that for every and every some has . So contains no set with for , of either sign of , which answers the question of Problem 120 yes for that . The proof first passes to a subsequence with for a constant (Lemma 2.1, p. 3, which uses no monotonicity). It then proves a periodic hitting statement (Proposition 2.2, p. 3): for every an open -periodic set of density at most meets for every real and every . Removing from the dilations , , of the sets that the proposition gives for , each of density at most , gives the theorem (pp. 3--4). The proposition is proved in Sections 3–5 by the finite-tree construction of the OpenAI release's dyadic and geometric manuscripts, with grids of mesh comparable to , index windows on the edges of a complete ordered tree, independent random selector and terminal tables, and an open-neighborhood repair of the exceptional centers. The lower ratio bound limits the number of representative dilations to an exponential in the window length (estimate (5.1)), and the upper bound separates the grids. Corollary 6.1 (p. 9) adds a single compact set of measure greater than for any countable family of such sequences at once, with infinitely many distinct points of every affine copy of each outside it. Section 6 (pp. 9–10) notes that the estimate (5.1) fails for , a limitation of the estimate and not a claim of universality, and that the whole conjecture would follow from the case of positive null sequences with . Read depth: the whole of v1; the proof was read through but not checked step by step.
Covers. The question, answered yes for every set with , and . It follows for every set that contains an affine copy of such a sequence, since the same avoids it; the paper's closing remark uses the same observation. Equivalently, the theorem reaches every set with an accumulation point approached from one side by points of whose distances to shrink by at most a fixed factor at each step. The geometric progressions , , are the case of constant ratio , with the same form of avoiding set: the release's geometric-progression claim and its instance , the accepted dyadic case, are special cases. So are the decreasing sequences with of Eigen and Falconer. Corollary 6.1 also gives one avoiding set for countably many ratios at once, which the release's geometric manuscript does not assert. Not covered: the bounded infinite sets at none of whose accumulation points such a sequence exists. Among positive null sequences these are exactly the sequences whose terms, listed in decreasing order, have , such as and ; by the paper's closing remark the whole conjecture reduces to these. Uncountable sets of that kind include the Cantor sets whose dissection ratios tend to . The general question is not addressed.
Depends on. Nothing in this wiki. The paper credits the finite-tree construction and the geometric case to the two release manuscripts, which it cites as unpublished, and writes the adapted proof out in full (p. 2), so the theorem does not rest on either release claim.
Standing. Claimed: an arXiv preprint with no refereed version, Lean proof or outside review known to this corpus. It appeared after erdosproblems.com froze proof claims on 6 October 2026, so no proof claim on erdosproblems.com records it; Bloom's catalog labeled the problem OPEN.