Chojecki 2026 note erdos problem 1201

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theorem_1: States that for every epsilon > 0 the upper density of the n for which every prime factor of n(n+1)...(n+h-1) is at most n^(1-epsilon) tends to 0 as h tends to infinity, so for every epsilon, eta > 0 some k gives the set of n with P^+(n(n+1)...(n+k)) > n^(1-epsilon) lower density at least 1 - eta.

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P. Chojecki, A note on Erdős Problem #1201. Preprint note (ulam.ai) (2026). No copyright or license line is printed in the file; the site that hosts it (https://www.ulam.ai/, read 2026-10-02) carries the footer "© 2017-2026 ULAM" and no license or terms-of-use statement, every other right reserved.

Theorem 1 proves that for every ε>0 the upper density of integers n with P^+(n(n+1)...(n+h-1)) <= n^{1-ε} tends to 0 as h → ∞, so for every ε,η > 0 some k makes the lower density of G_{ε,k} = {n : P^+(n(n+1)...(n+k)) > n^{1-ε}} at least 1-η. The method is a short specialization of the Matomaki-Radziwill theorem on multiplicative functions in short intervals (quoted as Theorem 2) applied to the smooth-number indicator f_X(m) = 1_{P^+(m) <= X^β} with β = 1-ε/2, combined with the Dickman-de Bruijn estimate for Ψ(x,y): since the mean of f_X over [X,2X) is ρ(1/β)+o(1) < 1, all but CX((log h)^{1/3}/(δ^2 h^{δ/25}) + 1/(δ^2 (log X)^{1/50})) integers n in [X,2X] have some n+j, 0 <= j < h, with a prime factor exceeding X^β, while every n in the bad set forces the short-interval average to equal 1. A dyadic decomposition passes the estimate to upper asymptotic density. Bearing on #1201: the note says it settles #1201 as stated on the website, and its own literature note observes the argument is an immediate but apparently unrecorded consequence of Matomaki-Radziwill. The deduction answers #1201 when the problem's density is read as lower density, the reading Sawin and Bloom gave in the site's thread; Tao held the problem technically open there, since the note does not show that the density of G_{ε,k} exists. The PDF carries no printed byline; the attribution rests on the site's thread, where P. Chojecki posted the note on 30 April 2026 as written by GPT-5.5 Pro.

Source: https://www.ulam.ai/research/erdos1201.pdf.

Bears on. #1201: Theorem 1 (p. 1) gives, for every ε,η > 0, a k with the lower density of G_{ε,k} at least 1-η, which is the problem's question with lower density in place of density; it does not show that the density of G_{ε,k} exists. The note says it settles the problem as stated on the site (p. 1).

Results. Pages are those of the five-page PDF named above.

• Theorem 1 (p. 1; proof pp. 2--4): for every ε>0, the upper density of the n with P^+(prod_{j<h}(n+j)) <= n^{1-ε} tends to 0 as h → ∞; consequently for every ε,η>0 some k gives lower density d(G_{ε,k}) >= 1-η.
• Theorem 2 (p. 2), Matomaki-Radziwill (external input, the note's half-open form of their Theorem 1): there are absolute constants C, C_0 > 0 such that for multiplicative f: N → [-1,1], 2 <= h <= X and δ > 0, the short-interval average over [x,x+h) is within δ + C_0 log log h / log h of the long average over [X,2X), for all but at most CX((log h)^{1/3}/(δ^2 h^{δ/25}) + 1/(δ^2 (log X)^{1/50})) integers x in [X,2X]; the constants are uniform in f, h, X and δ.
• Equation (1) (p. 2): Dickman-de Bruijn smooth-number count Ψ(tX, X^β) = tXρ(1/β) + o(X) as X → ∞, for fixed 0 < β < 1 and uniformly for t in [1,2], giving mean value r = ρ(1/β) < 1 for the smooth indicator.
• Equation (5) (p. 3): Dyadic bad-set bound: limsup_X |B_{ε,h} ∩ [X,2X]|/X <= C (log h)^{1/3} / (δ^2 h^{δ/25}), which tends to 0 as h → ∞.

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.