Wang 2026 proposed solution erdos problem 390

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Shouqiao Wang, A Proposed Solution to Erdős Problem 390. Preprint (github.com/ShouqiaoW/erdos) (2026); the held file is the revision of 2026-07-22, which followed uploads of 2026-07-18 and 2026-07-19. The file prints no notice of its own on pp. 1--2 or 116--117; the repository holding it, in its folder "390", declares the MIT License for the repository as a whole (https://github.com/ShouqiaoW/erdos, read 2026-10-02), and whether the author meant it to cover the manuscript PDF is not stated.

Theorem 1.1 (p. 3) claims that f(n), the smallest value the largest factor can take when n! (n >= 3) is written as a product of distinct integers each greater than n, satisfies f(n) = 2n + C0 n/log n + o(n/log n), with the explicit constant C0 = 4029639598/25970038185. The lower bound, Lemma 3.2 (f(n) >= 2n + (C0 - o(1)) n/log n), is the thirteen-layer valuation argument the paper credits to Mausberg (Mausberg 2026), preceded by Lemma 3.1, which excludes every M <= 2n. With Q(n,M) = M!/(n!)^2, f(n) <= M exactly when Q(n,M) is a product of distinct integers in (n, M]; each prime P of the thirteen layers divides Q(n,M) exactly once and forces a factor Pq with r+1 <= q <= 2r+1 in layer r, so q has a prime divisor at most 23, and counting these forced factors against the valuations of Q(n,M) at the primes up to 23 gives C0 as the ratio of sum_{r=1}^{13} 1/((r+1)(2r+1)) to sum_{p<=23} 1/(p-1). The upper bound is Theorem 10.3 (p. 115): for fixed c > C0 and M = 2n + ceil(c n/log n), every sufficiently large n has a set of distinct integers in (n, M] with product Q(n,M), whose complement in (n, M] gives f(n) <= M. Its construction runs through Sections 4--10: an exact cofactor allocation whose finite part (Lemma 4.1) is a rational certificate, central anchors that remove the central binomial coefficient, a precharged universal bank, a guarded rough-signature selector, marked friable counts with a smooth-row bridge, a finite-band tangent absorber, and column-sparse rounding with a final exactification; analytic inputs are de Bruijn–Saias smooth-number normalization (Theorem 2.1, Hildebrand–Tenenbaum–Saias), Mertens estimates, an interval Selberg sieve, a Poisson–Dickman bridge, an asymmetric local-lemma criterion, and Nagura's prime interval. A companion numerical_verifier.py is said to check the finite allocation certificate; the paper states the solution was found by GPT-5.6. For problem 390 it is the full proof claim posted on 2026-07-19 on the problem's erdosproblems.com proof-claims page: an exact asymptotic constant for f(n) - 2n. The paper rests the claim on its own argument and its Python checker; the Lean development added to the repository's folder 390/lean on 2026-07-28 is not described in the paper and has not been built here.

Source: https://github.com/ShouqiaoW/erdos/tree/main/390.

Bears on. #390

Results to transcribe.

• Theorem 1.1: Claimed exact asymptotic f(n) = 2n + C0 n/log n + o(n/log n) with C0 = 4029639598/25970038185.
• Lemma 3.2 (Section 3): f(n) >= 2n + (C0 - o(1)) n/log n, from thirteen disjoint prime layers with valuation exactly one, whose forced cofactors all have a prime factor at most 23; the paper credits the argument to Mausberg.
• Theorem 10.3: Upper-bound construction: for fixed c > C0 and M = 2n + ceil(c n/log n), every sufficiently large n has a set of distinct integers in (n, M] with product M!/(n!)^2, so that f(n) <= 2n + ceil(c n/log n).
• Theorem 2.1: Quoted Hildebrand-Tenenbaum-Saias smooth-number estimate used as the analytic normalization input.