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A Thirteen-Layer Lower Bound for Erdős Problem #390

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theorem_1: Mausberg's theorem that the least top factor f(n) in a factorization of n factorial into increasing factors above n satisfies liminf of (f(n) - 2n)/(n/log n) at least C_0 = 4029639598/25970038185, from the first thirteen large-prime layers and the primes up to 23.


Samuel Mausberg, A Thirteen-Layer Lower Bound for Erdős Problem #390, unpublished note, dated 2 May 2026, 4 pp.; its AI disclosure (p. 1) says it was prepared with assistance from GPT-5.5 Pro, the author taking responsibility for the final mathematical claims. No notice is printed in the four-page manuscript; it has no arXiv record (an arXiv author query on 2026-10-02 returned only an unrelated paper), and the Google Drive link the card names (https://drive.google.com/file/d/1gcFkDf-6PIIjpZAYC9rqfhWf5IKgL218/view) carries no terms; the term is unstated.

For f(n)f(n) the least mm with n!=a1⋯akn!=a_1\cdots a_k, n<a1<⋯<ak=mn<a_1<\cdots<a_k=m, Theorem 1 (p. 2) proves the unconditional bound lim inf⁡n→∞(f(n)−2n)/(n/log⁡n)≥C0\liminf_{n\to\infty}(f(n)-2n)/(n/\log n)\ge C_0 with $C_0=\bigl(\sum_{r=1}^{13}\frac{1}{(r+1)(2r+1)}\bigr)/\bigl(\sum_{p\le23}\frac{1}{p-1}\bigr) =4029639598/25970038185=0.15516494697830188\ldots$, so any asymptotic leading constant, if one exists, is at least C0C_0 (p. 1). The method is a finite large-prime obstruction in the complement formulation (Section 1, pp. 1--2): for an integer M>nM>n, f(n)≤Mf(n)\le M holds exactly when Q(n,M)=M!/(n!)2Q(n,M)=M!/(n!)^2 is a product of distinct integers from (n,M](n,M]. Two elementary valuation estimates, (1) M≥2n−O(log⁡n)M\ge2n-O(\log n) for every admissible MM, from the 2-adic valuation, and (2) vℓ(Q(n,2n+h))=h/(ℓ−1)+Oℓ(log⁡n)v_\ell(Q(n,2n+h))=h/(\ell-1)+O_\ell(\log n) for each fixed prime ℓ\ell whenever h=O(n/log⁡n)h=O(n/\log n) (p. 2), are combined over the layers Lr\mathcal L_r, 1≤r≤131\le r\le13, of primes PP with M/(2r+2)<P≤M/(2r+1)M/(2r+2)<P\le M/(2r+1) and n/(r+1)<P≤n/rn/(r+1)<P\le n/r (display (3), p. 2). Each such prime forces a factor PqPq with r+1≤q≤2r+1r+1\le q\le2r+1, and every such qq uses an exponent of a prime at most 23 (pp. 2--3). Section 3 (p. 4) recasts the count as linear-inequality bookkeeping, and Remark 1 (p. 4) says this relaxation is only a necessary condition, so Theorem 1 is a lower bound and not an asymptotic formula; the paper makes no upper-bound or full-asymptotic claim (p. 1).

The paper compares C0C_0 with the constant arbitrarily close to 1/91/9 that Erdős, Guy and Selfridge obtain in their proof of Theorem 3 (cited as [EGS82, p. 255]) and notes C0>1/9C_0>1/9. It presents its proof as a finite quantified version of an observation by Tao on the Erdős Problems forum, that the basic Erdős--Guy--Selfridge argument does not account for further prime intervals such as (2n/6,2n/5](2n/6,2n/5] and (2n/8,2n/7](2n/8,2n/7] (p. 1).

Source: https://drive.google.com/file/d/1gcFkDf-6PIIjpZAYC9rqfhWf5IKgL218/view.

Read status: claims checked for Theorem 1, (1), (2), the proof on pp. 2--3, the constant's arithmetic and Remark 1, read clause by clause on the page images. Nothing here is independently reviewed. Result page: theorem_1.

Bears on. #390: Theorem 1 (p. 2) proves lim inf⁡n→∞(f(n)−2n)/(n/log⁡n)≥C0≈0.15516\liminf_{n\to\infty}(f(n)-2n)/(n/\log n)\ge C_0\approx0.15516, so the constant cc the problem asks about is at least C0C_0 if it exists; the paper does not show that cc exists.

Results.

  • Theorem 1 (p. 2): lim inf⁡n→∞(f(n)−2n)/(n/log⁡n)≥C0=4029639598/25970038185\liminf_{n\to\infty}(f(n)-2n)/(n/\log n)\ge C_0=4029639598/25970038185, with the valuation estimates (1) and (2) (p. 2) it uses and the scope limitation of Remark 1 (p. 4).

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