Wiki
Wiki

Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

Updated


Source. Theorem 1, Section 1, p. 2 of arXiv:1607.02863v2 (30 July 2024; the paper is dated 29 July 2024), proof in Section 2, pp. 3--4; read on the PDF pages in the text layer. Preprint, not published in a journal (arXiv listing checked). Notation: Hn=cn/dnH_n=c_n/d_n in lowest terms and Dn=lcm(1,2,…,n)D_n=\mathrm{lcm}(1,2,\ldots,n).

Statement

Theorem 1 (p. 2). "For all n>1n>1, we have cn≠cn−1c_n\ne c_{n-1}. Also, each of the following holds for infinitely many nn:

(i) dn>dn−1d_n>d_{n-1}, (ii) dn=dn−1d_n=d_{n-1}, (iii) dn<dn−1d_n<d_{n-1}; (iv) cn>cn−1c_n>c_{n-1}, (v) cn<cn−1c_n<c_{n-1}."

Proof pointer and sketch (Section 2)

(i) p∣dpp\mid d_p for every prime pp, so dnd_n is unbounded. (ii) For n=2p>6n=2p>6, pp divides both dn−1d_{n-1} and dnd_n, and the two-term relations between cn/dnc_n/d_n and cn−1/dn−1c_{n-1}/d_{n-1} give dn∣dn−1d_n\mid d_{n-1} and dn−1∣dnd_{n-1}\mid d_n. (iii) For n=p(p−1)n=p(p-1), Hn−1=Hp−2/p+S0+⋯+Sp−2H_{n-1}=H_{p-2}/p+S_0+\cdots+S_{p-2} with Si=∑j=1p−11/(ip+j)S_i=\sum_{j=1}^{p-1}1/(ip+j): the p−2p-2 terms whose denominators are multiples of pp sum to Hp−2/pH_{p-2}/p, and each block SiS_i of the other terms has a reduced numerator divisible by pp; since p∣cp−1p\mid c_{p-1} (the pairing 1/j+1/(p−j)1/j+1/(p-j), display (1)), p∤dnp\nmid d_n while p∣dn−1p\mid d_{n-1}, and dn≤ndn−1/p2<dn−1d_n\le nd_{n-1}/p^2<d_{n-1}. (iv) follows from (i); (v) from (iii) by a short computation. The final paragraph shows cn=cn−1c_n=c_{n-1} would force dn−1≥3dnd_{n-1}\ge3d_n and Hn≥3Hn−1H_n\ge3H_{n-1}, impossible. The proofs are elementary and complete on pp. 3--4; they were read through here but not independently reviewed.

Dependencies and read depth

Elementary; the paper notes that Wolstenholme's theorem is not needed. Read depth: claims checked; the proof read through, not verified.

Relation to Problem 291

Part (iii) gives infinitely many nn with dn<dn−1d_n<d_{n-1}, and any such nn has dn<Dnd_n<D_n, that is (an,Ln)>1(a_n,L_n)>1 in the notation of Problem 291; this is a second route to the trivial half of that problem, beside the leading-digit observation the site records. The theorem says nothing about the open half, dn=Dnd_n=D_n infinitely often, which the paper states as a conjecture (the conjecture page). The non-monotonicity of dnd_n is also the a=1a=1 case of the denominator question of Problem 290, treated on that problem's page.

Bears on. #291 (part (iii) as a route to the trivial half); #290 (part (iii) answers the existence question for a=1a=1).