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

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Source. Theorem 2 and Conjecture 1, Section 1, printed p. 54 (PDF p. 3) of the retained publisher's PDF, C. R. Math. Acad. Sci. Paris 360 (2022), 53--57, DOI 10.5802/crmath.282; proof in Section 3, printed pp. 55--57. Read on the PDF pages in the text layer. Refereed journal article (received 11 August 2021, accepted 12 October 2021, published online 26 January 2022, as printed on p. 53 and in the Crossref record read). Notation: Hn=un/vnH_n=u_n/v_n in lowest terms; L\mathcal L is the set of n≥1n\ge1 with vn<lcm(1,…,n)v_n<\mathrm{lcm}(1,\ldots,n); dˉ(L)=lim sup⁡x→∞L(x)/x\bar d(\mathcal L)=\limsup_{x\to\infty}\mathcal L(x)/x.

Statement

Conjecture 1. If q1,…,qlq_1,\ldots,q_l are distinct primes, then 1/log⁡q1,…,1/log⁡ql1/\log q_1,\ldots,1/\log q_l are linearly independent over Q\mathbb Q. (The paper derives it from the weak Schanuel conjecture: for multiplicatively independent nonzero algebraic β1,…,βm\beta_1,\ldots,\beta_m the numbers log⁡βi\log\beta_i are algebraically independent.)

Theorem 2. Assuming Conjecture 1, dˉ(L)=1\bar d(\mathcal L)=1.

Proof pointer and sketch (Section 3)

Let pip_i be the iith prime and ai=∏2≤j≤i(1−1/pj)a_i=\prod_{2\le j\le i}(1-1/p_j), so ai→0a_i\to0 by Mertens' theorem (Lemma 3); fix kk with ak<ε/2a_k<\varepsilon/2. Conjecture 1 makes log⁡p2/log⁡pi\log p_2/\log p_i (2≤i≤k2\le i\le k) linearly independent over Q\mathbb Q, so Kronecker's theorem (Lemma 4) gives infinitely many qq and exponents sis_i with pisip_i^{s_i} within a factor pi±δp_i^{\pm\delta} of ai−1p2 qa_{i-1}p_2^{\,q} (display (2)). For nn in ((pi−1)pisi−1,pisi)((p_i-1)p_i^{s_i-1},p_i^{s_i}) the terms of HnH_n with denominators divisible by pisi−1p_i^{s_i-1} contribute Hpi−1/pisi−1H_{p_i-1}/p_i^{s_i-1}, whose numerator is a multiple of pip_i, so vpi(Hn)≥−(si−2)v_{p_i}(H_n)\ge-(s_i-2) while pisi−1∣lcm(1,…,n)p_i^{s_i-1}\mid\mathrm{lcm}(1,\ldots,n); hence these intervals lie in L\mathcal L (display (3)). Lemma 5 bounds the part of each (aip2 q,ai−1p2 q)(a_ip_2^{\,q},a_{i-1}p_2^{\,q}) outside the corresponding interval, and summing over ii gives L(p2 q)≥p2 q−εp2 q−3k\mathcal L(p_2^{\,q})\ge p_2^{\,q}-\varepsilon p_2^{\,q}-3k, so dˉ(L)≥1−ε\bar d(\mathcal L)\ge1-\varepsilon. The two-page proof was read through here, not verified.

Dependencies and read depth

Mertens' theorem and Kronecker's theorem (Hardy and Wright, Theorems 429 and 442), plus Conjecture 1 as an explicit hypothesis. Read depth: claims checked; the proof read through, not verified; nothing independently reviewed.

Relation to Problem 291

With ∑k≤n1/k=an/Ln\sum_{k\le n}1/k=a_n/L_n and Ln=lcm(1,…,n)L_n=\mathrm{lcm}(1,\ldots,n), vn=Ln/(an,Ln)v_n=L_n/(a_n,L_n), so n∈Ln\in\mathcal L exactly when (an,Ln)>1(a_n,L_n)>1. Theorem 2 says, conditionally, that this half of Problem 291 holds on a set of upper density 11; unconditionally that half is settled by the leading-digit criterion (a set of positive lower density). The theorem says nothing about the other half, (an,Ln)=1(a_n,L_n)=1 infinitely often, and Conjecture 1 remains open (a 2011 MathOverflow question asking for it had no answer on 2026-09-18).

Bears on. #291 (conditional density statement for the trivial half).