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Statement

Setting (p. 2). Let θ:N→R∪{∞}\theta:\mathbb N\to\mathbb R\cup\{\infty\} be an arithmetic function. B\mathcal B is the set of positive integers consisting of n=1n=1 and every n≥2n\ge2 with prime factorization n=p1α1⋯pkαkn=p_1^{\alpha_1}\cdots p_k^{\alpha_k}, p1<⋯<pkp_1<\cdots<p_k, such that pj+1≤θ(p1α1⋯pjαj)p_{j+1}\le\theta(p_1^{\alpha_1}\cdots p_j^{\alpha_j}) for 0≤j<k0\le j<k, the empty product being 11. The standing hypothesis is

θ:N→R∪{∞},θ(1)≥2,θ(n)≥P+(n)(n≥2),(3)\theta:\mathbb N\to\mathbb R\cup\{\infty\},\qquad \theta(1)\ge2,\qquad \theta(n)\ge P^+(n)\quad(n\ge2), \tag{3}

with P+(n)P^+(n) the largest prime factor of nn. B(x)B(x) counts the n≤xn\le x in B\mathcal B and χ\chi is the indicator function of B\mathcal B. With θ(n)=σ(n)+1\theta(n)=\sigma(n)+1, B\mathcal B is the set of practical numbers, by Sierpinski and Stewart (p. 2).

Lemma 1 (p. 2, recalled from the author's earlier papers). Under (3), for Re⁡(s)>1\operatorname{Re}(s)>1, ∑n≥1χ(n)n−s∏p≤θ(n)(1−p−s)=1\sum_{n\ge1}\chi(n)n^{-s}\prod_{p\le\theta(n)}(1-p^{-s})=1, and the equation also holds at s=1s=1 if B(x)=o(x)B(x)=o(x).

Lemma 2 (p. 2). Let θ\theta satisfy (3), let qq be prime and h∈Nh\in\mathbb N. For Re⁡(s)>1\operatorname{Re}(s)>1,

∑n≥1qh∥nχ(n)ns∏p≤θ(n)(1−1ps)=1−1/qsqsh∑n≥1θ(n)≥qχ(n)ns∏p≤θ(n)(1−1ps),\sum_{\substack{n\ge1\\ q^h\parallel n}}\frac{\chi(n)}{n^s} \prod_{p\le\theta(n)}\left(1-\frac1{p^s}\right) =\frac{1-1/q^s}{q^{sh}} \sum_{\substack{n\ge1\\ \theta(n)\ge q}}\frac{\chi(n)}{n^s} \prod_{p\le\theta(n)}\left(1-\frac1{p^s}\right),

and if B(x)=o(x)B(x)=o(x) the equation also holds at s=1s=1. Here pp runs over primes and qh∥nq^h\parallel n means that qhq^h divides nn and qh+1q^{h+1} does not.

The paper calls this a new identity (p. 1) and states it, with Lemma 1, in the general setting because it applies to other sets than the practical numbers (p. 2).

Proof pointer

Pp. 2--3. Every integer mm splits uniquely as m=nrm=nr with n∈Bn\in\mathcal B and every prime factor of rr above θ(n)\theta(n); summing m−sm^{-s} over the mm with qh∥mq^h\parallel m in two ways and dividing by ζ(s)\zeta(s) gives the identity for Re⁡(s)>1\operatorname{Re}(s)>1 after Lemma 1. At s=1s=1 the two sums are shown right-continuous, using the estimate ∏p≤θ(n)(1−p−s)≍s−1+1/log⁡θ(n)\prod_{p\le\theta(n)}(1-p^{-s})\asymp s-1+1/\log\theta(n) uniformly for 1≤s≤21\le s\le2 from the author's Math. Comp. paper.

Read depth

Claims checked: the setting, (3), Lemma 1 and Lemma 2 were read on the page images of pp. 2--3 of arXiv version 3, and the proof was followed at the level of the sketch above. Nothing here is independently reviewed.

Dependencies

Lemma 1 (p. 2), taken from A. Weingartner, On the constant factor in several related asymptotic estimates, Math. Comp. 88 (2019), 1883--1902, Lemma 1, and A. Weingartner, A sieve problem and its application, Mathematika 63 (2017), 213--229, Theorem 1. Lemma 2 feeds Lemma 3 (p. 3) and through it Theorem 1.

Source. Andreas Weingartner, The constant factor in the asymptotic for practical numbers, arXiv:1906.07819; the edition read is named on the source card.

Bears on

No Erdős problem directly; the lemma is a tool for Theorem 1.