Sums of divisors on arithmetic progressions

For each \(s\in \mathbb R\) and \(n\in \mathbb N\), let \(\sigma_s(n) = \sum_{d\mid n}d^s\). In this article, we give a comparison between \(\sigma_s(an+b)\) and \(\sigma_s(cn+d)\) where \(a\), \(b\), \(c\), \(d\), \(s\) are fixed, the vectors \((a,b)\) and \((c,d)\) are linearly independent over \(...

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Bibliographic Details
Published in:arXiv.org 2021-10
Main Author: Pongsriiam, Prapanpong
Format: Article
Language:English
Subjects:
Progressions
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Summary:For each \(s\in \mathbb R\) and \(n\in \mathbb N\), let \(\sigma_s(n) = \sum_{d\mid n}d^s\). In this article, we give a comparison between \(\sigma_s(an+b)\) and \(\sigma_s(cn+d)\) where \(a\), \(b\), \(c\), \(d\), \(s\) are fixed, the vectors \((a,b)\) and \((c,d)\) are linearly independent over \(\mathbb Q\), and \(n\) runs over all positive integers. For example, if \(|s|\leq 1\), \(a, b, c, d\in \mathbb N\) are fixed and satisfy certain natural conditions, then $$ \sigma_s(an+b) < \sigma_s(cn+d)\quad\text{ for all \(n\leq M\)} $$ where \(M\) may be arbitrarily large, but in fact \(\sigma_s(an+b) - \sigma_s(cn+d)\) has infinitely many sign changes. The results are entirely different when \(|s|>1\), where the following three cases may occur: \begin{itemize} \item[(i)] \(\sigma_s(an+b) < \sigma_s(cn+d)\) for all \(n\in \mathbb N\); \item[(ii)] \(\sigma_s(an+b) < \sigma_s(cn+d)\) for all \(n\leq M\) and \(\sigma_s(an+b) > \sigma_s(cn+d)\) for all \(n\geq M+1\); \item[(iii)] \(\sigma_s(an+b) - \sigma_s(cn+d)\) has infinitely many sign changes. \end{itemize} We also give several examples and propose some problems.
ISSN:2331-8422