ON A PROBLEM OF PONGSRIIAM ON THE SUM OF DIVISORS

Abstract For any positive integer n , let $\sigma (n)$ be the sum of all positive divisors of n . We prove that for every integer k with $1\leq k\leq 29$ and $(k,30)=1,$ $$ \begin{align*} \sum_{n\leq K}\sigma(30n)>\sum_{n\leq K}\sigma(30n+k) \end{align*} $$ for all $K\in \mathbb {N},$ which gives a positive answer to a problem posed by Pongsriiam [‘Sums of divisors on arithmetic progressions’, Period. Math. Hungar . 88 (2024), 443–460].

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Publication Details

Journal
Bulletin of the Australian Mathematical Society
Published
2024-09-13
DOI
https://doi.org/10.1017/s0004972724000492
Citations
1
Primary Topic
Analytic Number Theory Research
Type
article
Field-Weighted Citation Impact
1.27
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article

ON A PROBLEM OF PONGSRIIAM ON THE SUM OF DIVISORS

Rui-Jing Wang
1 citations
Bulletin of the Australian Mathematical Society
Analytic Number Theory Research
1.27
article

ON A PROBLEM OF PONGSRIIAM ON THE SUM OF DIVISORS

Rui-Jing Wang
article en
1 citations

Abstract

Abstract For any positive integer n , let $\sigma (n)$ be the sum of all positive divisors of n . We prove that for every integer k with $1\leq k\leq 29$ and $(k,30)=1,$ $$ \begin{align*} \sum_{n\leq K}\sigma(30n)>\sum_{n\leq K}\sigma(30n+k) \end{align*} $$ for all $K\in \mathbb {N},$ which gives a positive answer to a problem posed by Pongsriiam [‘Sums of divisors on arithmetic progressions’, Period. Math. Hungar . 88 (2024), 443–460].

Bulletin of the Australian Mathematical SocietyVol. 111(1)
Jiangsu Second Normal University (CN)
Openalex Percentile: Top 21%
Analytic Number Theory Research
1.27
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ON A PROBLEM OF PONGSRIIAM ON THE SUM OF DIVISORS — Rui-Jing Wang · Bulletin of the Australian Mathematical Society (2024) | TGRS Research Map | TGRS