{"title":"ON A PROBLEM OF PONGSRIIAM ON THE SUM OF DIVISORS","authors":"RUI-JING WANG","doi":"10.1017/s0004972724000492","DOIUrl":null,"url":null,"abstract":"<p>For any positive integer <span>n</span>, let <span><span><img data-mimesubtype=\"png\" data-type=\"\" src=\"https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240913031000660-0696:S0004972724000492:S0004972724000492_inline1.png\"><span data-mathjax-type=\"texmath\"><span>$\\sigma (n)$</span></span></img></span></span> be the sum of all positive divisors of <span>n</span>. We prove that for every integer <span>k</span> with <span><span><img data-mimesubtype=\"png\" data-type=\"\" src=\"https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240913031000660-0696:S0004972724000492:S0004972724000492_inline2.png\"><span data-mathjax-type=\"texmath\"><span>$1\\leq k\\leq 29$</span></span></img></span></span> and <span><span><img data-mimesubtype=\"png\" data-type=\"\" src=\"https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240913031000660-0696:S0004972724000492:S0004972724000492_inline3.png\"><span data-mathjax-type=\"texmath\"><span>$(k,30)=1,$</span></span></img></span></span> <span><img data-mimesubtype=\"png\" data-type=\"\" src=\"https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240913031000660-0696:S0004972724000492:S0004972724000492_eqnu1.png\"><span data-mathjax-type=\"texmath\"><span>$$ \\begin{align*} \\sum_{n\\leq K}\\sigma(30n)>\\sum_{n\\leq K}\\sigma(30n+k) \\end{align*} $$</span></span></img></span></p><p>for all <span><span><img data-mimesubtype=\"png\" data-type=\"\" src=\"https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240913031000660-0696:S0004972724000492:S0004972724000492_inline4.png\"><span data-mathjax-type=\"texmath\"><span>$K\\in \\mathbb {N},$</span></span></img></span></span> which gives a positive answer to a problem posed by Pongsriiam [‘Sums of divisors on arithmetic progressions’, <span>Period. Math. Hungar</span>. <span>88</span> (2024), 443–460].</p>","PeriodicalId":0,"journal":{"name":"","volume":null,"pages":null},"PeriodicalIF":0.0,"publicationDate":"2024-09-13","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"","FirstCategoryId":"100","ListUrlMain":"https://doi.org/10.1017/s0004972724000492","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
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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].
$\sigma (n)$ be the sum of all positive divisors of n. We prove that for every integer k with 
$1\leq k\leq 29$ and 
$(k,30)=1,$ 
$$ \begin{align*} \sum_{n\leq K}\sigma(30n)>\sum_{n\leq K}\sigma(30n+k) \end{align*} $$
$K\in \mathbb {N},$ 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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