{"title":"关于除数之和的庞氏难题","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":50720,"journal":{"name":"Bulletin of the Australian Mathematical Society","volume":"6 1","pages":""},"PeriodicalIF":0.6000,"publicationDate":"2024-09-13","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":"{\"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\":50720,\"journal\":{\"name\":\"Bulletin of the Australian Mathematical Society\",\"volume\":\"6 1\",\"pages\":\"\"},\"PeriodicalIF\":0.6000,\"publicationDate\":\"2024-09-13\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"\",\"citationCount\":\"0\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"Bulletin of the Australian Mathematical Society\",\"FirstCategoryId\":\"100\",\"ListUrlMain\":\"https://doi.org/10.1017/s0004972724000492\",\"RegionNum\":4,\"RegionCategory\":\"数学\",\"ArticlePicture\":[],\"TitleCN\":null,\"AbstractTextCN\":null,\"PMCID\":null,\"EPubDate\":\"\",\"PubModel\":\"\",\"JCR\":\"Q3\",\"JCRName\":\"MATHEMATICS\",\"Score\":null,\"Total\":0}","platform":"Semanticscholar","paperid":null,"PeriodicalName":"Bulletin of the Australian Mathematical Society","FirstCategoryId":"100","ListUrlMain":"https://doi.org/10.1017/s0004972724000492","RegionNum":4,"RegionCategory":"数学","ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"Q3","JCRName":"MATHEMATICS","Score":null,"Total":0}
引用次数: 0
摘要
对于任意正整数 n，让 $\sigma (n)$ 是 n 的所有正除数之和。我们证明，对于每一个整数k，只要有$1\leq k\leq 29$和$(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}, $ 这给了 Pongsriiam 提出的问题一个肯定的答案['Sums of divisors on arithmetic progressions', Period.Math.匈牙利。88 (2024), 443-460].
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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Bulletin of the Australian Mathematical Society aims at quick publication of original research in all branches of mathematics. Papers are accepted only after peer review but editorial decisions on acceptance or otherwise are taken quickly, normally within a month of receipt of the paper. The Bulletin concentrates on presenting new and interesting results in a clear and attractive way.
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$\sigma (n)$ 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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