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How can we get a hand on $\sum_{\substack{d|n\\d<\sqrt{n}}}d$

数论 Math StackExchange 0 票 0 回答 40 浏览 提问者: Marius S.L. 2026-06-23 22:13
number-theory arithmetic divisor-function

问题内容

I found the following statement (in different words with different functions) on another website: $$ \sigma(n)=2\left(n+\sum_{\substack{d|n\\d<\sqrt{n}}}d\right) -1 $$ if and only if $n=392.$

That $392$ is a solution is easy to check. Whether there are other solutions depends on "the first half" of the $\sigma$ function. The author had originally considered the function $$ \sum_{\substack{d|n\\d\ge\sqrt{n}}}d-\sum_{\substack{d|n\\d\le\sqrt{n}}}d $$ which makes the problem obvious: We have an additional equation with multiplicatively split sums.

My question is whether there are known formulas (expressions, relations) that allow us to tackle $$ \sum_{\substack{d|n\\d\le\sqrt{n}}}d\;? $$ It looks like a natural split to partition the divisors of $n,$ but I do not recall that I have seen it before. Does anybody else have?

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