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Consider the function $\sigma(n)/n$, where $\sigma$ is the usual sum-of-divisors function. I read somewhere that it is unknown what rational numbers are in fact values of this function (or at any rate that characterizing them is an open question). Well, that was a while ago, and I suspect it was in one of my older references. So:

What is the current status of this question - characterizing the $q\in \mathbb{Q}$ such that there exists $n\in \mathbb{N}$ with $\sigma(n)/n=q$?

I think there is a standard name for the function $\sigma(n)/n$. If I knew it, that would make things easier, so I apologize if this is easy to find once one knows that.

Edit after accepting answer: Of course, $\sigma(n)/n=\sigma_{-1}(n)$, but I don't know whether there is so much more information under that designation!

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This paper may help

MR2346095 (2008i:11005) Stanton, William G.; Holdener, Judy A. Abundancy "outlaws'' of the form $\frac{\sigma(N)+t}{N}$. (English summary) J. Integer Seq. 10 (2007), no. 9, Article 07.9.6, 19 pp. (electronic). 11A25 (11Y55 11Y70) PDF Clipboard Journal Article Make Link

The abundancy index of a positive integer $n$ is defined to be the rational number $I(n)=\sigma (n)/n$, where $\sigma$ is the sum of divisors function $\sigma(n)=\sum_{d|n}d$. A rational number $r/s$ greater than 1 is said to be an abundancy outlaw if $I(x)=r/s$ has no solution among the positive integers. By a theorem of Holdener [Math. Mag. 79 (2006), no. 5, 389--391] which provides conditions on $I(n)$ equivalent to the existence of an odd perfect number, it is useful to characterize those rational numbers in $(1, \infty)$ that are abundancy outlaws. In this paper, the authors consider rational numbers of the form ${(\sigma(N)+t)/N}$ and prove that under certain conditions such rationals are abundancy outlaws. As a result, 44 new abundancy outlaws with numerators $\leq 100$ are captured. As a way to visualize the distribution of abundancy outlaws, they include a list of the rationals with numerators $\leq 100$. Each rational number is colored according to its abundancy index/outlaw status. Reviewed by Zhenxiang Zhang

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  • $\begingroup$ 'Abundancy Index' - got it. Thank you! $\endgroup$
    – kcrisman
    Apr 15, 2011 at 18:37
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You can also look at http://upforthecount.com/math/abundance.html and http://upforthecount.com/math/black.html

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  • $\begingroup$ Yes, I think I found that site eventually at the time I was looking for this. Thanks. $\endgroup$
    – kcrisman
    Sep 22, 2011 at 3:32

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