For any positive integer $n$, we define $$\sigma(n) := \sum_{d \mid n} d,$$ and $$\delta(n) := \frac{\sigma(n)}{n} = \sum_{d \mid n} \frac{1}{d}.$$ Is there an (efficient) way to determine $\delta^{-1}$? In particular,

If $q \in \mathbb{Q}$ is given, can we determine whether $q \in \operatorname{im}(\delta)$?

And more precisely,

If $q \in \mathbb{Q}$ is given, can we find

allpositive integers $n$ such that $\delta(n) = q$? Is the set $\{ n \in \mathbb{N} \mid \delta(n) = q \}$ bounded? Can we at least find interesting restrictions on the possible values for $n$ with $\delta(n)=q$?

Of course, one obvious restriction is the fact that if $q = a/b$ with $\gcd(a,b)=1$, then any $n$ with $\delta(n)=q$ has to be a multiple of $b$.

all$n$ such that $\delta(n)=q$ seems indeed very difficult, but at least one can try to prove some qualitative results on $\delta$. For example, is there an integer $N$ such that $\{n \in \mathbb{N} | \delta(n) \in \frac{1}{N} \mathbb{Z}\}$ is infinite ? $\endgroup$ – François Brunault Feb 23 '11 at 10:11