Recovering n from sigma(n)/n 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 all positive 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 $b$.
 A: For the first question, note that the set $\Delta=\operatorname{im} (\delta)$ is not known very well understood. For example $5/3\in \Delta$ implies the existence of an odd perfect number. (C.W. Anderson showed that $\frac{\sigma(n)}{n}=\frac{5}{3}$ implies $5n$ is an odd perfect number.) It is also conjectured that $\Delta$ is recursive
A main reference is Carl Pomerance's "Multiply perfect numbers, Mersenne primes, and effective computability", where you can see Baker's method applied to such problems and references to other work. The second question is hard for most rationals $q > 1$. It is not known if $\delta^{-1}(q)$ is infinite for any $q$. On the other hand, Hornfeck and Wirsing have shown that the number of solutions to $\frac{\sigma(n)}{n}=q$ with $n\le x$ is $o(x^{\epsilon})$ for any $\epsilon > 0$ uniformly for all $q$. 
Another interesting theorem in this topic is Kanold's result that if there are infinitely many elements in $\delta^{-1}(q)$ which have a constant number of prime factors then there are infinitely many Mersenne primes. There also were some Monthly problems where one proves that there is a dense set of $q$ with $\delta^{-1}(q)=\emptyset$ and a dense set of $q$ with $|\delta^{-1}(q)|=1$.
A: I got a PARI/GP script that does this.
For instance, for 7/2
(05:45) gp > solveBA(7, 2, 10^10)
%1 = [4680, 26208, 4320, 197064960, 20427264, 57575890944, 21857648640]
