Define abundancy numbers as the rational numbers that are equal to the abundancy index of some integer (not to be confused with «abundant numbers», which are natural numbers with abundancyindex greater than 2). Abundancy index of $n$ is the function $\frac{\sigma(n)}{n}$, where $\sigma(n)$ is the sum of all divisors of $n$. Not all rational numbers are abundancy numbers. There are several additional properties, that abundancy numbers have to satisfy. For example:
- All abudancy numbers are not less than one (this condition is also not sufficient as there are some rational numbers greater than one that re proven not to be abundancy numbers (for example, $\frac{15840}{7921}$; proof of this fact you can find here: Are there any natural numbers $n$ that satisfy the condition $7921\sigma(n) = 15840n$?))
- All abundancy numbers are a product of numbers of the type $\frac{p^{n+1} - 1}{p(p-1)}$ for some integer $n$ and prime $p$, with different $p$-s (this is an equivalent characterisation of abundancy numbers due to multiplicativity of abundancy index, but this is not much informative, as in general it is quite hard to directly prove for a given rational number that it doesn’t satisfy this condition)
And what are other specific properties of abundancy numbers?
Why is it interesting: If one knows specific properties of abundancy numbers, one knows6 that the numbers that do not satisfy them are not abundancy numbers. And if $\frac{m}{n}$ is not an abundancy number, then the integer equation $n\sigma(x) = mx$ has no roots.
In specific cases:
- If integer $k$ is not an abundancy number, then there are no $k$-prefect numbers.
- If for an odd integer $k$ $\frac{k}{2}$ is not an abundancy number, then there are no $k$-hemiperfect numbers.
- If for a finite simple group $G$ $\frac{2|G|}{|G|+1}$ is not an abundancy number, then there are no immaculate group of the type $G\times C$ where $C$ is a finite cyclic group.