Subsection19.5.1Amicable Numbers
One other interesting idea is that of amicable numbers, such that \(\sigma(n)=\sigma(m)=m+n\). Clearly any perfect number is amicable with itself. The smallest pair of unequal amicable numbers is \((220,284)\); this was known to the ancient Greeks, cherished by some medieval Muslims, and apparently was not improved upon until the modern number-theoretic era.
Fermat, Descartes, and Euler all worked with this and found large examples, but it turns out that the next smallest pair was found by a sixteen-year old Italian boy in 1860!
Apparently he came up with this by trial and error, though no one knows for sure. Here is some of the most current data on these pairs.
There is a way to get as many amicable pairs as you like, discovered by Ibn Qurra and (later) Fermat, and used by Euler.
Algorithm19.5.1Get Amicable Numbers
- Make a list of numbers of the form \(p_n=3\cdot 2^n-1\) and \(q_n=9\cdot 2^{2n-1}-1\).
- Then check if \(p_{n-1}\), \(p_n\), and \(q_n\) are all prime.
- If so, then \(2^np_{n-1}p_n\) and \(2^nq_n\) are an amicable pair.
Proof
Since only primes and powers of two are involved, it's easy to calculate \(\sigma\) in this case, so proving it is left as an exercise.
Subsection19.5.2The abundancy index
Here's another largely open question which seems like it should be easy… which rational numbers can be gotten as \(\frac{\sigma(n)}{n}\)?
Definition19.5.2
The ratio \(\frac{\sigma(n)}{n}\) is sometimes called the abundancy index of \(n\).
There are some interesting theorems about this out there. For one thing, the abundancy index is also the same thing as \(\sigma_{-1}(n)\).
Fact19.5.3
\[\sigma_{-1}(n)=\frac{\sigma(n)}{n}\]Proof
- We have that \[\sigma(n)/n=\left(\sum_{d\mid n}d\right)/n\]
- But note that for every \(d\mid n\), the quotient is also an integer divisor \(d'\) of \(n\). So \[\sigma(n)/n=\sum_{d\mid n}\frac{1}{d'}\]
- This is the same list as the original divisor list, so reordering gives \[\sigma(n)/n=\sum_{d\mid n}\frac{1}{d}=\sigma_{-1}(n)\]
A more nuanced view will ask which rational numbers are possible for the abundancy. Clearly all such numbers are in the interval \(\left[1,\infty\right)\)! Here are some more facts.
Fact19.5.4
- If \(m\mid n\), then \(\sigma_{-1}(n)\geq \sigma_{-1}(m)\).
- If \(\sigma_{-1}(n)=\frac{a}{b}\) in lowest terms, then \(b\mid n\).
- If \(r\) is “caught” between \(\sigma(n)\) and \(n\) (such that \(n<r<\sigma(n)\)) and is relatively prime to \(n\), then \(r/n\) is not an abundancy index. (J. Holdener picturesquely calls rational numbers which are not abundancies abundancy outlaws.)
- The end of this paper has a nice list of which numbers thus far have been found, and which have not.
