A number is called **perfect** if it is the sum of its proper divisors, i.e. all divisors less than itself. For example, 28 is perfect because 1 + 2 + 4 + 7 + 14 = 28.

**Amicable** numbers are a sort of generalization of perfect numbers. Two numbers *a* and *b* are said to be amicable if *a* is the sum of *b*‘s proper divisors and vice versa.

The ancient Greeks knew of only one pair of amicable numbers: 220 and 284. Medieval mathematician Thâbit ibn Kurrah discovered two more pairs: (17296, 18416) and (9363584, 9437056). Leonard Euler (1707–1783) found 58 more pairs. Now over 12 million amicable number pairs have been found.

To generalize things further, start with a number *n* and compute the sum of its proper divisors, then the sum of the divisors of that number, etc. This sequence of numbers is called the aliquot sequence of *n*. If this sequence is periodic, *n* is called a **sociable** number.

If the aliquot sequence has period 1, *n* is a perfect number. If the sequence has period 2, *n* is part of an amicable number pair.

Are there numbers whose aliquot sequence has period 3? Not that we know of. Currently the only aliquot sequence periods that have been demonstrated are 4, 5, 6, 8, 9, and 28.

**Related posts**:

An alternative but equivalent definition of a perfect number is that the sum of all its divisors, including itself, is equal to twice itself. So if omega(n) is the sum of all n’s divisors, n is perfect when omega(n) = 2n.

What would happen if you generalized from this definition, sequences where omega(n_1) = 2 n_2, omega(n_2) = 2 n_3 …, omega(n_k) = 2 n_1? The pair (12,14) would have period 2, as would (48,62) and (112,124). These anecdotally appear to be easier to find than amicable numbers. Is there anything known about periods under this definition or is there some property of it that make it less interesting?

Replying to my own post, finding some 2-period pairs under this second definition I made up is trivial due to the fact that omega(n) is multiplicative for coprimes and equal to the product of the sums (1 + p^1 + p^2 + … p^k) where k is the maximum power of p which divides n, where the produce is over all prime divisors of n.

Thus, for powers of 2, omega(2^(m-1)) = 2^m – 1, and for primes omega(p) = p+1. So, for a Mersenne prime p = 2^m – 1, omega( (p+1)/2 ) = p and omega(p) = p+1. Close.

So, now take a pair of Mersenne primes p and q. Then, omega( (p+1)/2 * q ) = ( p * (q+1), half of that being p * (q + 1)/2, and omega( p * (q+1) / 2) = (p+1) * q, half of that being back where we started.