The search for perfect numbers loses its cachet after a while, when you realize that all the known perfect numbers are even and were found using a formula, and any outside the formula are exceedingly unlikely. We can interest ourselves in aliquot parts for a little while longer by turning our attention to abundant numbers. For one thing, they are, well, more abundant than perfect numbers. But more importantly, they yield to a little analytic number theory.

As you may recall, an *abundant number* is a number whose sum of proper divisors *exceeds* the number
itself:

`σ`_{0}(N) > N

In more modern terms using Euler's sigma function:

`σ(N) > 2N`

or

`σ(N)/N > 2`

Let's define

`η(n) = σ(N)/N`

If you recall the formula for the sigma function for a power of a prime,

`
σ (p`^{α}) = 1 + p + p^{2} + p^{3} + ... + p^{α}

= p^{α}(1 + 1/p + 1/p^{2} + 1/p^{3} + ... + 1/p^{α})

For a positive integer in general:

`σ (N) = σ(Πp`_{i}^{αi})

= Πσ(p_{i}^{αi})

= Πp_{i}^{αi}η(p_{i}^{αi})

= N * Πη(p_{i}^{αi})

where

`η(p`^{α}) = 1 + 1/p + 1/p^{2} + 1/p^{3} + ... +
1/p^{α}

This gives us a direct measurement of the ratio of `σ(N)` to `N`. This doesn't help us find perfect numbers. For that, we still need to find lucky combinations of factors of the single-prime sigmas. But the formula does limit the conditions under which we find perfect and abundant numbers.

How, you ask? The formulae for `σ(p`^{α}) and `η(p`^{α}) are both geometric series. However, the latter series converges (to `p/(p-1)`) where the former does not. So we know that however large we want to make an `α`_{i} for a given `p`_{i}, the prime
can contribute no more than `p/(p-1)` towards the final number's η. For the lowest primes:

p η(p) η(p^{2}) Lim (η(p^{i}), i -> ∞)
-- -------- -------- --------
2 1.5 1.75 2
3 1.333333 1.444444 1.5
5 1.2 1.24 1.25
7 1.142857 1.163265 1.166666
11 1.090909 1.099173 1.1
13 1.076923 1.082840 1.083333
17 1.058823 1.062283 1.0625
19 1.052631 1.055401 1.055555
23 1.043478 1.045368 1.045454
29 1.034482 1.035671 1.035714
31 1.032258 1.033298 1.033333
37 1.027027 1.027757 1.027777
41 1.024390 1.024985 1.025

As you can see from the table, the higher the prime is, the less it can contribute towards the η value of a number it divides. Since 2 has the highest η values, nearly all abundant numbers are even. The lowest odd abundant numbers are:

N prime factorization
==== ===================
945 3^{3}* 5* 7 σ(945)=1920=945*2.031746...
1575 3^{2}* 5^{2}* 7
2205 3^{2}* 5* 7^{2}
2835 3^{4}* 5* 7
3465 3^{2}* 5* 7* 11
4095 3^{2}* 5* 7* 13

As more and more of the lowest primes are excluded, more and more different prime factors are required to reach the
magical value of 2:

Exclude Lowest set of Smallest number of
primes prime factors different prime
required factors required
----------- ---------------- ----------------
2 3, 5, 7 3 1.5*1.25*1.16666=2.1875
2,3 5,7,11,13,17,19,23 7 1.25*1.16666*...*1.04545=2.0376
2,3,5 7 thru 61 15 1.16666*....*1.01666=2.014718
2,3,5,7 11 thru 113 27 (see note (1) below)
2 thru 11 13 thru 193 40
2 thru 13 17 thru 311 59
2 thru 17 19 thru 439 81
2 thru 19 23 thru 641 109
2 thru 23 29 thru 859 141
2 thru 29 31 thru 1093 174

An abundant number is guaranteed for the "lowest set" from the table above. However, the larger an exponent used for a given prime, the smaller the incremental effect on the η factor. For the "lowest sets" excluding primes greater than 2, the exponents required are very large, and you can get abundant numbers faster by using more primes. For example, the lowest abundant number not divisible by 2 or 3 is `5,391,411,025 = 5`^{2}*7*11*13*17*19*23*29. Calculating the lowest abundant number whose lowest prime factor is an even higher prime is an exercise left to the reader.

(1) Mathworld (at *http://mathworld.wolfram.com*) states that Catalan proved that 26 prime factors were required for an odd perfect number, and that Norton extended this to 27 in 1960. I have access to neither Catalan nor Norton to know what techniques they used. But I begin to see the power of analytic number theory.