lundy draft

Abundant and Deficient Numbers

We notice that perfect numbers take the form of $\sigma (n) = 2n$, but what about the numbers such that $\sigma (n) \neq 2n$?

Well, obviously, if $\sigma (n) \neq 2n$, then either $\sigma (n) < 2n$ or $\sigma (n) > 2n$. Actually, there are names given for these types of numbers.

We define deficient numbers as those numbers n such that $\sigma (n) < 2n$, and we define abundant numbers as those numbers n such that $\sigma (n) > 2n$.

There isn't much further classification for deficient numbers. If the sum of its divisors is less than $2n$, there is not much else to be determined. A deficient number's sigma function will always be strictly in between $n$ and $2n$.

An abundant number, however, can be further classified. The amount of divisors of a number can be very large, and the following definitions help to categorize how abundant a number is:

  • $n \in \mathbb{Z}$ is highly abundant $\iff$ $\sigma (n) > \sigma (m)$ $\forall m<n$.

An example of a highly abundant number is 6, since $1 + 2 + 3 + 6 = 12$ is greater than the sigma function of 1 through the sigma function of 5. (The only time the sigma function doesn't produce $n + 1$ is at 4, where it produces 7, which is still less than 12.

  • $n \in \mathbb{Z}$ is superabundant $\iff$ $\frac {\sigma (n)}{n} > \frac {\sigma (m)}{m}$ $\forall m<n$.

Notice that 6 is also superabundant.

  • $n \in \mathbb{Z}$ is colossally abundant $\iff \; \exists \epsilon >0 \; {s.t.} \; \forall k > 1 \;$ $\frac {\sigma (n)}{n^{1 + \epsilon}} \geq \frac {\sigma (k)}{k^{1 + \epsilon}}$.

This is more of a theoretical definition than the others, since it involves choosing an $\epsilon$ to make the statement true. Believe it or not, 6 is also a colossally abundant number. Other colossally abundant numbers include 2, 12, 60, and 120.

One should note that a number is always superabundant and highly abundant if it is colossally abundant. However, a highly abundant number is not necessarily superabundant or colossally abundant.

Amicable Numbers

Thabn ibt Qurra made the following proposition around the ninth century A.D.:

  • For p, q, and r primes defined as $p = 3 * 2^{n-1} - 1$, $q = 3 * 2^{n} - 1$, and $r = 9 * 2^{2n-1} - 1$, then $a = 2^{n}pq$ and $b = 2^{n}r$ form an amicable pair.

Proof: We want $\sigma (a) = \sigma (b) = a + b$, with $\sigma (a) = \sigma (2^{n}) \sigma (p) \sigma (q)$ and $\sigma (b) = \sigma (2^{n}) \sigma (r)$.

It suffices to demonstrate the following statements:

(1)
\begin{align} \sigma (p) \sigma (q) = \sigma (r) \end{align}
(2)
\begin{align} \sigma (2^{n}) \sigma (r) = 2^{n} (pq + r) \end{align}

Notice that statement (1) is derived from $\sigma (a) = \sigma (b)$ and that statement (2) is taken from the definition of a and b.

  • To show (1) holds:

$\sigma (p) \sigma (q) = (p + 1) (q + 1) = (3 * 2^{n-1}) (3 * 2^{n}) = 9 * 2^{2n-1} = r + 1 = \sigma (r)$.

  • To show (2) holds:

$\sigma (2^{n}) \sigma (r) = (2^{n+1} - 1) ( r + 1 )$. We can then break down $2^{n+1} - 1$ and $r + 1$ before we multiply:
$2^{n+1} - 1 = 2 * 2^{n} - 1 = 2 * 2^{n} - 2 * 2^{-1}$.
$r + 1 = 9 * 2^{2n-1} = 9 * 2^{2n} * 2^{-1} = 9 * 2^{n} * 2^{n} * 2^{-1}$
Multiplying those both through, we get $9 * 2^{n} * 2^{n} * 2^{n} * 2^{-1} - 9 * 2^{n} * 2^{n} * 2 * 2^{-1} * 2^{-1}$.
If we take a $2^{n}$ out of both sides and simplify, we're left with $\sigma (2^{n}) \sigma (r) = 2^{n} (9 * 2^{n} - 9 * 2^{n-1} )$.

Since (1) and (2) hold, we conclude that the proposition holds.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License