During this lecture we discussed many classifications of classical numbers. These numbers included perfect, amicable, and sociable numbers. We discussed many different types of perfect numbers, restrictions on even and odd perfect numbers, sociable numbers and aliquot sequences.

# Perfect Numbers

Recall: **Perfect numbers** are positive integers n such that $\sigma(n)=2n$ where $\sigma(n)$ is the sum of the divisors function

**Almost Perfect Numbers** are positive integers such that $\sigma(n)=2n-1$

- The only known almost perfect numbers are powers of two
- The first few almost perfect numbers are 1,2,4,8, and 16
- It is still open to show that a number is almost perfect if and only if it is of the form 2
^{n} - Example: $\sigma(16)=1+2+4+8+16=32-1=31$

**Superperfect Numbers** are numbers for which $\sigma(\sigma(n))=2n$

- Even superperfect numbers are just 2
^{p-1}where M_{p}=2^{p}-1 is a Mersenne Prime - Example: $\sigma(\sigma(16))=\sigma(31)=31+1=32$

**K-Perfect Numbers** are positive integers such that $\sigma(n)=kn$

- Example: 120 is 3-Perfect! $\sigma(120)$=$\sigma$(2
^{3}3*5)=15*4*6=360=3*120

**Multiplicative Perfect Numbers** are numbers for which the product of divisors is equal to n^{2}

- Example: 6 is a multiplicative perfect number. Its divisors are 1,2,3,and 6 and their product is equal to 36=6
^{2}

**Unitary Perfect Numbers** are numbers n such that the sum of its proper unitary divisors not including itself is equal to n

- a is a proper unitary divisor of b if $(a, \frac {b}{a})=1$
- Example: 6 is a unitary perfect number. The divisors of 6 are 1,2,3,6 and we should check if 1,2, and 3 are proper unitary divisors

$(1, \frac {6}{1})=1$ , $(2, \frac {6}{2})=1$ , $(3, \frac {6}{3})=1$ so they're all proper unitary divisors

so we have that the sum of proper unitary divisors of 6 =6

**Hyperperfect Numbers** are natural numbers n such that $n=1+k(\sigma(n)-n-1)$

- A number is perfect if it is 1-hyperperfect: $n=1+1(\sigma(n)-n-1)=\sigma(n)-n$
- Example: $6=1+1(\sigma(6)-6-1)$ where k=1

# 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.

# Even and Odd Perfect Numbers

At this time there are only even perfect numbers known and if an odd perfect number existed it must meet a multitude of conditions. The *Pomerance Hueristic* explains that it is unlikely that odd perfect numbers exist. *Sylvester's Web of Conditions* describes conditions that these odd perfect numbers must meet.

"… a prolonged meditation on the subject has satisfied me that the existence

of any one such — its escape, so to say, from the complex web of conditions

which hem it in on all sides — would be little short of a miracle."

-James Sylvester 1888

The following list are known restrictions on odd perfect numbers*

1. $N = p^{4λ+1}Q^2$ (Euler)

2. At least 9 distinct prime factors (Nielsen)

3. At least 75 prime factors, counting multiplicities. (Hare)

4. Greater than $10^300$(Brent, Cohen, the te Riele)

5. Less than $2^{4^k}$, k distinct primes (Nielsen)

6. Largest prime exceeds $10^8$ (Goto & Ohno)

7. Second largest prime exceeds 10,000 (Iannucci)

8. Third largest prime exceeds 100 (Iannucci)

9. Largest component exceeds $10^20$(Cohen)

10. Largest exponent at least 4 (Kanold)

*obtained from http://oddperfect.org/against.html

Another argument to refute the existence of odd perfect numbers is a heuristic argument put forth by Carl Pomerance.

Outline of the Pomerance Hueristic:

We know an odd perfect number must be of the form $N = pm^2$(Euler).

The idea is to consider an arbitrary m and the count the probability of making a perfect number with m. However, as m and subsequently N become large it becomes very unlikely that there are odd perfect numbers.

# Amicable Numbers

- An amicable number are numbers m and n such that $\sigma(m)=\sigma(n)=m+n$
- In every known case the numbers in an amicable pair are either both even or both odd, it is open to show that this is always the case
- Example: 1184 and 1210 form an amicable pair

$\sigma(1184)$=$\sigma$$(2^5*3*7)$=$\frac {(2^6)-1}{2-1}$ * $\frac {(37^2)-1}{37-1}$=(63)(38)=2394

$\sigma(1210)$=$\sigma$$(11^2*2*5)$=$\frac {(11^3)-1}{11-1}$ * $\frac {(2^2)-1}{2-1}$*$\frac {(5^2)-1}{5-1}$=(133)(3)(6)=2394

Therefore 1184 and 1210 are an amicable pair

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)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.

# Sociable Numbers

**Sociable Numbers** are numbers that result in a aliquot sequence of period greater than 2.

Restricted Divisor Function: $s(n)=\sigma(n)–n$

Aliquot Sequence: The aliquot sequence is a recursive function defined for any starting number $n \in \mathbb{Z}$ by repeatedly applying the restricted divisor function.

$a_{n+1} = s(a_{n}) = \sigma(a_{n}) – n$

Note: There are three different types of Aliquot Sequences

First, those for perfect, amicable and sociable numbers have a repeating period.

Second, those that terminate in 0.

A number is Perfect $\iff$ it has a periodic aliquot sequence of period 1.

Example:

6 is a perfect number so lets compute its Aliquot sequence

$a_{0}=6$

$a_{1} = s(6) = 1+2+3+6-6 = 6$

And so the Aliquot sequence looks like this ${6, 6, 6, 6, …}$ which is a repeating Aliquot sequence of period 1.

A number is Amicable $\iff$ it has a periodic aliquot sequence of period 2.

Example:

The numbers 1184 and 1210 are amicable numbers

$a_{0}=1184$

$a_[1} =s(1184)=1210$

$a_{2} =s(1210)=1184$

And so the Aliquot Sequence for both 1184 looks like ${1184,1210,1184,1210,…}$ which has a repeating period of 2.

A number is Sociable $\iff$ it has a periodic aliquot sequence of period 3 or greater.

The number 1264460 is a sociable number and its Aliquot Sequence looks like this

${1264460, 1547860, 1727636, 1305184, 1264460,…}$ and has a repeating period of length 4.

Aspiring numbers are numbers whose Aliquot sequence is not periodic but eventually becomes periodic.

Example 95

$s(95) = 1+5 +19 = 25$

$s(25) = 1 +5 = 6$

$s(6) = 1+2+3 = 6$

so this sequence eventually becomes the same as the sequence for the perfect number 6 but 95 is not perfect, amicable or sociable but it is aspiring.

# Suggested Homework

- From Ch 3 in Strayer
- 57, 58, 59

- From Ch 3 in Schumer
- Given m find n so that m and n form an amicable pair. Prove your result
- 12285
- 63020
- 66928

- Given m find n so that m and n form an amicable pair. Prove your result