k-Perfect, Abundant, and Deficient Numbers

# k-Perfect, Abundant, and Deficient Numbers

Recall from The Sum of Positive Divisors of an Integer page that if $n \in \mathbb{Z}$ then $\sigma(n)$ denotes the sum of all positive divisors of $n$ explicitly defined by:

(1)
\begin{align} \quad \sigma(n) = \sum_{d \mid n}_{d > 0} d \end{align}

We will now define some special types of numbers.

 Definition: A number $n \in \mathbb{N}$ is said to be Perfect if $\sigma(n) = 2n$ and are called $k$-Perfect if $\sigma (n) = kn$ for some $k \in \mathbb{N}$. $n$ is said to be Abundant if $\sigma (n) > 2n$ and Deficient if $\sigma(n) < 2n$.

We will now look at some simple results regarding these type of numbers.

 Proposition 1: If $p$ is a prime number then $p$ is deficient.
• Proof: We have that $\sigma (p) = 1 + p$. Since $p$ is a prime number we have $p > 1$, and thus:
(2)
\begin{align} \quad \sigma(p) = 1 + p < p +p \end{align}
• Thus $p$ is deficient. $\blacksquare$
 Corollary 2: There are infinitely many deficient numbers.
• Proof: There are infinitely many prime numbers and so by Proposition 1 there are infinitely many deficient numbers. $\blacksquare$
 Proposition 3: There are infinitely many abundant numbers.
• Proof: Suppose that $n = 2^a \cdot 3$ and $a \geq 2$. Hence there will be infinitely many abundant numbers for any $a \geq 2$ as shown below:
(3)
\begin{align} \sigma (n) & = \sigma (2^a) \sigma (3) \\ \sigma (n) & =(2^{a + 1} - 1)(3 + 1) \\ \sigma (n) & = \cdot 2^{a + 1} \cdot 3 + 2^{a + 1} - 4 \\ \sigma (n) & = 2 ( 2^a \cdot 3) + 2^{a + 1} - 4 \\ \sigma (n) & = 2n + 2^{a +1} - 4 \\ \sigma (n) & > 2n \end{align}