Abundant numbers

In this article, readers are expected to know how to sum the factors of any integer. If you are not familiar with that concept, please see the main article first: Sum of Factors.

An abundant number (also known as excessive numbers) is a positive integer such that the sum of its proper divisors is greater than the number itself. Or equivalently, a positive integer $n$ is said to be abundant if $\sigma_1 (n) > 2n$, where $\sigma_1(n)$ denotes the sum of factors of $n$.

12 is an abundant number, and this is because the sum of its proper divisors (1, 2, 3, 4, 6) is larger than 12. It can be shown by trial and error that 12 is the first abundant number (i.e. there are no abundant numbers smaller than 12).

The following is the list of the first 22 abundant numbers:

12, 18, 20, 24, 30, 36, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90, 96, 100.

In fact, the first 60 abundant numbers are all even numbers! But this does not mean that all abundant numbers are even numbers. Can you find the first odd abundant number? Submit your answer below.

As you might have already figured out from the exercise above, determining whether an integer is abundant is a rather tedious process and should be left to be solved by computers.

• There exist infinitely many even and odd abundant numbers.
• All multiples of an abundant number are also abundant.
• If $P$ is a perfect number, then $2P, 3P, 4P, \ldots$ are all abundant numbers.
• Every even integer greater than 46 can be written as the sum of 2 abundant numbers in at least 1 way.

Let $n > 46$ be an even number, so $n \bmod{12}$ has 6 possible values.

Note that 12, 20 and 40 are abundant numbers, and they will be used in the following 6 cases:

Case 1: If $n\equiv 0 \pmod{12}$, then it can be expressed as a sum of smaller multiples of 12 in at least 2 ways $($e.g $60 = 24+36 = 18+42).$

Case 2: If $n\equiv 2 \pmod{12}$, we can rewrite $n = 6m + 20$ for some odd $m>1$, so $6m$ is abundant.

Case 3: If $n\equiv 4 \pmod{12}$, we can rewrite $n = 12m + 20$ for some odd $m>1$, so $12m$ is abundant.

Case 4: If $n\equiv 6 \pmod{12}$, we can rewrite $n = 6m+12$.

Case 5: If $n\equiv 8 \pmod{12}$, we can rewrite $n = 12m + 20$.

Case 6: If $n\equiv 10 \pmod{12}$, we can rewrite $n = 6m + 40$ for some integer $m>1$.

We have considered all 6 cases, and thus the proof holds true. $_\square$

By applying the properties above, see if you can solve this exercise!

From above, we call a positive integer $n$ abundant if $\sigma_1 (n) > 2n$.
Likewise, we call a positive integer $n$ a perfect number if $\sigma_1 (n) > 2n$.
We call a positive integer $n$ deficient if $\sigma_1 (n) < 2n$.

If we call the abundancy of $n$ as $\text{abundancy}(n) = \dfrac{\sigma_{1}(n)}n$, then

• $\text{abundancy}(n) > 2$ if $n$ is abundant,

• $\text{abundancy}(n) = 2$ if $n$ is a perfect number,

• $\text{abundancy}(n) < 2$ if $n$ is deficient.

• Sum of Factors

• Amicable Numbers

• Perfect Numbers