Digital roots of even numbers whose abundance is of the specific form

Question

Let $n$ be an even number . Let $k$ denotes the abundance of $n$ and $dr(n)$ denotes a digital root of $n$ .

I have noticed the following pattern :

$\bullet$ If $k$ is of the form $18t+16$ then $dr(n)=1$ in almost all cases .

$\bullet$ If $k$ is of the form $18t+10$ then $dr(n)=4$ in almost all cases .

$\bullet$ If $k$ is of the form $18t+4$ then $dr(n)=7$ in almost all cases .

where $t \in \mathbb{Z}$ .

Here you can try it for yourself .

How to explain this phenomenon ?

Answer 1

I think what you are observing here is primarily that $\sigma(n)$ is often divisible by $9$.

Notice that, for $p$ prime, $\sigma(p^k)$ is divisible by $3$ when either

• $p \equiv 1 \,(\text{mod}\, 3)$ and $k \equiv 2 (\text{mod}\, 3)$, or
• $p \equiv 2 \,(\text{mod}\, 3)$ and $k$ is odd.

So there are lots of ways for factors of $3$ to end up in $\sigma(n)$. This is particularly true of even numbers: two-thirds of all even numbers pick up a factor of $3$ from the power of two in their prime factorization.

For example, if $n \equiv 1 (\text{mod}\, 9)$, then $\sigma(n)-2n \equiv \sigma(n)-2 \equiv \sigma(n)+7 (\text{mod}\, 9)$. Since $\sigma(n)$ is often congruent to $0$ modulo $9$, we can conclude that $\sigma(n)-2n$ is often congruent to $7$ modulo $9$ (corresponding to your $18t+16$ case). By choosing $n$ to be even, the phenomenon is even more common.