Even Integer with Abundancy Index greater than 9

Theorem

Let $n \in \Z_{>0}$ have an abundancy index greater than $9$.

Then $n$ has at least $35$ distinct prime factors.

We have for any prime $p$ and positive integer $k$:

$\ds \frac {\map {\sigma_1} {p^k} } {p^k}$

$=$

$\ds \frac {p^{k + 1} - 1} {p^k \paren {p - 1} }$

$\ds $

$=$

$\ds \frac {p - p^{-k} } {p - 1}$

$\ds $

$<$

$\ds \frac p {p - 1}$

as $p^{-k} > 0$

$\ds $

$=$

$\ds 1 + \frac 1 {p - 1}$

In fact this is the limit of the abundancy index of a prime power.

The greater the prime, the smaller this value is.

Therefore finding the least number of prime factors a number needs to have its abundancy index exceed $9$ is to find the least $m$ such that:

$\ds \prod_{i \mathop = 1}^m \frac {p_i} {p_i - 1} > 9$

where $p_i$ is the $i$th prime ordered by size.

The result can be verified by direct computation.

$\blacksquare$

Apr. 1986: Richard Laatsch: Measuring the Abundancy of Integers (Math. Mag. Vol. 59, no. 2: pp. 84 – 92) www.jstor.org/stable/2690424