Even Integer with Abundancy Index greater than 9

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Theorem

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


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


Proof

Since Sigma Function is Multiplicative, it follows easily that abundancy index is multiplicative as well.

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

\(\displaystyle \frac {\map \sigma {p^k} } {p^k}\) \(=\) \(\displaystyle \frac {p^{k + 1} - 1} {p^k \paren {p - 1} }\) Sigma Function of Power of Prime
\(\displaystyle \) \(=\) \(\displaystyle \frac {p - p^{-k} } {p - 1}\)
\(\displaystyle \) \(<\) \(\displaystyle \frac p {p - 1}\) as $p^{-k} > 0$
\(\displaystyle \) \(=\) \(\displaystyle 1 + \frac 1 {p - 1}\)

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:

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

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

The result can be verified by direct computation.

$\blacksquare$


Sources