Even Integer with Abundancy Index greater than 9

From ProofWiki
Jump to navigation Jump to search


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

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


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

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

\(\ds \frac {\map \sigma {p^k} } {p^k}\) \(=\) \(\ds \frac {p^{k + 1} - 1} {p^k \paren {p - 1} }\) Sigma Function of Power of Prime
\(\ds \) \(=\) \(\ds \frac {p - p^{-k} } {p - 1}\)
\(\ds \) \(<\) \(\ds \frac p {p - 1}\) 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:

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

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

The result can be verified by direct computation.