# 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 $k$:

\(\ds \frac {\map \sigma {p^k} } {p^k}\) | \(=\) | \(\ds \frac {p^{k + 1} - 1} {p^k \paren {p - 1} }\) | 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.

$\blacksquare$

## Sources

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

- 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $55$