Search This Blog


On the condition $n < \sigma(q^k)$ where $N = q^k n^2$ is an odd perfect number given in Eulerian form

Let $N = q^k n^2$ be an odd perfect number with Euler prime $q$.

Brown recently announced a proof for the inequality $q < n$, and a partial proof that $q^k < n$ holds "in many cases".  The inequality $q^k < n$ was conjectured by Dris in his M. Sc. thesis.

Note that when the proof for $q^k < n$ is completed, it would follow that the following implication holds:
$$n < \sigma(q^k) \Longrightarrow k > 1.$$
(This is because $q$ and $\sigma(q)=q+1$ are consecutive integers.)

In this post, we will derive bounds for the quantities
Notice that, already we have
$$\frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n}$$
$$I(q^k) < \frac{5}{4} < \bigg(\frac{8}{5}\bigg)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(n),$$
$$I(x) = \frac{\sigma(x)}{x}$$
is the abundancy index of $x \in \mathbb{N}$.

First, we dispose of the following lemmas:

Lemma 1.
$$\frac{\sigma(q^k)}{\sigma(n)} = \frac{\frac{\sigma(q^k)}{q^k} + \frac{\sigma(q^k)}{n}}{\frac{\sigma(n)}{q^k} + \frac{\sigma(n)}{n}}$$

Lemma 2.
$$\frac{q^k}{n} = \frac{\frac{q^k}{\sigma(q^k)} + \frac{q^k}{\sigma(n)}}{\frac{n}{\sigma(q^k)} + \frac{n}{\sigma(n)}}$$

The proofs of Lemmas 1 and 2 are trivial.

(This blog post is currently a WORK IN PROGRESS.)