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4.7.17

A curious inequality involving divisors of odd perfect numbers

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

Since $q^k n \mid N$ and $N$ is perfect, then $q^k n$ is deficient.  Therefore, $I(q^k n) < 2$, where $I(x)=\sigma(x)/x$ is the abundancy index of $x \in \mathbb{N}$.

It follows that
$$\frac{1}{2}\cdot\frac{\sigma(q^k)}{n} < \frac{q^k}{\sigma(n)}$$
and
$$\frac{1}{2}\cdot\frac{\sigma(n)}{q^k} < \frac{n}{\sigma(q^k)}.$$
Adding the last two inequalities, we get
$$\frac{\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}}{2} < \frac{q^k}{\sigma(n)} + \frac{n}{\sigma(q^k)}.$$

THIS BLOG POST IS CURRENTLY A WORK IN PROGRESS.