# If $q^k n^2$ is an odd perfect number with Euler prime $q$, is $\sigma(q^k)/n + \sigma(n)/q^k$ bounded from above?

STATEMENT OF THE PROBLEM

If $q^k n^2$ is an odd perfect number with Euler prime $q$, is $\sigma(q^k)/n + \sigma(n)/q^k$ bounded from above?

MOTIVATION

Let $\sigma=\sigma_{1}$ denote the classical sum-of-divisors function, and denote the abundancy index of $x \in \mathbb{N}$ by $I(x)=\sigma(x)/x$.

It is known that the inequality $$I(q^k) + I(n) < \frac{\sigma(q^k)}{n}+\frac{\sigma(n)}{q^k}$$ holds if and only if the biconditional $$q^k < n \iff \sigma(q^k) < \sigma(n)$$ is true. This biconditional is true if $\sigma(q^k)<n$, or if $\sigma(n) \leq q^k$. (I currently am not aware of any other conditions for which the biconditional holds.) Edit (August 11 2017): The biconditional is also true when $q^k < n$. (This follows from $I(q^k)<I(n)$.)

Note that if $\sigma(q^k)/n + \sigma(n)/q^k < C$ for some absolute constant $C$, then $$\sqrt{\frac{8}{5}}\frac{n}{C} < q^k < Cn,$$ so that $C > \sqrt[4]{8/5}$.

However, I know that $C > \sqrt[4]{8/5}$ is far from the truth, as I have recently been able to verify that either $$\frac{\sigma(q^k)}{n} < \sqrt{2} < \frac{\sigma(n)}{q^k}$$ or $$\frac{\sigma(n)}{q^k} < \sqrt{2} < \frac{\sigma(q^k)}{n}$$ is true. In the first case, $q^k < n\sqrt{2}$, while in the second case, we have $n < q^k$.

Of course, trivially we have $$I(q^k) + I(n) < I(q^k) + I(n^2) < 3.$$

Suppose to the contrary that $$\frac{\sigma(q^k)}{n}+\frac{\sigma(n)}{q^k}$$ is bounded from above.
Since $x < \sigma(x)$ for all $x > 1$, and $\sigma(y) < 2y$ for deficient $y$, $$\frac{q^k}{n}+\frac{n}{q^k}<\frac{\sigma(q^k)}{n}+\frac{\sigma(n)}{q^k}<2\cdot\bigg(\frac{q^k}{n}+\frac{n}{q^k}\bigg),$$ so that the biconditional $$\frac{\sigma(q^k)}{n}+\frac{\sigma(n)}{q^k} \text{ is bounded from above } \iff \frac{q^k}{n}+\frac{n}{q^k} \text{ is bounded from above }$$ holds.
But in general, we know that the function $$f(z) = z + \frac{1}{z}$$ is not bounded from above. (It suffices to let $z$ approach $0$ [from the right] and to let $z$ approach $\infty$.)
Consequently, the sum $$\frac{\sigma(q^k)}{n}+\frac{\sigma(n)}{q^k}$$ may not be bounded from above.
• This only shows that there is no trivial, i.e. purely analytic, argument proving that this expression is unbounded. But there might be only finitely many odd perfect numbers, or there might be some relation between $n$ and $q$, which implies that the ratio is bounded. – Jan-Christoph Schlage-Puchta Dec 9 '17 at 18:33