## If N = {q^k}{n^2} is an odd perfect number with Euler prime q, is sigma(q^k)/sigma(n^2) bounded from below?

A good day to everyone.

As most of you already know by now, I am currently researching the topic of odd perfect numbers (hereinafter abbreviated as OPN) and Sorli's conjecture, which predicts that, if $N = {q^k}{n^2}$ is an OPN with Euler prime $q$ (i.e. $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$), then $k = 1$.

A sufficient condition for Sorli's Conjecture is $n < q$. I currently do not know if this condition is also necessary for $k = 1$ to hold.

Let $I(x) = \frac{\sigma(x)}{x}$ denote the abundancy index of the positive integer $x$.

Currently, I am working on finding bounds for the quantity $$\frac{\sigma(q^k)}{\sigma(n^2)}.$$

It is easy to show that:

$$\frac{\sigma(q^k)}{n^2} \le \frac{2}{3} < 1 < I(q^k) < \frac{5}{4} < \frac{8}{5} < I(n^2) < 2 < 3 \leq \frac{\sigma(n^2)}{q^k}. \hspace{0.05in} (**)$$

By using the following equation:

$$\frac{\sigma(q^k)}{\sigma(n^2)} = \frac{\frac{\sigma(q^k)}{n^2} + I(q^k)}{I(n^2) + \frac{\sigma(n^2)}{q^k}},$$

we get the upper bound:

$$\frac{\sigma(q^k)}{\sigma(n^2)} < \frac{5}{12}.$$

Now here is a question that I have for which I currently have no satisfactory answer: Will it be possible to derive a lower bound for $\frac{\sigma(q^k)}{\sigma(n^2)}$, given the current state of knowledge on OPNs and (perhaps) further improvements (if any) to the bounds given in $(**)$?

I am thinking this can be done by fixing either $q$ or $k$ (or maybe, even both), but I am not entirely sure about this.

Finally, note that:

$$2 = I(q^k)I(n^2) = \left[\frac{\sigma(q^k)}{n^2}\right]\left[\frac{\sigma(n^2)}{q^k}\right] < \left[\frac{\sigma(q^k)}{n^2}\right]\left[\frac{[\sigma(n)]^2}{q^k}\right] = \left[\frac{\sigma(q^k)}{n}\right]\left[\frac{\sigma(n)}{q^k}\right]\cdot{I(n)},$$

and that:

$$\sigma(q^k) \neq n$$

because $\sigma(q^k) \equiv 2 \pmod 4$.

-
 Since $q$ is prime with $q \equiv 1 \pmod 4$, then $q \geq 5$. Consequently, we have the lower bound $$\frac{\sigma(q^k)}{\sigma(n^2)} \geq \frac{q^k + 1}{\sigma(n^2)} > \frac{q^k + 1}{2n^2} \geq \frac{5^1 + 1}{2n^2} = \frac{3}{n^2}.$$ I am interested in sharper lower bounds of the form $$\frac{\sigma(q^k)}{\sigma(n^2)} \geq \frac{a}{bn},$$ where $a, b \in \mathbb{R}$. – Jose Arnaldo Dris Jul 1 at 23:15