(Note: This has been cross-posted to MSE. However, I feel that it is more likely to receive a good answer here, because I believe that it *is* a research-level question. For the mathematicians who voted to put this question on hold, I invite you to peruse this preprint to give you an inkling of where I'm driving at. Thanks!)

Let $\sigma$ be the classical sum-of-divisors function.

A number is said to be *perfect* if $\sigma(N)=2N$.

If $q^k n^2$ is an odd perfect number with Euler prime $q$ (i.e., $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$), are the following statements known to hold in general?

$\bf{(a)}$ $\gcd(n^2, \sigma(n^2))$ is

large.$\bf{(b)}$ The deficiency $D(n^2) = 2n^2 - \sigma(n^2)$ is

large.$\bf{(c)}$ The

index$i(q^k) = \sigma(N/q^k)/q^k$ islarge.

Using the *trivial* relationships:

$$\gcd(n^2, \sigma(n^2)) = \frac{D(n^2)}{\sigma(q^{k-1})} = i(q^k) = \frac{2n^2}{\sigma(q^k)}$$

then if the Descartes-Frenicle-Sorli conjecture that $k = 1$ is true, it is possible to show that a lower bound for the quantities $\bf{(a)}$, $\bf{(b)}$ and $\bf{(c)}$ is given by $n/\sqrt{3}$. (Here, I have used Acquaah and Konyagin's estimate $q < n\sqrt{3}$. The inequality $q^k < n^2$ then gives the desired *large* numerical bound if we use known lower bounds for the odd perfect number $N = q^k n^2$, latest of which are by Ochem and Rao.)

What happens when $k > 1$? I do know that $\sigma(n^2)/q^k \geq 315$ by using a result of Broughan, Delbourgo, and Zhou.

Is it possible to do better than this, apart from attempting a proof of (*obviously*) $q^k < n$?