If $q$ is a prime with $q \equiv k \equiv 1 \pmod 4$ and $n^2$ is deficient, is $\frac{\sigma(q^k)}{\sigma(n^2)}$ bounded from below by a function of $n$?
Of course, an easy lower bound is $\frac{\sigma(q^k)}{\sigma(n^2)} \geq \frac{q^k + 1}{\sigma(n^2)} > \frac{q^k + 1}{2n^2} \geq \frac{5^1 + 1}{2n^2} = \frac{3}{n^2}.$
I am interested in lower bounds of the form $\frac{\sigma(q^k)}{\sigma(n^2)} \geq \frac{a}{bn},$ where $a, b \in \mathbb{R}$.