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On odd perfect numbers given in Eulerian form - Part 2

Let $N = q^k n^2$ be an odd perfect number given in Eulerian form.

The following equations can be easily derived:

$$N - (q^k + n^2) + 1 = \sigma(q^{k-1})(q-1)(n+1)(n-1)$$

$$\sigma(n^2) = {q^k}\cdot\left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)$$

$$n^2 = {\frac{\sigma(q^k)}{2}}\cdot\left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)$$

From the last two equations, it can be proved that
$$\gcd(n^2, \sigma(n^2)) = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})},$$
$$\sigma(n^2) - n^2 = \left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)\cdot\left(q^k - \frac{\sigma(q^k)}{2}\right)$$
$$= \left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)\cdot\left(\frac{q^k - \sigma(q^{k-1})}{2}\right)$$
$$= \gcd(n^2, \sigma(n^2))\cdot\left(\frac{q^k - \sigma(q^{k-1})}{2}\right).$$
This last equation expresses a relationship among the quantities
$$\sigma(n^2) - n^2,$$
$$\gcd(n^2, \sigma(n^2)),$$
Also, note that we actually have
$$\sigma(n^2) - n^2 = \frac{\left(2n^2 - \sigma(n^2)\right)\left(2q^k - \sigma(q^k)\right)}{2\sigma(q^{k-1})},$$
so that 
$$A(n^2) + \sigma(n^2) = \frac{{D(n^2)}\cdot{D(q^k)}}{\sigma(q^{k-1})},$$ 
where $A(x) = \sigma(x) - 2x$ is the abundance of $x$, and $D(x) = 2x - \sigma(x)$ is the deficiency of $x$.

In particular, we know that
$$\frac{\sigma(n^2)}{q^k} = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}$$
$$\frac{\sigma(q^k)}{n^2} = 2\cdot\left(\frac{\sigma(q^{k-1})}{2n^2 - \sigma(n^2)}\right).$$

Notice that, since $\sigma(q^k)\sigma(n^2) = \sigma(N) = 2N = 2{q^k}{n^2}$ and $\gcd(q^k,\sigma(q^k)) = 1$, then$$\frac{\sigma(n^2)}{q^k} = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}$$
is an odd integer.

The following papers obtain (ever-increasing) lower bounds for $\sigma(n^2)/q^k$:

(1)  J. A. B. Dris, The Abundancy Index of Divisors of Odd Perfect Numbers
(2)  J. A. B. Dris and F. Luca, A note on odd perfect numbers
(3)  F. J. Chen and Y. G. Chen, On Odd Perfect Numbers
(4)  K. A. Broughan, D. Delbourgo, and Q. Zhou, Improving the Chen and Chen result for odd perfect numbers,
(5)  F. J. Chen and Y. G. Chen, On the index of an odd perfect number

From this M. Sc. thesis, we know that
$$\frac{11}{3} \leq \frac{\sigma(q^k)}{n^2} + \frac{\sigma(n^2)}{q^k}.$$