Some further notes regarding Sorli's conjecture that $k = 1$ if $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form (i.e. $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$):
Suppose that $\sigma(q)/n = \sigma(n)/q$. Then we know that:
$$q\sigma(q) = n\sigma(n).$$
Since $\gcd(q, n) = 1$, then $q \mid \sigma(n)$ and $n \mid \sigma(q)$.
Therefore, it follows that $\sigma(q)/n$ and $\sigma(n)/q$ are equal positive integers.
This is a contradiction, as:
$$1 < \sigma(q)/q \leq 6/5 < \sqrt{5/3} < \sigma(n)/n < \sigma(qn)/qn < 2$$
which implies that:
$$1 < \sqrt{5/3} < [\sigma(q)/q][\sigma(n)/n] = [\sigma(q)/n][\sigma(n)/q] < 2$$
Therefore, either:
$$1 < \sigma(q)/q \leq 6/5 < \sqrt{5/3} < \sigma(n)/n < \sigma(qn)/qn < 2$$
which implies that:
$$1 < \sqrt{5/3} < [\sigma(q)/q][\sigma(n)/n] = [\sigma(q)/n][\sigma(n)/q] < 2$$
Therefore, either:
$\sigma(q)/n < \sigma(n)/q$, which implies $\sigma(q) < \sigma(n)$
OR
$\sigma(n)/q < \sigma(q)/n$, which implies $n < q$, further implying Sorli's conjecture that $k = 1$.
From previous posts, what remains is to dispose of the case:
$n < q < \sigma(q) \leq \sigma(n).$
Update: Brown has recently released a preprint where he claims a proof for $q < n$.
