Let $p^k m^2$ be an odd perfect number with special prime $p$.

This answer says something about the Descartes-Frenicle-Sorli Conjecture on odd perfect numbers (i.e. the prediction $k=1$).

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Assume that $p^k < 2m - 1$. We will show that $k \neq 1$.

Then
$$\frac{\sigma(p^k)}{2} = \frac{p^{k+1} - 1}{2(p - 1)} < \frac{2pm - p - 1}{2(p - 1)}$$
which is less than $m$ if $p > 2m - 1$. By assumption, we have
$$p \leq p^k < 2m - 1$$
which means that
$$m < \frac{\sigma(p^k)}{2}.$$
We thus obtain
$$p^k < 2m - 1 < \sigma(p^k) - 1 < \sigma(p^k)$$
which implies that $k \neq 1$ holds, under the assumption $p^k < 2m - 1$.

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Assume that $p^k > 2m - 1$. We will show that $k \neq 1$.

Then
$$\frac{\sigma(p^k)}{2} = \frac{p^{k+1} - 1}{2(p - 1)} > \frac{2pm - p - 1}{2(p - 1)}$$
which is more than $m$ if $p < 2m - 1$.

Note that
$$\frac{\sigma(p^k)}{2} \geq \frac{p^k + 1}{2} > m$$
does hold, under the assumption $p^k > 2m - 1$.

By assumption, we have
$$2m - 1 < p^k$$
which means that
$$p < 2m - 1 < p^k,$$
or $k \neq 1$ holds, under the assumption $p^k > 2m - 1$.