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On a biconditional for odd perfect numbers - Part 1

Let $N=q^k n^2$ be an odd perfect number with Euler prime $q$.  The Descartes-Frenicle-Sorli conjecture predicts that $k=1$.

In this preprint, it is shown that the divisibility condition $D(n^2) \mid n^2$ is equivalent to $k=1$, where $D(n^2)=2n^2 - \sigma(n^2)$ is the deficiency of the non-Euler part $n^2$.  (In particular, this shows that the Descartes-Frenicle-Sorli conjecture is true if and only if the non-Euler part $n^2$ is deficient-perfect.)

In this blog post, we investigate the range of possible values for the quotient ${n^2}/{D(n^2)}$.

First, we have the equation
from which it follows that
$$\frac{n^2}{D(n^2)} = \frac{\sigma(q^k)}{2\sigma(q^{k-1})}.$$
(When $k=1$, the RHS of the last equation is equal to $(q+1)/2$.)

We compute:
$$\frac{\sigma(q^k)}{2\sigma(q^{k-1})} = \frac{q^{k+1} - 1}{2\bigg(q^k - 1\bigg)} = \frac{q}{2} + \frac{1}{2}\cdot\frac{1}{\sigma(q^{k-1})}$$
Since we have the bounds
$$1 \leq I(q^{k-1}) < \frac{q}{q - 1}$$
where $I(x)=\sigma(x)/x$ is the abundancy index of $x$, then we have
$$\frac{q}{2} + \frac{q - 1}{2q^k} < \frac{\sigma(q^k)}{2\sigma(q^{k-1})} \leq \frac{q}{2} + \frac{1}{2q^{k-1}}.$$

Note that
$$\frac{n^2}{D(n^2)} = \frac{\sigma(q^k)}{2\sigma(q^{k-1})} \leq \frac{q+1}{2}.$$

Here is our conjecture:

Conjecture (March 8, 2017):
$$\frac{n^2}{D(n^2)} = \frac{q+1}{2}$$

Note that this conjecture is equivalent to the Descartes-Frenicle-Sorli conjecture.