If $N = q^k n^2$ is an odd perfect number and $n < q^{k+1}$, does it follow that $k > 1$? Let $\sigma(x)$ be the sum of the divisors of the positive integer $x$.
If $\sigma(M) = 2M$, then $M$ is said to be perfect.
Currently, as of December 2018, there are $51$ known examples of even perfect numbers -- on the other hand, we still do not know whether there are any odd perfect numbers.
Euler derived the general form that an odd perfect number $N$ must take:
$$N = {q^k}{n^2},$$
where $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n) = 1$.  We call $q$ the special or Euler prime of $N$.
Descartes, Frenicle and subsequently Sorli conjectured that $k = 1$.
In [Dris, 2012], it was shown that the implications
$$n < q \Longrightarrow k = 1$$
and
$$n < q^2 \Longrightarrow k = 1$$
are true.
Now, note that, since $q$ and $\sigma(q) = q + 1$ are consecutive integers, then the following implications are true.
Case 1: $q^k < n < \sigma(q^k) < \sigma(n) \Longrightarrow k > 1$
Case 2: $n < q^k < \sigma(n) < \sigma(q^k) \Longrightarrow k > 1$
The remaining cases to be considered are:
Case 3: $q^k < \sigma(q^k) < n < \sigma(n) \land k \geq 1$
Case 4: $n < \sigma(n) < q^k < \sigma(q^k) \land k \geq 1$
Case 5: $n < q^k \leq \sigma(n) < \sigma(q^k) \land k \geq 1$
Case 6: $n < q^k < \sigma(q^k) \leq \sigma(n) \land k \geq 1$ 
Am I right in thinking that it is possible to prove the following implication, given the above cases?


CLAIM: If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, then
    $$n < q^{k + 1} \Longrightarrow k \neq 1.$$


Easy to prove for Cases 1 and 2.  Should also be doable for the rest.  Anybody have any ideas?
Added September 12 2016

From one of the answers below, it can be shown that the condition $q^2 < n$ is equivalent to the truth of the implication
  $$n < q^{k + 1} \Longrightarrow k \neq 1.$$

 A: Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form.  Let
$$I(x) = \dfrac{\sigma(x)}{x}$$
be the abundancy index of $x$.
This is a partial answer to the original question, and proves the claim in the affirmative, subject to the validity of a recent proof claim by Patrick A. Brown that $q^k < n$ holds (in many cases).
First, we show the following lemmas:
Lemma 1.
$$I(q^k) + I(n) < \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} \implies \left(q^k < n \iff \sigma(q^k) < \sigma(n)\right)$$
Proof.
$$I(q^k) + I(n) < \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} \implies \left(n - q^k\right)\left(\sigma(q^k) - \sigma(n)\right) < 0 \implies \left(q^k < n \iff \sigma(q^k) < \sigma(n)\right)$$
QED
Lemma 2.
$$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} < I(q^k) + I(n) \implies \left(q^k < n \iff \sigma(n) < \sigma(q^k)\right)$$
Proof.
$$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} < I(q^k) + I(n) \implies 
\left(q^k - n\right)\left(\sigma(q^k) - \sigma(n)\right) < 0 \implies \left(n < q^k \iff \sigma(q^k) < \sigma(n)\right)$$
QED
Lemma 3.
$$I(q^k) + I(n) = \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} \iff \sigma(q^k) = \sigma(n)$$
Proof.
$$I(q^k) + I(n) = \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} \iff \left(q^k - n\right)\left(\sigma(q^k) - \sigma(n)\right) = 0 \iff \sigma(q^k) = \sigma(n)$$
since $q^k \neq n$.
QED
We are now ready to prove the following proposition:
Proposition 1.
$$q^k < n \iff \sigma(q^k) < \sigma(n) \iff \dfrac{\sigma(q^k)}{n} < \dfrac{\sigma(n)}{q^k}$$
Proof.
Since $I(q^k) < \sqrt[3]{2} < I(n)$ [Dris, 2012], then $\sigma(q^k) = \sigma(n) \implies n < q^k$.  So it suffices to consider the remaining case under Lemma 2.  It thus remains to consider $n < q^k < \sigma(q^k) < \sigma(n).$ But these are already ruled out, assuming Brown's proof for $q^k < n$ is successfully completed.
Therefore, we are left with the case under Lemma 1.  It follows that
$$q^k < n \iff \sigma(q^k) < \sigma(n) \iff \dfrac{\sigma(q^k)}{n} < \dfrac{\sigma(n)}{q^k}.$$
QED
Next, we prove the following proposition:
Proposition 2.
$$k = 1 \implies \sigma(q^k) < n$$
Proof.
We use Proposition 1 to list all possible permutations of the set $\left\{q^k, n, \sigma(q^k), \sigma(n)\right\}$.
So we have
$$\bf{A}: q^k < n < \sigma(q^k) < \sigma(n)$$
$$\bf{B}: q^k < \sigma(q^k) < n < \sigma(n)$$
$$\bf{C}: n < q^k < \sigma(n) < \sigma(q^k)$$
$$\bf{D}: n < \sigma(n) \leq q^k < \sigma(q^k)$$
(Notice that $\sigma(q^k) \neq n$ since $\sigma(q^k) \equiv k+1 \equiv 2 \pmod 4$ while $n$ is odd.)
Note that Brown's result rules out cases $\bf{C}$ and $\bf{D}$.
Now, $k = 1$ rules out case $\bf{A}$.
Therefore, under the assumption $k = 1$, we are left with case $\bf{B}$.
QED
We are left with proving the following lemma.
"Lemma 4".
$$n < \sigma(q^k) \iff n < q^{k+1}$$
"Proof".
$n < \sigma(q^k) < q^{k+1}$ since $I(q^k) < \dfrac{5}{4} < 5 \leq q$.
Hence, $n < \sigma(q^k) \implies n < q^{k+1}$.
It now remains to show that $n < q^{k+1} \implies n < \sigma(q^k)$.
"QED"
We now get the following "Theorem":
"Theorem".
$n < q^{k+1} \implies k \neq 1$
Proof.
Take the contrapositive of Proposition 2, and use "Lemma 4".
"QED"
In particular, by noting that $n < q^2 \implies k = 1$ is true, we obtain the following "Corollary":
"Corollary"
$q^2 < n$
Added September 08 2016 I just realized today that we actually have the following biconditional:
$$q^2 < n \iff \left\{n<q^{k+1} \implies k>1\right\}$$
A: Let me post another answer to this question, to collect some more thoughts regarding this problem.
The remaining cases to be considered are:
Case 3: $q^k < \sigma(q^k) < n < \sigma(n) \land k \geq 1$
Case 4: $n < \sigma(n) < q^k < \sigma(q^k) \land k \geq 1$
Case 5: $n < q^k \leq \sigma(n) < \sigma(q^k) \land k \geq 1$
Case 6: $n < q^k < \sigma(q^k) \leq \sigma(n) \land k \geq 1$
Since $q < n$ has already been proved by Brown, this means that $k > 1$ follows from Cases 4, 5, and 6.
Note that $k > 1$ also follows from Cases 1 and 2.
Let us now consider Case 3.  From Proposition 2 in the first answer, we have the following implication:
$$k=1 \implies \sigma(q^k)<n$$
Under Case 3, this implication is trivially true.  Additionally, if $k>1$, then the same implication is also vacuously true.
Since $k>1$ is sure to hold for all cases except Case 3, I am led to predict that:
Conjecture
If $N=q^k n^2$ is an odd perfect number with Euler prime $q$, then $k>1$ holds.
A: Solution-verification

Now, note that, since $q$ and $\sigma(q) = q + 1$ are consecutive integers, then the following implications are true.
Case 1: $q^k < n < \sigma(q^k) < \sigma(n) \Longrightarrow k > 1$
Case 2: $n < q^k < \sigma(n) < \sigma(q^k) \Longrightarrow k > 1$
The remaining cases to be considered are:
Case 3: $q^k < \sigma(q^k) < n < \sigma(n) \land k \geq 1$
Case 4: $n < \sigma(n) < q^k < \sigma(q^k) \land k \geq 1$
Case 5: $n < q^k \leq \sigma(n) < \sigma(q^k) \land k \geq 1$
Case 6: $n < q^k < \sigma(q^k) \leq \sigma(n) \land k \geq 1$

Note that Case 5 includes Case 2.
I would separate it into the following five cases :
Case 1 : $q^k < n < \sigma(q^k) < \sigma(n) \Longrightarrow k > 1$
Case 2 : $n < q^k < \sigma(n) < \sigma(q^k) \Longrightarrow k > 1$
Case 3 : $q^k\lt\sigma(q^k)\lt n\lt\sigma(n)\land k \geq 1$
Case 6 : $n\lt q^k\lt\sigma(q^k)\le\sigma(n)\land k \geq 1$
Case 7 : $n\lt \sigma(n)\le q^k\lt\sigma(q^k)\land k \geq 1$

Your proofs for Lemma 1, Lemma 2 and Lemma 3 are correct.
However, I think that the proof for Proposition 1 is not correct.
I don't think that you have proved
$$\sigma(q^k) < \sigma(n)\implies q^k \lt n$$


*

*It is true that $q^k\lt n\implies \sigma(q^k)\lt\sigma(n)$.


*It is true that $q^k\lt n\implies\frac{\sigma(q^k)}{n}\lt\frac{\sigma(n)}{q^k}$.


*Assuming that $q^k\lt n$, we see that it is true that $k = 1 \implies \sigma(q^k) < n$.


*Assuming that $q^k\lt n$, we see that the remaining case is Case 3.


*It is true that $q^2 < n \iff \left\{n<q^{k+1} \implies k>1\right\}$.
