The title says it all.
Question
If $N = qn^2$ is an odd perfect number with Euler prime $q$ and $\gcd(q,n)=1$, is it possible to have $q + 1 = \sigma(n)$?
Heuristic
From the Descartes spoof, with quasi-Euler prime $q_1$:
$$n_1 = 3003 < \sigma(n_1) = 5376 < q_1 = 22021$$
So it appears that it might be possible to prove that $q + 1 \neq \sigma(n)$.
Some Essential Estimates
Acquaah and Konyagin recently obtained the estimate $q < n\sqrt{3}$. We will use this estimate to obtain an upper bound for $\sigma(q)/n$.
Ochem and Rao recently obtained the lower bound $N > {10}^{1500}$ for the magnitude of an odd perfect number. Using this bound, together with the inequality $n < q$, gives $$I(q) < 1 + {10}^{-500}.$$
Motivation
We wish to prove the following proposition:
If $N = {q^k}{n^2}$ is an odd perfect number with Euler prime $q$, then $3 \nmid N$ implies that $q < n$.
If $q + 1 \neq \sigma(n)$, then it follows that $$I(q) + I(n) \neq \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q}$$ from which we obtain $$I(q) + I(n) < \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q},$$ since the reverse inequality $$\frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < I(q) + I(n)$$ will violate the inequality $I(q) < \sqrt[3]{2} < I(n)$ (see this paper). (Edit February 8 2016: Assuming $$\frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < I(q) + I(n)$$ is true, then $$\left(q < n\right) \land \left(\sigma(n) < \sigma(q)\right)$$ is false. However, I am currently unable to rule out $$\left(n < q\right) \land \left(\sigma(q) < \sigma(n)\right).$$ This particular case remains open.)
But the inequality $$I(q) + I(n) < \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q}$$ implies that the biconditional $$q < n \Longleftrightarrow \sigma(q) < \sigma(n)$$ holds.
This biconditional is then a key ingredient in the proof of the proposition mentioned earlier.
My method is able to rule out $\sigma(q) = q + 1 = \sigma(n)$ if $3 \nmid n$, since we obtain $$2.799 \approx 1 + 2^{\frac{\log(6/5)}{\log(31/25)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(5))}{\log(I(5^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$ $$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$$$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(q)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$ (where the smallest prime factor $u$ of $N$ satisfies $u \geq 5$), whence we arrive at a contradiction.
Further Considerations
If $\sigma(q) = q + 1 = \sigma(n)$ and $3 \mid n$, then the same method does not force a contradiction, because we then have $$2.7199 \approx 1 + 2^{\frac{\log(4/3)}{\log(13/9)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(3))}{\log(I(3^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$ $$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$$$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(q)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$ where $u$ is the smallest prime factor of $N$.
Added February 7 2016
If $q + 1 = \sigma(n)$, then $\sigma(n) \equiv 2 \pmod 4$, so that $n$ takes the form $$n = {p^r}{m^2}$$ where $p$ is a prime with $p \equiv r \equiv 1 \pmod 4$ and $\gcd(p, m) = 1$. If $3 \mid n$, then $p \neq 3$, so that $3 \mid m$.
So I have $$\sigma(n) = \sigma(p^r)\sigma(m^2)$$ where $\sigma(p^r) \equiv r + 1 \equiv 2 \pmod 4$.
Since $\gcd(q, q+1) = 1$, then $\gcd(q, \sigma(n)) = 1$, so that $$\gcd(q, \sigma(p^r)\sigma(m^2)) = 1.$$ Thus, $q \nmid \sigma(p^r)$ and $q \nmid \sigma(m^2)$.
However, note that $$q \mid \sigma(n^2) = \sigma(p^{2r})\sigma(m^4).$$
Alas here is where I get stuck.
