Hi all. My question today will be regarding what I consider to be a "stumbling block" while trying to research odd perfect numbers.
Let $N = {q^k}{n^2}$ be an odd perfect number with Euler prime $q$. Since $\gcd(q, n) = 1$, we know that $q \neq n$.
In 2008, I proved that $q^k < n^2$. This implies that, if $n < q$, then Sorli's conjecture that $k = {\nu}_q(N) = 1$ would follow.
I currently know that $I(q) \leq 6/5 < \sqrt{5/3} < I(n)$, where $I(x) = \sigma(x)/x$ is the abundancy index of $x$. In particular, this means that $$\frac{\sigma(q)}{\sigma(n)} < \frac{q}{n}$$.
Thus, if $q < n$, then $\sigma(q) < \sigma(n)$. (The contrapositive of this last implication is $\sigma(n) < \sigma(q)$ implies that $n < q$.)
Now, since $\sigma(q) = q + 1$, I believe we have three cases to consider:
Case 1: $q < \sigma(q) < n < \sigma(n)$
Case 2: $n < q < \sigma(q) \leq \sigma(n)$
Case 3: $n < \sigma(n) \leq q < \sigma(q)$
I also know that $$\frac{\sigma(q)}{n} \neq \frac{\sigma(n)}{q}.$$
My problem is: How do I dispose of Case 2? The motivation is that I want to establish an equivalence between the inequalities $\frac{\sigma(q)}{n} < \frac{\sigma(n)}{q}$ and $q < n$. This way, all it takes to prove Sorli's conjecture will be to show that $\frac{\sigma(n)}{q} < \frac{\sigma(q)}{n}$.
My idea is to show that $$\frac{\sigma(q)}{\sigma(n)} \leq 1 < \frac{q}{n}$$ cannot occur by considering two separate cases under Case 2:
Case 2A: $\frac{\sigma(q)}{n} < \frac{\sigma(n)}{q}$
Under this case:
$$\frac{\sigma(n)}{q} > \sqrt[4]{\frac{5}{3}}$$ $$2 > \frac{\sigma(n)}{n} > \sqrt{\frac{5}{3}}$$ $$1 < \frac{\sigma(q)}{q} \le \frac{6}{5}$$ $$\frac{\sigma(q)}{n} < \frac{2}{\sqrt[4]{\frac{5}{3}}}$$
$$\frac{\sigma(q)}{n} < \frac{\sigma(n)}{q}$$ $$\frac{\sigma(q)}{q} < \frac{\sigma(n)}{n}$$
Consequently:
$$\sigma(q)\left(\frac{1}{n} + \frac{1}{q}\right) < \sigma(n)\left(\frac{1}{q} + \frac{1}{n}\right)$$ $$\sigma(q) < \sigma(n)$$
Therefore:
$$\frac{\sigma(q)}{\sigma(n)} < 1$$
and:
$$\frac{n}{q} = \frac{\frac{\sigma(q)}{q}}{\frac{\sigma(q)}{n}} > \frac{\sqrt[4]{\frac{5}{3}}}{2} \approx 0.56811$$
Case 2B: $\frac{\sigma(n)}{q} < \frac{\sigma(q)}{n}$
Under this case:
$$\frac{\sigma(n)}{q} < \frac{2}{\sqrt[4]{\frac{5}{3}}}$$ $$\frac{\sigma(n)}{n} < 2$$ $$\frac{6}{5} \geq \frac{\sigma(q)}{q} > 1$$ $$\frac{\sigma(q)}{n} > \sqrt[4]{\frac{5}{3}}$$
$$\frac{\sigma(n)}{q} < \frac{\sigma(q)}{n}$$ $$\frac{\sigma(q)}{q} < \frac{\sigma(n)}{n}$$
Consequently:
$$\frac{1}{q}\left(\sigma(n) + \sigma(q)\right) < \frac{1}{n}\left(\sigma(q) + \sigma(n)\right)$$ $$n < q$$
Therefore:
$$\frac{\sigma(q)}{\sigma(n)} = \frac{\frac{\sigma(q)}{q} + \frac{\sigma(q)}{n}}{\frac{\sigma(n)}{q} + \frac{\sigma(n)}{n}} > \frac{1 + \sqrt[4]{\frac{5}{3}}}{\frac{2}{\sqrt[4]{\frac{5}{3}}} + 2} \approx 0.56811$$
and
$$\frac{n}{q} < 1$$
I was wondering if anybody out there would have some (better) ideas on how to improve on the bounds for $\frac{n}{q}$ in Case 2A and for $\frac{\sigma(q)}{\sigma(n)}$ in Case 2B. In particular, a sharp upper bound for $I(n) = \sigma(n)/n$ would be nice!
