In this post, we consider the problem of deriving bounds for the quantity
$$\frac{q^k}{n^2},$$
in terms of $n$, if $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form.
We will be using the bounds obtained in the post from July 2013.
Case I: $q^k < n$
Case I-A: $k = 1 \Longrightarrow q = q^k < n$
We have the chain of inequalities
$$\frac{1}{2}n < q^k < n < 2{q^k}$$
which implies that
$$\frac{1}{2n} < \frac{q^k}{n^2} < \frac{1}{n}.$$
Case I-B: $k \neq 1 \Longrightarrow q < q^k < n$
We have the chain of inequalities
$$\frac{n}{\sqrt{2}} < q^k < n < {q^k}\sqrt{2}$$
which implies that
$$\frac{1}{{\sqrt{2}}n} < \frac{q^k}{n^2} < \frac{1}{n}.$$
Case II: $n < q^k$
Case II-A:$k = 1 \Longrightarrow n < q = q^k$
We have the chain of inequalities
$$\frac{q^k}{\sqrt{3}} < n < \sqrt[4]{\frac{108}{125}}{q^k} < \sqrt[4]{\frac{108}{125}}{\sqrt{3}}{n}$$
which implies that
$$\sqrt[4]{\frac{125}{108}}\frac{1}{n} < \frac{q^k}{n^2} < \frac{\sqrt{3}}{n}.$$
Case II-B:$k \neq 1 \Longrightarrow q < n < q^k$
We have the chain of inequalities
$$\frac{q^k}{2} < n < \sqrt[4]{\frac{125}{128}}{q^k} < 2{\sqrt[4]{\frac{125}{128}}}{n}$$
which implies that
$$\sqrt[4]{\frac{128}{125}}\frac{1}{n} < \frac{q^k}{n^2} < \frac{2}{n}.$$
We have therefore proven the following theorem (as we already know, from the theorems $q^k < n^2$ [Dris, 2012] and $N = {q^k}{n^2} > {10}^{1500}$ [Ochem and Rao, 2012], that $n > {10}^{375}$):
Theorem: If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, then $q^k = \circ(n^2)$.
Some random thoughts on pseudo-random things
