## 2.7.13

### OPN Research - July 2013

If $N = {q^k}{n^2}$ is an odd perfect number (OPN) given in Eulerian form, then since $\gcd(q, n) = \gcd(q^k, n) = 1$, we either have $q^k < n$ or $n < q^k$.

As of July 2013, the author is able to obtain the following bounds:

Case 1:  $q^k < n$

If $k = 1$, then

$$\frac{1}{2} < \frac{\sigma(q)}{\sigma(n)} < \frac{q}{n} < \frac{\sigma(q)}{n} < 1$$
$$1 < \frac{n}{q} < \frac{\sigma(n)}{\sigma(q)} < \frac{\sigma(n)}{q} < 4$$

Otherwise, if $k > 1$ then

$$\frac{1}{2} < \frac{\sigma(q^k)}{\sigma(n)} < \sqrt{\frac{125}{128}}$$
$$\frac{1}{2} < \frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n} < 1$$
$$1 < \frac{\sigma(q^k)}{n} < \frac{n}{q^k} < \frac{\sigma(n)}{\sigma(q^k)} < \frac{\sigma(n)}{q^k} < 2$$
$$\sqrt{\frac{1}{2}} < \frac{q^k}{n} < 1 < \frac{n}{q^k} < \sqrt{2}$$
$$\sqrt{\frac{128}{125}} < \frac{\sigma(n)}{\sigma(q^k)} < \frac{\sigma(n)}{q^k} < 2$$
$$1 < \frac{\sigma(q^k)}{n} < I(q^k) < \frac{5}{4} < \sqrt{\frac{8}{5}} < I(n) < \frac{\sigma(n)}{q^k} < 2$$

Case 2:  $n < q^k$

If $k = 1$, then

$$\sqrt{\frac{1}{3}} < \frac{n}{q} < \frac{\sigma(n)}{\sigma(q)} < \frac{\sigma(n)}{q} < 1$$
$$1 < \frac{\sigma(q)}{\sigma(n)} < \frac{q}{n} < \sqrt{3}$$
$$1 < \frac{\sigma(q)}{\sigma(n)} < \frac{q}{n} < \frac{\sigma(q)}{n} < 2\sqrt{3}$$
$$\sqrt{\frac{1}{3}} < \frac{n}{q} < \sqrt[4]{\frac{108}{125}} < \sqrt[4]{\frac{125}{108}} < \frac{q}{n} < \sqrt{3}$$

Otherwise, if $k > 1$ then

$$\frac{1}{2} < \frac{n}{q^k} < \sqrt[4]{\frac{125}{128}} < \sqrt[4]{\frac{128}{125}} < \frac{q^k}{n} < 2$$
$$\frac{1}{2} < \frac{n}{q^k} < \frac{\sigma(n)}{\sigma(q^k)} < 1$$
$$1 < \frac{\sigma(n)}{q^k} < \frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n} < \frac{\sigma(q^k)}{n} < 2$$
$$\frac{4\left(1 + \sqrt{\frac{8}{5}}\right)}{13} < \frac{\sigma(n)}{\sigma(q^k)} < 1 < \frac{\sigma(q^k)}{\sigma(n)} < \frac{13}{4\left(1 + \sqrt{\frac{8}{5}}\right)}$$
$$1 < \frac{\sigma(n)}{q^k} < I(q^k) < \frac{5}{4} < \sqrt{\frac{8}{5}} < I(n) < \frac{\sigma(q^k)}{n} < 2$$