Antalan and Tagle showed that, if $M \neq 2^t$ ($t \geq 1$) and $\sigma(M) = 2M - 1$ (where $\sigma = \sigma_{1}$ is the (classical) sum-of-divisors function), then $M$ takes the form
$$M = 2^r b^2$$
where $b > 1$ is an odd composite.
In this preprint, Antalan and Dris proved the following bounds:
Theorem A. If $2^r b^2$ is an almost perfect number with $r \geq 1$, $\gcd(2, b) = 1$ and $b > 1$, then
$$\frac{\sigma(2^r)}{b} < 1 < \frac{\sigma(b)}{b} < \frac{4}{3} < \frac{3}{2} \leq \frac{\sigma(2^r)}{2^r} < 2 < \frac{\sigma(b)}{2^r}.$$
$$\frac{\sigma(2^r)}{b} < 1 < \frac{\sigma(b)}{b} < \frac{4}{3} < \frac{3}{2} \leq \frac{\sigma(2^r)}{2^r} < 2 < \frac{\sigma(b)}{2^r}.$$
In this blog post, we will prove the following series of inequalities:
Theorem B. If $2^r b^2$ is an almost perfect number with $r \geq 1$, $\gcd(2, b) = 1$ and $b > 1$, then
$$1 < \frac{b}{\sigma(2^r)} < \frac{\sigma(b)}{\sigma(2^r)} < \frac{b}{2^r} < \frac{\sigma(b)}{2^r} < \frac{b^2}{\sigma(2^r)} < \frac{\sigma(b^2)}{\sigma(2^r)} < \frac{b^2}{2^r} < \frac{\sigma(b^2)}{2^r}.$$
In other words, taking reciprocals, we should obtain:
$$\frac{2^r}{\sigma(b^2)} < \frac{2^r}{b^2} < \frac{\sigma(2^r)}{\sigma(b^2)} < \frac{\sigma(2^r)}{b^2} < \frac{2^r}{\sigma(b)} < \frac{2^r}{b} < \frac{\sigma(2^r)}{\sigma(b)} < \frac{\sigma(2^r)}{b} < 1.$$
Proof. Since
$$\frac{2^r}{\sigma(b^2)} < \frac{2^r}{b^2} < \frac{\sigma(2^r)}{\sigma(b^2)} < \frac{\sigma(2^r)}{b^2} < \frac{2^r}{\sigma(b)} < \frac{2^r}{b} < \frac{\sigma(2^r)}{\sigma(b)} < \frac{\sigma(2^r)}{b} < 1.$$
Proof. Since
$$\frac{\sigma(b)}{b} < \frac{\sigma(b^2)}{b^2} < \frac{4}{3},$$
by Theorem A, it suffices to prove the inequality in the middle, namely
$$\frac{\sigma(b)}{2^r} < \frac{\sigma(b^2)}{\sigma(2^r)}.$$
Suppose to the contrary that
$$\frac{\sigma(b^2)}{\sigma(2^r)} \leq \frac{\sigma(b)}{2^r}.$$
It follows that
$$\frac{\sigma(b^2)}{\sigma(b)} \leq \frac{\sigma(2^r)}{2^r} < 2,$$
by Theorem A. Since $\frac{\sigma(b)}{b} < \frac{4}{3}$ (again, by Theorem A), it follows that
$$\frac{3}{4}\cdot\left(\frac{\sigma(b^2)}{b}\right) < \left(\frac{b}{\sigma(b)}\right)\cdot\left(\frac{\sigma(b^2)}{b}\right) = \frac{\sigma(b^2)}{\sigma(b)} < 2.$$
Dividing through by $1 < b$, we obtain
$$\frac{3}{4}\cdot\left(\frac{\sigma(b^2)}{b^2}\right) < \frac{2}{b}.$$
By Theorem A, we get
$$\frac{3}{4} < \frac{3}{4}\cdot\left(\frac{\sigma(b^2)}{b^2}\right) < \frac{2}{b},$$
which finally implies that
$$b < \frac{8}{3}.$$
Since $1 < b$, this forces $b = 2$. A contradiction, as $b$ has to be odd.
We therefore conclude that
$$\frac{\sigma(b)}{2^r} < \frac{\sigma(b^2)}{\sigma(2^r)}.$$
The proof for
$$\frac{\sigma(b)}{2^r} < \frac{\sigma(b^2)}{\sigma(2^r)}.$$
The proof for
$$\frac{\sigma(b)}{2^r} < \frac{b^2}{\sigma(2^r)}$$
is very similar, and we are done.
is very similar, and we are done.
