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

$$\frac{\sigma(2^r)}{b} < 1 < \frac{\sigma(b)}{b} < \frac{4}{3} < \frac{3}{2} \leq \frac{\sigma(2^r)}{2^r} < 2 < \frac{\sigma(b)}{2^r}.$$

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
$$\frac{\sigma(b)}{b} < \frac{\sigma(b^2)}{b^2} < \frac{4}{3},$$

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.