(This question has been cross-posted from MSE to MO.)

Antalan and Tagle (in a 2004 preprint titled

Since $M$ is almost perfect, we have

$$\left(2^{r+1} - 1\right)\left(\sigma(b^2) - b^2\right) = \sigma(b^2) - 1$$

which further implies that

$$2^{r+1} = 1 + \frac{b^2 - 1}{\sigma(b^2) - b^2}.$$

Since $b$ is composite, then we have $\sigma(b^2) > b^2 + b + 1$, from which we obtain the inequality $2^{r + 1} < b$ using the last equation. This last inequality implies that

$$2^r < 2^{r+1} < b < \sigma(b)$$

and

$$\sigma(2^r) = 2^{r+1} - 1 < b - 1 < b$$

so that we have

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

Additionally, since $b^2$ is deficient, we can write $\sigma(b^2) = 2b^2 - c$, where we compute $c$ to be

$$c = b^2 - \frac{b^2 - 1}{\sigma(2^r)}$$

from which we obtain the upper bound

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

Lastly, since $r \geq 1$ and $2 \mid 2^r$, then

$$\frac{3}{2} = \frac{\sigma(2)}{2} \leq \frac{\sigma(2^r)}{2^r},$$

so that

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

Compare the results we have obtained for

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

Lastly, observe that, for the lone

$$\frac{\sigma(n_1)}{m} = \frac{5376}{22021} < 1 < \frac{22022}{22021} = \frac{\sigma(m)}{m} < \frac{5376}{3003} = \frac{\sigma(n_1)}{n_1} < \frac{\sigma(m)}{n_1} = \frac{22022}{3003},$$

where $m$ denotes the

My question is: Could there be a simple logical explanation for the

Antalan and Tagle (in a 2004 preprint titled

**) show that, if $M \neq 2^t$ is an even almost perfect number, then $M$ takes the form $M = {2^r}{b^2}$ where $r \geq 1$ and $b$ is an odd composite. We call $b^2$ the***Revisiting forms of almost perfect numbers***odd part**of the even almost perfect number $M$.Since $M$ is almost perfect, we have

$$\left(2^{r+1} - 1\right)\left(\sigma(b^2) - b^2\right) = \sigma(b^2) - 1$$

which further implies that

$$2^{r+1} = 1 + \frac{b^2 - 1}{\sigma(b^2) - b^2}.$$

Since $b$ is composite, then we have $\sigma(b^2) > b^2 + b + 1$, from which we obtain the inequality $2^{r + 1} < b$ using the last equation. This last inequality implies that

$$2^r < 2^{r+1} < b < \sigma(b)$$

and

$$\sigma(2^r) = 2^{r+1} - 1 < b - 1 < b$$

so that we have

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

Additionally, since $b^2$ is deficient, we can write $\sigma(b^2) = 2b^2 - c$, where we compute $c$ to be

$$c = b^2 - \frac{b^2 - 1}{\sigma(2^r)}$$

from which we obtain the upper bound

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

Lastly, since $r \geq 1$ and $2 \mid 2^r$, then

$$\frac{3}{2} = \frac{\sigma(2)}{2} \leq \frac{\sigma(2^r)}{2^r},$$

so that

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

Compare the results we have obtained for

*even almost perfect numbers other than powers of two*with the**conjectured**inequalities for the divisors of odd perfect numbers $N = {q^k}{n^2}$ given in Eulerian form (see this [link1] and [link2]):$$\frac{\sigma(q^k)}{n} < 1 < \frac{\sigma(q^k)}{q^k} < \frac{5}{4} < \sqrt{\frac{8}{5}} < \frac{\sigma(n)}{n} < \frac{\sigma(n)}{q^k}$$

Lastly, observe that, for the lone

**spoof**odd perfect number $D = m{n_1}^2 = 198585576189$ that we know of (see this [link3]), we actually have$$\frac{\sigma(n_1)}{m} = \frac{5376}{22021} < 1 < \frac{22022}{22021} = \frac{\sigma(m)}{m} < \frac{5376}{3003} = \frac{\sigma(n_1)}{n_1} < \frac{\sigma(m)}{n_1} = \frac{22022}{3003},$$

where $m$ denotes the

**quasi-Euler prime**of $D$.My question is: Could there be a simple logical explanation for the

**discrepancies**in the inequalities relating the divisors of*even almost perfect numbers other than powers of two*,*odd perfect numbers*, and*spoof odd perfect numbers*?