(This question has been cross-posted from MSE to MO.)
Antalan and Tagle (in a 2004 preprint titled Revisiting forms of almost perfect numbers) 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 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?
Antalan and Tagle (in a 2004 preprint titled Revisiting forms of almost perfect numbers) 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 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?
