(Note: This question has been cross-posted from MSE.)

Let $\sigma(a) = \sigma_{1}(a)$ be the sum of the divisors of the positive integer $a$.

A number $M$ is called almost perfect if $\sigma(M) = 2M - 1$. $N$ is called perfect if $\sigma(N) = 2N$.

$M_s = 2^s$ where $s \geq 0$ are almost perfect, with $M_0 = 1$ being the only

*odd almost perfect number*that is currently known. If $M \neq 2^t$ is an*even almost perfect number*, then Antalan and Tagle showed that $M$ must have the form
$$M = {b^2}{2^r}$$

where $r \geq 1$ and $b$ is an odd composite.

On the other hand, only $49$

*even perfect numbers*have been discovered, and they are of the form
$$N_e = 2^{p-1}\left(2^p - 1\right)$$

where $2^p - 1$ and $p$ are primes. It is currently unknown whether there are any

*odd perfect numbers*, but Euler showed that they are of the form
$$N_o = n^2 q^c$$

where $q$ is prime with $q \equiv c \equiv 1 \pmod 4$ and $\gcd(q,n) = 1$.

Notice the following:

__For numbers that we know exist__**Even Perfect Numbers**

Assuming $N_e \neq 6$ (because it is

*squarefree*),
$$\frac{\sigma(2^{(p-1)/2})}{2^p - 1} = \frac{2^{(p+1)/2} - 1}{2^p - 1}< 1 < 4 \le 2^{(p+1)/2} = \frac{\sigma(2^p - 1)}{2^{(p-1)/2}}.$$

Note that

$$\frac{\sigma(2^p - 1)}{2^p - 1} \leq \frac{8}{7} < \sqrt{\frac{7}{4}} < \frac{\sigma(2^{(p-1)/2})}{2^{(p-1)/2}}.$$

**Almost Perfect Numbers (Powers of Two)**

Since $s \geq 0$,

$$\frac{\sigma(\sqrt{1})}{2^s} \leq 1 \leq 2^{s+1} - 1 = \frac{\sigma(2^s)}{\sqrt{1}}.$$

Note that

$$\frac{\sigma(\sqrt{1})}{\sqrt{1}} = 1 \leq \frac{\sigma(2^s)}{2^s}.$$

In Descartes' example $D = km$, we have

$$m = 22021 = {{19}^2}\cdot{61}$$

and

$$\sqrt{k} = {3}\cdot{7}\cdot{11}\cdot{13}.$$

Note that

$$\frac{\sigma(\sqrt{k})}{m} = \frac{5376}{22021} < 1 < \frac{22022}{3003} < \frac{m+1}{\sqrt{k}}$$

and

$$\frac{m+1}{m} = \frac{22022}{22021} < \frac{5376}{3003} = \frac{\sigma(\sqrt{k})}{\sqrt{k}}.$$

__For numbers that are conjectured not to exist__**Even Almost Perfect Numbers (Other Than Powers of Two)**

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

Note that

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

"Some New Results On Even Almost Perfect Numbers Which Are Not Powers Of Two" - Theorem 2.2, page 5

**Odd Perfect Numbers**

The following inequalities are conjectured in New Results for Sorli's Conjecture on Odd Perfect Numbers - Part II:

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

Note that

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

(

**Added June 27 2016**- In fact, in a recent preprint, Brown claims a*partial proof*for $q^c < n$, which would be consistent with the conjecture here.)
Here are my questions:

**(1)**If $K = {x^2}{y^z}$ is a (hypothetical) number satisfying $\sigma(K) = 2K + \alpha$ (with $y$ prime, $\gcd(x,y)=1$, and where $\alpha$ could be zero or negative), might there be a

**specific reason**why the inequalities

$$\frac{\sigma(x)}{y^z} \leq 1 \leq \frac{\sigma(y^z)}{x}$$

seem to

*guarantee existence*of such numbers $K$?**(2)**If $L = {u^2}{v^w}$ is a (hypothetical) number satisfying $\sigma(L) = 2L + \beta$ (with $v$ prime, $\gcd(u,v)=1$ and where $\beta$ could be zero or negative), might there be a

**specific reason**why the inequalities

$$\frac{\sigma(v^w)}{u} < 1 < \frac{\sigma(u)}{v^w}$$

seem to

*predict nonexistence*of such numbers $L$?
Question

**(1)**is illustrated (as detailed above) in the case of even perfect numbers, almost perfect numbers which are powers of two, and spoof odd perfect numbers (otherwise known in the literature as Descartes numbers).
Question

**(2)**is illustrated (as detailed above) in the case of even almost perfect numbers which are not powers of two, and odd perfect numbers.