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

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

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

Almost Perfect Numbers (Powers of Two)

Since $s \geq 0$, $$\dfrac{\sigma(\sqrt{1})}{2^s} \leq 1 \leq 2^{s+1} - 1 = \dfrac{\sigma(2^s)}{\sqrt{1}}.$$

Note that $$\dfrac{\sigma(\sqrt{1})}{\sqrt{1}} = 1 \leq \dfrac{\sigma(2^s)}{2^s}.$$

Spoof Odd Perfect Numbers

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

For numbers that are conjectured not to exist

Even Almost Perfect Numbers (Other Than Powers of Two)

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

Note that $$\dfrac{\sigma(b)}{b} < \dfrac{4}{3} < \dfrac{3}{2} \leq \dfrac{\sigma(2^r)}{2^r}.$$ "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:

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

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

(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 $$\dfrac{\sigma(x)}{y^z} \leq 1 \leq \dfrac{\sigma(y^z)}{x}$$ 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 $$\dfrac{\sigma(v^w)}{u} < 1 < \dfrac{\sigma(u)}{v^w}$$ 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.

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  • $\begingroup$ I could be wrong, but isn't an odd almost perfect number known as a Descartes Number, denoted as $\mathscr{D}$? Thus $1$ is not the only odd almost perfect number known, assuming I am correct. I mean, they are the same...right? $\endgroup$ – Mr Pie Oct 27 '17 at 5:48
  • $\begingroup$ @user477343, no they are not the same. If $m$ is an odd almost perfect number, that is, $\sigma(m) = 2m - 1$ and $2m - 1$ is taken to be a "spoof" prime, then $n = m(2m - 1)$ is a Descartes number. $\endgroup$ – Jose Arnaldo Bebita-Dris Oct 27 '17 at 9:04
  • $\begingroup$ Aaaahhhh ok thanks for that :) $\endgroup$ – Mr Pie Oct 27 '17 at 9:44
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(1)

The inequalities do not guarantee existence of such numbers $K$.

Take $(x,y,z)=(3,2,2)$ where $y$ is prime with $\gcd(x,y)=1$.

Then, we get $$\frac{\sigma(x)}{y^z}=\frac{4}{4}\le 1\le \frac{7}{3}=\frac{\sigma(y^z)}{x}$$ and $$\alpha=\sigma(K)-2K=\sigma(36)-2\times 36=91-72=19$$ which is positive.


(2)

The inequalities do not predict nonexistence of such numbers $L$.

Take $(u,v,w)=(5,2,1)$ where $v$ is prime with $\gcd(u,v)=1$.

Then, we get $$\dfrac{\sigma(v^w)}{u}=\frac{3}{5} < 1 < \frac{6}{2}=\dfrac{\sigma(u)}{v^w}$$ and $$\beta=\sigma(L)-2L=\sigma(50)-2\times 50=93-100=-7$$ which is negative.

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    $\begingroup$ Well-said! I appreciate you taking the time out to answer (some of) my long-standing questions here at MSE. Again, my profuse thanks! +1! =) $\endgroup$ – Jose Arnaldo Bebita-Dris Oct 10 '18 at 11:27

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