## 6.12.15

### EAPN Research (October 2015) - Some updates on even almost perfect numbers other than the powers of two

Let $\sigma(x)$ be the sum of the divisors of $x$, and call the function

$$D(x) = 2x - \sigma(x)$$

as the deficiency of $x$.

If $D(n) = 1$, then $n$ is called an almost perfect number.

Antalan and Tagle (in a 2014 preprint titled "Revisiting forms of almost perfect numbers") show that, if $n \neq 2^k$ is an even almost perfect number, then $n$ takes the form

$$n = {2^r}{b^2}$$

where $b$ is an odd composite integer.

Using their result, we have

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

from which we obtain

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

From this equation, we obtain the following results (summarized from this paper):

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

Claim 2
$$c \geq \frac{2b^2 + 1}{3}$$

Claim 3
If ${2^r}{b^2}$ is an almost perfect number with $\gcd(2, b) = 1$ and $b > 1$, then $b^2$ is solitary.

Claim 4
If ${2^r}{b^2}$ is an almost perfect number with $\gcd(2, b) = 1$ and $b > 1$, then $\left(\sigma(b^2) - b^2\right) \mid \left(b^2 - 1\right)$.

Claim 5
Suppose that there exist at least two distinct even almost perfect numbers
$$M_1 = {2^{r_1}}{b_1}^2$$
and
$$M_2 = {2^{r_2}}{b_2}^2$$
with $\gcd(2, b_1) = \gcd(2, b_2) = 1$, $b_1 > 1$, $b_2 > 1$, and $r_1 \neq r_2$.  Then $b_1 \neq b_2$.

Claim 6
If ${2^r}{b^2}$ is an almost perfect number with $\gcd(2, b) = 1$ and $b > 1$, then $r < \log_{2}{b} - 1$.