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):

$$\sigma(b^2) = 2b^2 - c$$

where

$$c = b^2 - \frac{b^2 - 1}{2^{r + 1} - 1}$$

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

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

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

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 $b_1 \neq b_2$.

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

**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 $r < \log_{2}{b} - 1$.