Search limit for almost perfect numbers?

Question

The function $$\sigma(n)$$ denotes the sum of the divisors of $n$. Perfect numbers are positive integers $n$ with $$\sigma(n)=2n$$ almost perfect numbers are positive integers $n$ with $$\sigma(n)=2n-1$$ For the case $$\sigma(n)=2n+1$$ I found a search-limit on mathworld, but for "$2n-1$" I only found that it is conjectured that the only solutions are powers of $2$

Does anyone know a reference for a search limit for the almost perfect numbers ?

Answer 1

This answer is for the case of even almost perfect numbers (other than the powers of two).

(The following are taken from this preprint by Antalan and Dris.)

Antalan and Tagle showed that an even almost perfect number $n \neq 2^t$ must necessarily have the form $2^r b^2$ where $r \geq 1$, $\gcd(2,b)=1$, and $b$ is an odd composite.

Since $2^r b^2$ is almost perfect and $\gcd(2,b)=1$, then we have $$(2^{r+1} - 1)\sigma(b^2) = \sigma(2^r)\sigma(b^2) = \sigma(2^r b^2) = 2^{r+1} b^2 - 1$$ from which it follows that $$1=(1-2^{r+1})\sigma(b^2) + 2^{r+1} b^2$$ $$2^{r+1} (\sigma(b^2) - b^2) = \sigma(b^2) - 1$$ $$2^{r+1} = \frac{\sigma(b^2) - 1}{\sigma(b^2) - b^2} = 1 + \frac{b^2 - 1}{\sigma(b^2) - b^2}$$

This last equation gives the divisibility constraint in the following result:

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

Numbers $n$ such that $\sigma(n) − n$ divides $n − 1$ are listed in OEIS sequence A059046, the first $62$ terms of which are given below:

$$2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, 19, 23, 25, 27, 29, 31, 32, 37, 41, 43, 47, 49, 53, 59, 61,$$ $$64, 67, 71, 73, 77, 79, 81, 83, 89, 97, 101, 103, 107, 109, 113, 121, 125, 127, 128, 131, 137, 139,$$ $$149, 151, 157, 163, 167, 169, 173, 179, 181, 191, 193, 197, 199, 211.$$

Remark: Does OEIS sequence A059046 contain any odd squares $u^2$, with $\omega(u) \geq 2$? MSE user Charles checked and found that "there are no such squares with $u^2 < {10}^{22}$." (See this MSE question for Charles's answer.)