Is it possible to have an even superperfect number and an odd superperfect number whose product is an almost perfect number? A number $n \in \mathbb{N}$ is said to be superperfect if
$$\sigma(\sigma(n)) = 2n.$$
A number $m \in \mathbb{N}$ is said to be almost perfect if $$\sigma(m) = 2m - 1.$$
Here is my question:

Is it possible to have an even superperfect number and an odd superperfect number whose product is an almost perfect number?

By the results in a paper (by Antalan and Tagle) mentioned in this preprint, an even almost perfect number which is not a power of two must take the form $M = {2^r}{b^2}$ where $r \geq 1$ and $b$ is an odd composite.  Consequently, by the same preprint, we know that, if there exists an even almost perfect number which is not a power of two, then it must be a product of an even solitary number and an odd solitary number.  I wonder if a similar scenario holds for the product of two superperfect numbers of different parity.
 A: Guy's Unsolved Problems in Number Theory states that it is a result of Suryanarayana and Kanold that the even superperfect numbers are precisely the numbers $2^{n-1}$ where $2^n-1$ is a Mersenne prime. So let $r=n-1$. Since even almost perfect numbers which have an odd factor greater than one have the form $2^rb^2$ (with the stated conditions on $b$) by the result of Antalan and Tagle you mention, then suppose $2^r$ is the even superperfect number (so that $2^{r+1}-1$ is the Mersenne prime) and $b^2$ is the odd superperfect number. Then since as mentioned in the proof of Lemma 2.3 of [1],
$$
\sigma(b^2)=\frac{2^{r+1}b^2-1}{2^{r+1}-1},
$$
and
$$
\sigma\left(\frac{2^{r+1}b^2-1}{2^{r+1}-1}\right)=2b^2
$$
by the assumed superperfect property of $b^2$, then $\frac{2^{r+1}b^2-1}{2^{r+1}-1}$ must be of the form $q^km^2$ where $q\equiv k\equiv 1 \bmod 4$ and $gcd(q,m)=1$. Since $2^{r+1}-1$ is a Mersenne prime, either $r=1$ or $r$ is even. If $r$ is even, then any prime $p$ dividing $\frac{2^{r+1}b^2-1}{2^{r+1}-1}$ satisfies 
$$
\left(\frac{2}{p}\right) = 1
$$
as $2(2^{r/2}b)^2\equiv 1 \bmod p$, hence $p\equiv \pm 1 \bmod 8$. So when $r$ is even, the $q$ previously mentioned is $1\bmod 8$.
Letting $I(n)=\sigma(n)/n$, we also have
$$
I(\sigma(b^2)) = \frac{2^{r+1}-1}{2^r}+\frac{1}{2^r\sigma(b^2)}.
$$
In general, if one would ask the more general question of whether $n$ could both be a product of an even superperfect number and an odd superperfect number, and also a generalized quasi-perfect number, that is, for fixed nonzero integers $a_1, a_2$, $n$ satisfies $\sigma(n)=a_1n+a_2$, suppose $n=2^rb^2$ where $b^2$ is odd. Then for the case $r>1$, if $2^{r+1}-1$ is a Mersenne prime, a prime $p$ dividing $a_12^rb^2+a_2$ satisfies
$$
\left(\frac{-a_2a_1^{-1}}{p}\right)=1
$$
and this would be satisfied by $p$ in certain residue classes mod some modulus $D(a_1,a_2)$, and there are no solutions to the above general question for those $r$ mod the period of the powers of $2$ mod $D(a_1,a_2)$, which do not take the Mersenne prime $2^{r+1}-1$ to one of those residue classes.
[1] Antalan, Dris. "Some New Results on Even Almost Perfect Numbers Which Are Not Powers of Two", arXiv 1602.04248.
