Determine whether if $n$ is a primitive pseudoperfect (semiperfect) number, then $\sigma(n)<2^{\sigma_0(n)}$.

$\sigma_k(n)$ is the division function and $\sigma(n)=\sigma_1(n)$. A number is called pseudoperfect if it can be represented as a sum of some subset of distinct proper divisors of itself and primitive pseudoperfect if it also has no proper divisors with this property.

I've verified that the inequality holds for the first $10000$ terms.

As $\sigma(n)\geq 2n$ for every pseudoperfect number $n$, it is a weaker question to ask whether $n<2^{\sigma_0(n)-1}$.

It's easy to see that the strong form holds when $n=2^{p-1}(2^p-1)$ is an even perfect number, since $\sigma(n)=2n$ and $\sigma_0(n)=2p$, so $2^p(2^p-1)<2^{2p}$.

More generally, we can prove the strong form for primitive pseudoperfect numbers of the form $2^{\lfloor \log_2p \rfloor}p$, where $p$ is an odd prime (every such $n$ is primitive pseudoperfect):

We can see that

$$\sigma(n)=(2^{\lfloor \log_2p \rfloor+1}-1)(\dfrac{p^2-1}{p-1}),$$

and

$$\sigma_0(n)=2(\lfloor \log_2p \rfloor+1),$$

so we must confirm

$$\dfrac{p^2-1}{p-1}<\dfrac{2^{2(\lfloor \log_2p \rfloor+1)}}{2^{\lfloor \log_2p \rfloor+1}-1},$$

but since $2^{\lfloor \log_2p \rfloor+1}>p$, and $\dfrac{m^2-1}{m-1}=m+1$ is monotonically increasing,

$$\dfrac{p^2-1}{p-1}<\dfrac{2^{2(\lfloor \log_2p \rfloor+1)}-1}{2^{\lfloor \log_2p \rfloor+1}-1}<\dfrac{2^{2(\lfloor \log_2p \rfloor+1)}}{2^{\lfloor \log_2p \rfloor+1}-1},$$

and the proof is complete.

The general solution would probably have to be handled much differently, since it doesn't appear that the primitive pseudoperfect numbers have been completely classified, but I have noticed that many (though not all) other primitive pseudoperfect numbers are of the form $wp$, where $w$ is a weird number ($\sigma(n)>2n$ but $n$ is not pseudoperfect) and p is an odd prime, so I'll see if I can determine anything about these and the forms of the remaining terms. I would appreciate any insights you have into the full solution.