Eccoci.

For a positive integer $n$, let $\sigma (n)$ be the sum of all divisors of $n$. For example, $\sigma (6)=1+2+3+6=12$.

A perfect number, as you probably know, is a number with $\sigma (n)=2n$.

Let us define the perfection quotient of a positive integer as $p(n)={\displaystyle \frac{\sigma (n)}{n}}$.

Find the sum of all positive integers $n\le {10}^{18}$ for which $p(n)$ has the form $k+{\displaystyle \frac{1}{2}}$, where $k$ is an integer.