When is $f(n)=\sum\limits_{d\mid n}\sigma(d)$ prime?
Note, $f$ is multiplicative and $\sigma(n)>1, \;n>1$. Therefore $f(n)$ is prime only when $n=p^\alpha$, with $p$ prime, $\alpha\geq1$.
Furthermore, for $n=\prod\limits_{i=1}^k p_i^{n_i}$,we have that
$f(n)=\large\prod\limits_{i=1}^k \frac{p_i^{n_i+2}-(n_i-2)p_i+n_i+1}{(p_i-1)^2}=\prod\limits_{i=1}^k \left(\sum\limits_{j=0}^{n_i} (n_i-j+1)(p_i)^j\right)$.
So if $n$ is $p$ is an odd prime, then $\sum\limits_{j=0}^{n} (n-j+1)(p)^j\equiv 0\pmod{2}$ unless $n\equiv0,1 \pmod{4}$.
If $p=2$, Then $f(2^n)=2^{n+2}-2(n+2)+n+1=2^{n+2}-n+3\equiv0\pmod2$ unless $n\equiv0\pmod2$.
So consider $f(4^m)=4^{m+1}-2m+3\equiv 0 \pmod3$ if $m\equiv2\pmod3$.
Then we should consider $f(64^k)=4^{3k+1}-6k+3$ and $f(4^{3k+1})=4^{3k+2}-6k+1$. Now I'm not sure where to go.
To summarize, we need only consider the cases $2^{6k}$, $2^{6k+2}$ and $p^{4k},p^{4k+1}$.
Also it's prime for the least of a twin prime pair.