For a positive integer $n$, let $\sigma(n)$ denote the sum of the positive divisors of $n$.
Now, suppose that $n$ is a composite number. Is there an infinitely many $n$ composite such that $\sigma(\sigma(n)+n) = 2\cdot \sigma(n)$?
And in the case where there are an infinitely many $n$ composite such that $\sigma(\sigma(n)+n) = 2\cdot \sigma(n)$, it is conjectured that all numbers composite $n$ satisfying the equation $\sigma(\sigma(n)+n) = 2\cdot \sigma(n)$ are odd (except 2).
See the first 20 values of $n$ found by Peter (there are all odd, except 2).
I failed to prove this conjecture in my previous attempt, so I edited this thread.