Finding all $n$ such that $\sigma(\sigma(n))=2n$

Question

Find all $n$ such that $\sigma(\sigma(n))=2n$.

I know how to solve it for even number. Let $n=2^kt$, where $t\neq 1$. Than $\sigma(\sigma(n))=\sigma((2^{k+1}-1)\sigma(t))\geq (2^{k+1}-1)\sigma(t) + \sigma(t)>2^{k+1}t=2n$. So, if $n$ is even, than $n=2^k$. Than $\sigma(\sigma(n))=\sigma(2^{k+1}-1)$ and it must be equals to $2^{k+1}$. So, $2^{k+1}-1$ must be a prime number.

Answer 1

This is a well known problem and the $n$ that solve the equation $\sigma(\sigma(n)) = 2n$ are called

superperfect numbers

and have been defined by D. Suryanarayana in about 1969. Here is the sequence of solutions to your equation: A019279

You can find a lot of information here: "Superperfect Number." From MathWorld

To your question: All even superperfect numbers are $n = 2^{p-1}$ if $2^p-1$ is prime (Mersenne prime). Whether an odd superperfect number exists is still an open question in number theory.

If you are interested in this topic I suggest this papers:

• Cohen, G. L. and te Riele, J. J. "Iterating the Sum-of-Divisors Function." Experim. Math. 5, 93-100, 1996.
• Suryanarayana, D. "Super Perfect Numbers." Elem. Math. 24, 16-17, 1969. (I dont have a pdf)