When does :$\sigma(\sigma(2n))=\sigma(\sigma(n))$ and $\sigma(n)$ is sum divisors of the positive integer $n$? Is there someone who can show me When does: $$\sigma(\sigma(2n))=\sigma(\sigma(n))$$ where : $\sigma(n)$ denotes the sum of divisors of the positive integer $n$ ?
Note (1) : I came across this problem when I read some papers  about 
"Iterating of the sum divisors of sigma function ".
Note(2) :${\sigma}^{0}(n)=n$ and ${\sigma}^{m}(n)=\sigma({\sigma}^{m-1}(n))$ and $m \geq 1$
Thank you for any help!
 A: This is not a complete solution, but only a partial answer to the question.
Without loss of generality, let $n = {2^k}m$ where $k \geq 0$ and $\gcd(2,m)=1$.
Then we have:
$$\sigma(2n) = \sigma(2^{k+1})\sigma(m) = (2^{k+2} - 1)\sigma(m)$$
$$\sigma(\sigma(2n)) = \sigma((2^{k+2} - 1)\sigma(m))$$
$$\sigma(n) = \sigma(2^k)\sigma(m) = (2^{k+1} - 1)\sigma(m)$$
$$\sigma(\sigma(n)) = \sigma((2^{k+1} - 1)\sigma(m)).$$
Therefore, the equality
$$\sigma(\sigma(2n)) = \sigma(\sigma(n))$$
holds when there exist $k \geq 0$ and an odd $m$ such that
$$\sigma((2^{k+2} - 1)\sigma(m)) = \sigma((2^{k+1} - 1)\sigma(m)).$$
Lastly, using the inequalities
$$a\sigma(b) \leq \sigma(ab) \leq \sigma(a)\sigma(b)$$
we get
$$(2^{k+2} - 1)\sigma(\sigma(m)) \leq \sigma((2^{k+2} - 1)\sigma(m)) = \sigma((2^{k+1} - 1)\sigma(m)) \leq \sigma(\sigma(2^k))\sigma(\sigma(m)).$$
Hence,
$$2^{k+2} - 1 \leq \sigma(\sigma(2^k)).$$
We are therefore sure that
$$\sigma(2^k) = 2^{k+1} - 1$$
is not a (Mersenne) prime.
Additionally, we have the lower bound
$$3 < 4 - \frac{1}{2^k} \leq \frac{\sigma(\sigma(2^k))}{2^k},$$
so that $2^k$ is not an (even) superperfect number.  This agrees with our earlier finding that $2^{k+1} - 1$ is not prime.
A: $\sigma(2n)=3\times\sigma(n)$
if $m=\sigma(n)$
we should find $\sigma(3 m)=\sigma(m)$
$\sigma(3m)=4\times\sigma(m)$
then  $4\sigma(m)=\sigma(m)$
contradiction so there is no solution.
