When does :$\sigma(\sigma(2n))=\sigma(\sigma(n))$ and $\sigma(n)$ is sum divisors of the positive integer $n$?

Question

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!

Answer 1

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.

Answer 2

$\sigma(2n)=3\times\sigma(n)$

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.