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!

  • 1
    $\begingroup$ do you mean $\sigma_0$ or $\sigma_1$? $\endgroup$
    – pancini
    Jul 18, 2015 at 22:59
  • $\begingroup$ I added some necessary conditions about sigma function to define well the above equation!!! $\endgroup$ Jul 18, 2015 at 23:07
  • $\begingroup$ so the question for m=2 and it is \sigma_(2) $\endgroup$ Jul 18, 2015 at 23:15

2 Answers 2


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.



if $m=\sigma(n)$ we should find $\sigma(3 m)=\sigma(m)$


then $4\sigma(m)=\sigma(m)$

contradiction so there is no solution.

  • $\begingroup$ yes, thanks and i think it's has no solutions $\endgroup$ Jul 18, 2015 at 23:24
  • $\begingroup$ but your answer is not a proof , i think it showed by computational evidence that has no solution $\endgroup$ Jul 18, 2015 at 23:32
  • $\begingroup$ @zeraouliarafik Yes it do. $\endgroup$ Jul 18, 2015 at 23:35
  • 2
    $\begingroup$ The equations are only valid under some assumptions of relative primality. For example, $\sigma(2n)=3\times\sigma(n)$ if and only if $n$ is odd. $\endgroup$ Jul 19, 2015 at 0:20

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