For integers $m\geq 1$ let $\sigma(m)$ the sum of divisors function $\sum_{1\leq d\mid m}d$ and let $\psi(m)$ the Dedekind psi function (as reference I add the Wikipedia *Dedekind psi function*), then there exist integers $n\geq 1$ that satisfy $$\psi(\sigma(n))=2n.\tag{1}$$
I don't know it this equation is in the literature, compare this equation with the equation studied in the second page of [1] (or well from the last paragraph of the article *Totient Function* from the encyclopedia Wolfram MathWorld).

Up to $10^4$ these solutions are $n=2,3,4,16,64$ and $4096$. I believe that this sequence isn't in the OEIS, I've searched also the string *psi(sigma(n))*. It is easy to prove the following statement.

**Claim.** *If* $2^{\alpha+1}-1$ *is a Mersenne prime, then* $n=2^{\alpha}$ *is a solution of the equation* $(1)$.

Question.I would like to know if it is possible to do more work about the solutions of the equation $$\psi(\sigma(n))=2n.$$ What additional and reasonable* work can be done about it?Many thanks.

*I'm asking about if we can deduce more statements about the solutions (characterization of solutions and if there exist finitely/infinitely many solutions) of $(1)$.

**Remarks.** The functions $\sigma(n)$ and $\psi(n)$ are multuiplicative. It is unknown if there exist infinitely many Mersenne primes. As a side remark the integer $n=3$ also is a solution, for which $2n-1=5\in$*A175611* from the OEIS.

## References:

[1] L. Alaoglu and P. Erdös, *A conjecture in elementary number theory*, Bull. Amer. Math. Soc. Volume 50, Number 12 (1944), 881-882.

A175611versusA000668(Mersenne primes) from the OEIS. See theClaim. $\endgroup$ – user142929 Nov 22 '19 at 15:47Claimfor the equation $\psi(\sigma(n))=2n$ after I've known the comments for the sequenceA001229from the OEIS. $\endgroup$ – user142929 Nov 22 '19 at 16:40that if$n\geq 1$is an integer satisfying$2n\psi(n)=\sum_{1\leq d\mid n}\psi(dn)$then$n$is an even perfect number. The other direction $\Leftarrow$ is easy. I don't know if it is in the literature (I wrote also a variant involving the Euler's totient function) or it is easy to check. I hope don't disturb, isn't required a response of this comment and thanks again for your attention @MaxAlekseyev $\endgroup$ – user142929 Apr 1 '20 at 12:15