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



Solutions of the equation $\psi(\sigma(n))=2n$, where $\sigma(n)$ is the sum of divisors function and $\psi(n)$ the Dedekind psi function
There are no other solutions than $n=3$ and those from Claim: $n=2^{p-1}$ such that $2^p-1$ is prime. 


Consider several cases.


1. $n=2^ks$ is even (here $k\geqslant 1$ and $s$ is odd). Then $$\frac{\psi(\sigma(n))}{n}=\frac{\psi((2^{k+1}-1)\sigma(s))}{(2^{k+1}-1)\sigma(s)}\cdot \frac{2^{k+1}-1}{2^k}\cdot \frac{\sigma(s)}s.\quad(1)$$
Note that $$\frac{\psi(ab)}{ab}=\prod\_{p|ab}\left(1+\frac1p\right)\geqslant 
\prod\_{p|a}\left(1+\frac1p\right)=\frac{\psi(a)}a$$
for positive integers $a,b$, therefore 
$$
\frac{\psi((2^{k+1}-1)\sigma(s))}{(2^{k+1}-1)\sigma(s)}\geqslant 
\frac{\psi(2^{k+1}-1)}{2^{k+1}-1}\geqslant \frac{2^{k+1}}{2^{k+1}-1},\quad (2)
$$
the last inequality holds due to $k\geqslant 1$ and turns into equality
if and only if $2^{k+1}-1$ is a Mersenne prime. Substitute (2) to (1), we get
$$
\frac{\psi(\sigma(n))}{n}\geqslant \frac{2^{k+1}}{2^{k+1}-1}\cdot 
\frac{2^{k+1}-1}{2^k}\cdot \frac{\sigma(s)}s=2\frac{\sigma(s)}s\geqslant 2,
$$
with equality if and only if $s=1$ and $2^{k+1}-1$ is a Mersenne prime. So for even $n$ all solutions come from Claim.
2. $n$ is odd. Denote $\sigma(n)=p\_1^{\alpha\_1}\ldots p\_m^{\alpha\_m}$. Then $2n=\psi(\sigma(n))=(p\_1^{\alpha\_1}+p\_1^{\alpha\_1-1})\ldots (p\_m^{\alpha\_m}+p\_m^{\alpha\_m-1})$. Note that $p\_i^{\alpha\_i}+p\_i^{\alpha\_i-1}$
is even unless $p\_i^{\alpha\_i}=2$. But $2n$ is not divisible by 4, therefore either


(i) $m=1$ and $\sigma(n)=p^{\alpha}$; or 


(ii) $m=2$ and $\sigma(n)=2p^{\alpha-1}$ for odd $p$. 


In the case (i) we get $n=\frac{p+1}2 p^{\alpha-1}$. If $p=2$, then since $n$ should be odd we get $\alpha=2$, $n=3$. It is a solution. If $p$ is odd and $\alpha\geqslant 2$, then $p^\alpha=\sigma(n)=\sigma(\frac{p+1}2)\sigma(p^{\alpha-1})$, but $\sigma(p^{\alpha-1})$ is greater than 1 and not dividible by $p$. If $\alpha=1$, we get $\sigma(\frac{p+1}2)=p$. Then $n=\frac{p+1}2=q^s$ for a prime $q$ (otherwise $\sigma(\frac{p+1}2)$ is composite), $q$ is odd (since $q|n$) and
$$2q^s-1=p=\sigma(n)=q^s+q^{s-1}+\ldots+1$$
that fails modulo $q$. 


In the case (ii) we get $n=3\frac{p+1}2p^{\alpha-1}$ and $$\sigma(n)\geqslant n+\frac{n}3=2(p+1)p^{\alpha-1}>2p^\alpha,$$
a contradiction.


