In this post we denote the Dedekind psi function as $\psi(m)$ for integers $m\geq 1$. This is an important arithmetic fuction in several subjects of mathematics. As reference I add the Wikipedia Dedekind psi function, and [1]. On the other hand I add the reference that Wikipedia has the article Mersenne prime, and that I was inspired in the formula that defines the sequence A072868 from the On-Line Encyclopedia of Integer Sequences.
The Dedekind psi function can be represented for a positive integer $m>1$ as $$\psi(m)=m\prod_{\substack{p\mid m\\p\text{ prime}}}\left(1+\frac{1}{p}\right)$$
with the definition $\psi(1)=1$.
Claim. If we take $n=2^p$ with $2^p-1$ a Mersenne prime, then the equation
$$\psi(2(\psi(n)-n)-1)=n\tag{1}$$
holds.
Sketch of proof. Just direct computation using the mentioned representation for the Dedekind psi function.$\square$
I don't know if previous equation is in the literature, one can to state a similar equation than $(1)$ involving the sum of divisors function instead of the Dedekind psi function.
Question. I would like to know if it is possible to prove of refute that if an integer $n\geq 2$ satisfies $(1)$ then $n-1$ is a Mersenne prime. Many thanks.
With a Pari/GP script and for small segments of integers I have not found counterexamples. I'm asking what work can be done for previous question proving the conjecture, or if you can to find a counterexample, before I'm accepting an available answer.
References:
[1] Tom M. Apostol, Introduction to analytic number theory, Undergraduate Texts in Mathematics, New York-Heidelberg: Springer-Verlag (1976).
Best Answer
This is a partial answer.
This answer proves the following claims :
Claim 1 : There is no prime $p$ such that $p^3\mid 2\psi(n)-2n-1$.
Claim 2 : If $2\psi(n)-2n-1$ is a prime, then $n-1$ is a Mersenne prime.
Claim 3 : $n$ is not a prime.
Claim 4 : $n+2\le \psi(n)\lt\dfrac{3n+1}{2}$
Claim 5 : If $n$ is even, then $n-1$ is a Mersenne prime.
Claim 6 : $15\not \mid n$
Claim 7 : $21\not \mid n$
Claim 8 : If $n=3^aq^bc$ where $a,b$ are positive integers, $c$ is odd and $q\ge 11$ is a prime such that $\gcd(q,3)=\gcd(c,3q)=1$, then $c\gt \dfrac{16q+15}{q-8}$.
Claim 1 : There is no prime $p$ such that $p^3\mid 2\psi(n)-2n-1$.
Proof : Suppose that there is a prime $p$ such that $$p^3\mid 2\psi(n)-2n-1\tag2$$ then it follows from $(1)$ that $p^2\mid n$ and $p\mid\psi(n)$ from which we have $p\not\mid 2\psi(n)-2n-1$ which contradicts $(2)$. $\quad\blacksquare$
Claim 2 : If $2\psi(n)-2n-1$ is a prime, then $n-1$ is a Mersenne prime.
Proof : If $2\psi(n)-2n-1$ is a prime, then letting $n=\displaystyle\prod_{i=1}^{d}p_i^{e_i}$ where $p_1\lt p_2\lt\cdots\lt p_d$ are primes, we have $$(1)\implies 2\psi(n)-2n-1+1=n\implies 2\prod_{i=1}^{d}(p_i+1)=3\prod_{i=1}^{d}p_i$$
If $d\ge 2$, then we get $p_1=2$ and $\displaystyle\prod_{i=2}^{d}(p_i+1)=\displaystyle\prod_{i=2}^{d}p_i$ which is impossible.
If $d=1$, then $2(p_1+1)=3p_1$ implying $p_1=2$. So, $n$ is of the form $2^a$ where $a\ge 1$, and $n-1$ is a Mersenne prime. $\quad\blacksquare$
Claim 3 : $n$ is not a prime.
Proof : Suppose that $n$ is a prime. Then, it follows from $(1)$ that $n=1$ which contradicts that $n$ is a prime. $\quad\blacksquare$
Claim 4 : $n+2\le \psi(n)\lt\dfrac{3n+1}{2}$
Proof : Suppose that $2\psi(n)-2n-1=1$. Then, it follows from $(1)$ that $n=1$ for which $2\psi(n)-2n-1=1$ does not hold. So, we have $2\psi(n)-2n-1\ge 2$. Since $\psi(m)\gt m$ for $m\ge 2$, we have $n=\psi(2\psi(n)-2n-1)\gt 2\psi(n)-2n-1$, i.e. $\psi(n)\lt\dfrac{3n+1}{2}$. Also, since $m$ is a prime iff $\sigma(m)\le m+1$, we have $n+2\le \psi(n)$. $\quad\blacksquare$
Claim 5 : If $n$ is even, then $n-1$ is a Mersenne prime.
Proof : Suppose that there is an odd integer $q\gt 1$ such that $n=2^aq$ where $a\ge 1$. Then, $$\psi(n)\lt\frac{3n+1}{2}\implies q+2\le \psi(q)\lt q+\frac{1}{6}$$ which is impossible. So, we have $n=2^a$. Then, $(1)\implies \psi(2^{a}-1)=2^a$ which implies that $2^a-1$ is a prime. $\quad\blacksquare$
Claim 6 : $15\not \mid n$
Proof : Suppose that $n=3^a5^bc$ where $a,b$ are positive integers, and $c$ is odd such that $\gcd(c,15)=1$, then $$\psi(n)\lt\frac{3n+1}{2}\implies c+2\le\psi(c)\lt \frac{45c+1}{48}\implies c\lt -\frac{95}{3}$$ which contradicts that $c$ is positive. $\quad\blacksquare$
Claim 7 : $21\not \mid n$
Proof : Suppose that $n=3^a7^bc$ where $a,b$ are positive integers, and $c$ is odd such that $\gcd(c,21)=1$, then $$\psi(n)\lt\frac{3n+1}{2}\implies c+2\le\psi(c)\lt \frac{63c+1}{64}\implies c\lt -127$$which contradicts that $c$ is positive. $\quad\blacksquare$
Claim 8 : If $n=3^aq^bc$ where $a,b$ are positive integers, $c$ is odd and $q\ge 11$ is a prime such that $\gcd(q,3)=\gcd(c,3q)=1$, then $c\gt \dfrac{16q+15}{q-8}$.
Proof : We have $$\psi(n)\lt\frac{3n+1}{2}\implies c+2\le\psi(c)\lt \frac{9qc+1}{8(q+1)}\implies c\gt \frac{16q+15}{q-8}\quad\blacksquare$$