It is unknown whether $2^n-1$ is (a Mersenne) prime for infinitely many values of $n$. Such $n$ must necessarily be prime itself due to the factorization below $$(2^{ab}-1)=(2^a-1)(2^b+2^{b-1}+\cdots+2+1).$$

I am aware that there are related results on the size of the largest prime divisor of Mersenne numbers $M_n=2^n-1$ from this Mathoverflow question. But my question is somewhat different.

The question: How much do we need to weaken the question of primality to get a result that holds infinitely often. Is a statement such as $$ \varphi(2^n-1) \geq 2^n-1-K,\quad \textrm{for infinitely many }n\quad(1) $$ where $K$ is a bounded positive integer likely to hold, or proven? I think this is probably false.

Alternatively, how slowly growing a function of $n$, call it $K(n)$ can we specify so that $$ \varphi(2^n-1) \gg 2^n-1-K(n),\quad \textrm{for infinitely many }n\quad(2) $$ holds? In the light of Noam D. Elkies' comment, hopefully the following is better posed: Is it possible to find a constant $c \in (0,1)$ such that $$ \varphi(2^n-1) \geq c(2^n-1),\quad \textrm{for infinitely many }n\quad(3) $$ Experimentally, from Mathematica with $2\leq n\leq 220$ all but 2 $n$ satisfy (3) with $c=1/3$ and all but roughly 20% of $2\leq n \leq 220$ satisfy (3) with $c=1/2.$

It is unknown whether $2^n-1$ is (a Mersenne) prime for infinitely many values of $n$. Such $n$ must necessarily be prime itself due to the factorization below $$(2^{ab}-1)=(2^a-1)(2^b+2^{b-1}+\cdots+2+1).$$

I am aware that there are related results on the size of the largest prime divisor of Mersenne numbers $M_n=2^n-1$ from this Mathoverflow question. But my question is somewhat different.

The question: How much do we need to weaken the question of primality to get a result that holds infinitely often. Is a statement such as $$ \varphi(2^n-1) \geq 2^n-1-K,\quad \textrm{for infinitely many }n\quad(1) $$ where $K$ is a bounded positive integer likely to hold, or proven? I think this is probably false.

Alternatively, how slowly growing a function of $n$, call it $K(n)$ can we specify so that $$ \varphi(2^n-1) \gg 2^n-1-K(n),\quad \textrm{for infinitely many }n\quad(2) $$ holds? In the light of Noam D. Elkies' comment, hopefully the following is better posed: Is it possible to find a constant $c \in (0,1)$ such that $$ \varphi(2^n-1) \geq c(2^n-1),\quad \textrm{for infinitely many }n\quad(3) $$ Experimentally, from Mathematica with $2\leq n\leq 220$ all but 2 $n$ satisfy (3) with $c=1/3$ and all but roughly 20% of $2\leq n \leq 220$ satisfy (3) with $c=1/2.$

It is unknown whether $2^n-1$ is (a Mersenne) prime for infinitely many values of $n$. Such $n$ must necessarily be prime itself due to the factorization below $$(2^{ab}-1)=(2^a-1)(2^b+2^{b-1}+\cdots+2+1).$$

I am aware that there are related results on the size of the largest prime divisor of Mersenne numbers $M_n=2^n-1$ from this Mathoverflow question. But my question is somewhat different.

The question: How much do we need to weaken the question of primality to get a result that holds infinitely often. Is a statement such as $$ \varphi(2^n-1) \geq 2^n-1-K,\quad \textrm{for infinitely many }n\quad(1) $$ where $K$ is a bounded positive integer likely to hold, or proven? I think this is probably false.

Alternatively, how slowly growing a function of $n$, call it $K(n)$ can we specify so that $$ \varphi(2^n-1) \gg 2^n-1-K(n),\quad \textrm{for infinitely many }n\quad(2) $$ holds?

It is unknown whether $2^n-1$ is (a Mersenne) prime for infinitely many values of $n$. Such $n$ must necessarily be prime itself due to the factorization below $$(2^{ab}-1)=(2^a-1)(2^b+2^{b-1}+\cdots+2+1).$$

I am aware that there are related results on the size of the largest prime divisor of Mersenne numbers $M_n=2^n-1$ from this Mathoverflow question. But my question is somewhat different.

The question: How much do we need to weaken the question of primality to get a result that holds infinitely often. Is a statement such as $$ \varphi(2^n-1) \geq 2^n-1-K,\quad \textrm{for infinitely many }n\quad(1) $$ where $K$ is a bounded positive integer likely to hold, or proven? I think this is probably false.

Alternatively, how slowly growing a function of $n$, call it $K(n)$ can we specify so that $$ \varphi(2^n-1) \gg 2^n-1-K(n),\quad \textrm{for infinitely many }n\quad(2) $$ holds? In the light of Noam D. Elkies' comment, hopefully the following is better posed: Is it possible to find a constant $c \in (0,1)$ such that $$ \varphi(2^n-1) \geq c(2^n-1),\quad \textrm{for infinitely many }n\quad(3) $$ Experimentally, from Mathematica with $2\leq n\leq 220$ all but 2 $n$ satisfy (3) with $c=1/3$ and all but roughly 20% of $2\leq n \leq 220$ satisfy (3) with $c=1/2.$