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](https://mathoverflow.net/questions/38586/smoothness-in-mersenne-numbers). 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 $c=1/2.$ 



How often is $2^n-1$ a number with few divisors?
It is not so hard to prove that $\phi(2^p-1)/(2^p-1) \to 1$ as $p\to \infty$ along prime values. This follows from the usual formula for $\phi(n)/n$ as a product over primes along with the observation that $\sum\_{q\mid 2^p-1,~q\text{ prime}} 1/q \to 0$, as $p\to\infty$. This last fact can be seen by noting that --- from the maximal order of the distinct prime divisors function --- there are only $O(p/\log{p})$ distinct primes dividing $2^p-1$, while every prime dividing $2^p-1$ is congruent to $1$ modulo $p$, and so exceeds $p$.


In fact, $\phi(2^n-1)/(2^n-1)$ has a distribution function, in the sense of probabilistic number theory. As a consequence, given any $\epsilon > 0$, there is a $c >0$ such that $\phi(2^n-1)/(2^n-1) > c$ away from a set of $n$ of upper density $< \epsilon$. This distribution function result can be deduced from the general results of


Luca, Florian; Shparlinski, Igor E.(5-MCQR-CP)
Arithmetic functions with linear recurrence sequences. (English summary) 
J. Number Theory 125 (2007), no. 2, 459–472. 


(The full distribution function result is deeper than the consequence I noted above; that consequence follows, e.g., from a first moment argument.) 


