It seems that as $n$ increases, the ratio $$\frac{\varphi(2^n-1)}{2^n-1},$$ where $\varphi$ denotes the Euler totient function,
takes on values reasonably often in the interval $(.3,.4)$.


Is there anything known about $$\lim \inf\_{n \rightarrow \infty}\frac{\varphi(2^n-1)}{2^n-1}?$$



How Composite can $2^n-1$ be, infinitely often?
Let $n \in \mathbf N^+$ and $a \in \mathbf N\_{\ge 2}$. Every prime $\le n+1$ that doesn't divide $a$, is a divisor of $a^{n!} - 1$ (by Fermat's little theorem). So we have $$\frac{\varphi(a^{n!} - 1)}{a^{n!} - 1} = \prod\_{p \,\mid\, a^{n!} - 1} \left(1 - \frac{1}{p}\right) \le \prod\_{a < p \le n+1} \left(1 - \frac{1}{p}\right)\! \stackrel{n \to \infty}{\longrightarrow} 0,$$
where $p$ is always a prime and for the first equality we have used [Euler's product formula](https://en.wikipedia.org/wiki/Euler%27s_totient_function#Euler.27s_product_formula). In particular, this shows that the limit inferior in the OP is $0$.


