The $n$-th Mersenne number $M\_n$ is defined as
$$M\_n=2^n-1$$
A great deal of research focuses on Mersenne primes. What is known in the opposite direction about Mersenne numbers with only small factors (i.e. smooth numbers)? In particular, if we let $P\_n$ denote the largest prime factor of $M\_n$, are any results known of the form
$$\liminf\_{n\rightarrow \infty}\frac{P\_n}{f(n)}= 1$$
for some function $f$?


I've only come across two (fairly distant) bounds so far. If we consider even-valued $n$, then $M\_n=M\_{n/2}(M\_{n/2}+2)$, so:
$$\liminf\_{n\rightarrow \infty}\frac{P\_n}{2^{n/2}}\leq 1$$
In the other direction, [1] shows that $P\_n\geq 2n+1$ for $n>12$, so
$$\liminf\_{n\rightarrow \infty}\frac{P\_n}{2n}\geq 1$$


[1] A. Schinzel, On primitive prime factors of $a^n-b^n$, Proc. Cambridge Philos. Soc. 58
(1962), 555-562.



Smoothness in Mersenne numbers?
I can give you a slightly better upper bound. Recall that $2^n - 1 = \prod\_{d | n} \Phi\_d(2)$ where $\Phi\_d$ is a cyclotomic polynomial. Now,


$$\Phi\_d(2) = \prod\_{(k, d) = 1} (2 - \zeta\_d^k) \le 3^{\varphi(d)}$$


so that in particular the largest prime factor of $2^n - 1$ is at most $3^{\varphi(n)}$. By taking $n$ to be a product of the first $k$ primes and letting $k$ tend to infinity we have $\liminf\_{n \to \infty} \frac{\varphi(n)}{n} = 0$, hence


$$\liminf\_{n \to \infty} \frac{P\_n}{c^n} = 0$$


for any $c > 1$. In fact if $n$ is the product of the first $k$ primes then we should expect something like $3^{\varphi(n)} \approx 3^{ \frac{n}{\log k} }$ but this doesn't seem like a big improvement to me.


