TITLE: Smoothness in Mersenne numbers?
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: 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.