The $n$-th Mersenne number is $M\_n=2^n-1$. Write $M\_n=a\_n b\_n^2$ where $a\_n$ is positive and squarefree. 


**Question 1:** What lower bound can be proved for $a\_n$?


Let $A$ be the set of all possible $a\_n$. The **natural density** of $A$ is defined as
$$
\delta\_A=\lim\_{X \rightarrow \infty} \frac{\# \{a \in A | a \le X\}}{X}.
$$


**Question 2:** What can be proved about $\delta\_A$? Is it possible to show that $\delta\_A=0$?


**Note:** I am interested in unconditional answers to the above questions. It is easy to give answers conditional on the ABC conjecture. Indeed, the ABC conjecture shows that for any $\epsilon>0$ there is some $K\_\epsilon>0$ such that
$$
a\_n \ge K\_\epsilon \cdot 2^{n(1-\epsilon)}.
$$
Thus $\# \{ a \in A | a \le X\}=O(\log(X))$, which gives $\delta\_A=0$.



Squarefree parts of Mersenne numbers
Here's a simple proof using user43383's idea and a recent result of Andrew Granville: <http://arxiv.org/abs/1212.6306>


According to Theorem 1 of that paper, $2^n-1$ always has a primitive prime factor $p$ that occurs to an odd exponent, except when $n=1$ or $n=6$ (where there are no primitive prime factors at all). *Primitive* here means that $2$ has order $n$ modulo $p$. In particular, $p\equiv 1\pmod{n}$. Clearly, $p \mid a\_n$. So if $q\_n$ denotes the smallest prime congruent to $1$ modulo $n$, then $a\_n \geq q\_n$ for every $n \neq 1, 6$. And in fact, by a direct check, $a\_6 = 7 = q\_6$, so $a\_n \geq q\_n$ for every $n > 1$.


Trivially, $q\_n \geq n+1$ for every $n$. In fact, it is usually much larger. The simple fact that the primes have density zero implies that for every positive integer $K$, one has $q\_n > Kn$ apart from a set of $n$ of density zero. (To see this, note that if this inequality fails, then one of $n+1$, $2n+1$, $\dots$, or $(K-1)n+1$ is prime, and each of these conditions puts $n$ in a set of density zero.) 


Hence: $a\_n \geq n+1$ for all $n > 1$, and for every fixed $K$, $a\_n > Kn$ for all $n$ outside of a set of density zero. These two facts are enough to imply that $\{a\_n\}$ itself has density zero.


(Proof of the last bit: Fix $K$. Given a large $x$, if $a\_n \leq x$, then the inequality $a\_n \geq n+1$ forces $n \leq x$. If $n \leq x$ and $a\_n \leq x$, then either $n < x/K$ or $a\_n \leq Kn$. The former holds for at most $x/K$ values of $n$, and the latter holds only for $o(x)$ values of $n$, as $x\to\infty$. Thus, the number of distinct $a\_n$ with $a\_n \leq x$ is at most $x(1/K+o(1))$; so the upper density of $\{a\_n\}$ is at most $1/K$. But this holds for all $K$.) 


