Is it true that there exists $n \in \mathbb{N}$ with arbitrarily many prime factors such that $n$ divides $F\_n$, where $F\_n$ represents the n-th **Fibonacci number**?



Arbitrarily large $n$ divides $F_n$
Note that (see [here](http://oeis.org/A023172)):


* If $a$ and $b$ are in your sequence, then so is $\text{lcm}(a,b)$;
* If $n$ is in the sequence, then so is $F\_n$.


Now take any $n>12$ that belongs to the sequence, then $a\_1=\text{lcm}(n,F\_n)$, $a\_2=\text{lcm}(n,F\_n,F\_{F\_n})$, and so on. Each of $n$, $F\_n$, $F\_{F\_n}$, ... belongs to the sequence by the second fact. Their successive least common multiples do as well by the first fact. Moreover, one has $\omega(a\_k) \geq k-1+\omega(n)$ by [the theorem of Carmichael](http://en.wikipedia.org/wiki/Carmichael%27s_theorem), which answers your question affirmatively.


A more challenging version I didn't think about yet could be: for which $n \in \mathbb N$ is there an $N$ in the sequence such that $\omega(N)=n$?


