TITLE: Arbitrarily large $n$ divides $F_n$
QUESTION [5 upvotes]: 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?

REPLY [14 votes]: Note that (see here):

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, 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 $\omega(N)=n$?