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?
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1$\begingroup$ To respond to the original version which didn't ask for infinitely many prime factors, $F_{5(2n+1)} = 5 (5 F_{2n+1}^5 - 5 F_{2n+1}^3 + F_{2n+1})$ so we can show inductively that $5^k | F_{5^k}$. $\endgroup$– David E SpeyerCommented Aug 20, 2014 at 14:48
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$\begingroup$ That's the reason I changed the question, as I realized on the spot the many ways to answer that. $\endgroup$– shapiCommented Aug 20, 2014 at 15:00
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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 $\omega(N)=n$?
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4$\begingroup$ Just for the record, using the results from an upcoming work of mine plus some conjectures about primes in shifted multiply exponential sequences, I can answer the more challenging question affirmatively too. $\endgroup$– user41593Commented Aug 20, 2014 at 16:02
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2$\begingroup$ Here is the "upcoming work": arxiv.org/abs/1410.2489 $\endgroup$– user41593Commented Oct 10, 2014 at 6:36