It is not known if there are infinitely many prime Fibonacci numbers. But can one assert that there is no Fibonacci number >2 that is also highly composite (<https://en.wikipedia.org/wiki/Highly_composite_number>) - or that there are only finitely many such numbers?


**Remarks:** As given in <http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibtable.html>, every Fibonacci number bigger than 1 [except F(6)=8 and F(12)=144] has at least one prime factor that is *not* a factor of any earlier Fibonacci number. So, Fibonacci numbers tend to have large prime factors and it is quite conceivable that none of them are highly composite. However, a few are seen to be *semiprimes* (<https://en.wikipedia.org/wiki/Semiprime>). Not sure if the question of whether there are infinitely many Fibonacci semiprimes has been answered.



On Fibonacci numbers that are also highly composite
The largest highly composite Fibonacci number is $F\_{3} = 2$.


If $p$ is a prime number, then either $p \mid F\_{p-1}$ (if $p \equiv \pm 1 \pmod{5}$), $p \mid F\_{p}$ (if $p = 5$), or $p \mid F\_{p+1}$ (if $p \equiv \pm 2 \pmod{5}$). It follows that if $n > 12$ and $p$ is a prime that divides $F\_{n}$ and no previous Fibonacci number, then $p \geq n-1$. Assuming $F\_{n}$ is highly composite implies that all primes $\leq n-1$ divide $F\_{n}$. It follows that $F\_{n} \geq \prod\_{p \leq n-1} p$. This will lead to a contradiction for $n$ sufficiently large (which boils down to the fact that $\frac{1+\sqrt{5}}{2} < e$).


Let $\theta(x) = \sum\_{p \leq x} \log(p)$. The prime number theorem is equivalent to $\theta(x) \sim x$ and Rosser and Schoenfeld showed (see page 70 of their [1962 Illinois Journal of Mathematics paper](http://projecteuclid.org/euclid.ijm/1255631807)) that for $x \geq 41$, $\theta(x) \geq x \left(1 - \frac{1}{\log(x)}\right)$. This implies that for $n \geq 42$, we have
$$
 \log\left(\frac{1}{\sqrt{5}}\right) + n \log\left(\frac{1 + \sqrt{5}}{2}\right) \geq \log(F\_{n}) \geq \theta(n-1) \geq (n-1) - \frac{n-1}{\log(n-1)}.
$$
For $n \geq 22$, we have $(n-1) - \frac{(n-1)}{\log(n-1)} \geq \frac{2}{3} (n-1)$, which implies that the right hand side of the centered inequality above is greater than the left.


It suffices to check the prime factorization of $F\_{n}$ for $n \leq 42$ to verify that $F\_{n}$ is not highly composite for $4 \leq n \leq 42$. This can be done easily using the table of [Brillhart, Montgomery, and Silverman](https://www.ams.org/journals/mcom/1988-50-181/S0025-5718-1988-0917832-6/).


