On Fibonacci numbers that are also highly composite

Question

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.

Answer 1

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) 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.