Proof that the ratio between the logs of the product and LCM of the Fibonacci numbers converges to $\frac{\pi^2}{6}$

Question

I came across this amazing fact on Twitter. $$\lim_{n\to\infty} \frac{\log\left(F_1 \cdots F_n\right)}{\log \text{LCM}\left(F_1, \ldots, F_n\right)} = \frac{\pi^2}{6}$$

where $F_i$ is the $i$th Fibonacci number and LCM = Least Common Multiple. This is such an interesting link between the Fibonacci numbers (which are closely linked to the Golden Ratio $\varphi$) and $\pi$. I'm trying to prove why this is the case.

I know that, for the numerator, it'll become something like \begin{align} \log\left(F_1 \cdots F_n\right) &\sim \log\left(\varphi^1 \cdots \varphi^n\right) \\ &\sim \frac{\log \varphi}{2} n^2 \end{align}

Reverse engineering, that tells me that the denominator will turn out to be the same, but with an extra factor of $\frac{6}{\pi^2}$. What I'm wondering is, how to prove that this will be the case? Some relevant tools might be that $$\prod_i \left(1 - \frac{1}{p_i^2}\right) = \frac{6}{\pi^2}$$, where $p_i$ is the $i$th prime. Also, it's not hard to show that the LCM of the first $n$ natural numbers is roughly $e^n$. Finally, the probability that the $n$th Fibonacci is prime is roughly $\sim\frac{\log\varphi}{n}$ I think (though I'm not actually sure if this has been proven).

As an aside, even though it's cool that this is done with the Fibonacci sequence, judging by how the $\varphi$ factor just cleanly cancels on top and bottom, I have a feeling that this fact might be true for other linear integer recurrence relations that have an exploding dominant eigenvalue.

Answer 1

A full proof of this fact can be found in the paper A new Formula for $\pi$ by Yuri V. Matiyasevich and Richard K. Guy.

A brief summary: Notations I will use:

1. Let $\mu$ denote the Möbius function,
2. let $w_n:=\operatorname{LCM}(F_1,F_2,\dots,F_n)$.

As you noticed, we have (for large $n$) \begin{equation}\tag 1\label 1 \log(F_1\cdots F_n)\sim \frac{n^2 \ln\tau}{2}, \end{equation} where $\tau=\frac{1+\sqrt 5}{2}$ denotes the golden ratio.

So proving your statement is equivalent to proving that (for large $m$) \begin{equation}\tag 2\label 2 \ln w_m \sim 3\ln(\tau)\frac{m^2}{\pi^2}. \end{equation}

In the paper, \eqref{2} is called Chebyshev's form of the prime number theorem for Fibonacci numbers.

By rather lengthy applications of results on Arithmetic functions and the Möbius inversion formula, one gets

\begin{equation}\tag 3\label 3\begin{split} \ln w_m &= \sum_{d=1}^m \sum_{i \text{ such that } i | d} \mu\left(\frac di\right)\ln F_i \\ &= B(m)+\sum_{d=1}^m \sum_{i \text{ such that } i | d} \mu\left(\frac di\right) i \ln\tau, \end{split} \end{equation} where $0\le B(m)< 2m^\frac32$.

One can write (by elementary results on Euler's totient function $\phi$) \begin{equation}\tag 4\label 4 \sum_{d=1}^m \sum_{i \text{ such that } i | d} \mu\left(\frac di\right) i \ln\tau = \ln\tau\cdot\sum_{d=1}^m \phi(d)\sim 3\ln(\tau)\frac{m^2}{\pi^2} \end{equation} (the last similarity is a proven result from number Theory.)

So, as a result, $$\ln w_m \sim 3\ln(\tau)\frac{m^2}{\pi^2} + B(m),$$ where $0\le B(m)< 2m^\frac32$.

From this follows \eqref{2} and thus also your result.