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

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.

### 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.

• Fantastic! But from that last bit in (4), it seems like one interpretation is that each $F_n$ added on will increase the total LCM by roughly $\exp(\phi(n))$ on "average". Is there any rough intuition for why that might be the case? Also, I note here that the paper seems to confirm that this property is indeed not unique to the Fibonacci sequence, and other second-order recurrence sequences can be used. Oct 11, 2019 at 23:50
• Hi @chausies, the fact that you highlighted is beyond my intuition. It can be noted that the increase is "almost exact" (maybe "on average" is not the best description), in the sense that the error term $B$ grows very slowly. Also, you are right that this also works for other second-order recurrences Oct 15, 2019 at 6:48