# Limit of quotients of elements of special Fibonacci matrices

Let $$F_n$$ be the $$n$$-th Fibonacci number, started with $$F_0=0,F_1=1$$, and consider the matrices $$M_n=\pmatrix{F_{n+3} & F_{n+1} \\ F_{n+2} & F_{n}}.$$
Let $$\pmatrix{\alpha_n & \beta_n \\ \gamma_n & \delta_n}=M_1\cdot M_2\cdot \ldots \cdot M_n .$$

It is easy to see by computer, that the quotients $$\frac{\alpha_n}{\gamma_n},\frac{\beta_n}{\gamma_n},\frac{\delta_n}{\gamma_n}$$ are converging. I found that $$\lim_{n\to\infty} \frac{\delta_n}{\gamma_n}={\phi^2}$$ where $$\phi=\frac{\sqrt{5}-1}{2}$$. Unfortunately I can't find the other two limit, but numerically it seems to be that $$\lim_{n\to\infty} \frac{\alpha_n}{\gamma_n}\approx 1.3876267558043602953$$ $$\lim_{n\to\infty} \frac{\beta_n}{\gamma_n}\approx 0.53002625701851519880$$ Can anyone give me a "nice" description of these numbers? Alternatively, it would be enough, if someone can decide whether these numbers are algebraic or not.

(I remark that the limit of $$\alpha_n / \gamma_n$$ is the most interesting for me, since it has a continued fraction expansion: $$[1;2,1,1,2,1,1,1,2,...]$$ and so on, the number of ones between twos increasing by one. It is a non-periodic, badly approximable number.)

One can prove by induction that some of given ratios have nice continued fraction expansions: $$\begin{gather*} \frac{\alpha_n}{\beta_n }=[2;1^n,2,1^{n-1},\ldots,2,1,1,2,1]\to 1+\varphi=\varphi^2;\\ \frac{\gamma_n}{\delta_n }=[2;1^n,2,1^{n-1},\ldots,2,1,1,2]\to 1+\varphi=\varphi^2;\\ \frac{\alpha_n}{\gamma_n}=[1;2,1,1,2,1,1,1,2,\ldots,1^{n-1},2,1^n,2]\to \psi;\\ \frac{\beta_n}{\delta_n}=[1;2,1,1,2,1,1,1,2,\ldots,1^{n-1},2,1^n]\to \psi, \end{gather*}$$ where $$\psi=[1;2,1,1,2,1,1,1,2,\ldots]$$ is defined by its continued fraction expansion.
• So $\psi$ is certainly not a quadratic irrational and, if the standard conjectures are true, not an algebraic irrational but rather a transcendental number. Dec 1, 2018 at 4:00
• When the two's show up only every $2^k$ terms (rather than linearly) in the continued fraction for a number which is expanded in one and two, it is transcendental. This is in the introductory remarks here: web.williams.edu/Mathematics/sjmiller/public_html/book/papers/… Dec 1, 2018 at 6:38
• But there are numbers that are known to be transcendental and known not to satisfy Gauss-Kuz'min (for example, $e$), whereas there are no numbers known to be algebraic (of degree at least three) and known not to satisfy Gauss-Kuz'min. Dec 1, 2018 at 21:53