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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.)

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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.

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    $\begingroup$ So $\psi$ is certainly not a quadratic irrational and, if the standard conjectures are true, not an algebraic irrational but rather a transcendental number. $\endgroup$ – Gerry Myerson Dec 1 '18 at 4:00
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    $\begingroup$ 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/… $\endgroup$ – Josiah Park Dec 1 '18 at 6:38
  • $\begingroup$ Thank you for the answers! I wished, that it is an algebraic number of degree four, but this conjecture is disappointing. The paper of van der Poorten and Shallit is also very interesting, thank for it. It also suggest, that this is transcendental, but until the proof, I trust to be algebraic. $\endgroup$ – L.Remete Dec 1 '18 at 10:57
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    $\begingroup$ @L.Remete According to numerical experiments all algebraic numbers satisfy Gauss-Kuz'min statistics, while this number clearly not. $\endgroup$ – Alexey Ustinov Dec 1 '18 at 11:41
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    $\begingroup$ 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. $\endgroup$ – Gerry Myerson Dec 1 '18 at 21:53

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