Relation between two rational sequences approximating square root 2

Define the sequences $\,a_n := c_n/d_n\,$ where $\,c\,$ and $\,d\,$ are OEIS sequences A001333 and A000129. Consider the matrix $$ M := \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \tag{1}$$ whose powers are $$ M^n = \begin{pmatrix} c_n & d_n \\ 2d_n & c_n \end{pmatrix}. \tag{2}$$ Since $\,M^{n+1} = M^n\, M\,$ this explains the $\,a_n\,$ recursion.

Notice the algebraic matrix identity $$ \begin{pmatrix} c & d \\ 2d & c \end{pmatrix}^2 = \begin{pmatrix} c^2+2d^2 & 2cd \\ 4cd & c^2+2d^2 \end{pmatrix}. \tag{3}$$ Since $\,M^{2^{n+1}} = (M^{2^n})^2,\,$ this explains the $\,a_{2^n}\,$ recursion.

Note that the matrix $\,M\,$ is equivalent to $\,m:=1\pm\sqrt{2}.\,$ Thus equation $(2)$ is equivalent to $\,m^n = c_n\pm d_n\sqrt{2}\,$ and equation $(3)$ is equivalent to $\,(c\pm d\sqrt{2})^2 = (c^2+2d^2)\pm(2cd)\sqrt{2}.$

Here is an outline of a proof by induction that doesn't use the closed form expressions.

We are given $a_{n+1}=\dfrac{2+a_n}{1+a_n}$ and $a_1=1$.

First prove by induction on $m$ that $a_{n+m}=\dfrac{2+a_ma_n}{a_m+a_n}.$

It follows that $a_{2n}=\dfrac{2+a_n^2}{2a_n}$.

Now prove by induction that $b_{n+1}=a_{2^n}$.