Good morning. I was motivated by a question I answered here. I wanted to find a closed form for the continued fraction
$$[2;4,6,...,\overline{2n}]=2+\frac{1}{4+\frac{1}{6+...\frac{1}{2n+\frac{1}{2n+...}}}}$$
It makes sense first to find the closed form of the repeating part; i.e., the form of $[2n;\overline{2n}]$. Let $x_n=[2n;\overline{2n}]$.
$$x_n=2n+\frac{1}{2n+\frac{1}{2n+...}}\quad \Rightarrow \quad x_n=2n+\frac{1}{x_n}\quad \Rightarrow \quad x_n^2-2nx-1=0$$
Thus, $x_n=n+\sqrt{n^2+1}$
Now representing the c.f. as $y_n=[2;4,6,...,\overline{2n}]=[2;4,6,...,2(n-1),x_n]$, I started working backwards and trying to represent as a simple rational function of variable $x_n$. So far, looking at small cases, we have that
\begin{eqnarray*}y_0&=&x_0&=&\frac{x_0+0}{0x_0+1}\\y_1&=&2+\frac{1}{x_1}&=&\frac{2x_1+1}{x_1+0}\\y_2&=&2+\frac{1}{4+\frac{1}{x_2}}&=&\frac{9x_2+2}{4x_2+1}\\y_3&=&2+\frac{1}{4+\frac{1}{6+\frac{1}{x_3}}}&=&\frac{56x_3+9}{25x_3+4} \end{eqnarray*}
So I noticed that if I represent
$$y_n=\frac{a_nx_n+b_n}{c_nx_n+d_n} \qquad T_n=\begin{pmatrix}a_n & b_n\\c_n & d_n\end{pmatrix} \qquad \vec{x}=\begin{pmatrix}x_n\\1\end{pmatrix}$$
then we can define an operation such that
$$y_n=T_n\vec{x}=\frac{a_nx_n+b_n}{c_nx_n+d_n}$$
Then, if we let
$$A_n=\begin{pmatrix}2n & 1\\1 & 0\end{pmatrix}$$
we have the following.
\begin{eqnarray*}y_1&=&A_1\vec{x_1}\\y_2&=&(A_1A_2)\vec{x_2}\\y_3&=&(A_1A_2A_3)\vec{x_3}\\...\\y_n&=&\left(\prod_{k=1}^nA_k\right)\vec{x_n} \end{eqnarray*}
Certainly from here, using a CAS, it would be easy to determine the actual value given a certain $n$. First of all, is this correct? My calculations seem right. For a true closed form, though, should I try and solve for the individual $a_i$ to $d_i$? Are there any easier approaches as well?
This is a partial answer, some information is provided for potential answerers that need some numeric support. I cannot see a path to a closed formula. There is a small mathematical part, that puts order in the structure, a non-linear recursion relation of order two for the coefficients is extracted, and the asymmptotic is evaluated. There is also a small computer aided part with explicit data.
An invertible $2\times 2$ matrix $T$ with entries $a,b,c,d$ induces a momographic / Möbius transformation / action on $\Bbb C$ without pole given by $$ \underbrace{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} }_{=T}z :=\frac {az+b}{cz+d}\ . $$ Personally, i prefer to write this action as $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \left[\frac z1\right] := \left[\frac {az+b}{cz+d}\right]\ , $$ so that the action can be simpler seen with bare eyes, knowing that $2\times 2$ matrices act on $2\times 1$ vectors.
(The question was implicitly asking, if a formula has action character, yes.)
From this, we have using $$ \begin{aligned} A(k) &= \begin{bmatrix} k & 1\\ 1 & 0 \end{bmatrix} \ ,\\[2mm] T(n) &= A(2)\; A(4)\; \dots\; A(2n) \\ &= \begin{bmatrix} 2 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} 4 & 1\\ 1 & 0 \end{bmatrix} \dots \begin{bmatrix} 2n & 1\\ 1 & 0 \end{bmatrix} \\ &= \begin{bmatrix} t_n & t_{n-1}\\ u_n & u_{n-1} \end{bmatrix} \ , \end{aligned} $$ where the last form can be shown inductively.
The sequences $(t_n)$, $(u_n)$ are satisfying the parallel relations $$ \begin{aligned} t_n &= 2n \;t_{n-1} + t_{n-2}\ , \qquad & t_0 &=1\ , & t_{-1} &= 0\ ,\\ u_n &= 2n \; u_{n-1} + u_{n-2}\ , \qquad & u_0 &=0\ , & t_{-1} &= 1\ . \end{aligned} $$ These follow from $$ \begin{bmatrix} t_n & t_{n-1}\\ u_n & u_{n-1} \end{bmatrix} = \begin{bmatrix} t_{n-1} & t_{n-2} \\ u_{n-1} & u_{n-2} \end{bmatrix} \begin{bmatrix} 2n & 1\\ 1 & 0 \end{bmatrix} \ ,\ %T_1 = \begin{bmatrix} %2 & 1\\ 1 & 0 %\end{bmatrix}\ , %\ T_0 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\ . $$ An easy lower estimation is found by "forgetting the $t_{n-2}$ in the recursion", so we get for instance $t_n\ge (2n)!!=2^n\cdot n!$. So we have a rapid increasing sequence.
I could not find a closed form for the sequences $(t_n)$, $(u_n)$, appearing as entries in $(T(n))_{n\ge 0}$. And i also could not use the ideas of solving the differential equation $$ y'' = 2x\;y' + y $$ to get a "discrete idea" to proceed in our case. Note that the above differential equation can be used to get an asymptotic of $t(n)$. Using sage again for a quick typing:
(Result
y(x) = ...was rearrange manually.) $$ \left[y\left(x\right) = \frac{\left(\frac{1}{2}\right)^{\frac{1}{4}} \left(-x^{2}\right)^{\frac{1}{4}} K_{2} x I_{-\frac{1}{4}}(-\frac{1}{2} \, x^{2}) e^{\left(\frac{1}{2} \, x^{2}\right)}}{\left(-\frac{1}{2} i \, x^{2}\right)^{\frac{1}{4}} \sqrt{-x}} + \frac{K_{1} x Y_{-\frac{1}{4}}(-\frac{1}{2} i \, x^{2}) e^{\left(\frac{1}{2} \, x^{2}\right)}}{\sqrt{-x}}\right] $$Discretely, i am expecting a similar complexity. (As a human, i am completely ignoring all algebraic operations, and look at the word Bessel, this is a complicate function, research projects may be started if some starting idea is present. Else...)
I gave up also because this is only the first part in the computations of some "(real) surds" of a special shape that have a period of length one when written as continued fractions.
Here is a good place to insert some numerical data. We use sage. Please ignore this part if it is disturbing, some comments wanted explicit numerical data, here it is.
We can also check the continued fractions representations (easily and structurally, so let us do this). Again, please ignore if a theoretical interest predominates.