Closed form for $K(n)=[0;\overline{1,2,3,...,n}]$ I just started playing around with fairly simple periodic continued fractions, and I have a question. The fractions can be represented "linearly": for $n\in\Bbb N$,
$$K(n)=[0;\overline{1,2,3,...,n}]$$
I am seeking for a closed form for $K(n)$. I found the first few.
$n=1$:
$$K(1)=\frac1{1+K(1)}$$
$$\Rightarrow K(1)=\frac{-1\pm\sqrt{5}}2$$
$n=2$:
$$K(2)=\frac1{1+\frac1{2+K(2)}}$$
$$\Rightarrow K(2)=-1\pm\sqrt{3}$$
$n=3$:
$$K(3)=\frac1{1+\frac1{2+\frac1{3+K(3)}}}$$
$$\Rightarrow K(3)=\frac{-4\pm\sqrt{37}}3$$
$n=4$:
$$K(4)=\frac1{1+\frac1{2+\frac1{3+\frac1{4+K(4)}}}}$$
$$\Rightarrow K(4)=\frac{-9\pm2\sqrt{39}}5$$
As you may be able to tell, these results are all found by simplifying the fraction until one has a quadratic in $K(n)$, at which point the quadratic formula may be applied. 
I would be surprised if there wasn't a closed form expression for $K(n)$, as they can all be found the same way. I've failed to recognize any numerical patterns in the results, however. 
So, I have two questions:
$1)$: How does one express $K(n)$ in the $\operatorname{K}_{i=i_1}^\infty \frac{a_i}{b_i}$ notation? I was thinking something like 
$$K(n)=\operatorname{K}_{i\geq0}\frac1{1+\operatorname{mod}(i,n)}$$
$2)$: What is a closed form for $K(n)$?
Thanks.
Update:
I'm pretty sure that all the $\pm$ signs in the beginning of the question should be changed to a $+$ sign.
 A: Following up on Daniel Schepler's comment. Let $$P_n(x) = \frac{1}{1 + \frac{1}{2 + \ddots \frac{1}{n+x}}}.$$ This is basically the RHS of the recurrence equation for $K(n)$. Then:
\begin{align*}
P_1(x) &= \frac{1}{x+1} \\
P_2(x) &= \frac{x+2}{x+3} \\
P_3(x) &= \frac{2x+7}{3x+10} \\
P_4(x) &= \frac{7x+30}{10x+43} \\
P_5(x) &= \frac{30x+157}{43x+225} \\
P_6(x) &= \frac{157x+972}{225x+1393}.
\end{align*}
Note that $P_n(x) = P_{n-1}\left( \frac{1}{x+n}\right)$. Therefore, if $P_{n-1}(x) = \frac{ax+b}{cx+d}$, then
\begin{align*}P_n(x) &= \frac{\frac{a}{x+n} + b}{\frac{c}{x+n} + d} \\
&= \frac{bx + (a+bn)}{dx + (c+dn)}
\end{align*}
Thus in general, we may write $$P_n(x) = \frac{a_n x + a_{n+1}}{b_n x + b_{n+1}}$$
where $a$ and $b$ satisfy the recurrence $a_{n+1} = a_{n-1} + n a_n$ and likewise for $b$, with the initial conditions $a_1 = 0, a_2 = b_1 = b_2 = 1$. This recurrence gives the OEIS sequences linked by Jean-Claude Arbaut. $K(n)$ is a solution to $x - P_n(x) = 0$, or a root of the quadratic $$b_n x^2 + (b_{n+1} - a_n) x - a_{n+1} = 0.$$
