number theory - 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.

Answer

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 $$b_n x^2 + (b_{n+1} - a_n) x - a_{n+1} = 0.$$