There are some quadratic irrationals (like $\sqrt{2}, \sqrt{5},\sqrt{10}$, etc.) that have continued fractions with a period of one (e.g. $\sqrt{2}=[1;2,2,2,2,\dots]$). I know the period of the fraction ends whenever $a_i=2a_0$, but is there any pattern to the quadratic irrationals that exhibit this behavior?
The Golden Ratio, for example, also famously has a period of one, with its continued fraction being $\phi = [1;1,1,1,1,\dots]$.
On
$n \in \mathbb{Z^+}$ will have a (simple) continued fraction representation with a period of $1$ if and only if $(n-1)$ is a perfect square. See chapter 4 of http://www.ms.uky.edu/~sohum/ma330/files/Continued%20Fractions.pdf for details.
Note that $\sqrt{D} + a_1 = [\overline{(2\times a_1)}] = (2 \times a_1) \;+\; \frac{1}{\sqrt{D} + a_1} \;\Rightarrow$
$(\sqrt{D} - a_1)(\sqrt{D} + a_1) = 1.$
In a first step, the case where no irregular part is present:
$$[0,n,n,n,\cdots]=\cfrac{1}{n+\cfrac{1}{n+\cfrac{1}{n+\cdots}}}=\color{red}{\frac12\left(-n+\sqrt{4+n^2}\right)}\tag{1}$$
Formula (1) comes naturaly from the fact that if we denote by $x$ the continued fraction in (1),
$$x:=\cfrac{1}{n+\cfrac{1}{n+\cfrac{1}{n+\cdots}}}$$
we find back in the right hand side $x$ in this (classical) way:
$$x=\cfrac{1}{n+x}$$
giving rise to a quadratic equation whose positive root is the left hand side of (1)
The case $n=1$ gives in particular $\Phi-1$. One needs to add $1$ to get Golden Ratio $\Phi$.
More generally, all continued fractions with hopefuly an irregular part can be obtained in this way by eventually "prefixing" by the beginning of a continued fraction. For example $$[a,b,c,n,n,n,\cdots]=\cfrac{1}{a+\cfrac{1}{b+\cfrac{1}{c+\cfrac{1}{n+\frac12\left(-n+\sqrt{4+n^2}\right)}}}}$$
Examining this last form, one can see that, by successive multiplications par conjugate expressions, one can get an expression of the form indicated by richrow