I am interested in the continued fractions which $1$s are consecutive appears. For example, it is the following values.

$$ \sqrt{7} = [2;\overline{1,1,1,4}] \\ \sqrt{13} = [3;\overline{1,1,1,1,6}] $$ In this article, let us denote n consecutive $1$s as $1_n$. Applying this, the above numbers would be as follows.

$$ \sqrt{7} = [2;\overline{1_3,4}] \\ \sqrt{13} = [3;\overline{1_4,6}] $$

While investigating these numbers, the following pattern was found experimentally.

$$ \sqrt{F(n)^2m^2-(F(n)^2-L(n))m+\frac{(F(n)-1)(F(n)-3)}{4}-\frac{F(n-3)-1}{2}} =\left[F(n)m-\frac{F(n)-1}{2};\ \overline{1_{n-1},\ 2\left(F(n)m-\frac{F(n)-1}{2}\right)}\right] $$ ($m,n \in \mathbb{N},\ n\equiv\pm1\ (mod3),\ n > 3,\ $$F(n)$ is Fibonacci number, $L(n)$ is Lucas number)

I confirm that it works correctly when $n$ and $m$ are single digits. If you find a proof or counterexample, please let me know.

(2022/04/13 edit)

A general expression was derived. I think the expression I found is that special case. The condition is that the inside of the square root is always an integer.

Here are some concrete examples.

$$\begin{array}{|c|c|} \hline n & pattern \\ \hline 4 & \sqrt{9m^2-2m} = [3m-1;\overline{1_3,2(3m-1)}] \\ \hline 5 & \sqrt{25m^2-14m+2} = [5m-2;\overline{1_4,2(5m-2)}] \\ \hline 7 & \sqrt{169m^2-140m+29} = [13m-6;\overline{1_6,2(13m-6)}] \\ \hline 8 & \sqrt{441m^2-394m+88} = [21m-10;\overline{1_7,2(21m-10)}] \\ \hline \end{array}$$