Words of length $10$ in alphabet $\{a,b,c\}$ such that the letter $a$ is always doubled 
Compute the number of words of length $10$ in alphabet $\{a,b,c\}$ such that letter $a$ is always doubled (for example "$aabcbcbcaa$" is allowed but "$abcbcaabcc$" is forbidden).

I am looking for a quick/efficient way to resolve this problem. I thought of fixing "$aa$" in the beginning then draw a tree of the next possibilities but this tree will end up to be a whole forest. Can you help me ?

Here is a second approach: Let $x_n$ $(n\geq0)$ denote the number of admissible words having $n$ letters. Then
$$x_0=1,\quad x_1=2,\qquad x_n=2x_{n-1}+x_{n-2}\quad (n\geq2)\ .$$
The characteristic equation of the recursion is $\lambda^2-2\lambda-1=0$ with roots $\lambda=1\pm\sqrt{2}$. It follows that
$$x_n=c(1+\sqrt{2})^n+c'(1-\sqrt{2})^n\qquad(n\geq0)\ ,$$
where the constants $c$, $c'$ have to be determined from the initial conditions. The computation gives
$$c={2+\sqrt{2}\over4},\qquad c'={2-\sqrt{2}\over4}\ .$$
Now $\bigl|c'(1-\sqrt{2})^n\bigr|<{1\over2}$ for all $n\geq0$. We therefore can write
$$x_n={\tt round}\left({2+\sqrt{2}\over4}(1+\sqrt{2})^n\right)\qquad(n\geq0)\ .$$
This gives
$$x_{10}={\tt round}\bigl(5740.9999782267896753\bigr)=5741\ .$$
This coincides with the value obtained by ${\tt user10354138}$.
