I'm trying to calculate the closed-form expression of the recursive
$a_n = \frac{a_{n-1} + a_{n-2}}{3}$
Where $a_0 = 0, $ $a_1 = \frac {1}{3}$, and $a_2 = \frac {1}{9}$
I have tried to mimic the matrix method of calculating the fibonacci sequence (since this is so close to it), but I didn't understand how the eigenvalues and whatnot all were supposed to end up. I think many of the signs got messed up, but I based my method off of the article https://medium.com/@andrew.chamberlain/the-linear-algebra-view-of-the-fibonacci-sequence-4e81f78935a3
I also know the next 5 or so values, so if those would make the calculations easier, I can do that. Thank you.
The general method to solve linear recurrence relation is by using the associated characteristic equation.
First rewrite it: $\quad a_n-\frac 13\,a_{n-1}-\frac 13\,a_{n-2}=0$
Then the characteristic equation is: $\quad r^2-\frac 13r-\frac 13=0\iff 3r^2-r-1=0$
It has roots $r=\dfrac{1\pm\sqrt{13}}6$
Therefore the solution is given by $$a_n=\alpha\,r_1^n+\beta\,r_2^n$$ where $\alpha,\beta$ are calculatated to fit the initial values of the sequence.
The reason behind this is given here for instance: Recursively given sequence question about changing the sequence with exponent function
In our case $\begin{cases}a_0=\alpha+\beta=0\\a_1=\alpha r_1+\beta r_2=\frac 16(\alpha+\beta)+\frac{\sqrt{13}}6(\alpha-\beta)=\frac 13\end{cases}\iff \alpha=-\beta=\frac 1{\sqrt{13}}$
Finally $$a_n=\frac 1{\sqrt{13}}\left(\frac{1+\sqrt{13}}6\right)^n-\frac 1{\sqrt{13}}\left(\frac{1-\sqrt{13}}6\right)^n$$
Notice also that $\ b_n=3^na_n$
verifies $3^na_n=3^{n-1}a_{n-1}+3^{n-1}a_{n-2}\iff b_n=b_{n-1}+3b_{n-2}$
And since $b_0=0$ and $b_1=1$ are integers, then $b_n$ is an integer by induction.