Recurrence relation of binomial sum. I'm trying to find a closed-form solution to the sum
$$
a(n):= \sum_{k=0}^{\lfloor n/3 \rfloor} \binom{n}{3k}.
$$
In my attempt, I found the first few values of $a(n)$ and entered them into the OEIS and got a hit for sequence A024493. In the notes there I saw that there was a recurrence relation given, namely
$$
a(n) = 3a(n-1)-3a(n-2)+2a(n-3)
$$
or perhaps more illuminatingly
$$
a(n)-3a(n-1)+3a(n-2)-a(n-3) = a(n-3)
$$
where we can see that the coefficients on the right hand side are $(-1)^i \binom{3}{i}$ for $0\leq i \leq 3$.
I've tried proving this relation by induction, but the result seems to depend on the value of
$n\mod 3$ more than on the previous terms.
Any thoughts on how I can prove that $a(n)$ satisfies the given recursion?

Bearing in mind that for $ k > n$ or $ k < 0$, ${ n \choose k } =0 $, we can write $a_n = \sum_{k= - \infty } ^\infty { n \choose 3k}$. This allows us to avoid the "have to consider $n \pmod{3}$ cases".
Then use the identity $ { n\choose k } = { n-1 \choose k-1  } + { n - 1 \choose k }$ (which is still true when $k > n$ or $k < 0$) to iteratively reduce $a_n - 3 a_{n-1} + 3a_{n-2} + 2 a_{n-3}$, IE
$= \left[ \sum_{k} { n \choose 3k} \right] - 3 a_{n-1} + 3a_{n-2} - 2 a_{n-3}    $
$ = \left[ \sum_{k} { n-1 \choose 3k-1}  + {n-1 \choose 3k }\right] - 3 a_{n-1} + 3a_{n-2} - 2 a_{n-3}$
$ = \left[ \sum_{k} { n-1 \choose 3k-1}  - 2 {n-1 \choose 3k }\right]   + 3a_{n-2} - 2 a_{n-3}$
$ = \ldots $
Can you complete this to show that it equals 0?
