A recurrence with a hidden invariant

1. An invariant. Using $a_{n+3}=4a_{n+2}-a_{n+1}$ and then $4a_{n+1}-a_{n+2}=a_n$,
$$D_{n+1}=a_{n+1}a_{n+3}-a_{n+2}^2=a_{n+1}\bigl(4a_{n+2}-a_{n+1}\bigr)-a_{n+2}^2=a_{n+2}\bigl(4a_{n+1}-a_{n+2}\bigr)-a_{n+1}^2=a_{n+2}a_n-a_{n+1}^2=D_n.$$
So $D_n=D_0=a_0a_2-a_1^2=0\cdot4-1=-1$.

2. Closed form. The roots of $r^2-4r+1=0$ are $r_\pm=2\pm\sqrt3$. A general solution is $a_n=Ar_+^n+Br_-^n$. From $a_0=0$ we get $B=-A$, and $a_1=1$ gives $A(r_+-r_-)=2\sqrt3A=1$. So
$$a_n=\frac{(2+\sqrt3)^n-(2-\sqrt3)^n}{2\sqrt3}.$$

Checks. $n=2$: $a_2=4$ ✓. Also $r_+r_-=1$ makes part 1 transparent: $a_na_{n+2}-a_{n+1}^2=-\frac{(r_+r_-)^n(r_+-r_-)^2}{12}=-1$.

Remark. The numbers $a_n$ are the $y$-values of solutions of the Pell-type equation $x^2-3y^2=1$.

Lesson. For a linear recurrence, look for a conserved quantity (a determinant-like expression) by checking that it does not change when $n\to n+1$.