It was some time ago I studied recurrence relations and I came across this one that I cannot solve:
$a_{n+3}=-3a_{n+2}+4a_{n}$ with $a_{0}=2$ and $a_{1}=-5$
Ansatz: $a_{n}=r^{0}$ then I get
$r^{3}+3r^{2}-4=0$ with roots $r=1$ and $r=-2$ of double multiplicity
so the solution is
$a_{n}=A+B(-2)^{n}+Cn(-2)^{n}$ is this correct?
But I cannot find constants as I have 2 equations with 3 indeterminants...
What you’ve done is correct. There’s no way to pin down $A,B$, and $C$ with the available information: it’s a third-order recurrence, so you need three initial values. The best you can do is solve for $A,B$, and $C$ in terms of the missing value $a_2$.