In my research, I encounter a linear recursion of the form: $$a_n = (AB^n+C)a_{n-1}-AB^na_{n-2},$$ where $A,B,C$ are all positive (arbitrary) constants such that $B,C>1$.
Is it possible to get a non-recursive formula for $a_n$? It does not have to be closed-form, but I want to get rid of the recursion (to be able to estimate the convergence rate of $a_n$).
Hint:
Let $a_n=\int_{-\infty}^\infty B^{nt}K(t)~dt$ ,
Then $\int_{-\infty}^\infty B^{nt}K(t)~dt=(AB^n+C)\int_{-\infty}^\infty B^{(n-1)t}K(t)~dt-AB^n\int_{-\infty}^\infty B^{(n-2)t}K(t)~dt$
$\int_{-\infty}^\infty B^{nt}K(t)~dt=A\int_{-\infty}^\infty B^{n(t+1)}B^{-t}K(t)~dt+C\int_{-\infty}^\infty B^{nt}B^{-t}K(t)~dt-A\int_{-\infty}^\infty B^{n(t+1)}B^{-2t}K(t)~dt$
$\int_{-\infty}^\infty B^{nt}K(t)~dt-A\int_{-\infty}^\infty B^{nt}B^{-t+1}K(t-1)~dt-C\int_{-\infty}^\infty B^{nt}B^{-t}K(t)~dt+A\int_{-\infty}^\infty B^{nt}B^{-2t+2}K(t-1)~dt=0$
$\int_{-\infty}^\infty((1-CB^{-t})K(t)+A(B^{-2t+2}-B^{-t+1})K(t-1))B^{nt}~dt=0$
$\therefore(1-CB^{-t})K(t)+A(B^{-2t+2}-B^{-t+1})K(t-1)=0$
$(1-CB^{-t})K(t)=A(B^{-t+1}-B^{-2t+2})K(t-1)$
$(B^{2t}-CB^t)K(t)=A(B^{t+1}-B^2)K(t-1)$
$K(t)=\dfrac{A(B^{t+1}-B^2)}{B^{2t}-CB^t}K(t-1)$