I have three equations with three sequences, $a_n, b_n, c_n$. I tried to isolate $a_n$ with no luck.
$$a_n = 2b_{n-1}+c_{n-1}$$ $$b_n=2a_{n-1}+2b_{n-1}+c_{n-1}$$ $$c_n = 4a_{n-1}+4b_{n-1}$$
Is it even possible to get an expression based only on $a_n$ terms here?
Notice that the expression $2b_{n-1}+c_{n-1}$, which is equal to $a_n$, appears in the expression for $b_n$. So we can rewrite that as $$b_n = 2a_{n-1}+a_n$$ Thus $$c_n = 4a_{n-1} + 4b_{n-1} = 4a_{n-1} + 8a_{n-2} + 4a_{n-1} = 8a_{n-1} + 8a_{n-2}$$ And, finally, $$a_n = 2b_{n-1}+c_{n-1} = 4a_{n-2} + 2a_{n-1} + 8a_{n-2} + 8a_{n-3} = 2a_{n-1} + 12a_{n-2} + 8a_{n-3}$$
Your coefficient matrix is $$ M = \left( \begin{array}{ccc} 0 & 2 & 1 \\ 2 & 2 & 1 \\ 4 & 4 & 0 \\ \end{array} \right) $$ which satisfies (Cayley-Hamilton) $$ M^3 - 2 M^2 - 12 M - 8I = 0 $$ $$ a_{n+3} = 2 a_{n+2} + 12 a_{n+1} + 8 a_n \; . $$ You also get $$ b_{n+3} = 2 b_{n+2} + 12 b_{n+1} + 8 b_n \; , $$ $$ c_{n+3} = 2 c_{n+2} + 12 c_{n+1} + 8 c_n \; . $$
If we make the column vector $$ x_n = \left( \begin{array}{c} a_n \\ b_n \\ c_n \\ \end{array} \right) \; , $$ we find $x_{n+1} = M x_n \; ,$ then $x_{n+2} = M x_{n+1} = M^2 x_n \; ,$ finally $x_{n+3}= M x_{n+2} = M^3 x_n.$ Cayley Hamilton says $$ x_{n+3} = M^3 x_n = \left( 2M^2 + 12 M + 8 I \right)x_n = 2M^2 x_n + 12 M x_n + 8 I x_n = 2 x_{n+2} + 12 x_{n+1} + 8 x_n $$
