\(\begin{cases}a_0=3\\a_1=6\\a_n=a_{n-1}+6a_{n-2}&\text{for }n>2\end{cases}\)
Let the generating function for \(a_n\) be
\(A(x)=\displaystyle\sum_{n\ge0}a_nx^n\)
Multiplying both sides by \(x^{n-2}\)
\(a_nx^{n-2}=a_{n-1}x^{n-2}+6a_{n-2}x^{n-2}\)
Summing both sides over the non-negative integers greater than or equal to 2 gives
\(\displaystyle\sum_{n\ge2}a_nx^{n-2}=\sum_{n\ge2}a_{n-1}x^{n-2}+6\sum_{n\ge2}a_{n-2}x^{n-2}\)
\(\displaystyle\frac1{x^2}\sum_{n\ge2}a_nx^n=\frac1x\sum_{n\ge1}a_nx^n+6\sum_{n\ge0}a_nx^n\)
\(\displaystyle\frac1{x^2}\left(\sum_{n\ge0}a_nx^n-a_0-a_1x\right)=\frac1x\left(\sum_{n\ge0}a_nx^n-a_0\right)+6\sum_{n\ge0}a_nx^n\)
\(\displaystyle\frac1{x^2}\left(A(x)-3-6x\right)=\frac1x\left(A(x)-3\right)+6A(x)\)
\(A(x)=\dfrac{3x+3}{1-x-6x^2}\)
\(A(x)=\dfrac3{5(1+2x)}+\dfrac{12}{5(1-3x)}\)
For \(|x|<\dfrac13\), the two series converge to
\(\displaystyle A(x)=\frac35\sum_{n\ge0}(-2x)^n+\frac{12}5\sum_{n\ge0}(3x)^n\)
\(\implies A(x)=\displaystyle\sum_{n\ge0}\dfrac{3(-2)^n+12(3)^n}5x^n\)
\(a_n=\dfrac{3(-2)^n+12(3)^n}5=\dfrac{3(-2)^n+4(3)^{n+1}}5\)
