Here are two variants to derive $a_n$. The first one gives a closed form, the other one an explicit expression, which results in a nice binomial identity.
First variant: Partial fractions
In case it's easy to derive the zeros of the denominator of
\begin{align*}
G(x) = \frac{1}{1-x-4x^2-2x^3}
\end{align*}
the partial fraction decomposition is a convenient method. As @J.G. indicated is $x=-1$ a zero.
Omitting some intermediary calculations we obtain
\begin{align*}
G(x)&=\frac{1}{1-x-4x^2-2x^3}\\
&=\frac{1}{1+x}-\frac{2x}{2x^2+2x-1}\\
&=\frac{1}{1+x}-\frac{x}{(x+\frac{1}{2}(1+\sqrt{3}))(x+\frac{1}{2}(1-\sqrt{3}))}\\
&=\frac{1}{1+x}-\frac{1}{2\sqrt{3}}\cdot\frac{1+\sqrt{3}}{\left(x+\frac{1}{2}+\frac{\sqrt{3}}{2}\right)}
+\frac{1}{2\sqrt{3}}\cdot\frac{1-\sqrt{3}}{\left(x+\frac{1}{2}-\frac{\sqrt{3}}{2}\right)}\\
&=\frac{1}{1+x}-\frac{1}{\sqrt{3}}\cdot\frac{1}{1-(1-\sqrt{3})x}+\frac{1}{\sqrt{3}}\cdot\frac{1}{1-(1+\sqrt{3})x}\\
&=\sum_{n=0}^\infty\left[(-1)^n-\frac{1}{\sqrt{3}}(1-\sqrt{3})^n+\frac{1}{\sqrt{3}}(1+\sqrt{3})^n\right]x^n\tag{1}
\end{align*}
Second variant: Geometric series
We can also directly apply a geometric series expansion. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series and obtain
\begin{align*}
[x^n]G(x)&=[x^n]\frac{1}{1-x-4x^2-2x^3}\\
&=[x^n]\sum_{j=0}^\infty x^j(1+4x+2x^2)^j\\
&=\sum_{j=0}^n[x^{n-j}]\sum_{k=0}^j\binom{j}{k}(2x)^k(2+x)^k\tag{2}\\
&=\sum_{j=0}^n\sum_{k=0}^{\min\{j,n-j\}}\binom{j}{k}2^k[x^{n-j-k}](2+x)^k\\
&=\sum_{j=0}^n\sum_{k=0}^{\min\{j,n-j\}}\binom{j}{k}\binom{k}{n-j-k}2^{3k-n+j}\tag{3}
\end{align*}
Comment:
In (2) we use the linearity of the coefficient of operator and apply the rule
\begin{align*}
[x^{p-q}]A(x)=[x^p]x^qA(x)
\end{align*}
We also set the upper limit of the outer sum to $n$ since the exponent of $x^{n-j}$ is non-negative.
In (3) we select the coefficient of $x^{n-j-k}$.
Binomial identity: We derive from (1) and (3) the following binomial identity by changing the order of summation in the outer sum of (3), i.e. $j\rightarrow n-j$.
\begin{align*}
\sum_{j=0}^n&\sum_{k=0}^{\min\{j,n-j\}}\binom{n-j}{k}\binom{k}{j-k}2^{3k-j}\\
&=(-1)^n-\frac{1}{\sqrt{3}}(1-\sqrt{3})^n+\frac{1}{\sqrt{3}}(1+\sqrt{3})^n\qquad\qquad n\geq 0
\end{align*}