Suppose the following linear recurrence sequence
$$
C\_n:=C\_{n-2}+C\_{n-4}+C\_{n-6}\, .
$$ 
With the initial values 
$$
C\_0=0 \, , \, C\_1=1 \, , \, C\_2=0 \, , \, C\_3=0 \, , \, C\_4=1 \, , \, C\_5=1 \, , \, C\_6=1 \, .
$$
It can be proved that another form of the $C\_n$ sequence is as follows 
\begin{equation}
C\_n:=\left\{
\begin{array}{ccc}
C\_{n-3}+C\_{n-1} &\mbox{if}& n=1~\mbox{mod}~2,\\
\\
C\_{n-3}+C\_{n-2} & \mbox{if}&n=0~\mbox{mod}~2.
\end{array}
\right.
\end{equation}


With boundary conditions
$$
C\_0=0 \, , \, C\_1=1 \, , \, C\_2=0 \, .
$$


We can proof that the generating function of $C\_n$ sequence is in the following form


$$
C(x)=\frac{x-x^3+x^4}{1-x^2-x^4-x^6}\, .
$$


In addition, I found the combinatorial forms of even and odd terms of $C\_n$ sequence. For even terms, we have 


$$
C\_{2n}=\sum\_{(k\_1,k\_2,k\_3)} 
\left(
\begin{array}{c}
k\_1+k\_2+k\_3 \\
k\_1,k\_2 , k\_3
\end{array}
\right)
$$
where the summation is over non-negative integers satisfying
$$
k\_1+2\, k\_2+3\, k\_3=n-1 \, .
$$
and for odd terms, the following relation is obtained 
$$
C\_{2n+1}=\sum\_{(k\_1,k\_2,\cdots,k\_p)} \frac{k\_2+k\_3}{k\_1+k\_2+k\_3}\times
\left(
\begin{array}{c}
k\_1+k\_2+k\_3 \\
k\_1,k\_2 , k\_3
\end{array}
\right)
$$
where the summation is over non-negative integers satisfying
$$
k\_1+2\, k\_2+3\, k\_3+=n \, .
$$
My question is that how to find a closed-form expression for $C\_n$ sequence, based on the parameter $n$. 
I used the auxiliary equation method to find closed-form expression for $C\_n$ sequence but the auxiliary equation 
of $C\_n$ sequence have complex roots and it's closed-form expression is complicated.


I would greatly appreciate for any suggestions.


EDIT: I claimed that the closed-form expression of the $C\_n$ sequence ,by using auxiliary equation method, is complicated. I want to 
show it's complexity. In fact, I want to say why I want to find a closed-form expression with less complexity. 
The auxiliary equation of the $C\_n$ sequence is as follows
$$
x^6-x^4-x^2-1=0~.
$$
The roots of the above equation are
$$\left\{
\begin{array}{ccc}
x\_1&=1.356203066~,&\\
x\_2&=-1.356203066~,&\\
x\_3&=0.3985657592&+ \hspace{3mm} 0.7605905878\,i~,\\
x\_4&=0.3985657592&- \hspace{3mm} 0.7605905878\,i~,\\
x\_5&=-0.3985657592&+ \hspace{3mm} 0.7605905878\,i~,\\
x\_6&=-0.3985657592&- \hspace{3mm} 0.7605905878\,i~.\\
\end{array}
\right.$$
Using the Demorgan's law about complex number
$$Z=x+\,i y \Leftrightarrow Z=r(\cos(\theta)+\,i \sin(\theta))$$
and based on initial values of the $C\_n$ sequence that is defined at the first, the following 
closed-form expression for the $C\_n$ sequence is obtained
\begin{eqnarray}\nonumber
C\_n&=&0.1954392117(1.356203066)^n-0.01263567906(-1.356203066)^n \\ \nonumber
&&+{0.8586924398}^n(0.2878832995\cos(1.088116773n) \\ \nonumber
&&\hspace{32mm}- 0.06742448463\sin(1.088116773n) \\ \nonumber
&&\hspace{32mm}-0.4706868334\cos(2.053475881n) \\ \nonumber
&&\hspace{32mm}+0.6136685762\sin(2.053475881n))~.
\end{eqnarray}
In my research, the above closed-form is not applicable, just because of this, I asked this question. 



The closed-form expression for $C_n$ sequence
The roots of your denominator $x^6 - x^4 - x^2 - 1$ are $\pm \sqrt{r\_i}$ where $r\_1, \ldots, r\_3$ are the roots of $z^3 - z^2 - z - 1$, namely 
$$ \eqalign{r\_1 &= \dfrac{1}{3} + \dfrac{1}{3} (19 + 3 \sqrt{33})^{1/3} + \dfrac{4}{3} (19+3 \sqrt{33})^{-1/3}\cr
r\_2, r\_3 &= \dfrac{1}{3} - \dfrac{1}{6} (19 + 3 \sqrt{33})^{1/3} - \dfrac{2}{3} (19 + 3 \sqrt{33})^{-1/3}) \cr &\pm \frac{i \sqrt{3}}{6} \left((19 + 3 \sqrt{33})^{1/3} - 4 (19 + 3 \sqrt{33})^{-1/3}\right) }
 $$


