Evaluate:$S_{n}=\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{n-3}-\binom{n-3}{n-6}+.......$ If$$S_{n}=\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{n-3}-\binom{n-3}{n-6}+.......$$
Does $S_{n}$ have a closed form.
My Attempt
$$S_{n}=\binom{n}{0}-\binom{n-1}{n-2}+\binom{n-2}{2}-\binom{n-3}{3}+.......$$
$$S_{n}=\binom{n}{n}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+.......$$
$=$coefficient of $x^n$ in $\left\{(1+x)^n-x^2(1+x)^{n-1}+x^4(1+x)^{n-2}-x^6(1+x)^{n-3}+....\right\}$
After this not able to proceed

If $S_{n} = \sum_{j\geq 0} (-1)^{j}\binom{n-j}{j}$, then 
\begin{align}
S_{n+1} &= \sum_{j\geq 0} (-1)^{j} \binom{n+1-j}{j} \\
&= \sum_{j\geq 0} (-1)^{j} \left[\binom{n-j}{j} + \binom{n-j}{j-1}\right] \\
&= \sum_{j\geq 0} (-1)^{j} \binom{n-j}{j} - \sum_{j\geq 1} (-1)^{j-1}\binom{n-1-(j-1)}{j-1} \\
&= S_{n} - S_{n-1}
\end{align}
and $S_{1} = 1, S_{2} = 0$, so $S_{n}$ is 
$$
1, 0, -1, -1, 0, 1, 1, 0, -1, -1, 0, 1, \dots
$$
which is a periodic sequence. 
