Let $\zeta=e^{2\pi i/(n+1)}$, so $\zeta$ is a primitive complex root of unity of order $n+1$. Then there are constants $b_0,b_1,\dots,b_n$ such that for all $r$ the $r$th term in your sequence is given by the formula $b_0+b_1\zeta^r+b_2\zeta^{2r}+\cdots+b_n\zeta^{nr}$ You can find the $b_j$ by solving the system of $n+1$ linear equations in $n+1$ unknowns $a_r=b_0+b_1\zeta^r+b_2\zeta^{2r}+\cdots+b_n\zeta^{nr},\qquad r=0,1,\dots,n$ This isn't quite as bad as it looks, since the matrix of coefficients has a very special form and there are tricks based on that observation, but still doing it by hand even for $n=4$ could get very messy very fast.
Let's do a not-so-messy example: $(0,8,3,1,0,8,3,1,0,8,3,1,\dots)$ Here $n=3$, so $\zeta=i=\sqrt{-1}$, and the equations are $\eqalign{0&=b_0+b_1+b_2+b_3\cr8&=b_0+b_1i-b_2-b_3i\cr3&=b_0-b_1+b_2-b_3\cr1&=b_0-b_1i-b_2+b_3i\cr}$ Add the first and 3rd equations to get $b_0+b_2=3/2$, add the 2nd and 4th to get $b_0-b_2=9/2$, deduce $b_0=3$, $b_2=-3/2$. Then it's easy to get $b_1$ and $b_3$, stick 'em in the first displayed formula, and verify that it works.