Is there a large counterexample?

Observe that from the given equation, we can obtain the recursive relation $S_{n+3}+S_{n+2}+S_{n+1}=S_n,\ n\ge 1$ We can check it manually that the signs of $S_1,S_2,S_3$ are not the same. Now, let us assume that $\exists n\ge 1$ such that the signs of $S_{n+1},S_{n+2},S_{n+3}$ matches and let the sign be called $s \in \{-1,0,1\}$ . Then, from the recursive relation, it is evident that $S_n$ should also have the sign $s$ . Using the recursive relation and the fact that $S_{n+2},S_{n+1},S_{n}$ have the same sign let us further conclude that $S_{n-1}$ also have the same sign $s$ . Then by induction it will follow that all $S_k,\ 1\le k\le n+3$ have the same sign $s$ , which clearly simply is not true because of the sign mismatch of $S_1,S_2,S_3$ . Thus the answer is $\boxed{\mathrm{false}}$ .