It appears more than you think!

The observation that I found to be the key to solving the problem was that after some algebraic manipulations, the equation can be reduced to $x^5-5x-4=0$ This leads to the following recursive relation, $S_{n}=5S_{n-4}+4S_{n-5},\ n\ge 6$ Now, one can check it manually that $S_1=2,S_2=-2,S_3=2,S_4=18$ . This implies that every $S_n$ is an integer. Note that we do not need to calculate any more $S_n$ 's manually, and we can use the recursive relation to our advantage to tell which $n$ 's are going to give $|S_n|=2$ . Now, note that apart from $1,2,3$ , the only other $n$ 's for which $|S_n|=2$ , are going to be $6$ (from $\{1,2\}$ ), $7$ (from $\{2,3\}$ ), and $11$ (from $\{6,7\}$ ). This can be seen by an use of the recursive relation, as the relation implies that if both of $S_{n-4},\ S_{n-5}$ are positive, $S_n$ is going to be higher than $2$ . Since every $S_n$ , for $n\ge 12$ can be determined, using the recursion, from the $S_n$ 's with $n\ge 7$ , and since every $S_7>0$ , $S_n>2\ \forall n\ge 12$ . Thus the answer is $\boxed{30}$ .

The cubic $f(x) = x^3 - 2x^2 + 3x - 4$ has one real root $p$ and two complex roots $q,r = q^\star$ . Since $f(1.65) < 0 < f(1.66)$ , we deduce that $1.65 < p < 1.66$ . Also $4 = pqr = p|q|^2$ , and so $|q| = \tfrac{2}{\sqrt{p}}$ , so that $\tfrac{2}{\sqrt{1.66}} < |q| < \tfrac{2}{\sqrt{1.65}}$ . If we write $q = |q| e^{i\theta}$ then $S_n \; = \; p^n + q^n + r^n \;= \; p^n + 2|q|^n \cos n\theta \; = \; |q|^n\Big[\Big(\frac{p}{|q|}\Big)^n + 2\cos n\theta\Big]$ Since $\Big(\frac{p}{|q|}\Big)^n \; > \; \left(\frac{1.65}{\frac{2}{\sqrt{1.65}}}\right)^n \; = \; \Big(\frac{1.65^{1.5}}{2}\Big)^n$ and $\left(\frac{1.65^{1.5}}{2}\right)^{12} > 2$ , it is clear that $S_n > 0$ for all $n \ge 12$ , and moreover that $S_n \; \ge \; |q|^n\left[\left(\tfrac{p}{|q|}\right)^n - 2\right] \; \ge \; \left(\tfrac{2}{\sqrt{1.66}}\right)^n\left[\left(\frac{1.65^{1.5}}{2}\right)^n - 2\right] \qquad n \ge 12 \;.$ Hence we deduce that $S_n \ge \left(\tfrac{2}{\sqrt{1.66}}\right)^n\left[\left(\frac{1.65^{1.5}}{2}\right)^n - 2\right] \; \ge \; \left(\tfrac{2}{\sqrt{1.66}}\right)^{13}\left[\left(\frac{1.65^{1.5}}{2}\right)^{13} - 2\right] \; > \; 38$ for all $n \ge 13$ . Certainly, then, $|S_n| > 2$ for all $n \ge 13$ . It is now just a matter of checking that the recurrence relation $S_{n+3} - 2S_{n+2} + 3S_{n+1} - 4S_n = 0 \qquad \qquad S_{-1} = \tfrac34\,,\,S_0 = 3\,,\,S_1 = 2$ gives $S_1 = S_3 = S_7 = S_{11} = 2$ and $S_2 = S_6 = -2$ , and that no other values of $S_n$ for $1 \le n \le 12$ are equal to $2$ or $-2$ .

Thus the answer is $1+2+3+6+7+11 = \boxed{30}$