What's the longest sequence?

It is a simple induction that (here $F_n$ is the $n$ th Fibonacci number) $S_n \; = \; (-1)^n\big[F_{n-1}S_2 - 2018F_{n-2}\big] \hspace{2cm} n \ge 2$ and so the term $S_n$ will be non-negative provided that $S_2 \le 2018 \tfrac{F_{n-2}}{F_{n-1}}$ when $n$ is odd and provided that $S_2 \ge 2018 \tfrac{F_{n-2}}{F_{n-1}}$ when $n\ge2$ is even. The sequence $2018 \tfrac{F_{n-2}}{F_{n-1}}$ is increasing for odd $n$ and decreasing for even $n$ .

Since $2018 \tfrac{F_{n-2}}{F_{n-1}}$ is equal to $1246.41$ , $1247.49$ , $1247.08$ when $n = 10,11,12$ , we see that it is possible to have $S_1,S_2,...,S_{11}$ all nonegative, by choosing $S_2 = 1247$ , but that after that $S_{12}$ is negative. Any other value of $S_2$ yields a negative term in the sequence before $S_{12}$ . Thus the answer is $\boxed{12}$ .

The number of terms is maximised when $\frac{S_1}{S_2}$ is as close to the golden ratio $\phi$ as possible. In this case it is when $S_2=1247$ .

Then the sequence would go $2018 , 1247 , 771 , 476 , 295 , 181 , 114 , 67 , 47 , 20 , 27 , -7$ which is $12$ terms.