Inspired by Fibonacci et al
The numerators \(g(n)\) are given by \(g(n)=g(n-1)-4g(n-2)\) for \(n>2\), with \(g(1)=g(2)=1\).
If \(M\) and \(m\) are the supremum and the infimum, respectively, of the partial sums of the series above, find \(M+m\). As your answer, enter \(\lfloor 1000(M+m) \rfloor\).
Enter 2016 if you come to the conclusion that the supremum or the infimum fail to exist.
Notation: \( \lfloor \cdot \rfloor \) denotes the floor function.
This problem is a follow-up to this question.