We want to find the coefficients of $$f(x)={-x\over x^2+x-1},$$ that is, if $$f(x)=\sum_{i=0}^\infty F_ix^i,$$ we want a formula for $F_i$. We begin by finding the partial fraction decomposition of $-x/(x^2+x-1)$.
Let's check those answers to make sure we didn't do something stupid:
The partial fraction decomposition looks like this: $${-x\over x^2+x+1}={A\over x-a}+{B\over x-b}={A(x-b)+B(x-a)\over(x-a)(x-b)}.$$ We want to find $A$ and $B$ so that $A(x-b)+B(x-a)=-x$.
Now $$f(x)={A\over x-a}+{B\over x-b}=-{A\over a}{1\over 1-x/a}-{B\over b}{1\over 1-x/b}=-{A\over a}\sum_{i=0}^\infty (1/a)^ix^i-{B\over b}\sum_{i=0}^\infty (1/b)^ix^i=\sum_{i=0}^\infty \Bigl(-{A\over a}\bigl({1\over a}\bigr)^i-{B\over b}\bigl({1\over b}\bigr)^i\Bigr)x^i,$$ that is, the coefficient formula is $$-{A\over a}\bigl({1\over a}\bigr)^i-{B\over b}\bigl({1\over b}\bigr)^i.$$
Now let's look at the first few terms to check:
Let's look at the decimal values of $1/a$ and $1/b$.
Since $|1/b|<1$, $(1/b)^i$ goes to 0 as $i$ goes to infinity. This means that after a while, just the term $$-{A\over a}\bigl({1\over a}\bigr)^i$$ is very close to the true (integer) value of $F_i$. So let's look at just these terms, rounded to the nearest integer.
So in fact from the very beginning this gives the correct value.