Invert the formula $F(n)= (r^n - (1-r)^n)/sqrt(5) 1-r)^n)/\sqrt{5}$ where r=(1+sqrt(5))/2 $r=(1+\sqrt{5})/2$ by the Lagrange Inversion Formula (LIF). Let X=sqrt(5)*F(n) $X=\sqrt{5}*F(n)$ and s=1-r $s=1-r$ so $X=r^n - s^ns^n$. So X=sum from j=0 to infinity of ((ln(r))^j
$$X=\sum_{j=0}^\infty \frac{(\ln(r))^j - (ln(s))^j)/j! * n^j.\ln(s))^j}{j! n^j}$$
Then n = X + sum of X^j * sum over all sequences (b(2),b(3),....,) of nonnegative integers of (-1)^(sum of b(i) from i=2 to j) * ( (sum of b(i) from i=2 to j) + j-1)!/(j!) * product from i=2 to j of (((ln(r))^i - (ln(s))^i)/i!)^b(i) / (b(i)!) ) such that sum of (i-1)*b(i) from i=2 to j equals j-1 and pray for convergence everywhere.
I got this form of the LIF from page 264 of G.P. Egorychev's book, "Integral Representations of Combinatorial Sums"
