Invert the formula $F(n)= (r^n - (1-r)^n)/\sqrt{5}$ where $r=(1+\sqrt{5})/2$ by the 
[Lagrange Inversion Formula](http://en.wikipedia.org/wiki/Lagrange_inversion_theorem) (LIF).  Let $X=\sqrt{5}*F(n)$ and $s=1-r$ so $X=r^n - s^n$.
So 

$$X=\sum_{j=0}^\infty \frac{(\ln(r))^j - (\ln(s))^j}{j! n^j}$$

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"