This is the fastest algorithm I could come up with for the Fibonacci sequence in Python. After coding a solution with $O(\sqrt{n})$ time complexity, I managed to decrease it to just $O(\log{n})$ using identities of the Lucas sequence. However, for very large $n$, this algorithm is closer to having $O(\log^2{n})$ time complexity.
These identities serve as the basis for the algorithm.
$F_{2n} = F_nL_n$
$L_{2n} = L_n^2-2(-1)^n$
$2F_{n+1} = F_n+L_n$
$2L_{n+1} = 5F_n+L_n$
Explanation
Any Fibonacci number $F_n$ can be calculated if the previous Fibonacci and Lucas numbers $F_{n-1}$ and $L_{n-1}$ are known or if the middle Fibonacci and Lucas numbers $F_{\frac{n}{2}}$ and $L_{\frac{n}{2}}$ exist.
If $n$ is zero, return the base cases $F_0$ and $L_0$.
If $n$ is odd, make it even by calculating $F_{n-1}$ and $L_{n-1}$.
If $n$ is even, halve it by calculating $F_{\frac{n}{2}}$ and $L_{\frac{n}{2}}$.
Halving $n$ every step is what gives the algorithm its $O(\log{n})$ time complexity. Even when $n$ is in the form $2^k-1$, the algorithm ensures that the next $n$ will be even if it is currently odd, making it only twice as slow as the best case scenario when $n$ is in the form $2^k$.
These steps correlate with the binary representation of $n$. For example, $n = 100$ can also be written as $n = 1100100_2$. If the right bit is $0$, then $n$ is shifted to the right. Otherwise, it is set to $0$ before shifting to the right. This results in the following steps: