A fast recurrence for Fibonacci and Lucas numbers

Jeroen van de Graaf

We derive the double recurrence $e_n=\frac{1}{2}(a_{n-1}+5b_{n-1});f_n=\frac{1}{2}(a_{n-1}+b_{n-1})$ with $e_0=2;f_0=0$ for the Fibonacci numbers, leading to an extremely si…