He says that this is an inefficient way to generate the nth Fibonacci number, but since it actually gives you the first n Fibonacci numbers, thats not the fairest comparison. (Especially since the fastest methods for finding the nth Fibonacci number don't generate all the intermediates.) Just for curiosity sake, how does using this formula to generate all of the first n Fibonacci numbers compare to other ways of doing that?

(You get all n with the same formula by replacing the mod 2^(n+1) (or reading the last n+1 bits) by splitting the whole integer that is generated into n+1 bit chunks.