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&#x27;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?<p>(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.