Let argument exp be n. The number of steps for iterative process can be expressed by recurrency \(T(n)=T(n/2)+O(1)\), with solution \(O(\log{}n)\).

Louis replaced fast iterative process with recursive one, that splits in two branches for even n. This corresponds to different recurrency \(T(n)=2 T(n/2)+O(1)\), with solution \(O(n)\).