Reply To: Maths
I guess the point is that it is quicker to calculate S(v,p), and thereby determine if p is prime, by starting with S(v,p−1) and applying the difference −p(S(v/p,p−1)−S(p−1,p−1)). If you proceed incrementally you use the value of S(v, p) you have already obtained in the next step and only need to calculate the difference each time. That would be my take on the post.