We shall now examine the behavior of pn(k) as a function of k for a fixed n. We maintain that as k increases, pn(k) increases reaching a maximum for
k = kmax = [(n + 1)p], where the brackets mean the largest integer that does not exceed (n + 1)p. If (n + 1)p is an integer, then pn(k) is maximum for two consecutive values of k:

k = k1 = (n + 1)p and
k = k2 = k1 - 1 = np – q.