From the paper, this originates from: > ...Since i is a lower bound on j, j will equal the largest i for which P(j ≥ i), thus the largest i satisfying i ≤ (b+1

From the paper, this originates from: > ...Since i is a lower bound on j, j will equal the largest i for which P(j ≥ i), thus the largest i satisfying i ≤ (b+1) / r. Thus, by the definition of the floor function, j = floor((b+1) / r). It seems to me that since all the numbers are positive, you can safely use integer division if you like. AFAIK, floor(((double)a)/b) and a/b coincide for that case at least.