In
Xn+1 = (a*Xn + c) (mod m): If the offset
c is relatively prime with
m, you will generate a maximal-length sequence. Since
GotToBTru picked the first prime
≥ S/2, it will be maximal with length
S, and will guarantee that's there are at most 2 numbers from each period (ie,
Xn+1 may not have wrapped around
S, but
Xn+2 definitely will have).