Re: [math-fun] A binary sieve

"... then delete every j for which ..." A priori it seems possible that the order in which such j are deleted could change the outcome. —Dan Fred Lunnon wrote: ----- Choose an integer k >= 2 , let q' = 2^k - 1 , initialise set S = {1, ..., q'-1} ; then delete every j for which i < j exists with j == i*2^h (mod q') for some h . For example, when k = 5 , q' = 31 , finally S = {1, 3, 5, 7, 11, 15} . I conjecture that max S = (q'-1)/2 for all k ; but why? -----