8 Feb
2006
8 Feb
'06
2:05 p.m.
mreid>> (This may be asking way too much: I don't even know
how to prove that the floor of 2^n / n is odd for infinitely many values of n.)
Note that n = A090659 plus its conjectured continuation, 25, 91, 703, 1891, 12403, 38503,79003 ,88831 ,146611,188191, 218791, 269011, 286903, 385003, 497503, 597871, 736291, 765703, 954271, 1056331, 1314631, 1869211, 2741311, 3270403, 3913003, 4255903 ,4686391 ,5292631 gives all odds. --rwg