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
i didn't quite follow this. apart from the first term (25), the others are all of the form pq where p is a prime = 3 mod 4 , and q = 2 p - 1 . there is a comment in the online encyclopedia that later terms in the sequence are conjectured to be of this form (although it does not mention that p = 3 mod 4 ). for such numbers, we have 2^(pq) = 0 mod 2 , 2^(pq) = 2^q = 2 mod p , and 2^(pq) = 2^p = -2 mod q (using the fact that 2 is a quadratic non-residue mod q ). put these together to get 2^(pq) = (2p - 4) q - 2 mod 2pq . if p > 3 , then the remainder, (2p - 4) q - 2 , is in the range [pq, 2pq) , so indeed floor(2^(pq) / pq) is odd. (this also explains why 15 doesn't work.) conjecturely, there are infinitely many primes p = 3 mod 4 , for which 2p - 1 is also prime, but i don't believe it's been proven. nevertheless, these terms should grow slower (i.e. there are more of them) than those in the provably infinite "sequence" i gave, whose plateaus grow doubly exponentially. none of this speaks to jim's original question, which asked if, as N goes to infinity, floor(2^n / n) is even for asymptotically half the positive integers less than N . mike