OFFSET
1,1
COMMENTS
A student asked if the floor of 2^n / n was always even. He had a proof when n is prime. There is a shorter proof if you look at the binomial expansion of (1+1)^p.
There are infinitely many numbers in this sequence. (Because if n is even, then 2^n*12-n-2 is even, so 2^(2^n*12-n-2) is 4 (mod 6). Define x so that this is 6*x + 4, then dividing by 3 gives 2*x + (4/3), and the floor is an odd number.) - Jinyuan Wang, Oct 13 2018
MATHEMATICA
Select[ Range[300], OddQ[ Floor[2^# / # ]] & ]
PROG
(PARI) for(n=1, 1000, if((-1)^(floor(2^n/n))==-1+isprime(n), print1(n, ", ")))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
R. K. Guy, Jun 12 2002
EXTENSIONS
More terms from several correspondents, Jun 12, 2002
STATUS
approved