%I
%S 2,5,11,13,19,29,41,47,53,59,61,67,97,109,149,167,173,197,227,233,239,
%T 251,271,283,313,331,349,373,409,433,439,499,509,521,557,563,593,641,
%U 677,743,761,773,797,827,857,887,911,941,953,971,977,983,1013,1019,1021
%N Numbers n such that n and its 2's complement are both primes. In other words, n and 2^k  n (where k is the smallest power of 2 such that 2^k > n) are primes.
%C In the first 672509 primes, 64894 of them (about 9.65%) are 2'scomplement primes.
%H Charles R Greathouse IV, <a href="/A086081/b086081.txt">Table of n, a(n) for n = 1..10000</a>
%F If isPrime(p) And isPrime(2^(floor(Log(p, 2)) + 1)  p) then sequence.add(p)
%F If A(x) is the counting function of the terms a(n) <= x, then A(x) = O(xloglogx/(logx)^2) [From _Vladimir Shevelev_, Dec 04 2008]
%e a(5) = 19 because 19 is prime and (2^5  19) = (32  19) = 13 which is prime.
%e a(74) = 1777 because 1777 is prime and (2^11  1777) = (2048  1777) = 271 which is prime.
%t Join[{2}, Select[Prime[Range[250]], PrimeQ[BitXor[#, 2^Ceiling[Log[2, #]]  1] + 1] &]] (* _Alonso del Arte_, Feb 12 2013 *)
%o (PARI) select(n>isprime((2<<(log(n+.5)\log(2)))n), primes(100)) \\ _Charles R Greathouse IV_, Feb 13 2013
%Y Cf. A068811.
%K nonn,easy
%O 1,1
%A Chuck Seggelin (barkeep(AT)plastereddragon.com), Jul 08 2003
