%I #23 Apr 30 2016 19:47:51
%S 1,2,4,8,16,32,36,38,64,128,256,512,1024,2048,2056,2080,2088,2090,
%T 4096,8192,16384,16896,16900,16902,16928,18944,18952,18954,18988,
%U 32768,65536,131072,131328,131332,131334,131360,133376,133384,133386,133420,148224,148256,148258,150284
%N Numbers n such that the sum of the inverse of the exponents in the binary expansion of 2n is the inverse of an integer.
%C That is, numbers such that A116416(n) is equal to 1.
%C The powers of 2 (A000079) form a subsequence.
%H Peter Kagey, <a href="/A272034/b272034.txt">Table of n, a(n) for n = 1..354</a>
%e For n=36, 38_10=100100_2, and 1/3 + 1/6 = 1/2, the inverse of an integer.
%t Select[Range[2^18], IntegerQ[1/Total[1/# & /@ Flatten@ Position[Reverse@ IntegerDigits[#, 2], 1]]] &] (* _Michael De Vlieger_, Apr 18 2016 *)
%o (PARI) isok(n) = {my(b = Vecrev(binary(n))); numerator(sum(k=1, #b, b[k]/k)) == 1;}
%Y Cf. A000079, A116416, A116417, A272035, A272036, A272081.
%K nonn
%O 1,2
%A _Michel Marcus_, Apr 18 2016