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%I #14 Sep 26 2014 21:13:14
%S 4,6,7,11,16,18,20,26,33,39,41,43,45,46,53,55,61,63,65,67,68,69,71,74,
%T 76,82,89,97,100,106,108,110,113,114,115,116,120,126,128,130,135,137,
%U 150,157,159,163,164,171,174,178,180,183,188,191,195,206,209,212
%N Numbers k such that d(r,k) = 1 and d(s,k) = 1, where d(x,k) = k-th binary digit of x, r = {e}, s = {1/e}, and { } = fractional part.
%C Every positive integer lies in exactly one of these: A247542, A247543, A247544, A247545.
%H Clark Kimberling, <a href="/A247545/b247545.txt">Table of n, a(n) for n = 1..1000</a>
%e {e/1} has binary digits 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, ...
%e {1/e} has binary digits 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, ...
%e so that a(1) = 4 and a(2) = 6.
%t z = 400; r = FractionalPart[E]; s = FractionalPart[1/E];
%t u = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[r, 2, z]]
%t v = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[s, 2, z]]
%t t1 = Table[If[u[[n]] == 0 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t t2 = Table[If[u[[n]] == 0 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t t3 = Table[If[u[[n]] == 1 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t t4 = Table[If[u[[n]] == 1 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t Flatten[Position[t1, 1]] (* A247542 *)
%t Flatten[Position[t2, 1]] (* A247543 *)
%t Flatten[Position[t3, 1]] (* A247544 *)
%t Flatten[Position[t4, 1]] (* A247545 *)
%Y Cf. A247542, A247543, A247544.
%K nonn,easy,base
%O 1,1
%A _Clark Kimberling_, Sep 21 2014
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