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Numbers k such that d(r,k) != d(s,k), where d(x,k) = k-th binary digit of x, r = {e}, s = {1/e}, and { } = fractional part.
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%I #11 Sep 26 2014 21:13:32

%S 1,2,3,5,8,9,10,13,14,21,24,25,27,28,29,31,35,36,37,40,42,44,48,49,50,

%T 51,52,54,56,58,59,60,70,72,73,75,77,78,79,80,81,84,85,86,87,88,95,

%U 101,102,105,107,117,119,121,122,125,127,129,131,132,133,138

%N Numbers k such that d(r,k) != d(s,k), 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 the sequences A247546 and A247547.

%H Clark Kimberling, <a href="/A247547/b247547.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 = 200; 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 t = Table[If[u[[n]] == v[[n]], 1, 0], {n, 1, z}];

%t Flatten[Position[t, 1]] (* A247546 *)

%t Flatten[Position[t, 0]] (* A247547 *)

%Y Cf. A247546.

%K nonn,easy,base

%O 1,2

%A _Clark Kimberling_, Sep 21 2014