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 A247543 Numbers k such that d(r,k) = 0 and d(s,k) = 1, where d(x,k) = k-th binary digit of x, r = {e}, s = {1/e}, and { } = fractional part. 4

%I

%S 2,5,13,14,21,25,28,29,35,36,40,44,49,50,52,54,56,60,73,77,78,79,81,

%T 86,87,88,95,117,125,129,132,133,140,141,152,153,155,161,166,168,170,

%U 173,179,182,184,187,192,196,203,211,217,218,220,225,227,230,238

%N Numbers k such that d(r,k) = 0 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="/A247543/b247543.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) = 2 and a(2) = 5.

%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, A247544, A247545.

%K nonn,easy,base

%O 1,1

%A _Clark Kimberling_, Sep 21 2014

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