|
%I #14 Sep 26 2014 21:13:06
%S 1,3,8,9,10,24,27,31,37,42,48,51,58,59,70,72,75,80,84,85,101,102,105,
%T 107,119,121,122,127,131,138,139,142,143,144,148,151,158,160,165,169,
%U 172,177,181,186,190,193,198,199,200,201,210,222,226,228,233,236
%N Numbers k such that d(r,k) = 1 and d(s,k) = 0, 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="/A247544/b247544.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, A247543, A247545.
%K nonn,easy,base
%O 1,2
%A _Clark Kimberling_, Sep 21 2014
|
