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

%I #16 Sep 26 2014 21:12:47

%S 12,15,17,19,22,23,30,32,34,38,47,57,62,64,66,83,90,91,92,93,94,96,98,

%T 99,103,104,109,111,112,118,123,124,134,136,145,146,147,149,154,156,

%U 162,167,175,176,185,189,194,197,202,204,205,207,208,214,215,219

%N Numbers k such that d(r,k) = 0 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="/A247542/b247542.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) = 12 and a(2) = 15.

%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. A247543, A247544, A247545.

%K nonn,easy,base

%O 1,1

%A _Clark Kimberling_, Sep 21 2014