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a(n) = (1/2)*A293077(n).
4

%I #8 Oct 02 2017 07:58:16

%S 1,2,3,5,8,13,22,37,63,107,182,310,529,903,1541,2630,4489,7663,13081,

%T 22330,38119,65073,111086,189635,323727,552636,943408,1610498,2749292,

%U 4693335,8012024,13677380,23348748,39858806,68043238

%N a(n) = (1/2)*A293077(n).

%t z = 10; (* number of iterations *)

%t s = {0, 0}; u[0] = StringJoin[Map[ToString, s]]; w[0] = u[0];

%t u[n_] := u[n] = StringReplace[w[n - 1], {"00" -> "0010", "01" -> "010", "10" -> "010"}];

%t w[n_] := w[n] = If[OddQ[StringLength[u[n]]], StringDrop[u[n], -1], u[n]];

%t TableForm[Table[w[n], {n, 0, 8}]]

%t st = ToCharacterCode[w[z]] - 48 (* A293076 *)

%t p0 = Flatten[Position[st, 0]] (* A289036 *)

%t p1 = Flatten[Position[st, 1]] (* A289037 *)

%t v = Table[StringLength[w[n]], {n, 0, 34}] (* A293077 *)

%t v/2 (* A293078 *)

%Y Cf. A289035, A293076, A293077.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Sep 30 2017