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A299494 Solution a( ) of the complementary equation a(n) = b(n-1) + b(n-2) + b(n-3), where a(0) = 2, a(1) = 4, a(2) = 6; see Comments.

%I A299494 #5 Jan 05 2025 19:51:41
%S A299494 2,4,6,9,15,20,25,29,33,36,39,43,47,51,54,58,62,66,69,73,77,81,85,89,
%T A299494 93,97,101,106,110,115,119,123,127,131,135,139,143,147,151,155,160,
%U A299494 164,168,172,176,180,184,188,192,196,200,205,209,213,217,221,225,229
%N A299494 Solution a( ) of the complementary equation a(n) = b(n-1) + b(n-2) + b(n-3), where a(0) = 2, a(1) = 4, a(2) = 6; see Comments.
%C A299494 From the Bode-Harborth-Kimberling link:
%C A299494 a(n) = b(n-1) + b(n-2) + b(n-3) for n > 2;
%C A299494 b(0) = least positive integer not in {a(0),a(1),a(2)};
%C A299494 b(n) = least positive integer not in {a(0),...,a(n),b(0),...,b(n-1)} for n > 1.
%C A299494 Note that (b(n)) is strictly increasing and is the complement of (a(n)).
%C A299494 See A022424 for a guide to related sequences.
%H A299494 J-P. Bode, H. Harborth, C. Kimberling, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Papers1/45-3/bode.pdf">Complementary Fibonacci sequences</a>, Fibonacci Quarterly 45 (2007), 254-264.
%t A299494 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A299494 a[0] = 2; a[1] = 4; a[2] = 6; b[0] = 1; b[1] = 3; b[2] = 5;
%t A299494 a[n_] := a[n] = b[n - 1] + b[n - 2] + b[n - 3];
%t A299494 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A299494 Table[a[n], {n, 0, 100}]    (* A299494 *)
%t A299494 Table[b[n], {n, 0, 100}]    (* A299495 *)
%Y A299494 Cf. A022424, A299495.
%K A299494 nonn,easy
%O A299494 0,1
%A A299494 _Clark Kimberling_, Feb 21 2018