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A299545 Solution a( ) of the complementary equation a(n) = b(n-1) + 2*b(n-2) - b(n-3), where a(0) = 1, a(1) = 2, a(2) = 3; see Comments.

%I A299545 #5 Jan 05 2025 19:51:41
%S A299545 1,2,3,12,14,16,18,20,22,25,30,34,38,42,46,49,51,55,56,58,61,65,66,69,
%T A299545 73,74,77,81,82,85,89,90,93,97,99,104,107,108,112,119,121,123,127,128,
%U A299545 132,138,139,143,144,148,154,155,159,160,164,170,171,175,176
%N A299545 Solution a( ) of the complementary equation a(n) = b(n-1) + 2*b(n-2) - b(n-3), where a(0) = 1, a(1) = 2, a(2) = 3; see Comments.
%C A299545 From the Bode-Harborth-Kimberling link:
%C A299545 a(n) = b(n-1) + 2*b(n-2) - b(n-3) for n > 3;
%C A299545 b(0) = least positive integer not in {a(0),a(1),a(2)};
%C A299545 b(n) = least positive integer not in {a(0),...,a(n),b(0),...,b(n-1)} for n > 1.
%C A299545 Note that (b(n)) is strictly increasing and is the complement of (a(n)).
%C A299545 See A022424 for a guide to related sequences.
%H A299545 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 A299545 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A299545 a[0] = 1; a[1] = 2; a[2] = 3; b[0] = 4; b[1] = 5;
%t A299545 a[n_] := a[n] = b[n - 1] + 2 b[n - 2] - b[n - 3];
%t A299545 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A299545 Table[a[n], {n, 0, 100}]    (* A299545 *)
%t A299545 Table[b[n], {n, 0, 100}]    (* A299546 *)
%Y A299545 Cf. A022424, A299546.
%K A299545 nonn,easy
%O A299545 0,2
%A A299545 _Clark Kimberling_, Mar 01 2018