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A299497 Solution b( ) of the complementary equation a(n) = b(n-1) + b(n-2) + b(n-3), where a(0) = 3, a(1) = 4, a(2) = 5; see Comments.

%I A299497 #5 Jan 05 2025 19:51:41
%S A299497 1,2,3,7,8,9,10,11,13,14,15,16,17,19,20,21,22,23,25,26,28,29,31,32,33,
%T A299497 35,36,37,39,40,41,43,44,46,47,49,50,51,53,54,55,57,58,59,61,62,64,65,
%U A299497 67,68,69,71,72,73,75,76,77,78,80,81,82,84,85,86,87,89
%N A299497 Solution b( ) of the complementary equation a(n) = b(n-1) + b(n-2) + b(n-3), where a(0) = 3, a(1) = 4, a(2) = 5; see Comments.
%C A299497 From the Bode-Harborth-Kimberling link:
%C A299497 a(n) = b(n-1) + b(n-2) + b(n-3) for n > 2;
%C A299497 b(0) = least positive integer not in {a(0),a(1),a(2)};
%C A299497 b(n) = least positive integer not in {a(0),...,a(n),b(0),...,b(n-1)} for n > 1.
%C A299497 Note that (b(n)) is strictly increasing and is the complement of (a(n)).
%C A299497 See A022424 for a guide to related sequences.
%H A299497 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 A299497 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A299497 a[0] = 3; a[1] = 4; a[2] = 5; b[0] = 1; b[1] = 2; b[2] = 6;
%t A299497 a[n_] := a[n] = b[n - 1] + b[n - 2] + b[n - 3];
%t A299497 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A299497 Table[a[n], {n, 0, 100}]    (* A299496 *)
%t A299497 Table[b[n], {n, 0, 100}]    (* A299497 *)
%Y A299497 Cf. A022424, A299496.
%K A299497 nonn,easy
%O A299497 0,2
%A A299497 _Clark Kimberling_, Feb 21 2018