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A299418 Solution a( ) of the complementary equation a(n) = b(n-1) + b(n-2), where a(0) = 3, a(1) = 5; see Comments.

%I A299418 #10 Jan 05 2025 19:51:41
%S A299418 3,5,3,6,11,15,17,19,22,25,27,30,34,38,41,44,47,50,54,57,60,63,65,68,
%T A299418 71,73,76,79,82,85,88,91,94,97,100,103,105,108,111,114,117,120,123,
%U A299418 126,130,133,136,139,142,146,149,152,155,158,161,164,167,170,173
%N A299418 Solution a( ) of the complementary equation a(n) = b(n-1) + b(n-2), where a(0) = 3, a(1) = 5; see Comments.
%C A299418 a(n) = b(n-1) + b(n-2) for n > 2;
%C A299418 b(0) = least positive integer not in {a(0),a(1)};
%C A299418 b(n) = least positive integer not in {a(0),...,a(n),b(0),...b(n-1)} for n > 1.
%C A299418 Note that (b(n)) is strictly increasing and is the complement of (a(n)).
%C A299418 See A022424 for a guide to related sequences.
%H A299418 Clark Kimberling, <a href="/A299418/b299418.txt">Table of n, a(n) for n = 0..2000</a>
%H A299418 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 A299418 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A299418 a[0] = 3; a[1] = 5; b[0] = 1; b[1] = 2;
%t A299418 a[n_] := a[n] = b[n - 1] + b[n - 2];
%t A299418 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A299418 Table[a[n], {n, 0, 100}]  (* A299418 *)
%t A299418 Table[b[n], {n, 0, 100}]  (* A299419 *)
%Y A299418 Cf. A022424, A299419.
%K A299418 nonn,easy
%O A299418 0,1
%A A299418 _Clark Kimberling_, Feb 15 2018