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A299532 Solution b( ) of the complementary equation a(n) = 2*b(n-1) + b(n-2), where a(0) = 1, a(1) = 2; see Comments.

%I A299532 #5 Jan 05 2025 19:51:41
%S A299532 3,4,5,6,7,8,9,10,12,13,15,16,18,19,21,22,24,25,27,28,30,31,32,33,35,
%T A299532 36,37,39,40,41,42,44,45,46,48,49,50,51,53,54,55,57,58,59,60,62,63,64,
%U A299532 66,67,68,69,71,72,73,75,76,77,78,80,81,82,84,85,86,87
%N A299532 Solution b( ) of the complementary equation a(n) = 2*b(n-1) + b(n-2), where a(0) = 1, a(1) = 2; see Comments.
%C A299532 From the Bode-Harborth-Kimberling link:
%C A299532 a(n) = 2*b(n-1) + b(n-2) for n > 1;
%C A299532 b(0) = least positive integer not in {a(0),a(1)};
%C A299532 b(n) = least positive integer not in {a(0),...,a(n),b(0),...,b(n-1)} for n > 1.
%C A299532 Note that (b(n)) is strictly increasing and is the complement of (a(n)).
%C A299532 See A022424 for a guide to related sequences.
%H A299532 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 A299532 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A299532 a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
%t A299532 a[n_] := a[n] = 2*b[n - 1] + b[n - 2];
%t A299532 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A299532 Table[a[n], {n, 0, 100}]    (* A299531 *)
%t A299532 Table[b[n], {n, 0, 100}]    (* A299532 *)
%Y A299532 Cf. A022424, A299531.
%K A299532 nonn,easy
%O A299532 0,1
%A A299532 _Clark Kimberling_, Feb 21 2018