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%I #9 Mar 05 2018 13:14:18
%S 1,5,6,8,9,10,11,12,14,16,17,19,21,23,24,26,27,29,30,32,33,34,36,37,
%T 39,40,41,43,44,46,47,48,50,52,53,54,56,58,59,60,62,64,65,67,68,70,72,
%U 73,74,76,78,79,81,82,84,86,87,88,90,92,93,95,96,98,99,101
%N Solution b( ) of the complementary equation a(n) = b(n-1) + b(n-3), where a(0) = 2, a(1) = 3, a(2) = 4; see Comments.
%C From the Bode-Harborth-Kimberling link:
%C a(n) = b(n-1) + b(n-3) for n > 3;
%C b(0) = least positive integer not in {a(0),a(1),a(2)};
%C b(n) = least positive integer not in {a(0),...,a(n),b(0),...,b(n-1)} for n > 1.
%C Note that (b(n)) is strictly increasing and is the complement of (a(n)).
%C See A022424 for a guide to related sequences.
%H J-P. Bode, H. Harborth, C. Kimberling, <a href="https://www.fq.math.ca/Papers1/45-3/bode.pdf">Complementary Fibonacci sequences</a>, Fibonacci Quarterly 45 (2007), 254-264.
%t mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t a[0] = 2; a[1] = 3; a[2] = 4; b[0] = 1; b[1] = 5;
%t a[n_] := a[n] = b[n - 1] + b[n - 3];
%t b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t Table[a[n], {n, 0, 100}] (* A022437 *)
%t Table[b[n], {n, 0, 100}] (* A299538 *)
%Y Cf. A022424, A299537.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Feb 25 2018
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