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%I #9 Mar 05 2018 13:45:06
%S 1,2,5,6,8,9,10,12,13,15,16,18,20,21,23,24,26,27,29,30,32,33,35,36,37,
%T 39,40,42,43,45,46,48,49,51,52,54,55,57,58,60,61,63,64,66,67,69,70,72,
%U 74,75,77,78,80,81,83,84,86,87,89,90,92,93,95,96,98,99
%N Solution b( ) of the complementary equation a(n) = b(n-1) + b(n-2), where a(0) = 3, a(1) = 4; see Comments.
%C a(n) = b(n-1) + b(n-2) for n > 2;
%C b(0) = least positive integer not in {a(0),a(1)};
%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 Clark Kimberling, <a href="/A299417/b299417.txt">Table of n, a(n) for n = 0..2000</a>
%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] = 3; a[1] = 4; b[0] = 1; b[1] = 2;
%t a[n_] := a[n] = b[n - 1] + b[n - 2];
%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}] (* A299416 *)
%t Table[b[n], {n, 0, 100}] (* A299417 *)
%Y Cf. A022424, A299416.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Feb 15 2018
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