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%I #10 Jan 05 2025 19:51:41
%S 2,3,6,7,8,10,11,12,14,16,17,19,20,22,24,25,27,28,29,31,32,34,35,37,
%T 38,40,41,43,44,45,47,48,50,51,53,54,56,58,59,61,62,64,65,67,68,70,71,
%U 73,74,76,77,79,80,82,83,85,86,88,90,91,93,94,96,97,99,100
%N Solution b( ) of the complementary equation a(n) = b(n-1) + b(n-2), where a(0) = 1, 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="/A299407/b299407.txt">Table of n, a(n) for n = 0..2000</a>
%H 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 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t a[0] = 1; a[1] = 4; b[0] = 2; b[1] = 3;
%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}] (* A022425 *)
%t Table[b[n], {n, 0, 100}] (* A299407 *)
%Y Cf. A022424, A022425.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Feb 14 2018