%I #9 Jan 05 2025 19:51:41
%S 4,5,6,7,8,9,11,13,15,17,18,20,21,23,24,25,27,28,29,31,32,34,35,36,38,
%T 40,41,42,44,46,47,49,50,52,54,55,57,58,60,62,63,64,66,68,69,71,72,74,
%U 75,77,78,80,81,83,84,86,87,89,90,92,93,94,96,98,99,100
%N Solution b( ) of the complementary equation a(n) = b(n-1) + b(n-3), where a(0) = 1, a(1) = 2, a(2) = 3; 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://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] = 2; a[2] = 3; b[0] = 4; 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}] (* A022427 *)
%t Table[b[n], {n, 0, 100}] (* A299536 *)
%Y Cf. A022424, A022427 (complement).
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Feb 24 2018