%I #4 Feb 26 2018 09:11:35
%S 1,3,5,7,8,9,11,12,14,15,17,18,20,22,23,24,26,28,29,30,32,34,35,36,38,
%T 39,41,42,44,45,47,48,50,51,53,54,56,57,59,60,62,63,65,66,68,69,71,72,
%U 74,76,77,78,80,82,83,84,86,88,89,90,92,94,95,96,98,100
%N Solution b( ) of the complementary equation a(n) = b(n-1) + b(n-3), where a(0) = 2, a(1) = 4, a(2) = 6; 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] = 4; a[2] = 6; b[0] = 1; b[1] = 3;
%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}] (* A299541 *)
%t Table[b[n], {n, 0, 100}] (* A299542 *)
%Y Cf. A022424, A299541.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Feb 25 2018
|