A299487 - OEIS

%I #12 Mar 05 2018 13:47:21

%S 4,5,6,7,8,9,10,11,12,13,14,16,17,19,20,22,23,25,26,28,29,31,32,34,35,

%T 37,38,40,41,42,44,45,46,48,49,50,51,53,54,55,57,58,59,60,62,63,64,66,

%U 67,68,69,71,72,73,75,76,77,78,80,81,82,84,85,86,87,89

%N Solution b( ) of the complementary equation a(n) = b(n-1) + b(n-2) + b(n-3), where a(0) = 1, a(1) = 2, a(2) = 3; see Comments.

%C a(n) = b(n-1) + b(n-2) + b(n-3) for n > 2;

%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 Clark Kimberling, <a href="/A299487/b299487.txt">Table of n, a(n) for n = 0..1000</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] = 1; a[1] = 2; a[2] = 3; b[0] = 4; b[1] = 5; b[2] = 6;

%t a[n_] := a[n] = b[n - 1] + b[n - 2] + b[n - 3];

%t b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];

%t u = Table[a[n], {n, 0, 100}]    (* A299486 *)

%t v = Table[b[n], {n, 0, 100}]    (* A299487 *)

%Y Cf. A022424, A299486.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Feb 16 2018