A299492 - OEIS

%I #5 Jan 05 2025 19:51:41

%S 2,4,5,10,16,21,24,28,32,36,39,42,46,50,54,57,61,65,70,74,78,82,86,90,

%T 94,98,102,106,110,115,119,124,128,132,136,140,144,148,152,156,160,

%U 164,169,173,177,181,185,189,193,197,201,204,208,212,216,220,224

%N Solution a( ) of the complementary equation a(n) = b(n-1) + b(n-2) + b(n-3), where a(0) = 2, a(1) = 4, a(2) = 5; see Comments.

%C From the Bode-Harborth-Kimberling link:

%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 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] = 2; a[1] = 4; a[2] = 5; b[0] = 1; b[1] = 3; 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 Table[a[n], {n, 0, 100}]    (* A299492 *)

%t Table[b[n], {n, 0, 100}]    (* A299493 *)

%Y Cf. A022424, A299493.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Feb 20 2018