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%I #4 Feb 20 2018 21:11:23
%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://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
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