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Solution b( ) of the complementary equation a(n) = b(n-1) + b(n-2), where a(0) = 3, a(1) = 5; see Comments.
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%I #14 Jan 05 2025 19:51:41

%S 1,2,4,7,8,9,10,12,13,14,16,18,20,21,23,24,26,28,29,31,32,33,35,36,37,

%T 39,40,42,43,45,46,48,49,51,52,53,55,56,58,59,61,62,64,66,67,69,70,72,

%U 74,75,77,78,80,81,83,84,86,87,89,90,92,93,95,96,98,99

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

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

%C b(0) = least positive integer not in {a(0),a(1)};

%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="/A299419/b299419.txt">Table of n, a(n) for n = 0..2000</a>

%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] = 3; a[1] = 5; b[0] = 1; b[1] = 2;

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

%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}] (* A299418 *)

%t Table[b[n], {n, 0, 100}] (* A299419 *)

%Y Cf. A022424, A299418.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Feb 16 2018