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Solution b( ) of the complementary equation a(n) = b(n-1) + b(n-2), where a(0) = 2, a(1) = 3; see Comments.
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%I #9 Mar 05 2018 13:44:17

%S 1,4,6,7,8,9,11,12,14,16,18,19,21,22,24,25,27,28,29,31,32,33,35,36,38,

%T 39,41,42,44,45,47,48,50,51,53,54,56,58,59,61,62,64,66,67,69,70,72,73,

%U 75,76,78,79,81,82,84,85,87,88,90,91,93,94,96,97,99,100

%N Solution b( ) of the complementary equation a(n) = b(n-1) + b(n-2), where a(0) = 2, a(1) = 3; 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="/A299411/b299411.txt">Table of n, a(n) for n = 0..2000</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] = 2; a[1] = 3; b[0] = 1; b[1] = 4;

%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}] (* A022426 *)

%t Table[b[n], {n, 0, 100}] (* A299411 *)

%Y Cf. A022424, A022426.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Feb 14 2018