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 A294863 Solution of the complementary equation a(n) = a(n-2) + b(n-2) + 3, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences. 3

%I #4 Nov 18 2017 09:05:58

%S 1,2,7,9,15,18,26,31,40,46,56,63,75,83,97,106,121,131,147,158,175,188,

%T 206,220,239,255,275,292,313,331,353,372,395,416,440,462,487,510,537,

%U 561,589,614,643,669,699,726,757,786,818,848,881,912,946,979,1014

%N Solution of the complementary equation a(n) = a(n-2) + b(n-2) + 3, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

%C The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A294860 for a guide to related sequences.

%H Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.html">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.

%e a(0) = 1, a(1) = 2, b(0) = 3

%e b(1) = 4 (least "new number")

%e a(2) = a(0) + b(0) + 3 = 7

%e Complement: (b(n)) = (3, 4, 5, 6, 8, 10, 11, 12, 13, 14, 16, ...)

%t mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;

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

%t a[n_] := a[n] = a[n - 2] + b[n - 2] + 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, 18}] (* A294863 *)

%t Table[b[n], {n, 0, 10}]

%Y Cf. A294860, A294864.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Nov 16 2017

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