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A295366 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - b(n-3), where a(0) = 1, a(1) = 2, a[2] = 3, b(0) = 4, b(1) = 5, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A295366 #4 Nov 27 2017 12:49:47
%S A295366 1,2,3,12,23,44,77,132,221,367,604,987,1608,2613,4240,6873,11134,
%T A295366 18029,29186,47240,76453,123720,200201,323950,524181,848162,1372375,
%U A295366 2220570,3592979,5813584,9406599,15220220,24626857
%N A295366 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - b(n-3), where a(0) = 1, a(1) = 2, a[2] = 3,  b(0) = 4, b(1) = 5, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A295366 The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.
%H A295366 Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.html">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.
%F A295366 a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).
%e A295366 a(0) = 1, a(1) = 2, a[2] = 3, b(0) = 4, b(1) = 5, b(2) = 6, so that
%e A295366 b(2) = 7 (least "new number")
%e A295366 a(3) = a(2) + a(1) + b(2) + b (1) - b(0) = 12
%e A295366 Complement: (b(n)) = (4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, ...)
%t A295366 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A295366 a[0] = 1; a[1] = 2; a[2] = 3; b[0] = 4; b[1] = 5; b[2]=6;
%t A295366 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] - b[n - 3];
%t A295366 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A295366 z = 32; u = Table[a[n], {n, 0, z}]   (* A295366 *)
%t A295366 v = Table[b[n], {n, 0, 10}]  (* complement *)
%Y A295366 Cf. A001622, A295357.
%K A295366 nonn,easy
%O A295366 0,2
%A A295366 _Clark Kimberling_, Nov 22 2017