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A295361 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2) - b(n-3), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A295361 #4 Nov 21 2017 21:33:48
%S A295361 1,3,5,20,40,76,134,230,386,640,1052,1720,2802,4554,7390,11980,19408,
%T A295361 31429,50882,82357,133287,215694,349033,564781,913870,1478709,2392639,
%U A295361 3871410,6264113,10135589,16399770,26535429,42935271,69470774,112406121
%N A295361 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2) - b(n-3), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A295361 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 A295361 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 A295361 a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).
%e A295361 a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, so that
%e A295361 b(3) = 7 (least "new number")
%e A295361 a(3) = a(1) + a(0) + b(2) + 2*b(1) - b(0) = 20
%e A295361 Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...)
%t A295361 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A295361 a[0] = 1; a[1] = 3; a[2] = 5; b[0] = 2; b[1] = 4; b[2] = 6;
%t A295361 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + 2*b[n - 2] - b[n - 3];
%t A295361 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A295361 z = 32; u = Table[a[n], {n, 0, z}]   (* A295361 *)
%t A295361 v = Table[b[n], {n, 0, 10}]  (* complement *)
%Y A295361 Cf. A001622, A295357.
%K A295361 nonn,easy
%O A295361 0,2
%A A295361 _Clark Kimberling_, Nov 21 2017