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A295358 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) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A295358 #4 Nov 21 2017 21:33:25
%S A295358 1,3,5,16,30,55,95,161,268,442,724,1181,1921,3120,5061,8201,13283,
%T A295358 21506,34812,56342,91179,147547,238753,386328,625110,1011468,1636610,
%U A295358 2648112,4284756,6932903,11217695,18150635,29368368,47519042,76887450
%N A295358 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) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A295358 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 A295358 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 A295358 a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).
%e A295358 a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, so that
%e A295358 b(3) = 7 (least "new number")
%e A295358 a(3) = a(1) + a(0) + b(2) + b(1) - b(0) = 16
%e A295358 Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...)
%t A295358 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A295358 a[0] = 1; a[1] = 3; a[2] = 5; b[0] = 2; b[1] = 4; b[2] = 6;
%t A295358 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] - b[n - 3];
%t A295358 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A295358 z = 32; u = Table[a[n], {n, 0, z}]   (* A295358 *)
%t A295358 v = Table[b[n], {n, 0, 10}]  (* complement *)
%Y A295358 Cf. A001622, A295357.
%K A295358 nonn,easy
%O A295358 0,2
%A A295358 _Clark Kimberling_, Nov 21 2017