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A295136 Solution of the complementary equation a(n) = 3*a(n-1) + b(n-1) - 3, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A295136 #4 Nov 19 2017 10:44:11
%S A295136 1,2,7,23,72,221,669,2014,6050,18159,54487,163472,490428,1471297,
%T A295136 4413905,13241730,39725206,119175635,357526923
%N A295136 Solution of the complementary equation a(n) = 3*a(n-1) + b(n-1) - 3, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A295136 The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295053 for a guide to related sequences.
%H A295136 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 A295136 a(0) = 1, a(1) = 2, b(0) = 3
%e A295136 a(2) =3*a(1) + b(1) - 3 = 7
%e A295136 Complement: (b(n)) = (3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, ... )
%t A295136 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A295136 a[0] = 1; a[1] = 2; b[0] = 3;
%t A295136 a[n_] := a[n] = 3 a[n - 1] + b[n - 1] - 3;
%t A295136 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A295136 Table[a[n], {n, 0, 18}]  (* A295136 *)
%t A295136 Table[b[n], {n, 0, 10}]
%Y A295136 Cf. A295053.
%K A295136 nonn,easy
%O A295136 0,2
%A A295136 _Clark Kimberling_, Nov 19 2017