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A295620 Solution of the complementary equation a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4) + b(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A295620 #4 Nov 27 2017 12:50:42
%S A295620 1,2,3,4,12,20,49,85,177,304,578,979,1765,2953,5150,8538,14570,23997,
%T A295620 40352,66149,110094,179867,297172,484313,795934,1294823,2119684,
%U A295620 3443689,5621258,9123343,14860404,24100573,39192618,63526879,103182816,167177109,271286602
%N A295620 Solution of the complementary equation a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4) + b(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A295620 The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.
%H A295620 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 A295620 a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, so that
%e A295620 b(4) = 9 (least "new number")
%e A295620 a(4) = a(3) + 3*a(2) -2*a(1) - 2*a(0) + b(0) = 12
%e A295620 Complement: (b(n)) = (5, 6, 7, 8, 9, 10, 11, 13, 14, 15, ...)
%t A295620 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A295620 a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4;
%t A295620 b[0] = 5; b[1] = 6; b[2] = 7; b[3] = 8;
%t A295620 a[n_] := a[n] = a[n - 1] + 3*a[n - 2] - 2*a[n - 3] - 2 a[n - 4] + b[n - 4];
%t A295620 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A295620 z = 36;  Table[a[n], {n, 0, z}]   (* A295620 *)
%t A295620 Table[b[n], {n, 0, 20}]  (*complement *)
%Y A295620 Cf. A001622, A000045, A295619, A295621.
%K A295620 nonn,easy
%O A295620 0,2
%A A295620 _Clark Kimberling_, Nov 25 2017