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%I #6 Aug 27 2021 21:14:03
%S 2,4,6,15,33,68,130,237,417,716,1208,2013,3326,5461,8927,14547,23653,
%T 38400,62275,100920,163464,264678,428462,693487,1122324,1816215,
%U 2938973,4755653,7695123,12451307,20146996,32598905,52746540,85346122,138093378,223440256
%N Solution of the complementary equation a(n) = 2*a(n-1) - a(n-3) + b(n-1), where a(0) = 2, a(1) = 4, a(2) = 6, b(0) = 1, b(1) = 3, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
%C The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295613 for a guide to related sequences.
%C a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).
%H 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 a(0) = 2, a(1) = 4, a(2) = 6, b(0) = 1, b(1) = 3, b(2) = 5, so that
%e b(3) = 7 (least "new number")
%e a(3) = 2*a(2) - a(0) + b(2) = 15
%e Complement: (b(n)) = (1, 3, 5, 7, 8, 9, 10, 11, 12, 13, 14, 16, ...)
%t mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t a[0] = 2; a[1] = 4; a[2] = 6; b[0] = 1; b[1] = 3; b[2] = 5;
%t a[n_] := a[n] = 2 a[n - 1] - a[n - 3] + b[n - 1];
%t b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t Table[a[n], {n, 0, 30}] (* A295615 *)
%t Table[b[n], {n, 0, 20}] (* complement *)
%Y Cf. A001622, A000045, A295613.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Nov 25 2017
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