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A294560 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + 2*b(n-1) + 2*b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A294560 #4 Nov 15 2017 17:38:35
%S A294560 1,2,17,37,76,139,245,418,701,1161,1908,3119,5081,8258,13401,21727,
%T A294560 35202,57007,92291,149384,241765,391243,633106,1024451,1657663,
%U A294560 2682224,4340001,7022343,11362466,18384935,29747531,48132600,77880269,126013011,203893428
%N A294560 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + 2*b(n-1) + 2*b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A294560 The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A294532 for a guide to related sequences. Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).
%H A294560 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 A294560 a(0) = 1, a(1) = 2, b(0) = 3, so that
%e A294560 b(1) = 4 (least "new number")
%e A294560 a(2)  = a(0) + a(1) + 2*b(0) + 2*b(1) = 17
%e A294560 Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, ...)
%t A294560 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A294560 a[0] = 1; a[1] = 3; b[0] = 2;
%t A294560 a[n_] := a[n] = a[n - 1] + a[n - 2] + 2 b[n - 1] + 2 b[n - 2];
%t A294560 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A294560 Table[a[n], {n, 0, 40}]  (* A294560 *)
%t A294560 Table[b[n], {n, 0, 10}]
%Y A294560 Cf. A001622, A294532.
%K A294560 nonn,easy
%O A294560 0,2
%A A294560 _Clark Kimberling_, Nov 15 2017