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A295367 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)*b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A295367 #6 Nov 21 2017 21:34:24
%S A295367 1,2,15,37,82,161,299,532,921,1563,2616,4335,7133,11692,19097,31095,
%T A295367 50534,82009,132963,215434,348903,564889,914392,1479931,2395025,
%U A295367 3875712,6271549,10148131,16420610,26569733,42991399,69562254,112554843
%N A295367 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)*b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A295367 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 A295367 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 A295367 a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).
%e A295367 a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that
%e A295367 b(2) = 5 (least "new number")
%e A295367 a(2) = a(1) + a(0) + b(1)*b(0) = 15
%e A295367 Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 17, ...)
%t A295367 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A295367 a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
%t A295367 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1]*b[n - 2];
%t A295367 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A295367 z = 32; u = Table[a[n], {n, 0, z}]   (* A295367 *)
%t A295367 v = Table[b[n], {n, 0, 10}]  (* complement *)
%Y A295367 Cf. A001622, A295357.
%K A295367 nonn,easy
%O A295367 0,2
%A A295367 _Clark Kimberling_, Nov 21 2017