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%I #4 Nov 21 2017 21:34:01
%S 1,3,12,35,77,154,287,513,890,1513,2546,4241,6997,11478,18747,30531,
%T 49620,80531,130571,211564,342641,554757,897998,1453405,2352105,
%U 3806266,6159183,9966319,16126432,26093743,42221231
%N 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) = 3, b(0) = 2, b(1) = 4, 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 A295357 for a guide to related sequences.
%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.
%F a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).
%e a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e b(3) = 5 (least "new number")
%e a(2) = a(1) + a(0) + b(0)*b(1) = 12
%e Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, ...)
%t mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
%t a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1]*b[n - 2];
%t b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t z = 32; u = Table[a[n], {n, 0, z}] (* A295363 *)
%t v = Table[b[n], {n, 0, 10}] (* complement *)
%Y Cf. A001622, A295357.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Nov 21 2017
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