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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, a[2] = 5, b(0) = 2, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
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%I #4 Nov 21 2017 21:34:08

%S 1,3,5,32,79,167,318,575,1003,1710,2869,4761,7840,12841,20953,34100,

%T 55395,89875,145690,236027,382223,618802,1001625,1621077,2623404,

%U 4245237,6869453,11115560,17985943,29102526,47089591

%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, a[2] = 5, 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, a(2) = 5, b(0) = 2, b(1) = 4, so that

%e b(2) = 6 (least "new number")

%e a(3) = a(2) + a(1) + b(2)*b(1) = 32

%e Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, ...)

%t mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;

%t a[0] = 1; a[1] = 3; a[2] = 5; 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}] (* A295364 *)

%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