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%I #9 Nov 01 2017 20:54:25
%S 1,3,16,37,75,138,243,415,696,1153,1895,3098,5047,8203,13314,21587,
%T 34975,56640,91697,148423,240210,388727,629035,1017864,1647005,
%U 2664979,4312098,6977195,11289415,18266736,29556281,47823151,77379570,125202863,202582581
%N 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) = 3, b(0) = 2, b(1) = 4.
%C The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A294414 for a guide to related sequences.
%C Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio.
%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) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e a(2) = a(1) + a(0) + 2*b(1) + 2*b(0) = 16
%e Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17,...)
%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] + 2 b[n - 1] + 2 b[n - 2];
%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, 40}] (* A294419 *)
%t Table[b[n], {n, 0, 10}]
%Y Cf. A293076, A293765, A294414.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Oct 31 2017
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