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%I #7 Nov 02 2017 09:20:31
%S 1,3,11,22,42,74,127,213,353,581,950,1548,2516,4083,6619,10723,17364,
%T 28110,45498,73634,119159,192821,312009,504860,816900,1321792,2138725,
%U 3460551,5599311,9059898,14659246,23719182,38378467,62097689,100476197,162573928
%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 3, 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. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
%C Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio. See A293358 for a guide to related sequences.
%H Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.pdf">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) + b(1) + 3 = 11;
%e Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 12, 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] + 3;
%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}] (* A293766 *)
%t Table[b[n], {n, 0, 10}]
%Y Cf. A001622 (golden ratio), A293765.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Oct 29 2017
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