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%I #8 Nov 02 2017 09:20:06
%S 1,3,8,16,30,53,92,155,258,425,696,1135,1846,2998,4862,7879,12761,
%T 20661,33444,54128,87596,141749,229371,371147,600546,971722,1572299,
%U 2544053,4116385,6660472,10776892,17437400,28214329,45651767,73866135,119517942
%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1), 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 A293358: a(n) = a(n-1) + a(n-2) + b(n-1)
%C A293406: a(n) = a(n-1) + a(n-2) + b(n-1) + 1
%C A293765: a(n) = a(n-1) + a(n-2) + b(n-1) + 2
%C A293766: a(n) = a(n-1) + a(n-2) + b(n-1) + 3
%C A293767: a(n) = a(n-1) + a(n-2) + b(n-1) - 1
%C A294365: a(n) = a(n-1) + a(n-2) + b(n-1) + n
%C A294366: a(n) = a(n-1) + a(n-2) + b(n-1) + 2n
%C A294367: a(n) = a(n-1) + a(n-2) + b(n-1) + n - 1
%C A294368: a(n) = a(n-1) + a(n-2) + b(n-1) + n + 1
%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.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) = 8;
%e Complement: (b(n)) = (2, 4, 5, 6, 7, 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] + b[n - 1];
%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}] (* A293358 *)
%t Table[b[n], {n, 0, 10}]
%Y Cf. A001622 (golden ratio), A293076.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Oct 29 2017
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