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Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) - 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
2

%I #7 Nov 02 2017 09:19:31

%S 1,3,5,11,21,38,66,112,187,310,509,832,1355,2202,3573,5792,9383,15194,

%T 24598,39814,64435,104273,168733,273032,441792,714852,1156673,1871555,

%U 3028259,4899846,7928138,12828018,20756191,33584245,54340474,87924758,142265272

%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) - 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. See A293076 for a guide to related sequences.

%C Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio.

%e a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that

%e a(2) = a(1) + a(0) + b(0) -1 = 5;

%e a(3) = a(2) + a(1) + b(1) - 1 = 11.

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

%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 - 2] - 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}] (* A293317 *)

%t Table[b[n], {n, 0, 10}]

%Y Cf. A001622 (golden ratio), A293076.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Oct 28 2017