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A293058 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) + 3, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

%I A293058 #7 Nov 02 2017 09:19:24
%S A293058 1,3,9,19,36,64,110,185,308,507,830,1353,2200,3571,5790,9381,15192,
%T A293058 24596,39812,64433,104271,168731,273030,441790,714850,1156671,1871553,
%U A293058 3028257,4899844,7928136,12828016,20756189,33584243,54340472,87924756,142265270
%N A293058 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) + 3, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
%C A293058 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 A293058 Conjecture:  a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio.
%e A293058 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
%e A293058 a(2)  = a(1) + a(0) + b(0) + 3 = 9;
%e A293058 a(3) = a(2) + a(1) + b(1) + 1 = 19.
%e A293058 Complement: (b(n)) = (2,4,5,6,7,8,10,11,12,13,14,...)
%t A293058 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A293058 a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
%t A293058 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 2] + 3;
%t A293058 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A293058 Table[a[n], {n, 0, 40}]  (* A293316 *)
%t A293058 Table[b[n], {n, 0, 10}]
%Y A293058 Cf. A001622 (golden ratio), A293076.
%K A293058 nonn,easy
%O A293058 0,2
%A A293058 _Clark Kimberling_, Oct 28 2017