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A295140 Solution of the complementary equation a(n) = 3*a(n-2) - b(n-2) + 4, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A295140 #4 Nov 19 2017 19:05:20
%S A295140 1,3,5,9,13,24,35,66,98,190,284,559,840,1664,2506,4977,7502,14914,
%T A295140 22488,44723,67443,134147,202306,402417,606893,1207225,1820652,
%U A295140 3621647,5461927,10864911,16385749,32594700,49157213,97784065,147471603,293352158
%N A295140 Solution of the complementary equation a(n) = 3*a(n-2) - b(n-2) + 4, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A295140 The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295053 for a guide to related sequences.
%C A295140 The sequence a(n+1)/a(n) appears to have two convergent subsequences, with limits 1.40..., 2.13...
%H A295140 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 A295140 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
%e A295140 a(2) =3*a(0) - b(0) + 4 = 5
%e A295140 Complement: (b(n)) = (2, 4, 6, 7, 8, 10, 11, 12, 14, 15, 16, ...)
%t A295140 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A295140 a[0] = 1; a[1] = 3; b[0] = 2; b[1]=4;
%t A295140 a[n_] := a[n] = 3 a[n - 2] + b[n - 2] + 4;
%t A295140 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A295140 Table[a[n], {n, 0, 18}]  (* A295140 *)
%t A295140 Table[b[n], {n, 0, 10}]
%Y A295140 Cf. A295053.
%K A295140 nonn,easy
%O A295140 0,2
%A A295140 _Clark Kimberling_, Nov 19 2017