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A295139 Solution of the complementary equation a(n) = 3*a(n-2) + b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A295139 #4 Nov 19 2017 19:05:13
%S A295139 1,2,6,10,23,37,77,120,242,372,739,1130,2232,3406,6713,10236,20158,
%T A295139 30728,60495,92206,181509,276643,544553,829956,1633687,2489897,
%U A295139 4901091,7469722,14703305,22409199,44109949,67227632,132329883
%N A295139 Solution of the complementary equation a(n) = 3*a(n-2) + b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A295139 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 A295139 The sequence a(n+1)/a(n) appears to have two convergent subsequences, with limits 1.52..., 1.96...
%H A295139 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 A295139 a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4
%e A295139 a(2) =3*a(0) + b(0) = 6
%e A295139 Complement: (b(n)) = (3, 4, 5, 7, 8, 9, 11, 12, 13, 14, 15, ...)
%t A295139 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A295139 a[0] = 1; a[1] = 2; b[0] = 3; b[1]=4;
%t A295139 a[n_] := a[n] = 3 a[n - 2] + b[n - 2];
%t A295139 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A295139 Table[a[n], {n, 0, 18}]  (* A295139 *)
%t A295139 Table[b[n], {n, 0, 10}]
%Y A295139 Cf. A295053.
%K A295139 nonn,easy
%O A295139 0,2
%A A295139 _Clark Kimberling_, Nov 19 2017