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A295064 Solution of the complementary equation a(n) = 8*a(n-3) + b(n-1), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A295064 #4 Nov 19 2017 10:42:54
%S A295064 1,3,5,14,31,48,121,258,395,980,2077,3175,7856,16633,25418,62867,
%T A295064 133084,203365,502958,1064695,1626944,4023689,8517586,13015579,
%U A295064 32189540,68140717,104124662
%N A295064 Solution of the complementary equation a(n) = 8*a(n-3) + b(n-1), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A295064 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 A295064 The sequence a(n+1)/a(n) appears to have three convergent subsequences, with limits 1.52..., 2.11..., 2.47...
%H A295064 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 A295064 a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6
%e A295064 a(3)  = 8*a(0) + b(2) = 14
%e A295064 Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, ...)
%t A295064 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A295064 a[0] = 1; a[1] = 3; a[2] = 5; b[0] = 2;
%t A295064 a[n_] := a[n] = 8 a[n - 3] + b[n - 1];
%t A295064 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A295064 Table[a[n], {n, 0, 18}]  (* A295064 *)
%t A295064 Table[b[n], {n, 0, 10}]
%Y A295064 Cf. A295053.
%K A295064 nonn,easy
%O A295064 0,2
%A A295064 _Clark Kimberling_, Nov 19 2017