cp's OEIS Frontend

A295062 Solution of the complementary equation a(n) = 4*a(n-2) + b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A295062 #4 Nov 18 2017 20:55:18
%S A295062 1,3,6,16,29,71,124,293,506,1183,2036,4745,8158,18995,32649,75998,
%T A295062 130615,304012,522481,1216070,2089947,4864304,8359813,19457242,
%U A295062 33439279,77828996,133757146,311316015
%N A295062 Solution of the complementary equation a(n) = 4*a(n-2) + b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A295062 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 A295062 The sequence a(n+1)/a(n) appears to have two convergent subsequences, with limits 1.71... and 2.32... .
%H A295062 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 A295062 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
%e A295062 a(2) = 4*a(0) + b(0) = 6
%e A295062 Complement: (b(n)) = (2, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, ...)
%t A295062 mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t A295062 a[0] = 1; a[1] = 3; b[0] = 2; b[1]=4;
%t A295062 a[n_] := a[n] = 4 a[n - 2] + b[n - 2];
%t A295062 b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t A295062 Table[a[n], {n, 0, 18}]  (* A295062 *)
%t A295062 Table[b[n], {n, 0, 10}]
%Y A295062 Cf. A295053.
%K A295062 nonn,easy
%O A295062 0,2
%A A295062 _Clark Kimberling_, Nov 18 2017