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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.
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%I #4 Nov 18 2017 20:55:18

%S 1,3,6,16,29,71,124,293,506,1183,2036,4745,8158,18995,32649,75998,

%T 130615,304012,522481,1216070,2089947,4864304,8359813,19457242,

%U 33439279,77828996,133757146,311316015

%N 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 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 The sequence a(n+1)/a(n) appears to have two convergent subsequences, with limits 1.71... and 2.32... .

%H 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 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4

%e a(2) = 4*a(0) + b(0) = 6

%e Complement: (b(n)) = (2, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, ...)

%t mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;

%t a[0] = 1; a[1] = 3; b[0] = 2; b[1]=4;

%t a[n_] := a[n] = 4 a[n - 2] + b[n - 2];

%t b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];

%t Table[a[n], {n, 0, 18}] (* A295062 *)

%t Table[b[n], {n, 0, 10}]

%Y Cf. A295053.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Nov 18 2017