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 A295143 Solution of the complementary equation a(n) = 2*a(n-1) + a(n-2) + b(n-1), where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences. 5

%I #4 Nov 19 2017 19:05:44

%S 1,2,9,25,65,162,397,966,2340,5658,13669,33010,79704,192434,464589,

%T 1121630,2707868,6537386,15782661,38102730,91988144,222079042

%N Solution of the complementary equation a(n) = 2*a(n-1) + a(n-2) + b(n-1), where a(0) = 1, a(1) = 2, b(0) = 3, 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.

%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.

%F a(n+1)/a(n) -> 1 + sqrt(2).

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

%e a(2) =2*a(1) + a(0) + b(1) = 9

%e Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, ...)

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

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

%t a[n_] := a[n] = 2 a[ n - 1] + a[n - 2] + b[n - 1];

%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}] (* A295143 *)

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

%Y Cf. A295053, A295141, A295142, A295144.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Nov 19 2017

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