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A295361 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2) - b(n-3), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences. 2

%I #4 Nov 21 2017 21:33:48

%S 1,3,5,20,40,76,134,230,386,640,1052,1720,2802,4554,7390,11980,19408,

%T 31429,50882,82357,133287,215694,349033,564781,913870,1478709,2392639,

%U 3871410,6264113,10135589,16399770,26535429,42935271,69470774,112406121

%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2) - b(n-3), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, 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 A295357 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)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).

%e a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, so that

%e b(3) = 7 (least "new number")

%e a(3) = a(1) + a(0) + b(2) + 2*b(1) - b(0) = 20

%e Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...)

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

%t a[0] = 1; a[1] = 3; a[2] = 5; b[0] = 2; b[1] = 4; b[2] = 6;

%t a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + 2*b[n - 2] - b[n - 3];

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

%t z = 32; u = Table[a[n], {n, 0, z}] (* A295361 *)

%t v = Table[b[n], {n, 0, 10}] (* complement *)

%Y Cf. A001622, A295357.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Nov 21 2017

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Last modified April 23 15:11 EDT 2024. Contains 371914 sequences. (Running on oeis4.)