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Solution of the complementary equation a(n) = 2*a(n-1) - a(n-3) + b(n-2), where a(0) = 1, a(1) = 2, a(2) = 3, b(0) = 4, b(1) = 5, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.
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%I #6 Aug 27 2021 21:13:50

%S 1,2,3,10,24,52,102,189,337,584,992,1661,2753,4530,7416,12097,19683,

%T 31970,51864,84067,136187,220535,357029,577898,935289,1513578,2449288,

%U 3963318,6413090,10376925,16790566,27168077,43959265,71128001,115087963,186216700

%N Solution of the complementary equation a(n) = 2*a(n-1) - a(n-3) + b(n-2), where a(0) = 1, a(1) = 2, a(2) = 3, b(0) = 4, b(1) = 5, 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 A295613 for a guide to related sequences.

%C a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).

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

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

%e a(3) = 2*a(2) - a(0) + b(1) = 10

%e Complement: (b(n)) = (4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, ...)

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

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

%t a[n_] := a[n] = 2 a[n - 1] - a[n - 3] + 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, 30}] (* A295616 *)

%t Table[b[n], {n, 0, 20}] (* complement *)

%Y Cf. A001622, A000045, A295613.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Nov 25 2017