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A295616 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. 2
1, 2, 3, 10, 24, 52, 102, 189, 337, 584, 992, 1661, 2753, 4530, 7416, 12097, 19683, 31970, 51864, 84067, 136187, 220535, 357029, 577898, 935289, 1513578, 2449288, 3963318, 6413090, 10376925, 16790566, 27168077, 43959265, 71128001, 115087963, 186216700 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

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.

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

LINKS

Table of n, a(n) for n=0..35.

Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

EXAMPLE

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

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

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

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

MATHEMATICA

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

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

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

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

Table[a[n], {n, 0, 30}]  (* A295616 *)

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

CROSSREFS

Cf. A001622, A000045, A295613.

Sequence in context: A320812 A162034 A105286 * A059929 A123029 A103018

Adjacent sequences:  A295613 A295614 A295615 * A295617 A295618 A295619

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Nov 25 2017

STATUS

approved

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Last modified December 15 20:49 EST 2018. Contains 318154 sequences. (Running on oeis4.)