A324128 a(n) = 2*n*Fibonacci(n) + (-1)^n + 1.

2, 2, 6, 12, 26, 50, 98, 182, 338, 612, 1102, 1958, 3458, 6058, 10558, 18300, 31586, 54298, 93026, 158878, 270602, 459732, 779286, 1318222, 2225666, 3751250, 6312438, 10606572, 17797418, 29825282, 49922402, 83468678, 139411778, 232622148, 387796318, 645922550, 1074985346, 1787678458, 2970700846

Offset

0,1

Comments

This sequence is distantly related to the number of losing strings using a binary alphabet in the "same game" described by Burns and Purcell (2007) and Kurz (2001). - Petros Hadjicostas, Sep 01 2019

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Chris Burns and Benjamin Purcell, Counting the number of winning strings in the 1-dimensional same game, Fibonacci Quarterly, 45(3) (2007), 233-238.

Sascha Kurz, Polynomials in "same game", 2001.

Index entries for linear recurrences with constant coefficients, signature (2,2,-4,-2,2,1).

Formula

From Chai Wah Wu, Feb 20 2019: (Start)

a(n) = 2*a(n-1) + 2*a(n-2) - 4*a(n-3) - 2*a(n-4) + 2*a(n-5) + a(n-6) for n > 5.

G.f.: (2*x^5 - 2*x^4 - 4*x^3 + 2*x^2 + 2*x - 2)/((x - 1)*(x + 1)*(x^2 + x - 1)^2). (End)

From Petros Hadjicostas, Sep 01 2019: (Start)

a(n) = 2*A324129(n) for n >= 0.

a(n) = A309874(n) + 2*A099920(n-1) = 2^n - A035615(n) + 2*A099920(n-1) for n >= 2.[Here A309874 counts the losing strings while A035615 counts the winning strings using a binary alphabet in the "same game". See Burns and Purcell (2007) and Kurz (2001).]

(End)

Mathematica

A324128[n_]:=Fibonacci[n]2n+(-1)^n+1; Array[A324128, 50, 0] (* Paolo Xausa, Nov 15 2023 *)

Prog

(PARI) Vec(2*(1 - x - x^2 + 2*x^3 + x^4 - x^5) / ((1 - x)*(1 + x)*(1 - x - x^2)^2) + O(x^40)) \\ Colin Barker, Mar 03 2019

Crossrefs

Cf. A000045, A035615, A099920, A309874, A324129.

Sequence in context: A345988 A275439 A173392 * A217211 A035615 A115962

Adjacent sequences: A324125 A324126 A324127 * A324129 A324130 A324131

Keyword

nonn,easy

Author

N. J. A. Sloane, Feb 20 2019

Status

approved