A324129 a(n) = n*Fibonacci(n) + ((-1)^n + 1)/2.

1, 1, 3, 6, 13, 25, 49, 91, 169, 306, 551, 979, 1729, 3029, 5279, 9150, 15793, 27149, 46513, 79439, 135301, 229866, 389643, 659111, 1112833, 1875625, 3156219, 5303286, 8898709, 14912641, 24961201, 41734339, 69705889, 116311074, 193898159, 322961275

Offset: 0

Author

N. J. A. Sloane, Feb 20 2019

Keywords

• nonn
• easy

Comments

Equals A324128(n)/2.

This sequence is distantly related to (one-half) 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).

Crossrefs

Cf. A000045, A035615, A099920, A323812, A324128.

Programs

Magma

[n*Fibonacci(n)+((-1)^n+1)/2:n in [0..35]]; // Marius A. Burtea, Aug 29 2019

Mathematica

A324129[n_]:=Fibonacci[n]n+((-1)^n+1)/2;Array[A324129,50,0] (* Paolo Xausa, Nov 15 2023 *)

PARI

Vec((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

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

a(n) = A309874(n)/2 + A099920(n-1) = 2^(n-1) - A035615(n)/2 + A099920(n-1) = A323812(n) + A099920(n-1) for n >= 2. [Sequence 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).] - Petros Hadjicostas, Sep 01 2019