OFFSET
0,3
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).
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
MATHEMATICA
PROG
(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
(Magma) [n*Fibonacci(n)+((-1)^n+1)/2:n in [0..35]]; // Marius A. Burtea, Aug 29 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Feb 20 2019
STATUS
approved