%I #41 Jan 05 2025 19:51:41
%S 1,1,3,6,13,25,49,91,169,306,551,979,1729,3029,5279,9150,15793,27149,
%T 46513,79439,135301,229866,389643,659111,1112833,1875625,3156219,
%U 5303286,8898709,14912641,24961201,41734339,69705889,116311074,193898159,322961275
%N a(n) = n*Fibonacci(n) + ((-1)^n + 1)/2.
%C Equals A324128(n)/2.
%C 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
%H Colin Barker, <a href="/A324129/b324129.txt">Table of n, a(n) for n = 0..1000</a>
%H Chris Burns and Benjamin Purcell, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Papers1/45-3/burns.pdf">Counting the number of winning strings in the 1-dimensional same game</a>, Fibonacci Quarterly, 45(3) (2007), 233-238.
%H Sascha Kurz, <a href="/A035617/a035617.pdf">Polynomials in "same game"</a>, 2001.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-4,-2,2,1).
%F From _Chai Wah Wu_, Feb 20 2019: (Start)
%F 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.
%F G.f.: (x^5 - x^4 - 2*x^3 + x^2 + x - 1)/((x - 1)*(x + 1)*(x^2 + x - 1)^2). (End)
%F 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
%t A324129[n_]:=Fibonacci[n]n+((-1)^n+1)/2;Array[A324129,50,0] (* _Paolo Xausa_, Nov 15 2023 *)
%o (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
%o (Magma) [n*Fibonacci(n)+((-1)^n+1)/2:n in [0..35]]; // _Marius A. Burtea_, Aug 29 2019
%Y Cf. A000045, A035615, A099920, A323812, A324128.
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_, Feb 20 2019