A187273 - OEIS

%I #34 Feb 19 2024 10:20:31

%S 0,3,12,27,72,135,324,567,1296,2187,4860,8019,17496,28431,61236,98415,

%T 209952,334611,708588,1121931,2361960,3720087,7794468,12223143,

%U 25509168,39858075,82904796,129140163,267846264,416118303,860934420,1334448351,2754990144,4261625379,8781531084,13559717115,27894275208

%N a(n) = (n/4)*3^(n/2)*((1+sqrt(3))^2+(-1)^n*(1-sqrt(3))^2).

%H Vincenzo Librandi, <a href="/A187273/b187273.txt">Table of n, a(n) for n = 0..1000</a>

%H R. Kemp, <a href="http://dx.doi.org/10.1016/0012-365X(82)90123-6">On the number of words in the language {w in Sigma* | w = w^R }^2</a>, Discrete Math., 40 (1982), 225-234. See Lemma 1.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,6,0,-9).

%F From _Colin Barker_, Jul 24 2013: (Start)

%F a(n) = 6*a(n-2) - 9*a(n-4).

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

%F a(2*n) = 4*n*3^n, a(2*n+1) = (2*n+1)*3^(n+1). - _Andrew Howroyd_, Mar 28 2016

%p See A187272.

%t LinearRecurrence[{0,6,0,-9},{0,3,12,27},40] (* _Harvey P. Dale_, Apr 21 2014 *)

%t CoefficientList[Series[3 x (x + 1) (3 x + 1)/(3 x^2 - 1)^2, {x, 0, 40}], x] (* _Vincenzo Librandi_, Apr 23 2014 *)

%o (PARI) for(n=0,50, print1(round((n/4)*3^(n/2)*((1+sqrt(3))^2+(-1)^n*(1-sqrt(3))^2)), ", ")) \\ _G. C. Greubel_, Jul 08 2018

%o (Magma) [Round((n/4)*3^(n/2)*((1+Sqrt(3))^2+(-1)^n*(1-Sqrt(3))^2)): n in [0..50]]; // _G. C. Greubel_, Jul 08 2018

%o (Python)

%o def A187273(n): return n*3**(1+(n>>1)) if n&1 else (n<<1)*3**(n>>1) # _Chai Wah Wu_, Feb 19 2024

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_, Mar 07 2011