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a(n) = 6*a(n-1) - 9*a(n-2) + a(n-3), with a(0)=3, a(1)=6 and a(2)=18.
12

%I #24 Apr 29 2020 05:54:12

%S 3,6,18,57,186,621,2109,7251,25146,87726,307293,1079370,3798309,

%T 13382817,47191491,166501902,587670810,2074699233,7325660010,

%U 25869337773,91359785781,322660334739,1139593274178,4024976418198,14216179376325,50211881768346,177350652641349

%N a(n) = 6*a(n-1) - 9*a(n-2) + a(n-3), with a(0)=3, a(1)=6 and a(2)=18.

%C The Berndt-type sequence number 1 for the argument 2*Pi/9 (see also A215007, A215008) is connected with the following trigonometric identities: f(n;x)=g(n;x)=const for n=1,2 (and are equal to 6 and 18 respectively), f(n;x)+g(n;x)=const for n=3,4,5 (and are equal to 120, 420 and 1512 respectively). Moreover each of the functions f(3;x), g(3;x) and f(6;x)+g(6;x) is not the constant function. Here f(n;x) := (2*cos(x))^(2n) + (2*cos(x-Pi/3))^(2n) + (2*cos(x+Pi/3))^(2n), and g(n;x) := (2*sin(x))^(2n) + (2*cos(x-Pi/6))^(2n) + (2*cos(x+Pi/6))^(2n), for every n=1,2,..., and x in R (see Witula-Slota paper for details).

%H Andrew Howroyd, <a href="/A215455/b215455.txt">Table of n, a(n) for n = 0..500</a>

%H R. Witula and D. Slota, <a href="http://dx.doi.org/10.1016/j.jmaa.2005.12.020">On modified Chebyshev polynomials</a>, J. Math. Anal. Appl., 324 (2006), 321-343.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-9,1).

%F a(n) = c(1)^(2*n) + c(2)^(2*n) + c(4)^(2*n), where c(j) = 2*cos(Pi*j/9).

%F G.f.: 3*(1 - x)*(1 - 3*x)/(1 - 6*x + 9*x^2 - x^3).

%F a(n) = 3*A094831(n). - _Andrew Howroyd_, Apr 28 2020

%e From the identity c(j)^2 = 2 + c(2*j) we deduce that a(1)=6 is equivalent with c(2) + c(4) + c(8) = 0, where c(j) := 2*cos(Pi*j/9).

%t LinearRecurrence[{6,-9,1}, {3,6,18}, 50]

%o (PARI) Vec((3-12*x+9*x^2)/(1-6*x+9*x^2-x^3)+O(x^99)) \\ _Charles R Greathouse IV_, Sep 27 2012

%Y Cf. A094831, A215007, A215008.

%K nonn,easy

%O 0,1

%A _Roman Witula_, Aug 11 2012

%E Terms a(22) and beyond from _Andrew Howroyd_, Apr 28 2020