A215455 a(n) = 6*a(n-1) - 9*a(n-2) + a(n-3), with a(0)=3, a(1)=6 and a(2)=18.

3, 6, 18, 57, 186, 621, 2109, 7251, 25146, 87726, 307293, 1079370, 3798309, 13382817, 47191491, 166501902, 587670810, 2074699233, 7325660010, 25869337773, 91359785781, 322660334739, 1139593274178, 4024976418198, 14216179376325, 50211881768346, 177350652641349

Offset: 0

Author

Roman Witula, Aug 11 2012

Keywords

• nonn
• easy

Comments

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).

Examples

			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).
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..500
• R. Witula and D. Slota, On modified Chebyshev polynomials, J. Math. Anal. Appl., 324 (2006), 321-343.
• Index entries for linear recurrences with constant coefficients, signature (6,-9,1).

Crossrefs

Cf. A094831, A215007, A215008.

Programs

Mathematica

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

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

Formula

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

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

a(n) = 3*A094831(n). - Andrew Howroyd, Apr 28 2020

Extensions

Terms a(22) and beyond from Andrew Howroyd, Apr 28 2020