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Let M be the 3 X 3 Matrix [ -4 4 8 / 1 0 0 / 0 1 0], a(n) = absolute value of the center term of M^n * [1 1 1].
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%I #24 Apr 25 2024 09:18:04

%S 1,8,20,120,496,2304,10240,46208,207360,932352,4189184,18827264,

%T 84606976,380223488,1708703744,7678853120,34508439552,155079540736,

%U 696921096192,3131935031296,14074788184064,63251524091904

%N Let M be the 3 X 3 Matrix [ -4 4 8 / 1 0 0 / 0 1 0], a(n) = absolute value of the center term of M^n * [1 1 1].

%C The matrix is derived from the polynomial 8x^3 + 4x^2 - 4x - 1 shown on page 204 of "Advanced Trigonometry"; cos(2*Pi/7), cos(4*Pi/7), and cos(6*Pi/7) are roots of this polynomial.

%D C. V. Durell & A. Robson, "Advanced Trigonometry", Dover 2003, p. 204.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4, 4, -8).

%F a(n)/a(n-1) tends to 1/cos(3*Pi/7) = 4.4939592074...

%F From _Colin Barker_, Feb 01 2012: (Start)

%F a(n) = 4*a(n-1) + 4*a(n-2) - 8*a(n-3), n > 4.

%F G.f.: x*(1 + 4*x - 16*x^2 + 16*x^3)/(1 - 4*x - 4*x^2 + 8*x^3). (End)

%e a(3) = 20 since M^3 * [1 1 1] = [120 -20 8]. Take the absolute value of the center term.

%t Table[ Abs[ MatrixPower[{{-4, 4, 8}, {1, 0, 0}, {0, 1, 0}}, n].{1, 1, 1}][[2]], {n, 23}] (* _Robert G. Wilson v_, Apr 28 2004 *)

%K nonn,easy

%O 1,2

%A _Gary W. Adamson_, Apr 25 2004

%E More terms from _Robert G. Wilson v_, Apr 28 2004