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a(n) = 3*a(n-2) - a(n-3), with a(0)=3, a(1)=0, and a(2)=6.
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%I #37 Mar 10 2020 15:05:16

%S 3,0,6,-3,18,-15,57,-63,186,-246,621,-924,2109,-3393,7251,-12288,

%T 25146,-44115,87726,-157491,307293,-560199,1079370,-1987890,3798309,

%U -7043040,13382817,-24927429,47191491,-88165104,166501902,-311686803,587670810,-1101562311

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

%C The Berndt-type sequence number 5 for the argument 2Pi/9 defined by the first relation from the section "Formula" below. The respective sums with negative powers of the cosines form the sequence A215885. Additionally if we set b(n) = c(1)*c(2)^n + c(2)*c(4)^n + c(4)*c(1)^n and c(n) = c(4)*c(2)^n + c(1)*c(4)^n + c(2)*c(1)^n, where c(j):=2*cos(2*Pi*j/9), then the following system of recurrence equations holds true: b(n) - b(n+1) = a(n), a(n+1) - a(n) = c(n+1), a(n+2) - 2*a(n)=c(n). All three sequences satisfy the same recurrence relation: X(n+3) - 3*X(n+1) + X(n) = 0. Moreover we have a(n+1) + A215665(n) + A215666(n) = 0 since c(1) + c(2) + c(4) = 0, b(n)=A215665(n) and c(n)=A215666(n).

%C If X(n) = 3*X(n-2) - X(n-3), n in Z, with X(n) = a(n) for every n=0,1,..., then X(-n) = A215885(n) for every n=0,1,...

%C From initial values and the recurrence formula we deduce that a(n)/3 and a(3n+1)/9 are all integers. We have a(n)=3*(-1)^n *A188048(n) and a(2n)=A215455(n). Furthermore the following decomposition holds: (X - c(1)^n)*(X - c(2)^n)*(X - c(4)^n) = X^3 - a(n)*X^2 + ((a(n)^2 - a(2*n))/2)*X + (-1)^(n+1), which implies the relation (c(1)*c(2))^n + (c(1)*c(4))^n + (c(2)*c(4))^n = (-c(1))^(-n) + (-c(2))^(-n) + (-c(4))^(-n) = (a(n)^2 - a(2*n))/2.

%D D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the ninth order, (submitted, 2012).

%D R. Witula, Ramanujan type formulas for arguments 2Pi/7 and 2Pi/9, Demonstratio Math. (in press, 2012).

%H Michael De Vlieger, <a href="/A215664/b215664.txt">Table of n, a(n) for n = 0..3649</a>

%H Kai Wang, <a href="https://www.researchgate.net/publication/337943524_Fibonacci_Numbers_And_Trigonometric_Functions_Outline">Fibonacci Numbers And Trigonometric Functions Outline</a>, (2019).

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,-1).

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

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

%e We have c(1)^2 + c(2)^2 + c(4)^2 + 2*(c(1)^3 + c(2)^3 + c(4)^3) = 0 and 3*a(7) + a(8) = a(3).

%t LinearRecurrence[{0,3,-1}, {3,0,6}, 50].

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

%Y Cf. A215455, A215634, A215635, A215636, A215007, A214699, A214683.

%K sign,easy

%O 0,1

%A _Roman Witula_, Aug 20 2012