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%I #18 Jun 16 2016 23:27:48
%S 0,6,-15,45,-129,372,-1071,3084,-8880,25569,-73623,211989,-610398,
%T 1757571,-5060724,14571774,-41957751,120812529,-347865813,1001639688,
%U -2884106535,8304453792,-23911721688,68851058529,-198248721795,570834443697,-1643652272562
%N a(n) = -3*a(n-1) + a(n-3), with a(0)=0, a(1)=6, and a(2)=-15.
%C The Berndt-type sequence number 9 for the argument 2Pi/9 defined by the first relation from the section "Formula" below.
%C We have a(n) = 3*(-1)^(n+1)*A215448(n+1). From the recurrence formula for a(n) it follows that all a(3*n) are divisible by 9, a(3*n+1)/3 are congruent to 2 modulo 3, and a(3*n+2)/3 are congruent to 1 modulo 3. In the consequence also all sums a(n)+a(n+1)+a(n+2) are divisible by 9.
%C From general recurrence X(n) = -3*X(n-1) + X(n-3) the following formula can be deduced: 3*sum{k=2,..,n-1} X(k) = -X(n)-X(n-1)-X(n-2)+X(2)+X(1)+X(0). Hence, in the case of a(n) we obtain 3*sum{k=2,..,n-1} a(k) = -a(n)-a(n-1)-a(n-2)-9.
%C If we set X(n) = -3*X(n-1) + X(n-3), n in Z, with a(n) = X(n) for n=0,1,... then X(-n) = abs(A215666(n)) = (-1)^n*A215666(n), for every n=0,1,...
%C The following decomposition holds true (X - c(1)*(-c(4))^(-n))*(X - c(2)*(-c(1))^(-n))*(X - c(4)*(-c(2))^(-n)) = X^3 - a(n)*X^2 + (-1)^n*(A215665(n) - A215664(n))*X + 1.
%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 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-3, 0, 1).
%F a(n) = c(1)*(-c(4))^(-n) + c(2)*(-c(1))^(-n) + c(4)*(-c(2))^(-n), where c(j) := 2*cos(2*Pi*j/9).
%F a(n) = (-1)^n*(A215885(n+1) - A215885(n)).
%F G.f.: 3*x(x+2)/(1+3*x-x^3).
%p We have a(3) + 3*a(2) = 0, a(8) + 24*a(5) = 48 = a(3) + a(1)/2.
%t LinearRecurrence[{-3,0,1}, {0,6,-15}, 50].
%o (PARI) concat(0,Vec(3*(x+2)/(1+3*x-x^3)+O(x^99))) \\ _Charles R Greathouse IV_, Oct 01 2012
%Y Cf. A215448, A215885, A215664, A215665, A215666.
%K sign,easy
%O 0,2
%A _Roman Witula_, Aug 27 2012
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