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a(n) = 3*a(n-2) - a(n-3), with a(0)=0, a(1)=-3, and a(2)=6.
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%I #23 Jun 16 2016 23:27:48

%S 0,-3,6,-9,21,-33,72,-120,249,-432,867,-1545,3033,-5502,10644,-19539,

%T 37434,-69261,131841,-245217,464784,-867492,1639569,-3067260,5786199,

%U -10841349,20425857,-38310246,72118920,-135356595,254667006,-478188705,899357613

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

%C The Berndt-type sequence number 7 for the argument 2Pi/9 defined by the first relation from the section "Formula" below. Two sequences connected with a(n) (possessing the respective numbers 5 and 6) are discussed in A215664 and A215665 - for more details see comments to A215664 and Witula's reference. We have a(n) = A215664(n+2) - 2*A215664(n) and a(n+1) = A215664(n+1) - A215664(n).

%C From initial values and the title recurrence formula we deduce that a(n)/3 and a(3*n)/9 are all integers.

%C If we set X(n) = 3*X(n-2) - X(n-3), n in Z, with a(n) = X(n), for every n=0,1,..., then X(-n) = -abs(A215917(n)) = (-1)^n*A215917(n), for every n=0,1,...

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

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

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

%F a(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).

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

%e We have 8*a(3)+a(6)=5*a(6)+3*a(7)=0, a(5) + a(12) = 3000, and (a(30)-1000*a(10)-a(2))/10^5 is an integer. Further we obtain c(4)*cos(4*Pi/7)^7 + c(1)*cos(8*Pi/7)^7 + c(2)*c(2*Pi/7)^7 = -15/16.

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

%o (PARI) concat(0,Vec(-3*(1-2*x)/(1-3*x^2+x^3)+O(x^99))) \\ _Charles R Greathouse IV_, Oct 01 2012

%Y Cf. A215455, A215634, A215635, A215636, A215664, A214699, A215007, A214683.

%K sign,easy

%O 0,2

%A _Roman Witula_, Aug 20 2012