A215448 -id:A215448 - cp's OEIS frontend

A215917 a(n) = -3*a(n-1) + a(n-3), with a(0)=0, a(1)=6, and a(2)=-15.

0, 6, -15, 45, -129, 372, -1071, 3084, -8880, 25569, -73623, 211989, -610398, 1757571, -5060724, 14571774, -41957751, 120812529, -347865813, 1001639688, -2884106535, 8304453792, -23911721688, 68851058529, -198248721795, 570834443697, -1643652272562

Offset: 0

Roman Witula, Aug 27 2012

The Berndt-type sequence number 9 for the argument 2Pi/9 defined by the first relation from the section "Formula" below.
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.
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.
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,...
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. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the ninth order, (submitted, 2012).
R. Witula, Ramanujan type formulas for arguments 2Pi/7 and 2Pi/9, Demonstratio Math. (in press, 2012).

Index entries for linear recurrences with constant coefficients, signature (-3,0,1).

Cf. A215448, A215885, A215664, A215665, A215666.

We have a(3) + 3*a(2) = 0, a(8) + 24*a(5) = 48 = a(3) + a(1)/2.

LinearRecurrence[{-3,0,1}, {0,6,-15}, 50]

concat(0,Vec(3*(x+2)/(1+3*x-x^3)+O(x^99))) \\ Charles R Greathouse IV, Oct 01 2012

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).
a(n) = (-1)^n*(A215885(n+1) - A215885(n)).
G.f.: 3*x(x+2)/(1+3*x-x^3).

A052545 Expansion of (1-x)^2/(1-3*x+x^3).

Original entry on oeis.org

1, 1, 4, 11, 32, 92, 265, 763, 2197, 6326, 18215, 52448, 151018, 434839, 1252069, 3605189, 10380728, 29890115, 86065156, 247814740, 713554105, 2054597159, 5915976737, 17034376106, 49048531159, 141229616740, 406654474114

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

(1, 4, 11, 32, ...) = INVERT transform of (1, 3, 4, 5, 6, 7, ...).

G. C. Greubel, Table of n, a(n) for n = 0..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 481
Index entries for linear recurrences with constant coefficients, signature (3,0,-1).

Cf. A215448. First differences of A052536.

a:=[1,1,4];; for n in [4..40] do a[n]:=3*a[n-1]-a[n-3]; od; a; # G. C. Greubel, May 08 2019

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1-x)^2/(1-3*x+x^3) )); // G. C. Greubel, May 08 2019

spec := [S,{S=Sequence(Prod(Z,Union(Z,Sequence(Z)),Sequence(Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);

LinearRecurrence[{3,0,-1}, {1,1,4}, 40] (* G. C. Greubel, May 08 2019 *)

my(x='x+O('x^40)); Vec((1-x)^2/(1-3*x+x^3)) \\ G. C. Greubel, May 08 2019

TOP = 33
a = [1]*TOP
a[2]=4
for n in range(3,TOP):
    print(a[n-3], end=',')
    a[n] = 3*a[n-1] - a[n-3]
# Alex Ratushnyak, Aug 10 2012

((1-x)^2/(1-3*x+x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, May 08 2019

G.f.: (1-x)^2/(1-3*x+x^3).
a(n) = 3*a(n-1) - a(n-3), with a(0)=a(1)=1, a(2)=4.
a(n) = Sum_{alpha = RootOf(1-3*x+x^3)} (-1/9 * (-1+2*alpha^2-2*alpha) * alpha^(-1-n)).
a(n) = A076264(n) - 2*A076264(n-1) + A076264(n-2). - R. J. Mathar, Nov 28 2011

cp's OEIS Frontend

A215917 a(n) = -3*a(n-1) + a(n-3), with a(0)=0, a(1)=6, and a(2)=-15.

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