A215664 a(n) = 3*a(n-2) - a(n-3), with a(0)=3, a(1)=0, and a(2)=6.

3, 0, 6, -3, 18, -15, 57, -63, 186, -246, 621, -924, 2109, -3393, 7251, -12288, 25146, -44115, 87726, -157491, 307293, -560199, 1079370, -1987890, 3798309, -7043040, 13382817, -24927429, 47191491, -88165104, 166501902, -311686803, 587670810, -1101562311

Offset

0,1

Comments

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).

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,...

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.

References

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).

Links

Michael De Vlieger, Table of n, a(n) for n = 0..3649

Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).

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

Formula

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

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

Example

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).

Mathematica

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

Prog

(PARI) Vec(3*(1-x^2)/(1-3*x^2+x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 27 2012

Crossrefs

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

Sequence in context: A356195 A051021 A294777 * A088162 A133170 A062542

Adjacent sequences: A215661 A215662 A215663 * A215665 A215666 A215667

Keyword

sign,easy

Author

Roman Witula, Aug 20 2012

Status

approved