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.

Links

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

Barbara Smolen and Roman Witula, Two-parametric quasi-Fibonacci numbers, Silesian J. Pure Appl. Math. vol. 7, is. 1 (2017), 99-121.

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

Roman Witula, Ramanujan type trigonometric formulae, Demonstratio Math., Volume 45, Issue 4, May 2017.

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