login
A215917
a(n) = -3*a(n-1) + a(n-3), with a(0)=0, a(1)=6, and a(2)=-15.
7
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
0,2
COMMENTS
The Berndt-type sequence number 9 for the argument 2*Pi/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.
LINKS
Barbara Smolen and Roman Witula, Two-parametric quasi-Fibonacci numbers, Silesian J. Pure Appl. Math. vol. 7, is. 1 (2017), 99-121.
Roman Witula, Ramanujan type trigonometric formulae, Demonstratio Math., Volume 45, Issue 4, May 2017.
FORMULA
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).
EXAMPLE
We have a(3) + 3*a(2) = 0, a(8) + 24*a(5) = 48 = a(3) + a(1)/2.
MATHEMATICA
LinearRecurrence[{-3, 0, 1}, {0, 6, -15}, 50]
PROG
(PARI) concat(0, Vec(3*(x+2)/(1+3*x-x^3)+O(x^99))) \\ Charles R Greathouse IV, Oct 01 2012
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Roman Witula, Aug 27 2012
STATUS
approved