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