OFFSET
0,2
COMMENTS
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.
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
Index entries for linear recurrences with constant coefficients, signature (-3, 0, 1).
FORMULA
MAPLE
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