OFFSET
0,1
COMMENTS
The Berndt-type sequence number 3 for the argument 2Pi/9 defined by the relation: X(n) = a(n) + b(n)*sqrt(2), where X(n) := ((cos(Pi/24))^(2*n) + (cos(7*Pi/24))^(2*n) + ((cos(3*Pi/8))^(2*n))*(-4)^n. We have b(n)=A215636(n).
We note that above formula is the Binet form of the following recurrence sequence: X(n+3) + 6*X(n+2) + 9*X(n+1) + (2 + sqrt(2))*X(n) = 0, which is a special type of the sequence X(n)=X(n;g) defined in the comments to A215634 for g:=Pi/24. The sequences a(n) and b(n) satisfy the following system of recurrence equations: a(n) = -b(n+3)-6*b(n+2)-9*b(n+1)-2*b(n), 2*b(n) = -a(n+3)-6*a(n+2)-9*a(n+1)-2*a(n).
REFERENCES
R. Witula, D. Slota, On modified Chebyshev polynomials, J. Math. Anal. Appl., 324 (2006), 321-343.
LINKS
Index entries for linear recurrences with constant coefficients, signature (-12, -54, -112, -105, -36, -2).
FORMULA
G.f.:(3+30*x+108*x^2+168*x^3+105*x^4+18*x^5) / (1+12*x+54*x^2+112*x^3+105*x^4+36*x^5+2*x^6).
MATHEMATICA
LinearRecurrence[{-12, -54, -112, -105, -36, -2}, {3, -6, 18, -60, 210, -756}, 50]
PROG
(PARI) Vec((3+30*x+108*x^2+168*x^3+105*x^4+18*x^5) /(1+12*x+54*x^2+112*x^3+105*x^4+36*x^5+2*x^6)+O(x^99)) \\ Charles R Greathouse IV, Oct 01 2012
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Roman Witula, Aug 18 2012
STATUS
approved