OFFSET
0,1
COMMENTS
Ramanujan-type sequence number 4 for the argument 2*Pi/9 is defined by the following relation: 9^(1/3)*a(n)=(c(1)/c(2))^(n - 1/3) + (c(2)/c(4))^(n - 1/3) + (c(4)/c(1))^(n - 1/3), where c(j) := Cos(2Pi*j/9) - for the proof see Witula et al.'s papers. We have a(n)=bx(3n-1), where the sequence bx(n) and its two conjugate sequences ax(n) and cx(n) are defined in the comments to the sequence A214779. We note that ax(3n-1)=cx(3n-1)=0. Moreover we have ax(3n)=A214778(n), bx(3n)=cx(3n)=0 and cx(3n+1)=A214954(n), ax(3n+1)=bx(3n+1)=0.
REFERENCES
R. Witula, E. Hetmaniok, D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the Fifteenth International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012. (in review)
LINKS
Roman Witula, Ramanujan Type Trigonometric Formulae, Demonstratio Math. 45 (2012) 779-796.
Index entries for linear recurrences with constant coefficients, signature (3,6,1).
FORMULA
G.f.: (2-x-x^2)/(1-3*x-6*x^2-x^3).
a(n+1) - a(n) = A214778(n+1). - Roman Witula, Oct 06 2012
EXAMPLE
We have 2*9^(1/3) = (c(2)/c(1))^(1/3) + (c(4)/c(2))^(1/3) + (c(1)/c(4))^(1/3), 5*9^(1/3) = (c(1)/c(2))^(2/3) + (c(2)/c(4))^(2/3) + (c(4)/c(1))^(2/3), and 110*9^(1/3)=(c(1)/c(2))^(8/3) + (c(2)/c(4))^(8/3) + (c(4)/c(1))^(8/3). Moreover we obtain a(6)-a(2)-a(1)-a(0)=9500, a(12)-a(2)-a(1)-a(0)=70250000 and a(12)-a(6)=3^3*43*a(1)*a(3)^2. - Roman Witula, Oct 06 2012
MATHEMATICA
LinearRecurrence[{3, 6, 1}, {2, 5, 26}, 40] (* T. D. Noe, Jul 30 2012 *)
PROG
(PARI) Vec((2-x-x^2)/(1-3*x-6*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Oct 01 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Roman Witula, Jul 30 2012
STATUS
approved