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A214778
a(n) = 3*a(n-1) + 6*a(n-2) + a(n-3), with a(0) = 3, a(1) = 3, and a(2) = 21.
9
3, 3, 21, 84, 381, 1668, 7374, 32511, 143445, 632775, 2791506, 12314613, 54325650, 239656134, 1057236915, 4663973199, 20574997221, 90766067772, 400412159841, 1766407883376, 7792462676946, 34376247490935, 151649926417857, 668999726876127, 2951274986626458
OFFSET
0,1
COMMENTS
Ramanujan-type sequence number 3 for the argument 2Pi/9 is equal to the subsequence ax(3n) of the sequence ax(n), which (with its two conjugate sequences bx(n) and cx(n)) is defined in the comments to the sequence A214779 (we note that simultaneously we have bx(3n)=cx(3n)=0).
From example below follows that a(n) is equal to the sum of n-th powers of the roots of the polynomial x^3-3*x^2-6*x-1.
We note that all a(n) are divisible by 3 and a(n)/3 == 1 (mod 3). - Roman Witula, Oct 06 2012
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.
FORMULA
a(n) = (c(1)/c(2))^n + (c(2)/c(4))^n + (c(4)/c(1))^n, where c(j) := Cos(2*Pi*j/9).
G.f.: (3-6*x-6*x^2)/(1-3*x -6*x^2-x^3).
a(n+1) = A214951(n+1) - A214951(n). - Roman Witula, Oct 06 2012
EXAMPLE
From a(1)=3 (after squaring) and a(2)=21 the following equality follows c(1)/c(4) + c(4)/c(2) + c(2)/c(1) = -6, which implies the decomposition x^3 - 3*x^2 - 6*x - 1 =(x - c(1)/c(2))*(x - c(2)/c(4))*(x - c(4)/c(1)).
MATHEMATICA
LinearRecurrence[{3, 6, 1}, {3, 3, 21}, 40] (* T. D. Noe, Jul 30 2012 *)
PROG
(PARI) Vec((3-6*x-6*x^2)/(1-3*x -6*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Oct 01 2012
(PARI) polsym(x^3 - 3*x^2 - 6*x - 1, 30) \\ Charles R Greathouse IV, Jul 20 2016
CROSSREFS
Sequence in context: A369078 A209528 A369751 * A180754 A224091 A224751
KEYWORD
nonn,easy
AUTHOR
Roman Witula, Jul 28 2012
STATUS
approved