|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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.
|
|
LINKS
|
|
|
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).
|
|
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
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|