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A214699 a(n) = 3*a(n-2) - a(n-3) with a(0)=0, a(1)=3, a(2)=0. 19
0, 3, 0, 9, -3, 27, -18, 84, -81, 270, -327, 891, -1251, 3000, -4644, 10251, -16932, 35397, -61047, 123123, -218538, 430416, -778737, 1509786, -2766627, 5308095, -9809667, 18690912, -34737096, 65882403, -122902200, 232384305, -434589003, 820055115, -1536151314 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
All a(n) are divisible by 3.
The Ramanujan-type sequence number 1 for the argument 2*Pi/9 defined by the following identity:
3^(1/3)*a(n) = (c(1)/c(2))^(1/3)*c(1)^n + (c(2)/c(4))^(1/3)*c(2)^n + (c(4)/c(1))^(1/3)*c(4)^n = -( (c(1)/c(2))^(1/3)*c(2)^(n+1) + (c(2)/c(4))^(1/3)*c(4)^(n+1) + (c(4)/c(1))^(1/3)*c(1)^(n+1) ), where c(j) := 2*cos(2*Pi*j/9).
The definitions of other Ramanujan-type sequences, for the argument of 2*Pi/9 in one's, are given in the Crossrefs section.
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
Roman Witula, Full Description of Ramanujan Cubic Polynomials, Journal of Integer Sequences, Vol. 13 (2010), Article 10.5.7.
Roman Witula, Ramanujan Cubic Polynomials of the Second Kind, Journal of Integer Sequences, Vol. 13 (2010), Article 10.7.5.
Roman Witula, Ramanujan Type Trigonometric Formulae, Demonstratio Math. 45 (2012) 779-796.
FORMULA
G.f.: 3*x/(1 - 3*x^2 + x^3).
From Roman Witula, Oct 06 2012: (Start)
a(n+1) = 3*(-1)^n*A052931(n), which from recurrence relations for a(n) and A052931 can easily be proved inductively.
a(n) = -A214779(n+1) - A214779(n). (End)
EXAMPLE
We have a(2) = a(1) + a(4) = a(4) + a(7) + a(8) = -a(3) + a(5) + a(6) = 0, which implies
(c(1)/c(2))^(1/3)*c(1)^2 + (c(2)/c(4))^(1/3)*c(2)^2 + (c(4)/c(1))^(1/3)*c(4)^2 = (c(1)/c(2))^(1/3)*(c(1) + c(1)^4) + (c(2)/c(4))^(1/3)*(c(2) + c(2)^4) + (c(4)/c(1))^(1/3)*(c(4) + c(4)^4) = (c(1)/c(2))^(1/3)*(c(1)^4 + c(1)^7 + c(1)^8) + (c(2)/c(4))^(1/3)*(c(2)^4 + c(2)^7 + c(2)^8) + (c(4)/c(1))^(1/3)*(c(4)^4 + c(4)^7 + c(4)^8) = 0.
Moreover we have 3000*3^(1/3) = (c(1)/c(2))^(1/3)*c(1)^13 + (c(2)/c(4))^(1/3)*c(2)^13 + (c(4)/c(1))^(1/3)*c(4)^13. - Roman Witula, Oct 06 2012
MATHEMATICA
LinearRecurrence[{0, 3, -1}, {0, 3, 0}, 30]
CoefficientList[Series[3*x/(1 - 3*x^2 + x^3), {x, 0, 34}], x] (* James C. McMahon, Jan 09 2024 *)
PROG
(Magma) [n le 3 select 3*(1+(-1)^n)/2 else 3*Self(n-2) - Self(n-3): n in [1..40]]; // G. C. Greubel, Jan 08 2024
(SageMath)
def a(n): # a=A214699
if (n<3): return 3*(n%2)
else: return 3*a(n-2) - a(n-3)
[a(n) for n in range(41)] # G. C. Greubel, Jan 08 2024
CROSSREFS
Sequence in context: A021101 A154202 A352491 * A317825 A002346 A021327
KEYWORD
sign,easy
AUTHOR
Roman Witula, Jul 26 2012
STATUS
approved

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Last modified February 21 04:08 EST 2024. Contains 370219 sequences. (Running on oeis4.)