A214699 a(n) = 3*a(n-2) - a(n-3) with a(0)=0, a(1)=3, a(2)=0.

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

Offset: 0

Author

Roman Witula, Jul 26 2012

Keywords

• sign
• easy

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.

Examples

			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
		

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

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• 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.
• Index entries for linear recurrences with constant coefficients, signature (0,3,-1).

Crossrefs

Cf. A006053, A052931, A214683, A214778, A214779.

Cf. A214951, A214954, A217053, A217052, A217069.

Programs

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

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 *)

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

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)