A214699 - OEIS

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

(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

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).

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