A214778 -id:A214778 - cp's OEIS frontend

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

Roman Witula, Jul 26 2012

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.

			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

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

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).

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

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

[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

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 *)

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

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)

A019879 Decimal expansion of sine of 70 degrees.

Original entry on oeis.org

9, 3, 9, 6, 9, 2, 6, 2, 0, 7, 8, 5, 9, 0, 8, 3, 8, 4, 0, 5, 4, 1, 0, 9, 2, 7, 7, 3, 2, 4, 7, 3, 1, 4, 6, 9, 9, 3, 6, 2, 0, 8, 1, 3, 4, 2, 6, 4, 4, 6, 4, 6, 3, 3, 0, 9, 0, 2, 8, 6, 6, 6, 2, 7, 7, 4, 2, 2, 1, 2, 1, 0, 9, 9, 5, 8, 8, 9, 4, 5, 8, 9, 4, 9, 7, 4, 5, 8, 8, 9, 8, 3, 7, 9, 4, 8, 0, 6, 7

Offset: 0

N. J. A. Sloane

It is well known that the length sin 70° (cos 20°) is not constructible with ruler and compass, since it is a root of the irreducible polynomial 8x^3 - 6x - 1 and 3 fails to divide any power of 2. - Jean-François Alcover, Aug 10 2014 [cf. the Maxfield ref.]
A cubic number with denominator 2. - Charles R Greathouse IV, Aug 27 2017
From Peter Bala, Oct 21 2021: (Start)
The minimal polynomial of cos(Pi/9) is 8*x^3 - 6*x - 1 with discriminant (2^6)*(3^4), a square: hence the Galois group of the algebraic number field Q(sin(70°) over Q is the cyclic group of order 3.
The two other zeros of the minimal polynomial are cos(5*Pi/9) = - A019819 and cos(7*Pi/9) = - A019859. The mapping z -> 1 - 2*z^2 cyclically permutes the three zeros.  The inverse permutation is given by the mapping z -> 2*z^2 - z - 1. (End)

			0.93969262...

J. E. Maxfield and M. W. Maxfield, Abstract Algebra and Solution by Radicals, Dover Publications ISBN 0-486-67121-6, (1992), p. 197.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Michael Penn, Proof that cos(20°) is irrational, YouTube video, 2022.
Index entries for algebraic numbers, degree 3

Cf. A019819, A019844, A019859, A214778, A332437.

RealDigits[Sin[70 Degree],10,120][[1]] (* Harvey P. Dale, Aug 17 2012 *)

cos(Pi/9) \\ Charles R Greathouse IV, Aug 27 2017

Equals 2*A019844*A019864. - R. J. Mathar, Jan 17 2021
Equals cos(Pi/9) = (1/2)*A332437. - Peter Bala, Oct 21 2021
Equals 2F1(-1/6,1/6 ; 1/2; 3/4). - R. J. Mathar, Aug 31 2025

A332437 Decimal expansion of 2*cos(Pi/9).

Original entry on oeis.org

1, 8, 7, 9, 3, 8, 5, 2, 4, 1, 5, 7, 1, 8, 1, 6, 7, 6, 8, 1, 0, 8, 2, 1, 8, 5, 5, 4, 6, 4, 9, 4, 6, 2, 9, 3, 9, 8, 7, 2, 4, 1, 6, 2, 6, 8, 5, 2, 8, 9, 2, 9, 2, 6, 6, 1, 8, 0, 5, 7, 3, 3, 2, 5, 5, 4, 8, 4, 4, 2, 4, 2, 1, 9, 9, 1, 7, 7, 8, 9, 1, 7, 8, 9, 9, 4, 9, 1, 7, 7, 9, 6, 7, 5, 8, 9, 6, 1, 3, 4, 9

Offset: 1

Wolfdieter Lang, Mar 27 2020

This algebraic number called rho(9) of degree 3 = A055034(9) has minimal polynomial C(9, x) = x^3 - 3*x - 1 (see A187360).
rho(9) gives the length ratio diagonal/side of the smallest diagonal in the regular 9-gon.
The length ratio diagonal/side of the second smallest and the third smallest (or the largest) diagonal in the regular 9-gon are rho(9)^2 - 1 = A332438 - 1 and rho(9) + 1, respectively. - Mohammed Yaseen, Oct 31 2020

			rho(9) = 1.87938524157181676810821855464946293987241626852892926618...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 207.

Index entries for algebraic numbers, degree 3.

Cf. A019879, A055034, A056020, A156638, A187360, A214778, A215885, A332438.

RealDigits[2 * Cos[Pi/9], 10, 100][[1]] (* Amiram Eldar, Mar 27 2020 *)

2*cos(Pi/9) \\ Michel Marcus, Mar 28 2020

rho(9) = 2*cos(Pi/9).
Equals (-1)^(-1/9)*((-1)^(1/9) - i)*((-1)^(1/9) + i). - Peter Luschny, Mar 27 2020
Equals 2*A019879. - Michel Marcus, Mar 28 2020
Equals sqrt(A332438). - Mohammed Yaseen, Oct 31 2020
From Peter Bala, Oct 20 2021: (Start)
The zeros of x^3 - 3*x - 1 are r_1 = -2*cos(2*Pi/9), r_2 = -2*cos(4*Pi/9) and r_3 = -2*cos(8*Pi/9) = 2*cos(Pi/9).
The polynomial x^3 - 3*x - 1 is irreducible over Q (since it is irreducible mod 2) with discriminant equal to 3^4, a square. It follows that the Galois group of the number field Q(2*cos(Pi/9)) over Q is cyclic of order 3.
The mapping r -> 2 - r^2 cyclically permutes the zeros r_1, r_2 and r_3. The inverse cyclic permutation is given by r -> r^2 - r - 2.
The first differences r_1 - r_2, r_2 - r_3 and r_3 - r_1 are the zeros of the cyclic cubic polynomial x^3 - 9*x - 9 of discriminant 3^6.
First quotient relations:
r_1/r_2 = 1 + (r_3 - r_1); r_2/r_3 = 1 + (r_1 - r_2); r_3/r_1 = 1 + (r_2 - r_3);
r_2/r_1 = (r_3 - r_2) - 2; r_3/r_2 = (r_1 - r_3) - 2; r_1/r_3 = (r_2 - r_1) - 2;
r_1/r_2 + r_2/r_3 + r_3/r_1 = 3; r_2/r_1 + r_3/r_2 + r_1/r_3 = -6.
Thus the first quotients r_1/r_2, r_2/r_3 and r_3/r_1 are the zeros of the cyclic cubic polynomial x^3 - 3*x^2 - 6*x - 1 of discriminant 3^6. See A214778.
Second quotient relations:
(r_1*r_2)/(r_3^2) = 3*r_2 - 6*r_1 - 8, with two other similar relations by cyclically permuting the 3 zeros. The three second quotients are the zeros of the cyclic cubic polynomial x^3 + 24*x^2 + 3*x - 1 of discriminant 3^10.
(r_1^2)/(r_2*r_3) = 1 - 3*(r_2 + r_3), with two other similar relations by cyclically permuting the 3 zeros. (End)
Equals i^(2/9) + i^(-2/9). - Gary W. Adamson, Jun 25 2022
Equals Re((4+4*sqrt(3)*i)^(1/3)). - Gerry Martens, Mar 19 2024
From Amiram Eldar, Nov 22 2024: (Start)
Equals Product_{k>=1} (1 - (-1)^k/A056020(k)).
Equals 1 + Product_{k>=1} (1 + (-1)^k/A156638(k)). (End)

A214779 a(n) = 3*a(n-2) - a(n-3) with a(0)=-1, a(1)=1, a(2)=-4.

Original entry on oeis.org

-1, 1, -4, 4, -13, 16, -43, 61, -145, 226, -496, 823, -1714, 2965, -5965, 10609, -20860, 37792, -73189, 134236, -257359, 475897, -906313, 1685050, -3194836, 5961463, -11269558, 21079225, -39770137, 74507233, -140389636, 263291836, -495676141, 930265144

Offset: 0

Roman Witula, Jul 28 2012

Ramanujan-type sequence number 2 for argument 2Pi/9 is connected with the sequence A214699 (see also sequences A006053, A214683) - all have "similar" trigonometric description, for example in the case of a(n) the following formula hold true: 9^(1/3)*a(n) = (c(1)/c(4))^(1/3)*c(1)^n + (c(2)/c(1))^(1/3)*c(2)^n + (c(4)/c(2))^(1/3)*c(4)^n = -( (c(2)/c(1))^(1/3)*c(1)^(n+1) + (c(4)/c(2))^(1/3)*c(2)^(n+1) + (c(1)/c(4))^(1/3)*c(4)^(n+1) ), where c(j) := 2*Cos(2Pi*j/9) - for the proof see Witula et al.'s papers.
From a(0),A214699(0),a(2) and c(1)+c(2)+c(4)=0 we deduce
x^3 - 9^(1/3)*x - 1 = (x - (c(1)/c(2))^(1/3))*(x - (c(2)/c(4))^(1/3))*(x - (c(4)/c(1))^(1/3)), and
x^3 - 7*9^(1/3)*x - 1 = (x - (c(1)/c(2))^(1/3)*c(1)^2)*(x - (c(2)/c(4))^(1/3)*c(2)^2)*(x - (c(4)/c(1))^(1/3)*c(4)^2). We note that applying the Newton-Girard formulas to these polynomials two new sequences of real numbers can be discussed: X(n) := (c(1)/c(2))^(n/3) + (c(2)/c(4))^(n/3) + (c(4)/c(1))^(n/3), and Y(n) := ((c(1)/c(2))^(1/3)*c(1)^2)^n + ((c(2)/c(4))^(1/3)*c(2)^2)^n + ((c(4)/c(1))^(1/3)*c(4)^2)^n, where X(n)=9^(1/3)*X(n-2)+X(n-3), X(0)=3, X(1)=0, X(2)=2*9^(1/3), Y(n)=7*9^(1/3)Y(n-2)+Y(n-3), Y(0)=3, Y(1)=0, Y(2)=14*9^(1/3). It could be obtained the following decompositions: X(n) = ax(n) + 9^(1/3)*bx(n) + 81^(1/3)*cx(n), ax(0)=3, bx(0)=cx(0)=ax(1)=bx(1)=cx(1)=ax(2)=bx(2)=0, cx(2)=2, ax(n)=ax(n-3)+9*cx(n-2), bx(n)=bx(n-3)+ax(n-2), cx(n)=cx(n-3)+bx(n-2), and  Y(n) = ay(n) + 9^(1/3)*by(n) + 81^(1/3)*cy(n), ay(0)=3, by(0)=cy(0)=ay(1)=by(1)=cy(1)=ay(2)=cy(2)=0, by(2)=14, ay(n)=ay(n-3)+63*cy(n-2), by(n)=by(n-3)+7*ay(n-2), cy(n)=cy(n-3)+7*by(n-2). All these new sequence of positive integers ax(n),bx(n),...,cy(n) will be presented separately as A214778, A214951, A214954. - Roman Witula, Sep 27 2012
We note that all sums a(n+1) + a(n) are divisible by 3, which easily from recurrence formula for a(n) follows. Then it can be deduced the formula a(n+1) + a(n) = -A214699(n). - Roman Witula, Oct 06 2012

			From a(0)=-1 and A214699(0)=0 we obtain (c(1)/c(4))^(2/3) + (c(2)/c(1))^(2/3) + (c(4)/c(2))^(2/3) = 3*3^(1/3), whereas from a(1)=-1 and A214699(1)=3*3^(1/3) we get (c(1)/c(4))^(2/3)*2c(2) + (c(2)/c(1))^(2/3)*2c(4) + (c(4)/c(2))^(2/3)*2c(1) = 3*3^(1/3).

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.

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).

Cf. A214699, A214778, A214951, A214954, A217053, A217052, A217069.

LinearRecurrence[{0, 3, -1}, {-1, 1, -4}, 40] (* T. D. Noe, Jul 30 2012 *)

G.f.: -(1-x+x^2)/(1-3*x^2+x^3).

A214951 a(n) = 3a(n-1) + 6a(n-2) + a(n-3) with a(0)=2, a(1)=5, a(2)=26.

Original entry on oeis.org

2, 5, 26, 110, 491, 2159, 9533, 42044, 185489, 818264, 3609770, 15924383, 70250033, 309906167, 1367143082, 6031116281, 26606113502, 117372181274, 517784341115, 2284192224491, 10076654901437, 44452902392372, 196102828810229, 865102555686356, 3816377542312814

Offset: 0

Roman Witula, Jul 30 2012

Ramanujan-type sequence number 4 for the argument 2*Pi/9 is defined by the following relation: 9^(1/3)*a(n)=(c(1)/c(2))^(n - 1/3) + (c(2)/c(4))^(n - 1/3) + (c(4)/c(1))^(n - 1/3), where c(j) := Cos(2Pi*j/9) - for the proof see Witula et al.'s papers. We have a(n)=bx(3n-1), where the sequence bx(n) and its two conjugate sequences ax(n) and cx(n) are defined in the comments to the sequence A214779. We note that ax(3n-1)=cx(3n-1)=0. Moreover we have ax(3n)=A214778(n), bx(3n)=cx(3n)=0 and cx(3n+1)=A214954(n), ax(3n+1)=bx(3n+1)=0.

			We have 2*9^(1/3) = (c(2)/c(1))^(1/3) + (c(4)/c(2))^(1/3) + (c(1)/c(4))^(1/3), 5*9^(1/3) = (c(1)/c(2))^(2/3) + (c(2)/c(4))^(2/3) + (c(4)/c(1))^(2/3), and 110*9^(1/3)=(c(1)/c(2))^(8/3) + (c(2)/c(4))^(8/3) + (c(4)/c(1))^(8/3). Moreover we obtain a(6)-a(2)-a(1)-a(0)=9500, a(12)-a(2)-a(1)-a(0)=70250000 and a(12)-a(6)=3^3*43*a(1)*a(3)^2. - _Roman Witula_, Oct 06 2012

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. (in review)

Roman Witula, Ramanujan Type Trigonometric Formulae, Demonstratio Math. 45 (2012) 779-796.
Index entries for linear recurrences with constant coefficients, signature (3,6,1).

Cf. A214778, A214779, A214954, A214699, A217053, A217052, A217069.

LinearRecurrence[{3, 6, 1}, {2, 5, 26}, 40] (* T. D. Noe, Jul 30 2012 *)

Vec((2-x-x^2)/(1-3*x-6*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Oct 01 2012

G.f.: (2-x-x^2)/(1-3*x-6*x^2-x^3).

a(n+1) - a(n) = A214778(n+1). - Roman Witula, Oct 06 2012

A214954 a(n) = 3a(n-1) + 6a(n-2) + a(n-3), with a(0) = 0, a(1) = 2, and a(2) = 7.

Original entry on oeis.org

0, 2, 7, 33, 143, 634, 2793, 12326, 54370, 239859, 1058123, 4667893, 20592276, 90842309, 400748476, 1767891558, 7799007839, 34405121341, 151777302615, 669561643730, 2953753868221, 13030408769658, 57483311162030, 253586139972259, 1118688695658615

Offset: 0

Roman Witula, Jul 30 2012

Ramanujan-type sequence number 5 for the argument 2*Pi/9 is defined by the following relation: 81^(1/3)*a(n)=(c(1)/c(2))^(n + 1/3) + (c(2)/c(4))^(n + 1/3) + (c(4)/c(1))^(n + 1/3), where c(j) := Cos(2Pi*j/9) - for the proof see Witula's et al. papers. We have a(n)=cx(3n+1), where the sequence cx(n) and its two conjugate sequences ax(n) and bx(n) are defined in the comments to the sequence A214779. We note that ax(3n+1)=bx(3n+1)=0. Further we have ax(3n)=A214778(n), bx(3n)=cx(3n)=0 and bx(3n-1)=A214951(n), ax(3n-1)=cx(3n-1)=0.

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. (in review)

Roman Witula, Ramanujan Type Trigonometric Formulae, Demonstratio Math. 45 (2012) 779-796.
Index entries for linear recurrences with constant coefficients, signature (3,6,1).

Cf. A214779, A214778, A214951, A214699.

LinearRecurrence[{3, 6, 1}, {0, 2, 7}, 40] (* T. D. Noe, Jul 30 2012 *)
CoefficientList[Series[(2x+x^2)/(1-3x-6x^2-x^3),{x,0,30}],x] (* Harvey P. Dale, Sep 13 2021 *)

Vec((2*x+x^2)/(1-3*x-6*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Oct 01 2012

G.f.: (2*x+x^2)/(1-3*x-6*x^2-x^3).

A217052 a(n) = 3a(n-1) + 24a(n-2) + a(n-3), with a(0)=a(1)=1, and a(2)=19.

Original entry on oeis.org

1, 1, 19, 82, 703, 4096, 29242, 186733, 1266103, 8309143, 55500634, 367187437, 2441886670, 16193659132, 107553444913, 713750040577, 4738726458775, 31453733795086, 208804386436435, 1386041496850144, 9200883498819958, 61076450807299765, 405436597890428431

Offset: 0

Roman Witula, Sep 25 2012

The Ramanujan type sequence number 10 for the argument 2*Pi/9 defined by the relation a(n) = ((1/3 - c(1))^n + (1/3 - c(2))^n + (1/3 - c(4))^n)*3^(n-1), where c(j) := 2*cos(2*Pi*j/9). We note that c(4) = -cos(Pi/9). The conjugate with a(n) are sequences A217053 and A217069.
  For more informations about connections a(n) with these two sequences - see comments in A217053.
The 3-valuation of the sequence a(n) is equal to (1).

			We have a(4)=37*a(2) and a(5) = 2^(12), which implies (1/3 - c(1))^4 + (1/3 - c(2))^4 + (1/3 - c(4))^4 = (37/9)*((1/3 - c(1))^2 + (1/3 - c(2))^2 + (1/3 - c(4))^2) = (37/27)*19 = 703/27, (1/3 - c(1))^5 + (1/3 - c(2))^5 + (1/3 - c(4))^5 = (8/3)^4. Moreover we have a(10) = 676837*a(3).

Roman Witula, E. Hetmaniok, and D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the 15th International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012, in review.

Index entries for linear recurrences with constant coefficients, signature (3,24,1).

Cf. A217053, A217069, A214699, A214779, A214778, A214951, A214954, A215560, A214569, A215572.

LinearRecurrence[{3,24,1}, {1,1,19}, 30]

Vec((1-2*x-8*x^2)/(1-3*x-24*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Oct 01 2012

G.f.: (1-2*x-8*x^2)/(1-3*x-24*x^2-x^3).

A217053 a(n) = 3a(n-1) + 24a(n-2) + a(n-3), with a(0) = 2, a(1) = 5, and a(2) = 62.

Original entry on oeis.org

2, 5, 62, 308, 2417, 14705, 102431, 662630, 4460939, 29388368, 195890270, 1297452581, 8623112591, 57204089987, 379864424726, 2521114546457, 16737293922782, 111098495308040, 737511654617345, 4895636145167777, 32498286641627651, 215727639063526946

Offset: 0

Roman Witula, Sep 25 2012

The Ramanujan type sequence number 9 for the argument 2*Pi/9 defined by the following relation a(n)*3^(-n + 1/3) = ((-1)^n)*(c(1) - 1/3)^(n + 2/3) + ((-1)^n)*(c(2) - 1/3)^(n + 2/3) + (1/3 - c(4))^(n + 2/3), where c(j) := 2*cos(2*Pi*j/9). We note that c(4) = -cos(Pi/9). The conjugate with a(n) are the sequences A217052 and A217069.
In Witula's et al.'s paper the following sequence is discussed: S(n) := sum{j=0,1,2} (1/3 - c(2^j))^(n/3). It is proved that S(n+3) = (3^(1/3))*S(n+1) + S(n)/3, n=0,1,... Moreover the following decomposition holds true S(n) = x(n) + y(n)*3^(1/3) + z(n)*9^(1/3), which implies the system of recurrence relations: x(n+3) = 3*z(n+1) + x(n)/3, y(n+3) = x(n+1) + y(n)/3, z(n+3) = y(n+1) + z(n)/3, x(0)=3, y(0)=z(0)=x(1)=y(1)=z(1)=x(2)=z(2)=0, y(2)=2. Then it can be generated the relations X'(n+9) - 3*X'(n+6) - 24*X'(n+3) - X'(n) = 0, where X'(n) = X(n)*3^n for every X=x,y,z and n=0,1,..., from which we obtain that x(3*n+1)=x(3*n+2)=y(3*n)=y(3*n+1)=z(3*n)=z(3*n+2)=0 and a(n) = y(3*n+2)*3^n, A217052(n) = x(3*n)*3^(n-1), and A217069(n) = z(3*n+1)*3^(n-1).
Each a(n)+1 is divisible by 3 but which a(n)+1 are divisible by 9 - it is a question?

			We have 2*3^(1/3) = (c(1) - 1/3)^(2/3) + (c(2) - 1/3)^(2/3) + (1/3 - c(4))^(2/3), and 5*3^(-2/3) = -(c(1) - 1/3)^(5/3) - (c(2) - 1/3)^(5/3) + (1/3 - c(4))^(5/3).
Moreover we have 12*a(1) + a(0) = a(2), 5*a(2) = a(3) + a(0).

Roman Witula, E. Hetmaniok, and D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the 15th International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012, in review.

Index entries for linear recurrences with constant coefficients, signature (3,24,1).

Cf. A217052, A217069, A214699, A214778, A214779, A214951, A214954, A215560, A215569, A215572.

LinearRecurrence[{3,24,1}, {2,5,62}, 30]

Vec((2-x-x^2)/(1-3*x-24*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Oct 01 2012

G.f.: (2-x-x^2)/(1-3*x-24*x^2-x^3).

A217069 a(n) = 3a(n-1) + 24a(n-2) + a(n-3), with a(0)=0, a(1)=2, and a(2)=7.

Original entry on oeis.org

0, 2, 7, 69, 377, 2794, 17499, 119930, 782560, 5243499, 34631867, 230522137, 1527974718, 10151087309, 67355177296, 447219602022, 2968334148479, 19705628071261, 130804123379301, 868315777996646, 5763951923164423, 38262238564792074, 253989877628319020

Offset: 0

Roman Witula, Sep 26 2012

The Ramanujan sequence number 11 for the argument 2*Pi/9 defined by the relation a(n)*9^(1/3) = (3^(n-1))*(((-1)^(n-1))*(c(1) - 1/3)^(n + 1/3) + ((-1)^(n-1))*(c(2) - 1/3)^(n + 1/3) + (1/3 - c(4))^(n + 1/3)), where c(j) := 2*cos(2*Pi*j/9). The sequences A217052 and A217053 are conjugate with the sequence a(n). For more information on these connections - see Comments in A217053.

The 3-valuation of the sequence a(n) is equal to (0,2,1).

			We have a(4)-5*a(3)=32, 8*a(4)-a(5)=222, a(9)-a(6)=5226000. Furthermore from a(0)=0 we get (c(1) - 1/3)^( 1/3) + (c(2) - 1/3)^(1/3) = (1/3 - c(4))^(1/3), while from a(3)=69 we obtain 23*9^(-1/6) = (c(1) - 1/3)^(10/3) + (c(2) - 1/3)^(10/3) + (1/3 - c(4))^(10/3).

Roman Witula, E. Hetmaniok, and D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the 15th International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012, in review.

Index entries for linear recurrences with constant coefficients, signature (3,24,1).

Cf. A217052, A217053, A214699, A214779, A214778, A214951, A214954, A215560, A215569, A215572.

Mathematica
```
LinearRecurrence[{3,24,1}, {0,2,7}, 30]
```

Vec((2+x)/(1-3*x-24*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 27 2012

G.f.: x*(2+x)/(1-3*x-24*x^2-x^3).

cp's OEIS Frontend

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

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A019879 Decimal expansion of sine of 70 degrees.

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A332437 Decimal expansion of 2*cos(Pi/9).

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A214779 a(n) = 3*a(n-2) - a(n-3) with a(0)=-1, a(1)=1, a(2)=-4.

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A214951 a(n) = 3*a(n-1) + 6*a(n-2) + a(n-3) with a(0)=2, a(1)=5, a(2)=26.

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A214954 a(n) = 3*a(n-1) + 6*a(n-2) + a(n-3), with a(0) = 0, a(1) = 2, and a(2) = 7.

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A217052 a(n) = 3*a(n-1) + 24*a(n-2) + a(n-3), with a(0)=a(1)=1, and a(2)=19.

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A214951 a(n) = 3a(n-1) + 6a(n-2) + a(n-3) with a(0)=2, a(1)=5, a(2)=26.

A214954 a(n) = 3a(n-1) + 6a(n-2) + a(n-3), with a(0) = 0, a(1) = 2, and a(2) = 7.

A217052 a(n) = 3a(n-1) + 24a(n-2) + a(n-3), with a(0)=a(1)=1, and a(2)=19.

A217053 a(n) = 3a(n-1) + 24a(n-2) + a(n-3), with a(0) = 2, a(1) = 5, and a(2) = 62.

A217069 a(n) = 3a(n-1) + 24a(n-2) + a(n-3), with a(0)=0, a(1)=2, and a(2)=7.