A217069 -id:A217069 - 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)

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

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).

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

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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.

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.