A215829 -id:A215829 - cp's OEIS frontend

A215831 a(n) = 3^(-1-floor(n/3))*A215829(n).

1, -1, 9, -11, 59, -267, 433, -2041, 9753, -15443, 73547, -349875, 555073, -2641297, 12569673, -19938491, 94883195, -451526331, 716237425, -3408408073, 16219834713, -25728821507, 122437560587, -582652240611, 924236100865, -4398227463841, 20930155058697, -33200601349355

Offset: 0

Roman Witula, Aug 24 2012

The Berndt-type sequence number 8a for the argument 2Pi/9 (see A215829 for details).

From the recurrence relation for A215829 it can be proved that a(3*n+2) is divisible by 3, a(3*n) is congruent to 1 modulo 3, and a(3*n+1) is congruent to 2 modulo 3, which implies that a(3*n)+a(3*n+1) is divisible by 3.

			We have a(6)+3*a(3)=400, while a(30)+3*a(3) is divisible by 1000.

Index entries for linear recurrences with constant coefficients, signature (0,0,-33,0,0,105,0,0,1).

i:=28; I:=[3,-3,27]; A215829:=[m le 3 select I[m] else -3*Self(m-1)+9*Self(m-2)+3*Self(m-3): m in [1..i]]; [3^(-1-Floor((n-1)/3))*A215829[n]: n in [1..i]]; // Bruno Berselli, Oct 02 2012

LinearRecurrence[{0,0,-33,0,0,105,0,0,1},{1,-1,9,-11,59,-267,433,-2041,9753},30] (* Harvey P. Dale, Nov 29 2013 *)

G.f.: (3*x^8-11*x^7+35*x^6-30*x^5-26*x^4-22*x^3-9*x^2+x-1)/(x^9+105*x^6-33*x^3-1). [Colin Barker, Oct 28 2012]

A215634 a(n) = - 6a(n-1) - 9a(n-2) - 3*a(n-3) with a(0)=3, a(1)=-6, a(2)=18.

Original entry on oeis.org

3, -6, 18, -63, 234, -891, 3429, -13257, 51354, -199098, 772173, -2995218, 11619045, -45073827, 174857211, -678335958, 2631522330, -10208681991, 39603398850, -153636822171, 596016389349, -2312177133105, 8969825761002

Offset: 0

Roman Witula, Aug 18 2012

The Berndt-type sequence number 2 for the argument 2Pi/9 . Similarly like the respective sequence number 1 -- see A215455 -- the sequence a(n) is connected with the following general recurrence relation: X(n+3) + 6*X(n+2) + 9*X(n+1) + ((2*cos(3*g))^2)*X(n) = 0, X(0)=3, X(1)=-6, X(2)=18. The Binet formula for this one has the form: X(n) = (-4)^n*((cos(g))^(2*n) + cos(g+Pi/3))^(2*n) + cos(g-Pi/3))^(2*n)) - for details see Witula-Slota's reference and comments to A215455.

The characteristic polynomial of a(n) has the form x^3 + 6*x^2 + 9*x + 3 = (x + (2*cos(Pi/18))^2)*(x+(2*cos(5*Pi/18))^2)*(x+(2*cos(7*Pi/18))^2). We note that (2*cos(Pi/18))^2 = 2 - c(4), (2*cos(5*Pi/18))^2 = 2 - c(2), and (2*cos(7*Pi/18))^2 = 2 - c(1), where c(j) = 2*cos(2*Pi*j/9) - see trigonometric relations for A215455. Furthermore all numbers a(n)*3^(-ceiling((n+1)/3)) are integers.

R. Witula, On some applications of formulas for sums of the unimodular complex numbers, Wyd. Pracowni Komputerowej Jacka Skalmierskiego, Gliwice 2011 (in Polish).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
R. Witula, D. Slota, On modified Chebyshev polynomials, J. Math. Anal. Appl., 324 (2006), 321-343.
Index entries for linear recurrences with constant coefficients, signature (-6,-9,-3).

Cf. A215455, A215635, A215636, A215664, A215885, A215665, A215666, A215829, A215831, A215917, A215919, A215945, A216034, A215948, A216757.

I:=[3,-6,18]; [n le 3 select I[n] else -6*Self(n-1)-9*Self(n-2)-3*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Aug 30 2017

LinearRecurrence[{-6,-9,-3}, {3,-6,18}, 50]
CoefficientList[Series[(3 + 12 x + 9 x^2)/(1 + 6 x + 9 x^2 + 3 x^3), {x, 0, 33}], x] (* Vincenzo Librandi, Aug 30 2017 *)

Vec((3+12*x+9*x^2)/(1+6*x+9*x^2+3*x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 27 2012

a(n) = (-4)^n*((cos(Pi/18))^(2*n) + (cos(5*Pi/18))^(2*n) + (cos(7*Pi/18))^(2*n)).
G.f.: (3 + 12*x + 9*x^2)/(1 + 6*x + 9*x^2 + 3*x^3).
a(n)*(-1)^n = s(1)^(2*n) + s(2)^(2*n) + s(4)^(2*n), where s(j) := 2*sin(2*Pi*j/9) -- for the proof see Witula's book. The respective sums with odd powers of sines in A216757 are given. - Roman Witula, Sep 15 2012

A215636 a(n) = - 12a(n-1) - 54a(n-2) - 112a(n-3) - 105a(n-4) - 36a(n-5) - 2a(n-6) with a(0)=a(1)=a(2)=0, a(3)=-3, a(4)=24, a(5)=-135.

Original entry on oeis.org

0, 0, 0, -3, 24, -135, 660, -3003, 13104, -55689, 232500, -958617, 3916440, -15890355, 64127700, -257698347, 1032023136, -4121456625, 16421256420, -65301500577, 259259758056, -1027901275131, 4070632899300, -16104283594083, 63657906293520, -251447560563465, 992593021410900

Offset: 0

Roman Witula, Aug 18 2012

The Berndt-type sequence number 4 for the argument 2*Pi/9 defined by the relation: X(n) = b(n) + a(n)*sqrt(2), where X(n) := ((cos(Pi/24))^(2*n) + (cos(7*Pi/24))^(2*n) + (cos(3*Pi/8))^(2*n))*(-4)^n. We have b(n) = A215635(n) (see also section "Example" below). For more details - see comments to A215635, A215634 and Witula-Slota's reference.

			We have X(1)=-6, X(2)=18 and X(3)=-60-3*sqrt(2), which implies the equality: (cos(Pi/24))^6 + (cos(7*Pi/24))^6 + (cos(3*Pi/8))^6 = (60+3*sqrt(2))/64.

Roman Witula and D. Slota, On modified Chebyshev polynomials, J. Math. Anal. Appl., 324 (2006), 321-343.
Index entries for linear recurrences with constant coefficients, signature (-12,-54,-112,-105,-36,-2).

Cf. A215455, A215634, A215635, A215664, A215885, A215665, A215666, A215829, A215831, A215917, A215919, A215945, A216034, A215948, A216757.

LinearRecurrence[{-12,-54,-112,-105,-36,-2}, {0,0,0,-3,24,-135}, 50]

G.f.: (-3*x^3-12*x^4-9*x^5)/(1+12*x+54*x^2+112*x^3+105*x^4+36*x^5+2*x^6).

A215945 a(n) = - 3^nA(2n+1), where A(n) = 3*A(n-1) + A(n-2) - A(n-3)/3, with A(0)=A(1)=3, A(2)=11.

Original entry on oeis.org

-3, -105, -3387, -108945, -3504051, -112702329, -3624894315, -116589061665, -3749904995427, -120609834867081, -3879226882922139, -124769271310005681, -4013008656895890963, -129072153032843014809, -4151404124161560449739

Offset: 0

Roman Witula, Aug 28 2012

The Berndt-type sequence number 11 for the argument 2Pi/9 connecting with the following sequence T(n) := t(1)^n + (-t(2))^n + t(4)^n, where t(j) := tan(2*Pi*j/9). More precisely we have sqrt(3)*a(n) = T(2*n+1) = (-sqrt(3))^n*W(n;2*i/sqrt(3)), where W(n;d) := (1 + 2*i*d*s(1))^n
+ (1 - 2*i*d*s(2))^n + (1 + 2*i*d*s(4))^n, n=0,1,..., d in C. For example we have W(0;d)=W(1;d)=3, W(2;d)=3-6*d^2. The characteristic polynomial of W(n;d) has the form ((x-1)-2*i*d*s(1))*((x-1)-2*i*d*s(2))*((x-1)-2*i*d*s(4)) = (x-1)^3 + 3*d^2*(x-1) - sqrt(3)*i*d^3 = x^3 - 3*x^2 + 3*(1+d^2)*x - 1 - 3*d^2 - sqrt(3)*i*d^3, which implies the recurrence form of W(n;d) = 3*W(n-1;d) - 3*(1+d^2)*W(n-2;d) + (1+3*d^2+sqrt(3)*i*d^3)*W(n-3;d). In consequence we obtain the title recurrence relation for
A(n) := W(n;2*i/sqrt(3)). The polynomials W(n;d) are equivalent of the big omega function with index n of argument d defined and discussed in Witula-Slota's paper (Section 6) and in comments to A215794.
We note that all numbers a(n)/3 and 3^(-1+floor((n+1)/3))*A(n) = A216034(n) are integers.
The following decomposition hold true: (X - t(1)^n)*(X - (-t(2))^n)*(X - t(4)^n) = X^3 - (-sqrt(3))^n*A(n)*X^2 + A215829(n)*X - sqrt(3)^n. Moreover we have A215829(n) = (T(n)^2 - T(2*n))/2, which implies A215829(2*n+1) = (3*a(n)^2 - A215948(2*n+1))/2 and A215829(2*n) = (A215948(n)^2 - A215948(2*n))/2.

			We have 35*(t(1) - t(2) + t(3)) =  t(1)^3 - t(2)^3 + t(4)^3, t(1)^7 - t(2)^7 + t(4)^7 = -5*81*269*sqrt(3) and t(1)^9 - t(2)^9 + t(4)^9 = -9*389339*sqrt(3).

D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the ninth order, (submitted, 2012).
R. Witula, Ramanujan type formulas for arguments 2Pi/7 and 2Pi/9, Demonstratio Math. (in press, 2012).

Vincenzo Librandi, Table of n, a(n) for n = 0..600
Roman Witula and Damian Slota, New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7, Journal of Integer Sequences, Vol. 10 (2007), Article 07.5.6
Index entries for linear recurrences with constant coefficients, signature (33,-27,3).

Cf. A215948, A216034, A215829, A215794, A215575.

m:=17; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(-3*(1+x)^2/(1-33*x+27*x^2-3*x^3))); // Bruno Berselli, Aug 29 2012

I:=[-3, -105, -3387]; [n le 3 select I[n] else 33*Self(n-1)-27*Self(n-2)+3*Self(n-3): n in [1..20]]; // Vincenzo Librandi, Mar 19 2013

LinearRecurrence[{33, -27, 3}, {-3, -105, -3387}, 17] (* Bruno Berselli, Aug 29 2012 *)
CoefficientList[Series[-3 (1 + x)^2/(1 - 33 x + 27 x^2 - 3 x^3), {x, 0, 20}], x] (* Vincenzo Librandi, Mar 19 2013 *)

sqrt(3)*a(n) = t(1)^(2*n+1) - t(2)^(2*n+1) + t(4)^(2*n+1) = (-sqrt(3) + 4*s(1))^(2*n+1) + (-sqrt(3) - 4*s(2))^(2*n+1) + (-sqrt(3) + 4*s(4))^(2*n+1), where t(j) := tan(2*Pi*j/9) and s(j) := sin(2*Pi*j/9).
a(n) = 33*a(n-1) - 27*a(n-2) + 3*a(n-3).
G.f.: -3*(1+x)^2/(1-33*x+27*x^2-3*x^3). - Bruno Berselli, Aug 29 2012

A215948 a(n) = 3^nA(2n), where A(n) = 3*A(n-1) + A(n-2) - A(n-3)/3 with A(0)=A(1)=3, A(2)=11.

Original entry on oeis.org

3, 33, 1035, 33273, 1070163, 34420113, 1107069147, 35607149289, 1145248319907, 36835122733569, 1184744167018155, 38105444942752473, 1225602095969542131, 39419576386041628017, 1267869080483024344443, 40779027899804588036553, 1311593714249667872790339

Offset: 0

Roman Witula, Aug 28 2012

The Berndt-type sequence number 12 for the argument 2*Pi/9 defined by the first trigonometric relations from the section "Formula" below (it is the complement of the sequence A215945). For more information see comments to A215945. We note that all a(n)/3 and 3^(-1 + floor((n+3)/3))*A(n) = A216034(n) are integers.

			We have t(1)^4 + t(2)^4 + t(4)^4 = 1035 = (345/11)*(t(1)^2 + t(2)^2 + t(4)^2) and (1 - 4*s(1)/sqrt(3))^4 + (1 + 4*s(2)/sqrt(3))^4 + (1 - 4*s(4)/sqrt(3))^4 = 115. Moreover we get a(2)/a(1) = 31,(36), a(3)/a(1) = 1008,(27), a(4)/a(1) = 32429,(18).

D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the ninth order, (submitted, 2012).
R. Witula, Ramanujan type formulas for arguments 2Pi/7 and 2Pi/9, Demonstratio Math. (in press, 2012).

Roman Witula and Damian Slota, New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7, Journal of Integer Sequences, Vol. 10 (2007), Article 07.5.6.
Index entries for linear recurrences with constant coefficients, signature (33,-27,3).

Cf. A215945, A216034, A215829, A215794, A215575.

LinearRecurrence[{33,-27,3}, {3,33,1035}, 50]

a(n) = t(1)^(2*n) + t(2)^(2*n) + t(4)^(2*n) = (-sqrt(3) + 4*s(1))^(2*n) + (sqrt(3) + 4*s(2))^(2*n) + (-sqrt(3) + 4*s(4))^(2*n), where t(j) := tan(2*Pi*j/9) and s(j) := sin(2*Pi*j/9). For the respective sums of odd powers - see A215945.
a(n) = 33*a(n-1) - 27*a(n-2) + 3*a(n-3).
G.f.: 3*(1-22*x+9*x^2)/(1-33*x+27*x^2-3*x^3).
a(n) = cot(Pi/18)^(2*n) + cot(5*Pi/18)^(2*n) + cot(7*Pi/18)^(2*n). - Greg Dresden, Oct 01 2020

A216034 a(n) = 3^(-1+floor((n+1)/3))A(n), where A(n) = 3A(n-1) + A(n-2) - A(n-3)/3 with A(0)=A(1)=3, A(2)=11.

Original entry on oeis.org

1, 1, 11, 35, 115, 1129, 3697, 12105, 118907, 389339, 1274819, 12522481, 41002561, 134255345, 1318783307, 4318113395, 14138868147, 138885370201, 454754601649, 1489010307001, 14626471197755, 47891689912619, 156812530628611, 1540361374197601

Offset: 0

Roman Witula, Aug 30 2012

The Berndt-type sequence number 11a for the argument 2Pi/9 - see A215945 for more details.

Index entries for linear recurrences with constant coefficients, signature (0,0,105,0,0,33,0,0,-1).

Cf. A215945, A215948, A215829, A215794, A215575.

i:=24; I:=[3,3,11]; A:=[m le 3 select I[m] else 3*Self(m-1)+Self(m-2)-Self(m-3)/3: m in [1..i]]; [3^(-1+Floor(n/3))*A[n]: n in [1..i]]; // Bruno Berselli, Oct 02 2012

G.f.: (1+x+11*x^2-70*x^3+10*x^4-26*x^5-11*x^6-3*x^7-x^8)/(1-105*x^3-33*x^6+x^9). [Bruno Berselli, Oct 02 2012]

A217336 a(n) = 3^(-1+floor(n/2))A(n), where A(n) = 3A(n-1) + A(n-2) - A(n-3)/3 with A(0)=A(1)=3, A(2)=11.

Original entry on oeis.org

1, 1, 11, 35, 345, 1129, 11091, 36315, 356721, 1168017, 11473371, 37567443, 369023049, 1208298105, 11869049763, 38863020555, 381749439969, 1249968331809, 12278374244523, 40203278289027, 394914722339385, 1293075627640713

Offset: 0

Roman Witula, Oct 01 2012

The Berndt-type sequence number 14 for the argument 2Pi/9 defined by the relation: A(n)*(-sqrt(3))^n = t(1)^n + (-t(2))^n + t(4)^n = (-sqrt(3) + 4*s(1))^n + (-sqrt(3) - 4*s(2))^n + (-sqrt(3) + 4*s(4))^n, where s(j) := sin(2*Pi*j/9) and t(j) := tan(2*Pi*j/9).
The definitions of the other Berndt-type sequences for the argument 2Pi/9 like A215945, A215948, A216034 in Crossrefs are given.
We note that all a(2*n), n=2,3,..., are divisible by 3, and it is only when n=5 that a(2*n) is divisible by 9.

			Note that A(0)=A(1)=3, a(0)=a(1)=1, A(2)=a(2)=11, A(3)=a(3)=35, A(4)=115, a(4)=345 and A(5) = 1129/3, which implies the equality  3387*sqrt(3) = -t(1)^5 + t(2)^5 - t(4)^5.

D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the ninth order, (submitted, 2012).
R. Witula, Ramanujan type formulas for arguments 2Pi/7 and 2Pi/9, Demonstratio Math. (in press, 2012).

Paolo Xausa, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,33,0,-27,0,3).

Cf. A215455, A215634-A215636, A215664, A215885, A215665, A215666, A215829, A215831, A215917, A215919, A215945, A216034, A215948, A216757.

/* By definition: */ i:=22; I:=[3,3,11]; A:=[m le 3 select I[m] else 3*Self(m-1)+Self(m-2)-Self(m-3)/3: m in [1..i]]; [3^(-1+Floor((n-1)/2))*A[n]: n in [1..i]]; // Bruno Berselli, Oct 02 2012

LinearRecurrence[{0, 33, 0, -27, 0, 3}, {1, 1, 11, 35, 345, 1129},25] (* Paolo Xausa, Feb 23 2024 *)

G.f.: (1+x-22*x^2+2*x^3+9*x^4+x^5)/(1-33*x^2+27*x^4-3*x^6). - Bruno Berselli, Oct 01 2012

cp's OEIS Frontend

A215831 a(n) = 3^(-1-floor(n/3))*A215829(n).

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A215634 a(n) = - 6*a(n-1) - 9*a(n-2) - 3*a(n-3) with a(0)=3, a(1)=-6, a(2)=18.

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A215636 a(n) = - 12*a(n-1) - 54*a(n-2) - 112*a(n-3) - 105*a(n-4) - 36*a(n-5) - 2*a(n-6) with a(0)=a(1)=a(2)=0, a(3)=-3, a(4)=24, a(5)=-135.

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A215945 a(n) = - 3^n*A(2*n+1), where A(n) = 3*A(n-1) + A(n-2) - A(n-3)/3, with A(0)=A(1)=3, A(2)=11.

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A215948 a(n) = 3^n*A(2*n), where A(n) = 3*A(n-1) + A(n-2) - A(n-3)/3 with A(0)=A(1)=3, A(2)=11.

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A216034 a(n) = 3^(-1+floor((n+1)/3))*A(n), where A(n) = 3*A(n-1) + A(n-2) - A(n-3)/3 with A(0)=A(1)=3, A(2)=11.

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A217336 a(n) = 3^(-1+floor(n/2))*A(n), where A(n) = 3*A(n-1) + A(n-2) - A(n-3)/3 with A(0)=A(1)=3, A(2)=11.

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A215634 a(n) = - 6a(n-1) - 9a(n-2) - 3*a(n-3) with a(0)=3, a(1)=-6, a(2)=18.

A215636 a(n) = - 12a(n-1) - 54a(n-2) - 112a(n-3) - 105a(n-4) - 36a(n-5) - 2a(n-6) with a(0)=a(1)=a(2)=0, a(3)=-3, a(4)=24, a(5)=-135.

A215945 a(n) = - 3^nA(2n+1), where A(n) = 3*A(n-1) + A(n-2) - A(n-3)/3, with A(0)=A(1)=3, A(2)=11.

A215948 a(n) = 3^nA(2n), where A(n) = 3*A(n-1) + A(n-2) - A(n-3)/3 with A(0)=A(1)=3, A(2)=11.

A216034 a(n) = 3^(-1+floor((n+1)/3))A(n), where A(n) = 3A(n-1) + A(n-2) - A(n-3)/3 with A(0)=A(1)=3, A(2)=11.

A217336 a(n) = 3^(-1+floor(n/2))A(n), where A(n) = 3A(n-1) + A(n-2) - A(n-3)/3 with A(0)=A(1)=3, A(2)=11.