A215455 -id:A215455 - cp's OEIS frontend

A215575 a(n) = 7*(a(n-1) - a(n-2) - a(n-3)), with a(0)=3, a(1)=7, a(2)=35.

3, 7, 35, 175, 931, 5047, 27587, 151263, 830403, 4560871, 25054435, 137642127, 756187747, 4154438295, 22824258947, 125395430335, 688917131651, 3784882096583, 20793986742179, 114241312597615, 627637106311971, 3448212648805239, 18944339609269571

Offset: 0

Roman Witula, Aug 16 2012

The Berndt-type sequence number 8 for the argument 2Pi/7 defined by trigonometric relations from "Formula" below.
We note that the following decompositions hold true: (X-cot(2*Pi/7)^n)*(X-cot(4*Pi/7)^n)*(X-cot(8*Pi/7)^n) = X^3 - sqrt(7)^(-n)*a(n)*X^2 + (-sqrt(7))^(-n)*B(n)*X
  - (-sqrt(7))^(-n), and (X-tan(2*Pi/7)^n)*(X-tan(4*Pi/7)^n)*(X-tan(8*Pi/7)^n) = X^3 - B(n)*X^2  + (-1)^n*a(n)*X - (-sqrt(7))^n, where B(n) := tan(2*Pi/7)^n + tan(4*Pi/7)^n + tan(8*Pi/7)^n = (-sqrt(7) + 4*sin(2*Pi/7))^n + (-sqrt(7) + 4*sin(4*Pi/7))^n + (-sqrt(7) + 4*sin(8*Pi/7))^n. Moreover we have 2*(-1)^n*B(n) = 7^(-n/2)*(a(n)^2 - a(2*n)). For the proof of these decompositions see Witula-Slota's (Section 6) and Witula's (Remark 11) reference.
We note that the numbers a(n)*7^(-ceiling(n/3)) are all integers. Moreover from the recurrence relation: a(n+3)+7*a(n+1)=7*(a(n+2)-a(n)) it can be easily obtained the following summations formulas: 8*sum{k=1,..,n} a(2*k) = 7*(a(2*n+1)-2)-a(2*n+2), which also means that the result is divisible by 8 for every n=1,2,..., and 8*sum{k=1,..,n} a(2*k-1) = 7*(a(2*n)-2)-a(2*n+1), which implies that 7*(a(n)-2)-a(n+1) is divisible by 8 for each n=0,1,...

			We have  cot(2*Pi/7)^2 + cot(4*Pi/7)^2 + cot(8*Pi/7)^2 = 5,  cot(2*Pi/7)^4 + cot(4*Pi/7)^4 + cot(8*Pi/7)^4 = 19, but cot(2*Pi/7)^6 + cot(4*Pi/7)^6 + cot(8*Pi/7)^6 = 563/7. Similarly the numbers sqrt(7)*(cot(2*Pi/7)^n + cot(4*Pi/7)^n + cot(8*Pi/7)^n) are integers for n=1,3,5,7 (equal to 7, 25, 103, 441, respectively), whereas for n=9 we obtain the rational value 13297/7.

E Hetmaniok, P Lorenc, S Damian, et al., Periodic orbits of boundary logistic map and new kind of modified Chebyshev polynomials in R. Witula, D. Slota, W. Holubowski (eds.), Monograph on the Occasion of 100th Birthday Anniversary of Zygmunt Zahorski. Wydawnictwo Politechniki Slaskiej, Gliwice 2015, pp. 325-343.

G. C. Greubel, Table of n, a(n) for n = 0..1350
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
Roman Witula, Ramanujan Type Trigonometric Formulas: The General Form for the Argument 2*Pi/7, Journal of Integer Sequences, Vol. 12 (2009), Article 09.8.5.
R. Witula, P. Lorenc, M. Rozanski, M. Szweda, Sums of the rational powers of roots of cubic polynomials, Zeszyty Naukowe Politechniki Slaskiej, Seria: Matematyka Stosowana z. 4, Nr. kol. 1920, 2014.
Index entries for linear recurrences with constant coefficients, signature (7, -7, -7).

Cf. A215007, A215008, A215143, A215493, A215494, A215510, A215512, A215455.

I:=[3,7,35]; [n le 3 select I[n] else 7*Self(n-1) - 7*Self(n-2) - 7*Self(n-3): n in [1..30]]; // G. C. Greubel, Apr 25 2017

LinearRecurrence[{7,-7,-7}, {3,7,35}, 50]

polsym(x^3 - 7*x^2 + 7*x + 7, 30) \\ Charles R Greathouse IV, Jul 20 2016

a(n) = (sqrt(7)^n)*(cot(2*Pi/7)^n + cot(4*Pi/7)^n + cot(8*Pi/7)^n) = (3 + 4*cos(2*Pi/7))^n + (3 + 4*cos(4*Pi/7))^n + (3 + 4*cos(8*Pi/7))^n = (-tan(2*Pi/7)*tan(4*Pi/7))^n + (-tan(2*Pi/7)*tan(8*Pi/7))^n + (-tan(4*Pi/7)*tan(8*Pi/7))^n.
G.f.: (3-14*x+7*x^2)/(1-7*x+7*x^2+7*x^3).
a(n) = x1^n + x2^n + x3^n, where x1, x2, x3 are the roots of the polynomial x^3 - 7*x^2 + 7*x + 7, that is, x1 = sqrt(7)/tan(Pi/7), x2 = sqrt(7)/tan(2*Pi/7), x3 = sqrt(7)/tan(4*Pi/7). - Kai Wang, Jul 19 2016

A215664 a(n) = 3*a(n-2) - a(n-3), with a(0)=3, a(1)=0, and a(2)=6.

Original entry on oeis.org

3, 0, 6, -3, 18, -15, 57, -63, 186, -246, 621, -924, 2109, -3393, 7251, -12288, 25146, -44115, 87726, -157491, 307293, -560199, 1079370, -1987890, 3798309, -7043040, 13382817, -24927429, 47191491, -88165104, 166501902, -311686803, 587670810, -1101562311

Offset: 0

Roman Witula, Aug 20 2012

The Berndt-type sequence number 5 for the argument 2Pi/9 defined by the first relation from the section "Formula" below. The respective sums with negative powers of the cosines form the sequence A215885. Additionally if we set b(n) = c(1)*c(2)^n + c(2)*c(4)^n + c(4)*c(1)^n and c(n) = c(4)*c(2)^n + c(1)*c(4)^n + c(2)*c(1)^n, where c(j):=2*cos(2*Pi*j/9), then the following system of recurrence equations holds true: b(n) - b(n+1) = a(n), a(n+1) - a(n) = c(n+1), a(n+2) - 2*a(n)=c(n). All three sequences satisfy the same recurrence relation: X(n+3) - 3*X(n+1) + X(n) = 0. Moreover we have a(n+1) + A215665(n) + A215666(n) = 0 since c(1) + c(2) + c(4) = 0, b(n)=A215665(n) and c(n)=A215666(n).
If X(n) = 3*X(n-2) - X(n-3), n in Z, with X(n) = a(n) for every n=0,1,..., then X(-n) = A215885(n) for every n=0,1,...
From initial values and the recurrence formula we deduce that a(n)/3 and a(3n+1)/9 are all integers. We have a(n)=3*(-1)^n *A188048(n) and a(2n)=A215455(n). Furthermore the following decomposition holds: (X - c(1)^n)*(X - c(2)^n)*(X - c(4)^n) = X^3 - a(n)*X^2 + ((a(n)^2 - a(2*n))/2)*X + (-1)^(n+1), which implies the relation (c(1)*c(2))^n + (c(1)*c(4))^n + (c(2)*c(4))^n = (-c(1))^(-n) + (-c(2))^(-n) + (-c(4))^(-n) = (a(n)^2 - a(2*n))/2.

			We have c(1)^2 + c(2)^2 + c(4)^2 + 2*(c(1)^3 + c(2)^3 + c(4)^3) = 0 and 3*a(7) + a(8) = a(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).

Michael De Vlieger, Table of n, a(n) for n = 0..3649
Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).
Index entries for linear recurrences with constant coefficients, signature (0,3,-1).

Cf. A215455, A215634, A215635, A215636, A215007, A214699, A214683.

Mathematica
```
LinearRecurrence[{0,3,-1}, {3,0,6}, 50]
```

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

a(n) = c(1)^n + c(2)^n + c(4)^n, where c(j) := 2*cos(2*Pi*j/9).

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

A215512 a(n) = 5a(n-1) - 6a(n-2) + a(n-3), with a(0)=1, a(1)=3, a(2)=8.

Original entry on oeis.org

1, 3, 8, 23, 70, 220, 703, 2265, 7327, 23748, 77043, 250054, 811760, 2635519, 8557089, 27784091, 90213440, 292919743, 951102166, 3088205812, 10027335807, 32558546329, 105716922615, 343260670908, 1114560365179, 3618954723062, 11750672095144, 38154192502527

Offset: 0

Roman Witula, Aug 14 2012

The Berndt-type sequence number 7 for the argument 2Pi/7 defined by the relation: sqrt(7)*a(n) = s(1)*c(4)^(2*n) + s(2)*c(1)^(2*n) + s(4)*c(2)^(2*n), where c(j):=2*cos(2*Pi*j/7) and s(j):=2*sin(2*Pi*j/7). If we additionally defined the following sequences:
sqrt(7)*b(n) = s(2)*c(4)^(2*n) + s(4)*c(1)^(2*n) + s(1)*c(2)^(2*n),
sqrt(7)*c(n) = s(4)*c(4)^(2*n) + s(1)*c(1)^(2*n) + s(2)*c(2)^(2*n), and
sqrt(7)*a1(n) = s(1)*c(4)^(2*n+1) + s(2)*c(1)^(2*n+1) + s(4)*c(2)^(2*n+1),
sqrt(7)*b1(n) = s(2)*c(4)^(2*n+1) + s(4)*c(1)^(2*n+1) + s(1)*c(2)^(2*n+1),
sqrt(7)*c1(n) = s(4)*c(4)^(2*n+1) + s(1)*c(1)^(2*n+1) + s(2)*c(2)^(2*n+1), then the following simple relationships between elements of these sequences hold true: a(n)=c1(n), c(n+1)=a1(n), -a(n)-b(n)=b1(n), which means that the sequences a1(n), b1(n), and c1(n) are completely and in very simple way determined by the sequences a(n), b(n) and c(n). However the last one's satisfy the following system of recurrence equations: a(n+1) = 2*a(n) + b(n), b(n+1) = a(n) + 2*b(n) - c(n), c(n+1) = c(n) - b(n). We have b(n)=A215694(n) and c(n)=A215695(n).
We note that a(n)=A000782(n) for every n=0,1,...,4 and A000782(5)-a(5)=2.
From general recurrence relation: a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3), i.e. a(n) = 5*(a(n-1)-a(n-2)) + (a(n-3)-a(n-2)) the following summation formula can be easily obtained: sum{k=3,..,n} a(k) = 5*a(n-1)-a(n-2)+a(0)-5*a(1). Hence in discussed sequence it follows that: sum{k=3,..,n} a(k) = 5*a(n-1) - a(n-2) - 14.

			We have a(6) = 10*a(4)+a(1), a(5) = 11*(a(3)-a(1)), a(10)-a(4)+a(3)+a(1)+a(0) = 77*10^3, and a(11)-a(4)+a(3)-a(2)+a(0) = 25*10^4 = (5^6)*(2^4).

G. C. Greubel, Table of n, a(n) for n = 0..1000
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 (5, -6, 1).

Cf.A215694, A215695, A215007, A215008, A215143, A215493, A215494, A215510, A215575, A215455, A214683, A214699.

I:=[1,3,8]; [n le 3 select I[n] else 5*Self(n-1) - 6*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Apr 23 2018

Mathematica
```
LinearRecurrence[{5,-6,1}, {1,3,8}, 50]
```

x='x+O('x^30); Vec((1-2*x-x^2)/(1-5*x+6*x^2-x^3)) \\ G. C. Greubel, Apr 23 2018

G.f.: (1-2*x-x^2)/(1-5*x+6*x^2-x^3).

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

A215665 a(n) = 3*a(n-2) - a(n-3), with a(0)=0, a(1)=a(2)=-3.

Original entry on oeis.org

0, -3, -3, -9, -6, -24, -9, -66, -3, -189, 57, -564, 360, -1749, 1644, -5607, 6681, -18465, 25650, -62076, 95415, -211878, 348321, -731049, 1256841, -2541468, 4501572, -8881245, 16046184, -31145307, 57019797, -109482105, 202204698, -385466112, 716096199

Offset: 0

Roman Witula, Aug 20 2012

The Berndt-type sequence number 6 for the argument 2Pi/9 defined by the first relation from the section "Formula" below. Two sequences connected with a(n) (possessing the respective numbers 5 and 7) are discussed in A215664 and A215666 - for more details see comments to A215664 and Witula's reference. We have a(n) - a(n+1) = A215664(n).
From initial values and the recurrence formula we deduce that a(n)/3 are all integers.
We note that a(10) is the first element of a(n) which is positive integer and all (-1)^n*a(n+10) are positive integer, which can be obtained from the title recurrence relation.
The following decomposition holds (X - c(1)*c(2)^n)*(X - c(2)*c(4)^n)*(X - c(4)*c(1)^n) = X^3 - a(n)*X^2 - A215917(n-1)*X + (-1)^n.
If X(n) = 3*X(n-2) - X(n-3), n in Z, with X(n) = a(n) for every n=0,1,..., then X(-n) = abs(A215919(n)) = (-1)^n*A215919(n) for every n=0,1,...

			We have a(1)=a(2)=a(8)=-3, a(3)=a(6)=-9, a(4)+a(11)=-10*a(10), and 47*a(5)=2*a(11).

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

Index entries for linear recurrences with constant coefficients, signature (0,3,-1).

Cf. A215455, A215634, A215635, A215636, A215664, A214699, A215007, A214683.

LinearRecurrence[{0,3,-1}, {0,-3,-3}, 50]

concat(0,Vec(-3*(1+x)/(1-3*x^2+x^3)+O(x^99))) \\ Charles R Greathouse IV, Oct 01 2012

a(n) = c(1)*c(2)^n + c(2)*c(4)^n + c(4)*c(1)^n, where c(j) := 2*cos(2*Pi*j/9).

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

A215666 a(n) = 3*a(n-2) - a(n-3), with a(0)=0, a(1)=-3, and a(2)=6.

Original entry on oeis.org

0, -3, 6, -9, 21, -33, 72, -120, 249, -432, 867, -1545, 3033, -5502, 10644, -19539, 37434, -69261, 131841, -245217, 464784, -867492, 1639569, -3067260, 5786199, -10841349, 20425857, -38310246, 72118920, -135356595, 254667006, -478188705, 899357613

Offset: 0

Roman Witula, Aug 20 2012

The Berndt-type sequence number 7 for the argument 2Pi/9 defined by the first relation from the section "Formula" below. Two sequences connected with a(n) (possessing the respective numbers 5 and 6) are discussed in A215664 and A215665 - for more details see comments to A215664 and Witula's reference. We have a(n) = A215664(n+2) - 2*A215664(n) and a(n+1) = A215664(n+1) - A215664(n).
From initial values and the title recurrence formula we deduce that a(n)/3 and a(3*n)/9 are all integers.
If we set X(n) = 3*X(n-2) - X(n-3), n in Z, with a(n) = X(n), for every n=0,1,..., then X(-n) = -abs(A215917(n)) = (-1)^n*A215917(n), for every n=0,1,...

			We have 8*a(3)+a(6)=5*a(6)+3*a(7)=0, a(5) + a(12) = 3000, and (a(30)-1000*a(10)-a(2))/10^5 is an integer. Further we obtain  c(4)*cos(4*Pi/7)^7 + c(1)*cos(8*Pi/7)^7 + c(2)*c(2*Pi/7)^7 = -15/16.

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

Index entries for linear recurrences with constant coefficients, signature (0,3,-1).

Cf. A215455, A215634, A215635, A215636, A215664, A214699, A215007, A214683.

LinearRecurrence[{0,3,-1}, {0,-3,6}, 50]

concat(0,Vec(-3*(1-2*x)/(1-3*x^2+x^3)+O(x^99))) \\ Charles R Greathouse IV, Oct 01 2012

a(n) = c(4)*c(2)^n + c(1)*c(4)^n + c(2)*c(1)^n, where c(j) := 2*cos(2*Pi*j/9).

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

A215635 a(n) = - 12a(n-1) - 54a(n-2) - 112a(n-3) - 105a(n-4) -36a(n-5) - 2a(n-6), with a(0)=3, a(1)=-6, a(2)=18, a(3)=-60, a(4)=210, a(5)=-756.

Original entry on oeis.org

3, -6, 18, -60, 210, -756, 2772, -10296, 38610, -145860, 554268, -2116296, 8112462, -31201644, 120347532, -465328200, 1803025410, -6999149124, 27213719148, -105960069864, 413078158350, -1612098272460, 6297409350492, -24620247483624, 96324799842498, -377102656201956, 1477141800784668

Offset: 0

Roman Witula, Aug 18 2012

The Berndt-type sequence number 3 for the argument 2*Pi/9 defined by the relation: X(n) = a(n) + b(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) = A215636(n).
We note that above formula is the Binet form of the following recurrence sequence: X(n+3) + 6*X(n+2) + 9*X(n+1) + (2 + sqrt(2))*X(n) = 0, which is a special type of the sequence X(n)=X(n;g) defined in the comments to A215634 for g:=Pi/24. The sequences a(n) and b(n) satisfy the following system of recurrence equations: a(n) = -b(n+3)-6*b(n+2)-9*b(n+1)-2*b(n), 2*b(n) =  -a(n+3)-6*a(n+2)-9*a(n+1)-2*a(n).
There exists an amazing relation: (-1)^n*a(n)=3*A000984(n) for every n=0,1,...,11 and 3*A000984(12)-a(12)=6.

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, A215636, A215007, A214699, A214683.

LinearRecurrence[{-12,-54,-112,-105,-36,-2}, {3,-6,18,-60,210,-756}, 50]

Vec((3+30*x+108*x^2+168*x^3+105*x^4+18*x^5) /(1+12*x+54*x^2+112*x^3+105*x^4+36*x^5+2*x^6)+O(x^99)) \\ Charles R Greathouse IV, Oct 01 2012

G.f.: (3+30*x+108*x^2+168*x^3+105*x^4+18*x^5) / (1+12*x+54*x^2+112*x^3+105*x^4+36*x^5+2*x^6).

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).

A215829 a(n) = -3a(n-1) + 9a(n-2) + 3*a(n-3), with a(0)=3, a(1)=-3, a(2)=27.

Original entry on oeis.org

3, -3, 27, -99, 531, -2403, 11691, -55107, 263331, -1250883, 5957307, -28339875, 134882739, -641835171, 3054430539, -14535159939, 69169849155, -329162695299, 1566411248475, -7454188455651, 35472778517331, -168806797907427, 803312835011307

Offset: 0

Roman Witula, Aug 24 2012

The Berndt-type sequence number 8 for the argument 2*Pi/9 defined by the trigonometric relations from the Formula section below.
From the general recurrence relation: b(n) = -3*b(n-1) + 9*b(n-2) + 3*b(n-3), i.e., b(n) - b(n-2) = 8*b(n-2) + 3(b(n-3) - b(n-1)) the following summation formulas can be easily deduced: b(2*n+1) + 3*b(2*n) - 3*b(0) - b(1) = 8*Sum_{k=1..n} b(2*k-1) and b(2*n+2) + 3*b(2*n+1) - b(2) - 3*b(1) = 8*Sum_{k=1..n} b(2*k). Hence it follows that (a(2*n+1) + 3*a(2*n))/2 are all integers congruent to 3 modulo 4, and (a(2*n+2) + 3*a(2*n+1))/2 are all integers congruent to 1 modulo 4.
We note that all numbers 3^(-1-floor(n/3))*a(n) = A215831(n) and 3^(-n-2)*a(3*n+2) are integers.
The following decomposition holds true: (X - k(1)^n)*(X - (-k(2))^n)*(X - k(3)^n) = X^3 - sqrt(3)^(-n)*a(n)*X^2 + sqrt(3)^(-n)*T(n) - sqrt(3)^(-n), where T(2*n+1) = sqrt(3)*A215945(n) and T(2*n) = A215948(n). [Roman Witula, Aug 30 2012]

			We have k(1)^3 - k(2)^3 + k(4)^3 = -11*sqrt(3).

D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the nine order, (submitted, 2012).

Roman Witula, Ramanujan Type Trigonometric Formulas: The General Form for the Argument 2*Pi/7, Journal of Integer Sequences, Vol. 12 (2009), Article 09.8.5
Index entries for linear recurrences with constant coefficients, signature (-3, 9, 3).

Cf. A215945, A215948, A216034, A215455, A215634, A215635, A215636, A215664, A215665, A215666, A215831, A215575, A108716, A215794, A215828.

LinearRecurrence[{-3, 9, 3}, {3, -3, 27}, 50]

a(n) = (k(1)^n + (-k(2))^n + k(4)^n)*(sqrt(3))^n = (-1+4*c(1))^n + (-1+4*c(2))^n + (-1+4*c(4))^n, where k(j) := cot(2*Pi*j/9) and c(j) := cos(2*Pi*j/9).

G.f.: (3 + 6*x - 9*x^2)/(1 + 3*x - 9*x^2 - 3*x^3). [corrected by Georg Fischer, May 10 2019]

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

A215575 a(n) = 7*(a(n-1) - a(n-2) - a(n-3)), with a(0)=3, a(1)=7, a(2)=35.

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A215664 a(n) = 3*a(n-2) - a(n-3), with a(0)=3, a(1)=0, and a(2)=6.

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A215512 a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3), with a(0)=1, a(1)=3, a(2)=8.

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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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A215665 a(n) = 3*a(n-2) - a(n-3), with a(0)=0, a(1)=a(2)=-3.

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A215666 a(n) = 3*a(n-2) - a(n-3), with a(0)=0, a(1)=-3, and a(2)=6.

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A215512 a(n) = 5a(n-1) - 6a(n-2) + a(n-3), with a(0)=1, a(1)=3, a(2)=8.

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

A215635 a(n) = - 12a(n-1) - 54a(n-2) - 112a(n-3) - 105a(n-4) -36a(n-5) - 2a(n-6), with a(0)=3, a(1)=-6, a(2)=18, a(3)=-60, a(4)=210, a(5)=-756.

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.

A215829 a(n) = -3a(n-1) + 9a(n-2) + 3*a(n-3), with a(0)=3, a(1)=-3, a(2)=27.

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.