A215664 -id:A215664 - cp's OEIS frontend

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

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).

A215885 a(n) = 3*a(n-1) - a(n-3), with a(0) = 3, a(1) = 3, and a(2) = 9.

Original entry on oeis.org

3, 3, 9, 24, 69, 198, 570, 1641, 4725, 13605, 39174, 112797, 324786, 935184, 2692755, 7753479, 22325253, 64283004, 185095533, 532961346, 1534601034, 4418707569, 12723161361, 36634883049, 105485941578, 303734663373, 874569107070, 2518221379632, 7250929475523

Offset: 0

Roman Witula, Aug 25 2012

The Berndt-type sequence number 5a for the argument 2Pi/9 defined by the first relation from the section "Formula". We see that a(n) is equal to the sum of the n-th negative powers of the c(j) := 2*cos(2*Pi*j/9), j=1,2,4 (the A215664(n) is equal to the respective n-th positive powers, further both sequences can be obtained from the two-sided recurrence relation: X(n+3) = 3*X(n+1) - X(n), n in Z, with X(-1) = X(0) = 3, and X(1) = 0).
From the last formula in Witula's comments to A215664 it follows that 2*(-1)^n*a(n) = A215664(n)^2 - A215664(2*n).
The following decomposition holds true: (X - c(1)^(-n))*(X - c(2)^(-n))*(X - c(4)^(-n)) = X^3 - a(n)*X^2 - (-1)^n*A215664(n)*X - (-1)^n.
For n >= 1, a(n) is the number of cyclic (0,1,2)-compositions of n that avoid the pattern 110 provided the positions of the parts of the composition on the circle are fixed. (Similar comments hold for the pattern 012 and for the pattern 001.) - Petros Hadjicostas, Sep 13 2017
See the Maple program by Edlin and Zeilberger for counting the q-ary cyclic compositions of n that avoid one or more patterns provided the positions of the parts of the composition are fixed on the circle. The program is located at D. Zeilberger's personal website (see links). For the sequence here, q=3 and the pattern is A=110. - Petros Hadjicostas, Sep 13 2017

			For n=3, we have a(3) = 3^3 - 3 = 24 ternary cyclic compositions of n=3 (with fixed positions on the circle for the parts) that avoid 110 because we have to exclude 110, 101, and 011. - _Petros Hadjicostas_, Sep 13 2017

Michael De Vlieger, Table of n, a(n) for n = 0..1999
A. E. Edlin and D. Zeilberger, The Goulden-Jackson cluster method for cyclic words, Adv. Appl. Math. 25 (2000), 228-232.
A. E. Edlin and D. Zeilberger, Maple program; Local copy
Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).
Roman Witula, Ramanujan Type Trigonometric Formulae, Demonstratio Math. 45 (2012), 779-796.
Index entries for linear recurrences with constant coefficients, signature (3,0,-1).

Cf. A130880, A147704, A215664, A215665, A215666, A274018.

Mathematica
```
LinearRecurrence[{3,0,-1}, {3,3,9}, 50]
```

my(x='x+O('x^30)); Vec(3*(1-2*x)/(1-3*x+x^3)) \\ Altug Alkan, Sep 13 2017

a(n) = 3*A147704(n).
a(n) = c(1)^(-n) + c(2)^(-n) + c(4)^(-n) = (-c(1)*c(2))^n + (-c(1)*c(4))^n + (-c(2)*c(4))^n, where c(j) := 2*cos(2*Pi*j/9).
G.f.: Sum_{n>=0} a(n)*x^n = 3-3*x*(x^2-1)/(1-3*x+x^3) = 3*(1-2*x)/(1-3*x+x^3).
G.f. of Edlin and Zeilberger (2000): 1+Sum_{n>=1} a(n)*x^n = 1-3*x*(x^2-1)/(1-3*x+x^3) = (1-2*x^3)/(1-3*x+x^3). - Petros Hadjicostas, Sep 13 2017
a(n) = ceiling(r^n) for n >= 1, where r = 1/A130880 is the largest root of x^3 - 3*x^2 + 1. - Tamas Lengyel, Feb 20 2022

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]

A215917 a(n) = -3*a(n-1) + a(n-3), with a(0)=0, a(1)=6, and a(2)=-15.

Original entry on oeis.org

0, 6, -15, 45, -129, 372, -1071, 3084, -8880, 25569, -73623, 211989, -610398, 1757571, -5060724, 14571774, -41957751, 120812529, -347865813, 1001639688, -2884106535, 8304453792, -23911721688, 68851058529, -198248721795, 570834443697, -1643652272562

Offset: 0

Roman Witula, Aug 27 2012

The Berndt-type sequence number 9 for the argument 2Pi/9 defined by the first relation from the section "Formula" below.
We have a(n) = 3*(-1)^(n+1)*A215448(n+1). From the recurrence formula for a(n) it follows that all a(3*n) are divisible by 9, a(3*n+1)/3 are congruent to 2 modulo 3, and a(3*n+2)/3 are congruent to 1 modulo 3. In the consequence also all sums a(n)+a(n+1)+a(n+2) are divisible by 9.
From general recurrence X(n) = -3*X(n-1) + X(n-3) the following formula can be deduced: 3*Sum_{k=2..n-1} X(k) = -X(n)-X(n-1)-X(n-2)+X(2)+X(1)+X(0). Hence, in the case of a(n) we obtain 3*Sum_{k=2..n-1} a(k) = -a(n)-a(n-1)-a(n-2)-9.
If we set X(n) = -3*X(n-1) + X(n-3), n in Z, with a(n) = X(n) for n=0,1,... then X(-n) = abs(A215666(n)) = (-1)^n*A215666(n), for every n=0,1,...
The following decomposition holds true (X - c(1)*(-c(4))^(-n))*(X - c(2)*(-c(1))^(-n))*(X - c(4)*(-c(2))^(-n)) = X^3 - a(n)*X^2 + (-1)^n*(A215665(n) - A215664(n))*X + 1.

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).

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

Cf. A215448, A215885, A215664, A215665, A215666.

We have a(3) + 3*a(2) = 0, a(8) + 24*a(5) = 48 = a(3) + a(1)/2.

LinearRecurrence[{-3,0,1}, {0,6,-15}, 50]

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

a(n) = c(1)*(-c(4))^(-n) + c(2)*(-c(1))^(-n) + c(4)*(-c(2))^(-n), where c(j) := 2*cos(2*Pi*j/9).
a(n) = (-1)^n*(A215885(n+1) - A215885(n)).
G.f.: 3*x(x+2)/(1+3*x-x^3).

A094649 An accelerator sequence for Catalan's constant.

Original entry on oeis.org

4, 1, 7, 4, 19, 16, 58, 64, 187, 247, 622, 925, 2110, 3394, 7252, 12289, 25147, 44116, 87727, 157492, 307294, 560200, 1079371, 1987891, 3798310, 7043041, 13382818, 24927430, 47191492, 88165105, 166501903, 311686804, 587670811, 1101562312

Offset: 0

Paul Barry, May 18 2004

From L. Edson Jeffery, Apr 03 2011: (Start)
Let U be the unit-primitive matrix (see [Jeffery])
U = U_(9,1) =
(0 1 0 0)
(1 0 1 0)
(0 1 0 1)
(0 0 1 1).
Then a(n) = Trace(U^n). (End)
a(n)==1 (mod 3), a(3*n+1)==1 (mod 9). - Roman Witula, Sep 14 2012

			We have a(0)+a(3)=a(1)+a(2)=8, a(3)+a(4)=a(2)+a(5)=23, and a(7)+a(8)=a(9)+a(3)=247. - _Roman Witula_, Sep 14 2012

Michael De Vlieger, Table of n, a(n) for n = 0..3649
A. Akbary and Q. Wang, On some permutation polynomials over finite fields, International Journal of Mathematics and Mathematical Sciences, 2005:16 (2005) 2631-2640.
A. Akbary and Q. Wang, A generalized Lucas sequence and permutation binomials, Proceeding of the American Mathematical Society, 134 (1) (2006), 15-22, sequence a(n) with l=9.
David M. Bradley, A Class of Series Acceleration Formulae for Catalan's Constant, The Ramanujan Journal, Vol. 3, Issue 2, 1999, pp. 159-173.
David M. Bradley, A Class of Series Acceleration Formulae for Catalan's Constant, arXiv:0706.0356 [math.CA], 2007.
Russell A. Gordon, Lucas Type Sequences and Sums of Binomial Coefficients, Integers (2023) Vol 23, Art. No. A84. See p. 21.
L. E. Jeffery, Unit-primitive matrices
Genki Shibukawa, New identities for some symmetric polynomials and their applications, arXiv:1907.00334 [math.CA], 2019.
Q. Wang, On generalized Lucas sequences, Contemp. Math. 531 (2010) 127-141, Table 2 (k=4)
Index entries for linear recurrences with constant coefficients, signature (1,3,-2,-1).

Cf. A000032, A094648, A094650.

LinearRecurrence[{1, 3, -2, -1}, {4, 1, 7, 4}, 34] (* Jean-François Alcover, Sep 21 2017 *)

Vec((4-3*x-6*x^2+2*x^3)/(1-x-3*x^2+2*x^3+x^4)+O(x^66)) /* Joerg Arndt, Apr 08 2011 */

G.f.: ( 4-3*x-6*x^2+2*x^3 ) / ( (x-1)*(x^3+3*x^2-1) )
a(n) = 1+(2*cos(Pi/9))^n+(-2*sin(Pi/18))^n+(-2*cos(2*Pi/9))^n.
a(n) = 2^n*Sum_{k=1..4} cos((2*k-1)*Pi/9)^n. - L. Edson Jeffery, Apr 03 2011
a(n) = 1 + (-1)^n*A215664(n), which is compatible with the last two formulas above. - Roman Witula, Sep 14 2012
a(n) = 3*a(n-2) + a(n-3) - 3, with a(0)=4, a(1)=1, and a(2)=7. - Roman Witula, Sep 14 2012

A188048 Expansion of (1 - x^2)/(1 - 3*x^2 - x^3).

Original entry on oeis.org

1, 0, 2, 1, 6, 5, 19, 21, 62, 82, 207, 308, 703, 1131, 2417, 4096, 8382, 14705, 29242, 52497, 102431, 186733, 359790, 662630, 1266103, 2347680, 4460939, 8309143, 15730497, 29388368, 55500634, 103895601, 195890270, 367187437, 691566411, 1297452581

Offset: 0

L. Edson Jeffery, Mar 19 2011

Sequence is related to rhombus substitution tilings.

Robert Israel, Table of n, a(n) for n = 0..3286
L. Edson Jeffery, Unit-primitive matrix
Index entries for linear recurrences with constant coefficients, signature (0,3,1).

Cf. A052931.

I:=[1,0,2,1]; [n le 4 select I[n] else Self(n-1)+3*Self(n-2)-2*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jun 22 2015

F:= gfun:-rectoproc({a(n)=3*a(n-2)+a(n-3),a(0)=1,a(1)=0,a(2)=2},a(n),remember):
map(F, [$0..100]); # Robert Israel, Jun 21 2015

CoefficientList[Series[(1-x^2)/(1-3x^2-x^3),{x,0,40}],x]  (* Harvey P. Dale, Mar 31 2011 *)
LinearRecurrence[{0,3,1}, {1,0,2}, 50] (* Roman Witula, Aug 20 2012 *)

abs(polsym(1-3*x+x^3,66)/3) /* Joerg Arndt, Aug 19 2012 */

G.f.: (1 - x^2)/(1 - 3*x^2 - x^3).
a(n) = 3*a(n-2)+a(n-3), for n>=3, with a(0)=1, a(1)=0, a(2)=2.
a(n) = a(n-1)+3*a(n-2)-2*a(n-3)-a(n-4), for n>=4, with {a(k)}={1,0,2,1}, k=0,1,2,3.
a(n) = A187497(3*n+1).
a(n) = m_(3,3), where (m_(i,j)) = (U_1)^n, i,j=1,2,3,4 and U_1 is the tridiagonal unit-primitive matrix [0, 1, 0, 0; 1, 0, 1, 0; 0, 1, 0, 1; 0, 0, 1, 1].
3*(-1)^n*a(n) = A215664(n). - Roman Witula, Aug 20 2012
a(2n) = A094831(n); a(2n+1) = A094834(n). - John Blythe Dobson, Jun 20 2015
a(n) = A052931(n)-A052931(n-2). - R. J. Mathar, Nov 03 2020
a(n) = (2^n/3)*(cos^n(Pi/9) + cos^n(5*Pi/9) + cos^n(7*Pi/9)). - Greg Dresden, Sep 24 2022

A215919 a(n) = -3*a(n-1) + a(n-3), with a(0)=0, a(1)=-3, a(2)=12.

Original entry on oeis.org

0, -3, 12, -36, 105, -303, 873, -2514, 7239, -20844, 60018, -172815, 497601, -1432785, 4125540, -11879019, 34204272, -98487276, 283582809, -816544155, 2351145189, -6769852758, 19493014119, -56127897168, 161613838746, -465348502119, 1339917609189, -3858138988821

Offset: 0

Roman Witula, Aug 27 2012

The Berndt-type sequence number 10 for the argument 2Pi/9 defined by the first trigonometric relation from the section "Formula" below. The sequence a(n) is connected with sequences A215917 and A215885 - see the respective formula.

We have A035045(n)=abs(a(n+1)/3) for every n=0,1,...,5 and A035045(7) + a(7)/3 = 1, A035045(8) - a(8)/3 = 10, A035045(9) + a(9)/3 = 63, and A035045(10) - a(10)/3 = 320 - all these four results-numbers are in A069269.

			We have a(2)=-4*a(1), a(3)=-3*a(2), a(6)/a(3) = -24.25, and a(9) = 579*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).

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

Cf. A215917, A215885, A215664.

LinearRecurrence[{-3, 0, 1}, {0, -3, 12}, 50]

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).
a(n) = A215917(n+1) + A215917(n) - 2*(-1)^n*A215885(n).
G.f.: -3*x*(1-x)/(1+3*x-x^3).

cp's OEIS Frontend

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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A215885 a(n) = 3*a(n-1) - a(n-3), with a(0) = 3, a(1) = 3, and a(2) = 9.

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

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A215917 a(n) = -3*a(n-1) + a(n-3), with a(0)=0, a(1)=6, and a(2)=-15.

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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.

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