A215666 -id:A215666 - cp's OEIS frontend

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

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).

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).

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).

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

A218489 The sequence of coefficients of cubic polynomials p(x+n), where p(x) = x^3 - 3*x + 1.

Original entry on oeis.org

1, 0, -3, 1, 1, 3, 0, -1, 1, 6, 9, 3, 1, 9, 24, 19, 1, 12, 45, 53, 1, 15, 72, 111, 1, 18, 105, 199, 1, 21, 144, 323, 1, 24, 189, 489, 1, 27, 240, 703, 1, 30, 297, 971, 1, 33, 360, 1299, 1, 36, 429, 1693, 1, 39, 504, 2159, 1, 42, 585, 2703, 1, 45, 672, 3331

Offset: 0

Roman Witula, Oct 30 2012

sign

We note that p(x) =  (x - s(1))*(x + c(1))*(x - c(2)),
p(x+1) = x^3 + 3*x^2 -1 = (x + s(1)*c(1))*(x - s(1)*c(2))*(x + c(1)*c(2)), p(x+2) = x^3 + 6*x^2 + 9*x + 3 = (x + c(1/2)^2)*(x + s(2)^2)*(x + s(4)^2), and p(x + n) = (x + n - 2 + c(1/2)^2)*(x + n - 2 + s(2)^2)*(x + n - 2 + s(4)^2), n = 2,3,..., where c(j) := 2*cos(Pi*j/9) and s(j) := 2*sin(Pi*j/18). These one's are characteristic polynomials many sequences A... - see crossrefs.
A218332 is the sequence of coefficients of polynomials p(x-n).

Cf. A214699, A214779, A215634, A215664, A215666, A215917, A215919.

We have a(4*k) = 1, a(4*k + 1) = 3*k, a(4*k + 2) = 3*k^2 - 3, and a(4*k + 3) = k^3 - 3*k + 1. Moreover we obtain
b(k+1) = b(k) + 3, c(k+1) = 2*b(k) + c(k) + 3, d(k+1) = b(k) + c(k) + d(k) + 1, where p(x + k) = x^3 + b(k)*x^2 + c(k)*x + d(k).
Empirical g.f.: -(3*x^15-3*x^13+x^12-13*x^11+9*x^10+6*x^9-3*x^8+5*x^7-12*x^6-3*x^5+3*x^4-x^3+3*x^2-1) / ((x-1)^4*(x+1)^4*(x^2+1)^4). - Colin Barker, May 17 2013

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

Formula

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.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

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.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

Formula

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

Views

Author

Keywords

Comments

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.

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.