A215634 -id:A215634 - 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).

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]

A216757 a(n) = 6a(n-1) - 9a(n-2) + 3*a(n-3).

Original entry on oeis.org

0, -3, -15, -63, -252, -990, -3861, -15012, -58293, -226233, -877797, -3405564, -13211910, -51254775, -198838152, -771371667, -2992450959, -11608875207, -45035307612, -174709321686, -677764787229, -2629310751036, -10200109386213, -39570153919641, -153507871295037

Offset: 1

Roman Witula, Sep 15 2012

a(n) = X(2*n-1)/sqrt(3), where X(n) = 3*X(n-2) - sqrt(3)*X(n-3), with X(0)=3, X(1)=0, and X(2)=6.
The Berndt-type sequence number 13 for the argument 2*Pi/9 defined by the relation a(n)*sqrt(3) = s(1)^(2*n-1) - s(2)^(2*n-1) + s(4)^(2*n-1), where s(j) := 2*sin(2*Pi*j/9). For the respective sums with the even powers of sines - see A215634.
We note that X(n) = s(1)^n + (-s(2))^n + s(4)^n -- see Witula's book for details. Moreover the numbers of the form a(n)*3^(-1-floor((n-1)/3)) are integers.
The following summation formulas hold: Sum_{k=3..n} a(k) = 3*(2*a(n-1) - a(n-2) + 1), X(2*n+1) - X(1)*3^n = X(2*n+1) = -sqrt(3)*Sum_{k=1..n} X(2*(n-k))*3^(k-1), and X(2*n) - X(0)*3^n = X(2*n) - 3^(n+1) = -sqrt(3)*Sum{k=1..n} X(2*(n-k))*3^(k-1).

			We have s(1)^5 - s(2)^5 + s(4)^5 = 5*(s(1)^3 - s(2)^3 + s(4)^3) = -15*sqrt(3), s(1)^9 - s(2)^9 + s(4)^9 = 4*(s(1)^7 - s(2)^7 + s(4)^7) = -252*sqrt(3),
39*(s(1)^11 - s(2)^11 + s(4)^11) = 10*(s(1)^13 - s(2)^13 + s(4)^13) = -38610*sqrt(3),
s(1)^7 - s(2)^7 + s(4)^7 = 4*(s(1)^5 - s(2)^5 + s(4)^5) + (s(1)^3 - s(2)^3 + s(4)^3) = -63*sqrt(3), and  s(1)^15 - s(2)^15 + s(4)^15 = 1000*(s(1)^5 - s(2)^5 + s(4)^5) + 4*(s(1)^3 - s(2)^3 + s(4)^3) = -15012*sqrt(3).
We note that a(6) = 3*(a(5) + a(4) + a(3)).

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

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (6,-9,3).

Cf. A215634.

LinearRecurrence[{6,-9,3}, {0,-3,-15}, 30]
CoefficientList[Series[-3*x^2*(1 - x)/(1 - 6*x + 9*x^2 - 3*x^3), {x, 0,5 0}], x] (* G. C. Greubel, Apr 17 2017 *)

concat(0,Vec(-3*(1-x)/(1-6*x+9*x^2-3*x^3)+O(x^99))) \\ Charles R Greathouse IV, Oct 01 2012

a(n) = 6*a(n-1) - 9*a(n-2) + 3*a(n-3).
G.f.: -3*x^2*(1 - x)/(1 - 6*x + 9*x^2 - 3*x^3).
a(n) = Sum_{k=0..n} 3*(-1)^k*(binomial(2*n-1, n+9*k+7) - binomial(2*n-1, n+9*k+1)). - Greg Dresden, Jan 28 2023

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

A218332 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, 3, 1, -6, 9, -1, 1, -9, 24, -17, 1, -12, 45, -51, 1, -15, 72, -109, 1, -18, 105, -197, 1, -21, 144, -321, 1, -24, 189, -487, 1, -27, 240, -701, 1, -30, 297, -969, 1, -33, 360, -1297, 1, -36, 429, -1691, 1, -39, 504, -2157, 1, -42, 585, -2701

Offset: 0

Roman Witula, Nov 02 2012

sign

We note that p(x) =  (x - s(1))*(x + c(1))*(x - c(2)),
p(x-1) = x^3 - 3*x^2 + 3 = (x - s(2)*c(1/2))*(x - s(4)*c(1/2))*(x + s(2)*s(4)), p(x-2) = x^3 - 6*x^2 + 9*x - 1 = (x - c(1)^2)*(x - c(2)^2)*(x - c(4)^2), and p(x - n - 2) = (x - n - c(1)^2)*(x - n -c(2)^2)*(x - n - c(4)^2), n = 1,2,..., 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.
A218489 is the sequence of coefficients of polynomials p(x+n).

Cf. A214699, A214779, A215455, A215634, A215635, A215664, A215665.

We have a(4*k) = 1, a(4*k+1) = -3*k, a(4*k+2) = 3*k^2 - 3, a(4*k+3) = -k^3 + 3*k + 1. Moreover we obtain the relations b(k+1) = b(k) - 3, c(k+1) = c(k) - 2*b(k) + 3, b(k) - c(k) + d(k) - 1, whenever p(x-k) = x^3 + b(k)*x^2 + c(k)*x + d(k).

Empirical g.f.: (x^15-3*x^13-x^12-7*x^11-9*x^10+6*x^9+3*x^8-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

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.

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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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A215635 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)=3, a(1)=-6, a(2)=18, a(3)=-60, a(4)=210, a(5)=-756.

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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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A216757 a(n) = 6*a(n-1) - 9*a(n-2) + 3*a(n-3).

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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.

A216757 a(n) = 6a(n-1) - 9a(n-2) + 3*a(n-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.