cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A033304 Expansion of (2 + 2*x - 3*x^2) / (1 - 2*x - x^2 + x^3).

Original entry on oeis.org

2, 6, 11, 26, 57, 129, 289, 650, 1460, 3281, 7372, 16565, 37221, 83635, 187926, 422266, 948823, 2131986, 4790529, 10764221, 24186985, 54347662, 122118088, 274396853, 616564132, 1385407029, 3112981337, 6994805571, 15717185450
Offset: 0

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Author

Keywords

Comments

From L. Edson Jeffery, Mar 22 2011: (Start)
Let A be the unit-primitive matrix (see [Jeffery])
A=A_(7,2)=
(0 0 1)
(0 1 1)
(1 1 1).
Let B={b(n)} be this sequence shifted to the right one place and setting b(0)=3. Then B=(3,2,6,11,26,...) with generating function (3-4*x-x^2)/(1-2*x-x^2+x^3) and b(n)=Trace(A^n). (End)
The following identity hold true (a(n)^2 - a(2n+2))/2 = A094648(n+1) = (-1)^(n+1)*A096975(n+1) - for the proof see Witula et al.'s papers - Roman Witula, Jul 25 2012
We note that the joined sequences (-1)^(n+1)*a(n) and A094648(n) form a two-sided sequence defined either by the recurrence formula x(n+3) + x(n+2) - 2x(n+1) - x(n) = 0, n in Z, x(0)=3, x(-1)=-2, x(1)=-1, or by the following trigonometric identities: x(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) = (s(2)/s(1))^n + (s(4)/s(2))^n + (s(1)/s(4))^n, for n in Z, where c(j) := 2*cos(2Pi*j/7) and s(j) := sin(2*Pi*j/7) - for the proof see Witula's and Witula et al.'s papers. - Roman Witula, Jul 25 2012
We have 4*a(n+2) - a(n) = 7*A077998(n+2). - Roman Witula, Aug 13 2012
Two very intriguing identities of trigonometric nature hold: (-1)^n*(a(n)-a(n-1)) = c(1)*c(2)^(-n) + c(2)*c(4)^(-n) + c(4)*c(1)^(-n), and (-1)^(n+1)*(a(n-1)-a(n+1)) = c(1)*c(4)^(-n-1) + c(2)*c(1)^(-n-1) + c(4)*c(2)^(-n-1), where a(-1):=3 and c(j) is defined as above. For the proof see Remark 6 in the first Witula's paper. - Roman Witula, Aug 14 2012
With respect to the form of the trigonometric formulas describing a(n), we call this sequence the Berndt-type sequence number 20 for the argument 2Pi/7. The A-numbers of other Berndt-type sequences numbers are given in below. - Roman Witula, Sep 30 2012

References

  • R. P. Stanley, Enumerative Combinatorics I, p. 244, Eq. (36).

Crossrefs

Programs

  • Magma
    I:=[2,6,11]; [n le 3 select I[n] else 2*Self(n-1) +Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Apr 19 2018
  • Mathematica
    CoefficientList[Series[(2+2x-3x^2)/(1-2x-x^2+x^3),{x,0,50}], x]  (* Harvey P. Dale, Mar 14 2011 *)
    LinearRecurrence[{2, 1, -1}, {2, 6, 11}, 29] (* Jean-François Alcover, Sep 27 2017 *)
  • PARI
    {a(n)=if(n<0, n=-n; polsym(x^3-x^2-2*x+1,n-1)[n], n+=2; polsym(1-x-2*x^2+x^3,n-1)[n])} /* Michael Somos, Aug 03 2006 */
    
  • PARI
    x='x+O('x^99); Vec((2+2*x-3*x^2)/(1-2*x-x^2+x^3)) \\ Altug Alkan, Apr 19 2018
    

Formula

a(-1-n) = A096975(n).
a(n) = (1-2*cos(1/7*Pi))^(n+1)+(1+2*cos(2/7*Pi))^(n+1)+(1-2*cos(3/7*Pi))^(n+1). - Vladeta Jovovic, Jun 27 2001
a(n) = trace of (n+1)-th power of the 3 X 3 matrix (in the example of A066170): [1 1 1 / 1 1 0 / 1 0 0]. Alternatively, the sum of the (n+1)st powers of the roots of the corresponding characteristic polynomial: x^3 - 2*x^2 - x + 1 = 0. a(n) = A006356(n) + A006356(n-1) + 2*A006356(n-2). E.g., a(3) = 26 = the trace of M^4. The characteristic polynomial of this matrix (see A066170) is x^3 - 2*x^2 - x + 1 and the roots are 2.24697960372..., -0.8019377358... and 0.55495813208... = a, b, c. Then Sum(a^4 + b^4 + c^4) = 26. - Gary W. Adamson, Feb 01 2004
(-1)^(n+1)*a(n) = (c(1))^(-n-1) + (c(2))^(-n-1) + (c(3))^(-n-1) = (c(1)c(2))^(n+1) + (c(1)c(4))^(n+1) + (c(2)c(4))^(n+1) = (s(1)/s(2))^(n+1) + (s(2)/s(4))^(n+1) + (s(4)/s(1))^(n+1), where c(j) := 2*cos(2*Pi*j/7) and s(j) := sin(2*Pi*j/7) - for the proof see Witula's and Witula et al.'s papers. - Roman Witula, Jul 25 2012
a(n) = 3*A077998(n+1) - A006054(n+2) - A006054(n+1). - Roman Witula, Aug 13 2012
a(n)*(-1)^(n+1) = (A094648(n+1)^2 - A094648(2*(n+1)))/2. - Roman Witula, Sep 30 2012

A094648 An accelerator sequence for Catalan's constant.

Original entry on oeis.org

3, -1, 5, -4, 13, -16, 38, -57, 117, -193, 370, -639, 1186, -2094, 3827, -6829, 12389, -22220, 40169, -72220, 130338, -234609, 423065, -761945, 1373466, -2474291, 4459278, -8034394, 14478659, -26088169, 47011093, -84708772, 152642789, -275049240
Offset: 0

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Author

Paul Barry, May 18 2004

Keywords

Comments

The pair A094648 and the alternating sequence A033304 when joined form a two-sided sequence defined by the recurrence formula x(n+3) + x(n+2) - 2x(n+1) - x(n) = 0, n in Z, x(-1)=-2, x(0)=3, x(1)=-1 - for details see Witula's comments to A033304. - Roman Witula, Jul 25 2012
From Roman Witula, Aug 09 2012: (Start)
There exist two interesting subsequences b(n) and c(n) of the given above sequence x(n) defined by the following relations: b(n)=a(2^n) and c(n)=x(-2^n). These subsequences satisfy the following system of recurrence equations:
b(n+1)=b(n)^2-2*c(n), and c(n+1)=c(n)^2-2*b(n),
which easily follow from the general identity: x(n)^2=x(2*n)-2*x(-n), n in Z. We note that b(0)=-1, b(1)=5, b(2)=13, b(3)=117, c(0)=-2, c(1)=6, c(2)=26, c(3)=650. From the above system we deduce that all b(n) are odd, whereas all c(n) are even. Moreover we obtain c(n+1)-b(n+1)=(c(n)-b(n))*(b(n)+c(n)+2), which yields b(n+1)-c(n+1)=product{k=1,..,n}(b(k)+c(k)+2)=13*product{k=2,..,n}(b(k)+c(k)+2)=13^2*41*product{k=3,..,n}(b(k)+c(k)+2). It follows that b(n)-c(n) is divisible by 13^2*41 for every n=3,4,..., and after using the above system again each b(n) and c(n), for n=2,3,..., is divisible by 13. (End)
If we set W(n):=3*A077998(n)-A006054(n+1)-A006054(n), n=0,1,..., then a(n)=(W(n)^2-W(2*n))/2 and W(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 = (-1-c(1))^n + (-1-c(2))^n + (-1-c(4))^n, where c(j):=2*cos(2*Pi*j/7) - for the proof see Witula-Slota-Warzynski's paper. Moreover it follows from the comment at the top and from comments to A033304 that W(n+1)=A033304(n)=(-1)^(n+1)*x(-n-1). - Roman Witula, Aug 11 2012
The following trigonometric type identitities hold true: (1) -a(n-1)-a(n) = c(1)*c(2)^n + c(2)*c(4)^n + c(4)*c(1)^n and (2) a(n)-a(n+2) = c(4)*c(2)^(n+1) + c(1)*c(4)^(n+1) + c(2)*c(1)^(n+1), where a(-1)=-2 and c(j) is defined as above (see also the respective comment to A033304). For the proof see Remark 6 in Witula's paper. - Roman Witula, Aug 14 2012
It can be proved that A033304(n-1)*(-1)^n = (a(n)^2 - a(2*n))/2, n=1,2,... - Roman Witula, Sep 30 2012
With respect to the form of the trigonometric formulas describing a(n), we call this sequence the Berndt-type sequence number 19 for the argument 2*Pi/7. The A-numbers of other Berndt-type sequences numbers are given in below. - Roman Witula, Sep 30 2012

Examples

			We have a(17) = a(19) + 50000, a(4) + a(5) = -3, 2*a(7) + a(8) = 3, and 2*a(9) + a(10) = a(5). - _Roman Witula_, Sep 14 2012
		

Crossrefs

Programs

  • Magma
    I:=[3,-1,5]; [n le 3 select I[n]  else -Self(n-1)+2*Self(n-2)+Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jul 25 2015
    
  • Mathematica
    CoefficientList[ Series[(3 + 2x - 2x^2)/(1 + x - 2x^2 - x^3), {x, 0, 33}], x] (* Robert G. Wilson v, May 24 2004 *)
    a[n_] := Round[(2Sin[3Pi/14])^n + (-2Sin[Pi/14])^n + (-2Cos[Pi/7])^n]; Table[ a[n], {n, 0, 33}] (* Robert G. Wilson v, May 24 2004 *)
    LinearRecurrence[{-1,2,1}, {3,-1,5}, 50] (* Roman Witula, Aug 09 2012 *)
  • PARI
    x='x+O('x^30); Vec((3+2*x-2*x^2)/(1+x-2*x^2-x^3)) \\ G. C. Greubel, May 09 2018

Formula

G.f.: (3+2*x-2*x^2)/(1+x-2*x^2-x^3);
a(n) = (2*sin(3*Pi/14))^n+(-2*sin(Pi/14))^n+(-2*cos(Pi/7))^n.
a(p) == -1 mod(p), p prime. - Philippe Deléham, Oct 03 2009
a(n) = (2*cos(2*Pi/7))^n + (2*cos(4*Pi/7))^n + (2*cos(8*Pi/7))^n, which is equivalent to the formula given above (for analogous sums with sines see A215493 and A215494). Moreover we have a(n+3) + a(n+2) - 2a(n+1) - a(n) = 0 - for the proof see Witula-Slota's paper. - Roman Witula, Jul 24 2012
a(n) = 3*(-1)^n*A006053(n+2) +2*A078038(n-1). - R. J. Mathar, Nov 03 2020

A215008 a(n) = 7*a(n-1) - 14*a(n-2) + 7*a(n-3), a(0)=0, a(1)=1, a(2)=5.

Original entry on oeis.org

0, 1, 5, 21, 84, 329, 1274, 4900, 18767, 71687, 273371, 1041348, 3964051, 15083082, 57374296, 218205281, 829778397, 3155194917, 11996903828, 45614046737, 173428037986, 659377938380, 2506951364015, 9531364676687, 36237879209259, 137774708539300, 523812203582283, 1991504659990594
Offset: 0

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Author

Roman Witula, Jul 31 2012

Keywords

Comments

The Berndt-type sequence number 2 for argument 2*Pi/7 is defined by the following relation: a(n) = -(2^(2*n-1)/sqrt(7))*((s(1))^(2*n)/s(2) + (s(4))^(2*n)/s(1) + (s(2))^(2*n)/s(4)), where s(j) := sin(2*Pi*j/7) - see also sequence A215007. This sequence was motivated by Berndt's et al. papers.
We note that a(n) = A002054(n) for n=0,1,...,4, and A002054(5) - a(5) = 1. Moreover, we have a(n+1)=A026027(n) for n=0,...,6, and A026027(7) - a(8) = 1. The characteristic polynomial of a(n) has the form x^3 -7*x^2 +14*x -7 = (x-(2*s(1))^2)*(x-(2*s(2))^2)*(x-(2*s(4))^2) and was known to Johannes Kepler (1571-1630) - see Witula's book and Savio-Suryanarayan's paper.

Examples

			We have a(6)<a(8), but the following amazing equality holds:
  (s(1))^6/s(2) + (s(4))^6/s(1) + (s(2))^6/s(4) = (s(1))^8/s(2) + (s(4))^8/s(1) + (s(2))^8/s(4) = -21*sqrt(7)/32.
It can be also proved that
  (s(1))^3/s(2) + (s(4))^3/s(1) + (s(2))^3/s(4) = (s(1))^5/s(2) + (s(4))^5/s(1) + (s(2))^5/s(4) = (s(1))^7/s(2) + (s(4))^7/s(1) + (s(2))^7/s(4).
		

References

  • R. Witula, Complex numbers, Polynomials and Fractial Partial Decompositions, T.3, Silesian Technical University Press, Gliwice 2010 (in Polish).
  • R. Witula, E. Hetmaniok and D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the Fifteenth International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012.

Crossrefs

Cf. A215007.

Programs

  • GAP
    a:=[0,1,5];; for n in [4..30] do a[n]:=7*(a[n-1]-2*a[n-2]+a[n-3]); od; a; # G. C. Greubel, Oct 03 2019
  • Magma
    I:=[0,1,5]; [n le 3 select I[n] else 7*(Self(n-1) -2*Self(n-2) + Self(n-3)): n in [1..30]]; // G. C. Greubel, Feb 01 2018
    
  • Maple
    seq(coeff(series(x*(1-2*x)/(1-7*x+14*x^2-7*x^3), x, n+1), x, n), n = 0..30); # G. C. Greubel, Oct 03 2019
  • Mathematica
    LinearRecurrence[{7,-14,7},{0,1,5},30]
    CoefficientList[Series[x (1-2x)/(1-7x+14x^2-7x^3),{x,0,30}],x] (* Harvey P. Dale, Jul 01 2021 *)
  • PARI
    concat([0], Vec((x-2*x^2)/(1-7*x+14*x^2-7*x^3)+O(x^30))) \\ Charles R Greathouse IV, Sep 27 2012
    
  • Sage
    def A215008_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P(x*(1-2*x)/(1-7*x+14*x^2-7*x^3)).list()
    A215008_list(30) # G. C. Greubel, Oct 03 2019
    

Formula

G.f.: x*(1-2*x)/(1-7*x+14*x^2-7*x^3).
a(n+1) - 2*a(n) = (1/sqrt(7))*Sum_{k=0,1,2} cot(2^k * alpha) * (2*sin(2^k * alpha))^(2n), where alpha = 2*Pi/7. - Roman Witula, May 16 2014
a(n) = A217274(n) - 2*A217274(n-1). - R. J. Mathar, Feb 05 2020

A094429 Given the 3 X 3 matrix M = [0 1 0 / 0 0 1 / 7 -14 7], a(n) = (-) rightmost term of M^n * [1 1 1].

Original entry on oeis.org

0, 7, 42, 196, 833, 3381, 13377, 52136, 201341, 773122, 2958032, 11291903, 43042727, 163918671, 623875840, 2373568575, 9028148962, 34334213564, 130560389505, 496440779373, 1887579497489, 7176808297736, 27286630574917
Offset: 1

Views

Author

Gary W. Adamson, May 02 2004

Keywords

Comments

M is derived from the Lucas polynomial: x^3 - 7*x^2 + 14*x - 7 with a root (and eigenvalue of the matrix): 3.801377358... = (2*sin(3*Pi/7))^2, the convergent of the sequence.
From Roman Witula, Sep 29 2012: (Start)
The Berndt-type sequence number 16 for the argument 2*Pi/7 (see Formula section and Crossrefs for other Berndt-type sequences for the argument 2*Pi/7 - for numbers from 1 to 18 without 16).
Note that all numbers of the form a(n)*7^(-1 - floor((n-1)/3)) are integers and even a(10) and a(11) are divisible by 7^5. (End)

Examples

			a(5) = 833. M^5 * [1 1 1] = [ -42 -196 -833].
We have 4*a(4) - a(5) = 4*a(5) - a(6) = 7*a(2) = 49, 88*a(10) = 23*a(11), and a(3) = 6*a(2), which implies the equalities c(4)*(s(1))^6 + c(2)*(s(4))^6 + c(1)*(s(2))^6 = 6*(c(4)*(s(1))^4 + c(2)*(s(4))^4 + c(1)*(s(2))^4) and
s(2)*(s(1))^8 + s(4)*(s(2))^8 + s(1)*(s(4))^8 = 6*( s(2)*(s(1))^6 + s(4)*(s(2))^6 + s(1)*(s(4))^6). - _Roman Witula_, Sep 29 2012
		

Crossrefs

Programs

  • Magma
    I:=[0,7,42]; [n le 3 select I[n] else 7*Self(n-1) -14*Self(n-2) +7*Self(n-3): n in [1..30]]; // G. C. Greubel, May 09 2018
  • Mathematica
    Table[(MatrixPower[{{0, 1, 0}, {0, 0, 1}, {7, -14, 7}}, n].{-1, -1, -1})[[3]], {n, 23}] (* Robert G. Wilson v, May 08 2004 *)
    LinearRecurrence[{7,-14,7}, {0,7,42}, 50] (* Roman Witula, Aug 13 2012 *)
  • PARI
    x='x+O('x^30); concat([0], Vec(7*x^2*(1-x)/(1-7*x+14*x^2-7*x^3))) \\ G. C. Greubel, May 09 2018
    
  • PARI
    a(n) = -(([0, 1, 0; 0, 0, 1; 7, -14, 7]^n)*[1,1,1]~)[3]; \\ Michel Marcus, May 10 2018
    

Formula

From Colin Barker, Jun 19 2012: (Start)
a(n) = 7*a(n-1) - 14*a(n-2) + 7*a(n-3).
G.f.: 7*x^2*(1-x)/(1 - 7*x + 14*x^2 - 7*x^3). (End)
a(n) = c(4)*(s(1))^(2*n) + c(2)*(s(4))^(2*n) + c(1)*(s(2))^(2*n) = (-1/sqrt(7))*(c(1)*(s(1))^(2*n+3) + c(2)*(s(2))^(2*n+3) + c(3)*(s(3))^(2*n+3)) = (-1/sqrt(7))*(s(2)*(s(1))^(2*n+2) + s(4)*(s(2))^(2*n+2) + s(1)*(s(4))^(2*n+2)), where c(j) := 2*cos(2*Pi*j/7) and s(j) := 2*sin(2*Pi*j/7) (for the sums of the respective odd powers see A217274, see also A215493 and comments to A215494). For the proof of these formulas see Witula-Slota's paper. - Roman Witula, Jul 24 2012

Extensions

More terms from Robert G. Wilson v, May 08 2004

A094430 a(n) is the rightmost term of M^n * [1 0 0], where M is the 3 X 3 matrix [0 1 0 / 0 0 1 / 7 -14 7].

Original entry on oeis.org

7, 49, 245, 1078, 4459, 17836, 69972, 271313, 1044435, 4002467, 15294370, 58337097, 222255768, 846131608, 3219700183, 12247849145, 46582062709, 177142452214, 673583231587, 2561162729076, 9737971026812, 37024601601729
Offset: 1

Views

Author

Gary W. Adamson, May 02 2004

Keywords

Comments

In A094429 the multiplier is [1 1 1] instead of [1 0 0]. The matrix M is derived from the 3rd-order Lucas polynomial x^3 - 7x^2 + 14x - 7, with a convergent of the series = 3.801937735... = (2 sin 3*Pi/7)^2; (an eigenvalue of the matrix and a root of the polynomial).
From Roman Witula, Sep 29 2012: (Start)
This sequence is the Berndt-type sequence number 17 for the argument 2*Pi/7 (see Formula section and Crossrefs for other Berndt-type sequences for the argument 2*Pi/7 - for numbers from 1 to 18 without 17).
Note that all numbers of the form a(n)*7^(-floor((n+4)/3)) are integers. (End)

Examples

			a(4) = 1078 since M^4 * [1 0 0] = [49 245 1078] = [a(2), a(3), a(4)].
We have a(2)=7*a(1), a(3)=5*a(2), 22*a(3)=5*a(4), and a(6)=4*a(5), which implies s(2)*s(1)^15 + s(4)*s(2)^15 + s(1)*s(4)^15 = 4*(s(2)*s(1)^13 + s(4)*s(2)^13 + s(1)*s(4)^13). - _Roman Witula_, Sep 29 2012
		

Crossrefs

Programs

  • Magma
    I:=[49,245,1078]; [7] cat [n le 3 select I[n] else 7*Self(n-1) -14*Self(n-2) + 7*Self(n-3): n in [1..30]]; // G. C. Greubel, May 09 2018
  • Mathematica
    Table[(MatrixPower[{{0, 1, 0}, {0, 0, 1}, {7, -14, 7}}, n].{1, 0, 0})[[3]], {n, 22}] (* Robert G. Wilson v, May 08 2004 *)
    Join[{7}, LinearRecurrence[{7,-14,7}, {49,245,1078}, 50]] (* Roman Witula, Aug 13 2012 *)(* corrected by G. C. Greubel, May 09 2018 *)
  • PARI
    x='x+O('x^30); Vec(7*x/(1-7*x+14*x^2-7*x^3)) \\ G. C. Greubel, May 09 2018
    
  • PARI
    a(n) = (([0, 1, 0; 0, 0, 1; 7, -14, 7]^n)*[1,0,0]~)[3]; \\ Michel Marcus, May 10 2018
    

Formula

From Colin Barker, Jun 19 2012: (Start)
a(n) = 7*a(n-1)-14*a(n-2)+7*a(n-3).
G.f.: 7*x/(1-7*x+14*x^2-7*x^3). (End)
-a(n) = s(2)*s(1)^(2*n+3) + s(4)*s(2)^(2*n+3) + s(1)*s(4)^(2*n+3), where s(j) := 2*sin(2*Pi*j/7); for the proof see A215494 and the Witula-Slota paper. This formula and the respective recurrence also give a(0)=a(-1)=0. - Roman Witula, Aug 13 2012

Extensions

More terms from Robert G. Wilson v, May 08 2004
Name edited by Michel Marcus, May 10 2018

A217444 a(n) = A(n)*7^(-floor(n+1)/3), where A(n) = 7*A(n-1) - 14*A(n-2) + 7*A(n-3) with A(0)=0, A(1)=1, A(2)=7.

Original entry on oeis.org

0, 1, 1, 5, 22, 13, 52, 204, 113, 435, 1667, 910, 3471, 13224, 7192, 27367, 104105, 56563, 215098, 817909, 444276, 1689212, 6422529, 3488381, 13262821, 50424942, 27387681, 104126704, 395884336, 215018609, 817488295, 3108041875, 1688083894, 6417991803, 24400809980
Offset: 0

Views

Author

Roman Witula, Oct 03 2012

Keywords

Comments

The Berndt-type sequence number 18a for the argument 2Pi/7, which is closely connected with the sequence A217274. Definitions other Berndt-type sequences for the argument 2Pi/7 like A215575, A215877, A033304 in sequences from Crossrefs are given.

Crossrefs

Programs

  • Magma
    i:=35; I:=[0, 1, 7]; A:=[m le 3 select I[m] else 7*Self(m-1)-14*Self(m-2)+7*Self(m-3): m in [1..i]]; [7^(-Floor(n/3))*A[n]: n in [1..i]]; // Bruno Berselli, Oct 03 2012
    
  • Mathematica
    CoefficientList[Series[x*(1+x+5*x^2+12*x^3+3*x^4+2*x^5+x^6)/(1 - 10*x^3 + 17*x^6 - x^9), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 15 2012 *)
    LinearRecurrence[{0,0,10,0,0,-17,0,0,1}, {0, 1, 1, 5, 22, 13, 52, 204, 113}, 50] (* G. C. Greubel, Apr 23 2018 *)
  • PARI
    x='x+O('x^30); concat([0], Vec(x*(1+x+5*x^2+12*x^3+3*x^4 +2*x^5 +x^6)/(1- 10*x^3+17*x^6-x^9))) \\ G. C. Greubel, Apr 23 2018

Formula

G.f.: x*(1+x+5*x^2+12*x^3+3*x^4+2*x^5+x^6)/(1-10*x^3+17*x^6-x^9). - Bruno Berselli, Oct 03 2012
a(n) = 10*a(n-3) - 17*a(n-6) + a(n-9). - G. C. Greubel, Apr 23 2018

A319512 a(n) = 7*a(n-1) - 14*a(n-2) + 7*a(n-3), a(0) = 1, a(1) = 3, a(2) = 11.

Original entry on oeis.org

1, 3, 11, 42, 161, 616, 2352, 8967, 34153, 129997, 494606, 1881355, 7154980, 27208132, 103456689, 393367835, 1495638123, 5686513994, 21620239081, 82199944512, 312521862408, 1188195487255, 4517461948657, 17175149855885, 65298950120782, 248262786503683
Offset: 0

Views

Author

Kai Wang, Dec 10 2018

Keywords

Comments

Let {X,Y,Z} be the roots of the cubic equation
t^3 + at^2 + bt + c = 0
where {a, b, c} are integers. Let {u, v, w} be three numbers such that {u + v + w, u*X + v*Y + w*Z, u*X^2 + v*Y^2 + w*Z^2} are integers. Then
{p(n) = u*X^n + v*Y^n + w*Z^n | n = 0, 1, 2, ...}
is an integer sequence with the recurrence relation:
p(n) = -a*p(n-1) - b*p(n-2) - c*p(n-3).
This sequence has (a, b, c) = (-7, 14, -7), (u, v, w) = (1/(sqrt(7)*tan(4*(Pi/7))), 1/(sqrt(7)*tan(8*(Pi/7))), 1/(sqrt(7)*tan(2*(Pi/7)))).

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{7, -14, 7}, {1, 3, 11}, 30] (* Amiram Eldar, Dec 10 2018 *)
    CoefficientList[Series[(1-2x)^2/(1-7x+14x^2-7x^3),{x,0,30}],x] (* Harvey P. Dale, Oct 08 2023 *)
  • PARI
    Vec((1 - 2*x)^2 / (1 - 7*x + 14*x^2 - 7*x^3) + O(x^40)) \\ Colin Barker, Dec 11 2018

Formula

(X, Y, Z) = (4*sin^2(2*(Pi/7)), 4*sin^2(4*(Pi/7)), 4*sin^2(8*(Pi/7)));
a(n) = 7*a(n-1) - 14*a(n-2) + 7*a(n-3), a(0) = 1, a(1) = 3, a(2) = 11.
G.f.: (1 - 2*x)^2 / (1 - 7*x + 14*x^2 - 7*x^3). - Colin Barker, Dec 11 2018

Extensions

More terms from Felix Fröhlich, Dec 10 2018
Showing 1-7 of 7 results.