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 1-8 of 8 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

A215494 a(n) = 7*a(n-1) - 14*a(n-2) + 7*a(n-3) with a(1)=7, a(2)=21, a(3)=70.

Original entry on oeis.org

7, 21, 70, 245, 882, 3234, 12005, 44933, 169099, 638666, 2417807, 9167018, 34790490, 132119827, 501941055, 1907443237, 7249766678, 27557748813, 104759610858, 398257159370, 1514069805269, 5756205681709, 21884262613787, 83201447389466, 316323894905207
Offset: 1

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Author

Roman Witula, Aug 13 2012

Keywords

Comments

The Berndt-type sequence number 5 for the argument 2*Pi/7; see also A215007, A215008, A215143, A215493 and A215510.
We note that if we set:
x(n) := s(2)*s(1)^n + s(4)*s(2)^n + s(1)*s(4)^n,
y(n) := s(4)*s(1)^n + s(1)*s(2)^n + s(2)*s(4)^n,
z(n) := s(1)^(n+1) + s(2)^(n+1) + s(4)^(n+1),
for every n=0,1,..., where s(j) := 2*sin(2*Pi*j/7), then the following system of recurrence equations holds true:
x(n+2)=2*x(n)-y(n), y(n+2)=2*y(n)-x(n)+z(n), z(n+2)=y(n)+3*z(n).
Moreover we have a(n)=z(2*n-1), A215493(n)=z(2*n-2), A094429(n)=y(2n-1)-x(2n-1)=-x(2*n+2)/sqrt(7), A094430(n)=-x(2*n+3), y(2*n-2)=sqrt(7)*A215143(n), y(2*n-1)=A215510(n) and x(11)=-(y(10)+z(10))/sqrt(7)=-1078.
We can also deduce the following relations:
x(n-1) = c(1)*s(1)^n + c(2)*s(2)^n + c(4)*s(4)^n,
-y(n-1)-z(n-1) = c(2)*s(1)^n + c(4)*s(2)^n + c(1)*s(4)^n,
y(n-1)-x(n-1) = c(4)*s(1)^n + c(1)*s(2)^n + c(2)*s(4)^n,
for every n=1,2,..., where x(0)=y(0)=z(0)=sqrt(7), and c(j) := 2*cos(2*Pi*j/7).
All these sequences satisfy the following recurrence equation: Z(n+6)-7*Z(n+4)+14*Z(n+2)-7*Z(n)=0. The characteristic polynomial of this equation (after rescaling) has the form (X-s(1)^2)*(X-s(2)^2)*(X-s(3)^2)=X^3-7*X^2+14*X-7 and was recognized by Johannes Kepler (1571-1630); see the Savio-Suryanarayan paper.
We also have the following decomposition: (X-s(1)^(n+1))*(X-s(2)^(n+1))*(X-s(4)^(n+1)) = X^3 - z(n)*X^2 + (1/2)*(z(n)^2-z(2n+1))*X - (-sqrt(7))^(n+1).
Further we have a(n)=A146533(n) for n=1,...,6, and A146533(7)-a(7)=7. We note that all numbers 7^(-1-floor(n/3))*a(n) are integers.

Examples

			We have a(3)=5*7^2 and a(6)=5*7^4, which implies that s(1)^12 + s(2)^12 + s(4)^12 = 49*(s(1)^6 + s(2)^6 + s(4)^6). We also have a(9) = (a(1) + a(3))*7^49.
		

Crossrefs

See A122068.

Programs

  • Magma
    I:=[7,21,70]; [n le 3 select I[n] else 7*Self(n-1)-14*Self(n-2)+7*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Dec 01 2016
    
  • Mathematica
    LinearRecurrence[{7,-14,7}, {7,21,70}, 50]
  • PARI
    polsym(x^3 - 7*x^2 + 14*x - 7, 30) \\ (includes a(0)=3) Joerg Arndt, May 31 2017
    
  • PARI
    x='x+O('x^30); Vec((7-28*x+21*x^2)/(1-7*x+14*x^2-7*x^3)) \\ G. C. Greubel, Apr 23 2018

Formula

Equals 7*A122068. - M. F. Hasler, Aug 25 2012
a(n) = s(1)^(2n) + s(2)^(2n) + s(4)^(2n), where s(j) := 2*Sin(2*Pi*j/7) (for the sums of the respective odd powers see A215493, see also A094429, A115146). For the proof of these formula see Witula-Slota's paper.
G.f.: (7 - 28*x + 21*x^2)/(1 - 7*x + 14*x^2 - 7*x^3) = -d(log(1 - 7*x + 14*x^2 - 7*x^3))/dx.

A215510 a(n) = 7*a(n-1) - 14*a(n-2) + 7*a(n-3) with a(0)=0, a(1)=7, a(2)=35.

Original entry on oeis.org

0, 7, 35, 147, 588, 2303, 8918, 34300, 131369, 501809, 1913597, 7289436, 27748357, 105581574, 401620072, 1527436967, 5808448779, 22086364419, 83978326796, 319298327159, 1213996265902, 4615645568660, 17548659548105, 66719552736809, 253665154464813
Offset: 0

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Author

Roman Witula, Aug 14 2012

Keywords

Comments

The Berndt-type sequence number 6 for the argument 2Pi/7 (see A215007, A215008, A215143, A215493 and A215494 for the respective sequences numbers 1-5) is defined by the following relation: a(n) = s(1)*s(2)^(2n+1) + s(2)*s(4)^(2n+1) + s(4)*s(1)^(2n+1), where s(j) := 2*sin(2*Pi*j/7). For the respective sums with even powers see A215143.
We note that a(4)=49*sqrt(7)*(s(1)*s(4)^(-6) + s(2)*s(4)^(-6) + s(4)*s(1)^(-6)) - see the respective value of the sequence y*(n) in Witula-Slota's paper.

Examples

			We have  (1-7*x+14*x^2-7*x^3)*(a(1)*x + a(3)*x^2 + a(5)*x^3 + ...) = b(1)*x - b(2)*x^2 + b(3)*x^3 - b(4)*x^4 + (b(5)-2b(2))*x^5 + ..., where b(n)=A094430(n) for n=1,...,5.
		

Programs

  • Magma
    I:=[0,7,35]; [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, Apr 23 2018
  • Mathematica
    LinearRecurrence[{7,-14,7}, {0,7,35}, 50]
  • PARI
    x='x+O('x^30); concat([0], Vec((7*x-14*x^2)/(1-7*x+14*x^2-7*x^3))) \\ G. C. Greubel, Apr 23 2018
    

Formula

G.f.: (7*x-14*x^2)/(1-7*x+14*x^2-7*x^3).
a(n) = 7*A215008(n). - R. J. Mathar, Nov 07 2015

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

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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

A217274 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, 7, 35, 154, 637, 2548, 9996, 38759, 149205, 571781, 2184910, 8333871, 31750824, 120875944, 459957169, 1749692735, 6654580387, 25306064602, 96226175941, 365880389868, 1391138718116, 5289228800247, 20109822277181, 76457523763621, 290689756066542
Offset: 0

Views

Author

Roman Witula, Sep 29 2012

Keywords

Comments

This is the Berndt-type sequence number 18 for the argument 2*Pi/7 defined by the relation
a(n)*sqrt(7) = c(4)*s(1)^(2n+1) + c(2)*s(4)^(2n+1) + c(1)*s(2)^(2n+1) = (1/s(4))*s(1)^(2n+2) + (1/s(2))*s(4)^(2n+2) + (1/s(1))*s(2)^(2n+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 even powers see A094429). For the proof of this formula see the Witula/Slota and Witula references.
The definitions of the other Berndt-type sequences for the argument 2*Pi/7 (with numbers from 1 to 17) are in the cross references.
We note that all numbers of the form a(n)*7^(-floor((n+1)/3)) = A217444(n) are integers.
It can be proved that Sum_{k=2..n}a(k) = 7*(a(n-1) - a(n-2)).

Examples

			Writing c(j) as cj and s(k) as sk,
we have 7*sqrt(7) = c4*s1^5 + c2*s4^5 + c1*s2^5
and c4*s1^13 + c2*s4^13 + c1*s2^13 = 4(c4*s1^11 + c2*s4^11 + c1*s2^11).
We note that a(9) = 87*a(3)*a(2)^2 and a(11) = 2*a(3)*a(5)*a(2)^2.
		

Crossrefs

Programs

  • Magma
    I:=[0,1,7]; [n le 3 select I[n] else 7*Self(n-1)-14*Self(n-2)+7*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jul 26 2015
  • Mathematica
    LinearRecurrence[{7,-14,7}, {0,1,7}, 30]
    CoefficientList[Series[x/(1 - 7*x + 14*x^2 - 7*x^3), {x,0,50}], x] (* G. C. Greubel, Apr 16 2017 *)
  • Maxima
    a[0]:0$
    a[1]:1$
    a[2]:7$
    a[n]:=7*a[n-1] - 14*a[n-2] + 7*a[n-3];
    makelist(a[n], n, 0, 25); /* Martin Ettl, Oct 11 2012 */
    
  • PARI
    concat(0, Vec(x/(1-7*x+14*x^2-7*x^3) + O(x^40))) \\ Michel Marcus, Jul 25 2015
    

Formula

G.f.: x/(1-7*x+14*x^2-7*x^3).

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 1-8 of 8 results.