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.

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A108716 a(n) = tan(Pi/14)^(-2n) + tan(3*Pi/14)^(-2n) + tan(5*Pi/14)^(-2n).

Original entry on oeis.org

3, 21, 371, 7077, 135779, 2606261, 50028755, 960335173, 18434276035, 353858266965, 6792546291251, 130387472704741, 2502874814474531, 48044357383337973, 922243598852422035, 17703083191185355397
Offset: 0

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Author

Philippe Deléham, Jun 20 2005

Keywords

Comments

The Berndt-type sequence number 11 for the argument 2*Pi/7 defined by the relation a(n) = t(1)^(2*n) + t(2)^(2*n) + t(4)^(2*n) = (-sqrt(7) + 4*s(1))^(2*n) + (-sqrt(7) + 4*s(2))^(2*n) + (-sqrt(7) + 4*s(4))^(2*n), where t(j) = tan(2*Pi*j/7) and s(j) = sin(2*Pi*j/7) (the respective sum with odd powers are discussed in A215794). See also A215007, A215008, A215143, A215493, A215494, A215510, A215512, A215694, A215695, A215828 and especially A215575, where a(n) = B(2n) for the function B(n) defined in the comments. - Roman Witula, Aug 23 2012
The sequence a(n+1)/a(n) is increasing and convergent to (t(2))^2 = 19,195669... (we note that the sequence A215794(n+1)/A215794(n) is decreasing and converges to the same limit). - Roman Witula, Aug 24 2012
Let L(p) be the total length of all sides and diagonals of a regular p-sided polygon inscribed in a unit circle. Then (L(p)/p)^2 = cot(Pi/(2p))^2 is the largest root of the equation: C(p,k)-C(p,2+k)*x+C(p,4+k)*x^2-C(p,6+k)*x^3+ ... +(-1)^q*x^q = 0, where k=1 if p is odd, k=0 if p is even, q = floor(p/2), and where C denotes the binomial coefficient. The complete set of roots is: x(i) = cot((2*i-1)*Pi/(2p))^2, i=1,2,...,q. Then a(n) = x(1)^n+x(2)^n+...x(q)^n for p=7. - Seppo Mustonen, Mar 25 2014
Sum_{k=1..(m-1)/2} tan^(2n) (k*Pi/m) is an integer when m >= 3 is an odd integer (see AMM link and formula); this sequence is the particular case m = 7. All terms are odd. - Bernard Schott, Apr 22 2022

Crossrefs

Similar to: A000244 (m=3), 2*A165225 (m=5), this sequence (m=7), A353410 (m=9), A275546 (m=11), A353411 (m=13).

Programs

  • Maple
    A:= gfun:-rectoproc({-a(n+3)+21*a(n+2)-35*a(n+1)+7*a(n), a(0) = 3, a(1) = 21, a(2) = 371},a(n), remember):
    seq(A(n),n=0..20); # Robert Israel, Aug 23 2015
  • Mathematica
    Table[ Round[ Cot[Pi/14]^(2n) + Cot[3Pi/14]^(2n) + Cot[5Pi/14]^(2n)], {n, 0, 12}] (* Robert G. Wilson v, Jun 21 2005 *)
    RecurrenceTable[{a[0]== 3, a[1]== 21, a[2]==371, a[n]== 21*a[n-1] - 35*a[n-2] + 7*a[n-3]}, a, {n,30}] (* G. C. Greubel, Aug 22 2015 *)
  • PARI
    a(n)=round(tan(Pi/14)^(-2*n) + tan(3*Pi/14)^(-2*n) + tan(5*Pi/14)^(-2*n)); \\ Anders Hellström, Aug 22 2015

Formula

a(n) = 7^n*A(2n), where A(n) := A(n-1) + A(n-2) + A(n-3)/7, with A(0)=3, A(1)=1, and A(2)=3. - see Witula-Slota's (Section 6) and Witula's (Remark 11) papers for the proofs and details. In these papers A(n) denotes the value of the big omega function with index n for the argument 2*i/sqrt(7) (see also A215512). - Roman Witula, Aug 23 2012
Conjecture: a(n) = 21*a(n-1)-35*a(n-2)+7*a(n-3). G.f.: -(35*x^2-42*x+3) / (7*x^3-35*x^2+21*x-1). - Colin Barker, Jun 01 2013
To verify conjecture, note that the roots of 7*x^3-35*x^2+21*x-1 are tan(Pi/14)^2, tan(3*Pi/14)^2 and tan(5*Pi/14)^2. - Robert Israel, Aug 23 2015
E.g.f.: exp((tan(Pi/7))^2*x) + exp((cot(Pi/14))^2*x) + exp((cot(3*Pi/14))^2*x). - G. C. Greubel, Aug 22 2015
a(n) = A275195(2*n)/(7^n). - Kai Wang, Aug 02 2016
a(n) = (tan(1*Pi/7))^(2*n) + (tan(2*Pi/7))^(2*n) + (tan(3*Pi/7))^(2*n). - Bernard Schott, Apr 22 2022

Extensions

More terms from Robert G. Wilson v, Jun 21 2005

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

Views

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

Views

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

A215575 a(n) = 7*(a(n-1) - a(n-2) - a(n-3)), with a(0)=3, a(1)=7, a(2)=35.

Original entry on oeis.org

3, 7, 35, 175, 931, 5047, 27587, 151263, 830403, 4560871, 25054435, 137642127, 756187747, 4154438295, 22824258947, 125395430335, 688917131651, 3784882096583, 20793986742179, 114241312597615, 627637106311971, 3448212648805239, 18944339609269571
Offset: 0

Views

Author

Roman Witula, Aug 16 2012

Keywords

Comments

The Berndt-type sequence number 8 for the argument 2Pi/7 defined by trigonometric relations from "Formula" below.
We note that the following decompositions hold true: (X-cot(2*Pi/7)^n)*(X-cot(4*Pi/7)^n)*(X-cot(8*Pi/7)^n) = X^3 - sqrt(7)^(-n)*a(n)*X^2 + (-sqrt(7))^(-n)*B(n)*X
- (-sqrt(7))^(-n), and (X-tan(2*Pi/7)^n)*(X-tan(4*Pi/7)^n)*(X-tan(8*Pi/7)^n) = X^3 - B(n)*X^2 + (-1)^n*a(n)*X - (-sqrt(7))^n, where B(n) := tan(2*Pi/7)^n + tan(4*Pi/7)^n + tan(8*Pi/7)^n = (-sqrt(7) + 4*sin(2*Pi/7))^n + (-sqrt(7) + 4*sin(4*Pi/7))^n + (-sqrt(7) + 4*sin(8*Pi/7))^n. Moreover we have 2*(-1)^n*B(n) = 7^(-n/2)*(a(n)^2 - a(2*n)). For the proof of these decompositions see Witula-Slota's (Section 6) and Witula's (Remark 11) reference.
We note that the numbers a(n)*7^(-ceiling(n/3)) are all integers. Moreover from the recurrence relation: a(n+3)+7*a(n+1)=7*(a(n+2)-a(n)) it can be easily obtained the following summations formulas: 8*sum{k=1,..,n} a(2*k) = 7*(a(2*n+1)-2)-a(2*n+2), which also means that the result is divisible by 8 for every n=1,2,..., and 8*sum{k=1,..,n} a(2*k-1) = 7*(a(2*n)-2)-a(2*n+1), which implies that 7*(a(n)-2)-a(n+1) is divisible by 8 for each n=0,1,...

Examples

			We have  cot(2*Pi/7)^2 + cot(4*Pi/7)^2 + cot(8*Pi/7)^2 = 5,  cot(2*Pi/7)^4 + cot(4*Pi/7)^4 + cot(8*Pi/7)^4 = 19, but cot(2*Pi/7)^6 + cot(4*Pi/7)^6 + cot(8*Pi/7)^6 = 563/7. Similarly the numbers sqrt(7)*(cot(2*Pi/7)^n + cot(4*Pi/7)^n + cot(8*Pi/7)^n) are integers for n=1,3,5,7 (equal to 7, 25, 103, 441, respectively), whereas for n=9 we obtain the rational value 13297/7.
		

References

  • E Hetmaniok, P Lorenc, S Damian, et al., Periodic orbits of boundary logistic map and new kind of modified Chebyshev polynomials in R. Witula, D. Slota, W. Holubowski (eds.), Monograph on the Occasion of 100th Birthday Anniversary of Zygmunt Zahorski. Wydawnictwo Politechniki Slaskiej, Gliwice 2015, pp. 325-343.

Crossrefs

Programs

  • Magma
    I:=[3,7,35]; [n le 3 select I[n] else 7*Self(n-1) - 7*Self(n-2) - 7*Self(n-3): n in [1..30]]; // G. C. Greubel, Apr 25 2017
  • Mathematica
    LinearRecurrence[{7,-7,-7}, {3,7,35}, 50]
  • PARI
    polsym(x^3 - 7*x^2 + 7*x + 7, 30) \\ Charles R Greathouse IV, Jul 20 2016
    

Formula

a(n) = (sqrt(7)^n)*(cot(2*Pi/7)^n + cot(4*Pi/7)^n + cot(8*Pi/7)^n) = (3 + 4*cos(2*Pi/7))^n + (3 + 4*cos(4*Pi/7))^n + (3 + 4*cos(8*Pi/7))^n = (-tan(2*Pi/7)*tan(4*Pi/7))^n + (-tan(2*Pi/7)*tan(8*Pi/7))^n + (-tan(4*Pi/7)*tan(8*Pi/7))^n.
G.f.: (3-14*x+7*x^2)/(1-7*x+7*x^2+7*x^3).
a(n) = x1^n + x2^n + x3^n, where x1, x2, x3 are the roots of the polynomial x^3 - 7*x^2 + 7*x + 7, that is, x1 = sqrt(7)/tan(Pi/7), x2 = sqrt(7)/tan(2*Pi/7), x3 = sqrt(7)/tan(4*Pi/7). - Kai Wang, Jul 19 2016

A215143 a(n) = 7*a(n-1) -14*a(n-2) +7*a(n-3), with a(0)=1, a(1)=2, a(2)=7.

Original entry on oeis.org

1, 2, 7, 28, 112, 441, 1715, 6615, 25382, 97069, 370440, 1411788, 5375839, 20458921, 77833217, 296038498, 1125816895, 4281011812, 16277915640, 61891962377, 235320000363, 894697938743, 3401649302758, 12933013979445, 49170893188704, 186945601728004, 710757805310287
Offset: 0

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Author

Roman Witula, Aug 04 2012

Keywords

Comments

The Berndt-type sequence number 3 for the argument 2Pi/7 (see A215007 and A215008 for the respective sequences numbers 1 and 2) is defined by the following relations: sqrt(7) *a(n) = s(1)*s(2)^(2n) + s(2)*s(4)^(2n) + s(4)*s(1)^(2n) = s(4)*s(1)^(2n) + s(1)*s(2)^(2n) + s(2)*s(4)^(2n), where s(j) := 2*sin(2*Pi*j/7).

References

  • R. Witula, Complex numbers, Polynomials and Fractial Partial Decompositions, T.3, Silesian Technical University Press, Gliwice 2010 (in Polish).

Crossrefs

Programs

  • Magma
    I:=[1,2,7]; [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 19 2018
  • Mathematica
    LinearRecurrence[{7,-14,7},{1,2,7},40]
  • PARI
    Vec((1-5*x+7*x^2)/(1-7*x+14*x^2-7*x^3) + O(x^30)) \\ Michel Marcus, Apr 19 2016
    

Formula

G.f.: (1-5*x+7*x^2)/(1-7*x+14*x^2-7*x^3).

A215493 a(n) = 7*a(n-1) - 14*a(n-2) + 7*a(n-3) with a(0)=0, a(1)=1, a(2)=4.

Original entry on oeis.org

0, 1, 4, 14, 49, 175, 637, 2352, 8771, 32928, 124166, 469567, 1779141, 6749211, 25623472, 97329337, 369821228, 1405502182, 5342323441, 20307982135, 77201862045, 293497548512, 1115812645899, 4242135876440, 16128056932078, 61317184775679, 233122447515741
Offset: 0

Views

Author

Roman Witula, Aug 13 2012

Keywords

Comments

The Berndt-type sequence number 4 for the argument 2Pi/7 - see also A215007, A215008, A215143 and A215494.
We have a(n)=A079309(n) for n=1..6, and A079309(7)-a(7)=1.

Programs

  • Magma
    I:=[0,1,4]; [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}, {0,1,4}, 50]
  • PARI
    x='x+O('x^30); concat([0], Vec(x*(1-3*x)/(1-7*x+14*x^2-7*x^3))) \\ G. C. Greubel, Apr 23 2018
    

Formula

a(n)*sqrt(7) = s(1)^(2n-1) + s(2)^(2n-1) + s(4)^(2n-1), where s(j) := 2*Sin(2*Pi*j/7) (for the sums of the respective even powers see A215494, see also A094429, A115146). For the proof of these formula see Witula-Slota's paper.
G.f.: x*(1-3*x)/(1-7*x+14*x^2-7*x^3).
a(n) = A275830(2*n-1)/(7^n). - Kai Wang, May 25 2017

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

Views

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

Views

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

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

Original entry on oeis.org

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

Views

Author

Roman Witula, Aug 20 2012

Keywords

Comments

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.

Examples

			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).
		

References

  • 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).

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{0,3,-1}, {3,0,6}, 50]
  • PARI
    Vec(3*(1-x^2)/(1-3*x^2+x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 27 2012

Formula

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).
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