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.

Previous Showing 11-18 of 18 results.

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

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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

A062882 a(n) = (1 - 2*cos(Pi/9))^n + (1 + 2*cos(Pi*2/9))^n + (1 + 2*cos(Pi*4/9))^n.

Original entry on oeis.org

3, 9, 18, 45, 108, 270, 675, 1701, 4293, 10854, 27459, 69498, 175932, 445419, 1127763, 2855493, 7230222, 18307377, 46355652, 117376290, 297206739, 752553261, 1905530913, 4824972522, 12217257783, 30935180610, 78330624264
Offset: 1

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Author

Vladeta Jovovic, Jun 27 2001

Keywords

Comments

From L. Edson Jeffery, Apr 05 2011: (Start)
Let U be the matrix (see [Jeffery])
U = U_(9,2) =
(0 0 1 0)
(0 1 0 1)
(1 0 1 1)
(0 1 1 1).
Then a(n) = Trace(U^n).
(End)
We note that all numbers of the form a(n)*3^(-floor((n+4)/3)) are integers. - Roman Witula, Sep 29 2012

Examples

			We have a(2)=3*a(1), a(4)/a(3) = a(6)/a(5) = a(7)/a(6) = 5/2, a(6)=6*a(4), a(7)=15*a(4), and (1 + c(1))^8 + (1 + c(2))^8 + (1 + c(4))^8 = 7*3^5. - _Roman Witula_, Sep 29 2012
		

Crossrefs

Programs

  • Maple
    Digits := 1000:q := seq(floor(evalf((1-2*cos(1/9*Pi))^n+(1+2*cos(2/9*Pi))^n+(1+2*cos(4/9*Pi))^n)),n=1..50);
  • Mathematica
    LinearRecurrence[{3,0,-3},{3,9,18},25] (* Georg Fischer Feb 02 2019 *)
  • PARI
    { default(realprecision, 200); for (n=1, 200, a=(1 - 2*cos(1/9*Pi))^n + (1 + 2*cos(2/9*Pi))^n + (1 + 2*cos(4/9*Pi))^n; write("b062882.txt", n, " ", round(a)) ) } \\ Harry J. Smith, Aug 12 2009
    
  • PARI
    Vec((3-9*x^2)/(1-3*x+3*x^3)+O(x^66)) /* Joerg Arndt, Apr 08 2011 */

Formula

G.f.: x*(3 - 9*x^2)/(1 - 3*x + 3*x^3). The terms in parentheses in the definition are the roots of x^3-3*x^2+3. - Ralf Stephan, Apr 10 2004
a(n) = 3*(a(n-1) - a(n-3)) for n >= 4 - Roman Witula, Sep 29 2012

Extensions

More terms from Sascha Kurz, Mar 24 2002
Adapted formula, denominator of g.f. and modified g.f. (and offset) to accommodate added initial term a(0)=4. - L. Edson Jeffery, Apr 05 2011
a(0) = 4 removed, g.f. and programs adapted by Georg Fischer, Feb 02 2019

A321173 a(n) = -2*a(n-1) + a(n-2) + a(n-3), a(0) = -1, a(1) = 3, a(2) = -9.

Original entry on oeis.org

-1, 3, -9, 20, -46, 103, -232, 521, -1171, 2631, -5912, 13284, -29849, 67070, -150705, 338631, -760897, 1709720, -3841706, 8632235, -19396456, 43583441, -97931103, 220049191, -494446044, 1111010176, -2496417205, 5609398542, -12604204113, 28321389563, -63637584697
Offset: 0

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Author

Kai Wang, Jan 10 2019

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).
Let k = Pi/7.
This sequence has (a, b, c) = (2, -1, -1), (u, v, w) = (2*cos(4k), 2*cos(8k), 2*cos(2k)).
A033304, A274975: (a, b, c) = (2, -1, -1), (u, v, w) = (2*cos(2k), 2*cos(4k), 2*cos(8k)).
A321174 : (a, b, c) = (2, -1, -1), (u, v, w) = (2*cos(8k), 2*cos(2k), 2*cos(4k)).
X = sin(2k)/sin(4k), Y = sin(4k)/sin(8k), Z = sin(8k)/sin(2k).

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{-2,1,1},{-1,3,-9},50] (* Stefano Spezia, Jan 11 2019 *)
  • PARI
    Vec(-(1 - x + 2*x^2) / (1 + 2*x - x^2 - x^3) + O(x^30)) \\ Colin Barker, Jan 11 2019

Formula

G.f.: -(1 - x + 2*x^2) / (1 + 2*x - x^2 - x^3). - Colin Barker, Jan 11 2019

A321174 a(n) = -2*a(n-1) + a(n-2) + a(n-3), a(0) = -1, a(1) = -4, a(2) = 5.

Original entry on oeis.org

-1, -4, 5, -15, 31, -72, 160, -361, 810, -1821, 4091, -9193, 20656, -46414, 104291, -234340, 526557, -1183163, 2658543, -5973692, 13422764, -30160677, 67770426, -152278765, 342167279, -768842897, 1727574308, -3881824234, 8722379879, -19599009684, 44038575013
Offset: 0

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Author

Kai Wang, Jan 10 2019

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).
Let k = Pi/7.
This sequence has (a, b, c) = (2, -1, -1), (u, v, w) = (2*cos(8k), 2*cos(2k), 2*cos(4k)).
A033304, A274975: (a, b, c) = (2, -1, -1), (u, v, w) = (2*cos(2k), 2*cos(4k), 2*cos(8k)).
A321173 : (a, b, c) = (2, -1, -1), (u, v, w) = (2*cos(4k), 2*cos(8k), 2*cos(2k)).
X = sin(2k)/sin(4k), Y = sin(4k)/sin(8k), Z = sin(8k)/sin(2k).

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{-2,1,1},{-1,-4,5},50] (* Stefano Spezia, Jan 11 2019 *)
  • PARI
    Vec(-(1 + 6*x + 2*x^2) / (1 + 2*x - x^2 - x^3) + O(x^30)) \\ Colin Barker, Jan 11 2019

Formula

G.f.: -(1 + 6*x + 2*x^2) / (1 + 2*x - x^2 - x^3). - Colin Barker, Jan 11 2019

A287396 a(n) = (7*(csc(2*Pi/7))^2)^n + (7*(csc(4*Pi/7))^2)^n + (7*(csc(8*Pi/7))^2)^n.

Original entry on oeis.org

3, 56, 1568, 53312, 1931776, 71300096, 2645479424, 98305622016, 3654656065536, 135885355483136, 5052615982317568, 187873377732526080, 6985794697679601664, 259756778648305139712, 9658687473893481906176, 359144636249686988029952, 13354285908291066433372160
Offset: 0

Views

Author

Kai Wang, May 24 2017

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{56,-784,3136},{3,56,1568},30] (* Harvey P. Dale, Aug 08 2017 *)
  • PARI
    Vec((3 - 28*x)*(1 - 28*x) / (1 - 56*x + 784*x^2 - 3136*x^3) + O(x^30)) \\ Colin Barker, May 25 2017
    
  • PARI
    polsym(x^3 - 56*x^2 + 784* x - 3136, 20) \\ Joerg Arndt, May 26 2017

Formula

a(n) = x1^n + x2^n + x3^n, where x1, x2, x3 are the roots of x^3 - 56*x^2 + 784* x - 3136, x1 = 7*(csc(2*Pi/7))^2, x2 = 7*(csc(4*Pi/7))^2, x3 = 7*(csc(8*Pi/7))^2.
a(n) = 56*a(n-1) - 784*a(n-2) + 3136*a(n-3) for n>2, a(0) = 3, a(1) = 56, a(2) = 1568.
G.f.: (3 - 28*x)*(1 - 28*x) / (1 - 56*x + 784*x^2 - 3136*x^3). - Colin Barker, May 25 2017

A287405 a(n) = (7*(cot(1*Pi/7))^2)^n + (7*(cot(2*Pi/7))^2)^n + (7*(cot(4*Pi/7))^2)^n.

Original entry on oeis.org

3, 35, 931, 27587, 830403, 25054435, 756187747, 22824258947, 688917131651, 20793986742179, 627637106311971, 18944339609269571, 571808137046942019, 17259221092289630307, 520945214725090792931, 15723995613526902256387, 474606601742375424297731
Offset: 0

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Author

Kai Wang, May 24 2017

Keywords

Comments

a(n) = x1^n + x2^n + x3^n, where x1, x2, x3 are the roots of x^3 - 35*x^2 + 147*x - 49, x1 = 7*(cot(1*Pi/7))^2, x2 = 7*(cot(2*Pi/7))^2, x3 = 7*(cot(4*Pi/7))^2.

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{35,-147,49},{3,35,931},30] (* Harvey P. Dale, Mar 15 2018 *)
  • PARI
    Vec((3 - 7*x)*(1 - 21*x) / (1 - 35*x + 147*x^2 - 49*x^3) + O(x^30)) \\ Colin Barker, May 26 2017
    
  • PARI
    polsym(x^3 - 35*x^2 + 147*x - 49, 20) \\ Joerg Arndt, May 26 2017

Formula

a(n) = 35*a(n-1) - 147*a(n-2) + 49*a(n-3), a(0) = 3, a(1) = 35, a(2) = 931.
Bisection of A215575: a(n) = A215575(2*n).
G.f.: (3 - 7*x)*(1 - 21*x) / (1 - 35*x + 147*x^2 - 49*x^3). - Colin Barker, May 26 2017

A322455 Sum of n-th powers of the roots of x^3 - 20*x^2 - 9*x - 1.

Original entry on oeis.org

3, 20, 418, 8543, 174642, 3570145, 72983221, 1491970367, 30499826474, 623497246004, 12745935328713, 260560681614770, 5326550547499821, 108888803019858063, 2225975576006274419, 45504837297851710768, 930239414944110543194, 19016557810138882535211
Offset: 0

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Author

Kai Wang, Dec 09 2018

Keywords

Comments

Let A = sin(2*Pi/7), B = sin(4*Pi/7), C = sin(8*Pi/7).
In general, for integer h, k let
X = (B^h*C^k)/A^(h+k),
Y = (C^h*A^k)/B^(h+k),
Z = (A^h*B^k)/C^(h+k).
then X, Y, Z are the roots of a monic equation
t^3 + a*t^2 + b*t + c = 0
where a, b, c are integers and c = 1 or -1.
Then X^n + Y^n + Z^n, n = 0, 1, 2, ... is an integer sequence.
This sequence has (h,k) = (1,3) and its other half is A320918.

Crossrefs

Similar sequences with (h,k) values: A033304 (0,1), A215076 (1,1), A274032 (1,2).

Programs

  • Mathematica
    CoefficientList[Series[(3 - 40*x - 9*x^2) / (1 - 20*x - 9*x^2 - x^3) , {x, 0, 50}], x] (* Amiram Eldar, Dec 09 2018 *)
  • PARI
    Vec((3 - 40*x - 9*x^2) / (1 - 20*x - 9*x^2 - x^3) + O(x^20)) \\ Colin Barker, Dec 09 2018
    
  • PARI
    polsym(x^3 - 20*x^2 - 9*x - 1, 25) \\ Joerg Arndt, Dec 17 2018

Formula

a(n) = (B*C^3/A^4)^n + (C*A^3/B^4)^n + (A*B^3/C^4)^n.
a(n) = 20*a(n-1) + 9*a(n-2) + a(n-3) for n > 2.
G.f.: (3 - 40*x - 9*x^2) / (1 - 20*x - 9*x^2 - x^3). - Colin Barker, Dec 09 2018

A218664 Coefficients of cubic polynomials p(x+n), where p(x) = x^3 + x^2 - 2*x - 1.

Original entry on oeis.org

1, 1, -2, -1, 1, 4, 3, -1, 1, 7, 14, 7, 1, 10, 31, 29, 1, 13, 54, 71, 1, 16, 83, 139, 1, 19, 118, 239, 1, 22, 159, 377, 1, 25, 206, 559, 1, 28, 259, 791, 1, 31, 318, 1079, 1, 34, 383, 1429, 1, 37, 454, 1847, 1, 40, 531, 2339, 1, 43, 614, 2911, 1, 46, 703, 3569, 1, 49, 798, 4319
Offset: 0

Views

Author

Roman Witula, Nov 04 2012

Keywords

Comments

We have p(x) = (x - c(1))*(x - c(2))*(x - c(4)), where c(j) := 2*cos(2*Pi*j/7). We note that c(4) = c(3) = -c(1/2), c(1) = s(3) and c(2) = -s(1), where s(j) := 2*sin(Pi*j/14). Moreover we obtain -p(-x) = x^3 - x^2 - 2*x + 1 = (x + c(1))*(x + c(2))*(x + c(4)), q(x) := -x^3*p(1/x) = x^3 + 2*x^2 + x - 1 = (x - c(1)^(-1))*(x - c(2)^(-1))*(x - c(4)^(-1)), and -q(-x) = x^3 - 2*x^2 + x + 1 = (x + c(1)^(-1))*(x + c(2)^(-1))*(x + c(4)^(-1)).
We also have p(x+2) = x^3 + 7*x^2 + 14*x + 7 = (x + s(2)^2)*(x + s(4)^2)*(x + s(6)^2). The polynomial -p(-x-2) = x^3 - 7*x^2 + 14*x - 7 = (x - s(2)^2)*(x - s(4)^2)*(x - s(6)^2) is known as Johannes Kepler's cubic polynomial (see Witula's book).
Let us set r(x) := p(x+1). It can be verified that -x^3*r(1/x) = x^3 - 3*x^2 - 4*x - 1 = (x - c(1)/c(4))*(x - c(4)/c(2))*(x - c(2)/c(1)); for example, we have c(1)^3 + c(1)^2 - 2*c(1) - 1 = 0 which implies that c(1)^2 + 2*c(1) = 1/(c(1) - 1), and then c(1)^2 + 2*c(1) = c(4)/c(2) since c(4)/c(2) = (c(1)^4 - 4*c(1)^2 + 2)/(c(1)^2 - 2).
The polynomials p(x+n) and the ones obtained as above (i.e., after simple algebraic transformations) are the characteristic polynomials of many sequences in the OEIS; see crossrefs.

References

  • R. Witula, Complex Numbers, Polynomials and Partial Fraction Decomposition, Part 3, Wydawnictwo Politechniki Slaskiej, Gliwice 2010 (Silesian Technical University publishers).

Crossrefs

Formula

We have a(4*k) = 1, a(4*k + 1) = 3*k + 1, a(4*k + 2) = 3*k^2 + 2*k - 2, a(4*k + 3) = k^3 + k^2 - 2*k - 1. Further, the following relations hold true: b(k+1) = b(k) + 3, c(k+1) = 2*b(k) -2*c(k) + 3, d(k+1) = b(k) - 2*c(k) - d(k) + 1, where p(x + k) = x^3 + b(k)*x^2 + c(k)*x + d(k).
Empirical g.f.: -(x^15 - x^14 - 2*x^13 + x^12 - 5*x^11 + 10*x^10 + 3*x^9 - 3*x^8 - 3*x^7 - 11*x^6 + 3*x^4 + x^3 + 2*x^2 - x - 1) / ((x-1)^4*(x+1)^4*(x^2+1)^4). - Colin Barker, May 17 2013
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