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 1-4 of 4 results.

A080891 Period 5: repeat [0, 1, -1, -1, 1].

Original entry on oeis.org

0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0, 1, -1, -1, 1, 0
Offset: 0

Views

Author

N. J. A. Sloane, Sep 23 2003

Keywords

Comments

a(n) = (5/n), where (k/n) is the Kronecker symbol.
L(1;5) (Dirichlet L-series) is the integral from 0 to 1 of the g.f. of a(n+1). Partial sums are A092202. - Paul Barry, Apr 01 2005
From R. J. Mathar, Jul 15 2010, simplified Jul 27 2010: (Start)
The sequence is the real non-principal Dirichlet character mod 5. (The principal character mod 5 is A011558.)
Associated Dirichlet L-functions are, for example, L(1,chi) = Sum_{n>=1} a(n)/n = A086466 or L(2,chi) = Sum_{n>=1} a(n)/n^2 = 0.7062114... = 4*Pi^2/(25*sqrt(5)). (End)
This sequence {a(n)} appears in the formula 2*exp(2*Pi*n*i/5) = (A(n) + a(n)*phi) + (C(n) + D(n)*phi)*sqrt(2 + phi)*i, with the golden section phi, i = sqrt(-1) and A(n) = A164116(n+5), C(n) = A156174(n+4) and D(n) = A010891(n+3) for n >= 0. See a comment on A164116. - Wolfdieter Lang, Feb 26 2014
In Gil and Robins 2003 on page 33 the g.f. is denoted by f_{4, 4}(x). - Michael Somos, Sep 04 2015

Examples

			G.f. = x - x^2 - x^3 + x^4 + x^6 - x^7 - x^8 + x^9 + x^11 - x^12 - x^13 + ...
		

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1986, page 139, k=5, Chi_2(n).
  • H. Cohn, Advanced Number Theory, Dover Publications, Inc., 1962, p. 173.

Crossrefs

Programs

  • Magma
    &cat [[0, 1, -1, -1, 1]^^30]; // Wesley Ivan Hurt, Dec 26 2016
  • Maple
    A080891 := proc(n) numtheory[jacobi](n,5) ; end proc: seq(A080891(n),n=0..100) ; # R. J. Mathar, Jul 29 2010
  • Mathematica
    a[ n_] := Mod[n^2 + 1, 5] - 1; (* Michael Somos, May 24 2015 *)
    a[ n_] := KroneckerSymbol[ n, 5]; (* Michael Somos, May 24 2015 *)
    a[ n_] := {1, -1, -1, 1, 0}[[Mod[n, 5, 1]]]; (* Michael Somos, May 24 2015 *)
    PadRight[{},120,{0,1,-1,-1,1}] (* Harvey P. Dale, Nov 30 2023 *)
  • MuPAD
    numlib::jacobi(n,5)$ n=0..100 // Zerinvary Lajos, May 13 2008
    
  • PARI
    a(n)=kronecker(5,n) /* Also, a(n)=kronecker(n,5) */
    
  • PARI
    {a(n) = (n^2 + 1)%5 - 1}; /* Michael Somos, Dec 01 2004 */
    

Formula

If n == 0 (mod 5) a(n)=0; if n == 1 or 4 (mod 5) a(n)=1; if n == 2 or 3 (mod 5) a(n)=-1.
G.f.: x*(1-x^2)/(1+x+x^2+x^3+x^4). - Paul Barry, Apr 01 2005
G.f.: x * (1 - x) * (1 - x^2) / (1 - x^5). a(n) = a(-n) = a(n+5) for all n in Z. - Michael Somos, Jun 17 2005
Euler transform of length 5 sequence [-1, -1, 0, 0, 1]. - Michael Somos, Jun 17 2005
Transform of the Fibonacci numbers by the Riordan array A102587. - Paul Barry, Jul 14 2005
a(n) = -1 + floor(12002/99999*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 04 2013
a(n) = -1 + floor(137/242*3^(n+1)) mod 3. - Hieronymus Fischer, Jan 04 2013
|A011558(n)| = |a(n)| = |A100047(n)|. - Michael Somos, May 24 2015
a(n) is completely multiplicative with a(p) = Kronecker(5, p). - Michael Somos, Jun 17 2015
From Wesley Ivan Hurt, Dec 26 2016: (Start)
a(n) = a(n-5) for n > 4.
a(n) + a(n-1) + a(n-2) + a(n-3) + a(n-4) = 0 for n > 3.
a(n) = 1 + 2*floor((n-4)/5) - 2*floor((n-2)/5) + floor((n-1)/5) - floor(n/5). (End)
a(n) = 2*(cos(2*n*Pi/5) - cos(4*n*Pi/5))/sqrt(5). - Wesley Ivan Hurt, Sep 26 2018
a(n) = a(n-1)*a(n-4) - a(n-2)*a(n-3) for n > 3. - Nicolas Bělohoubek, May 21 2024
a(n) = n^2 - 5*floor((n^2+1)/5). - Aaron J Grech, Aug 28 2024

Extensions

Name specified by Wolfdieter Lang, Feb 26 2014

A008732 Molien series for 3-dimensional group [2,n] = *22n.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 9, 11, 13, 15, 18, 21, 24, 27, 30, 34, 38, 42, 46, 50, 55, 60, 65, 70, 75, 81, 87, 93, 99, 105, 112, 119, 126, 133, 140, 148, 156, 164, 172, 180, 189, 198, 207, 216, 225, 235, 245, 255, 265
Offset: 0

Views

Author

Keywords

Examples

			From _Philippe Deléham_, Apr 05 2013: (Start)
Stored in five columns:
    1   2   3   4   5
    7   9  11  13  15
   18  21  24  27  30
   34  38  42  46  50
   55  60  65  70  75
   81  87  93  99 105
  112 119 126 133 140
(End)
		

Crossrefs

Cf. A130520.

Programs

  • GAP
    List([0..50], n-> Int((n+3)*(n+4)/10)); # G. C. Greubel, Jul 30 2019
  • Magma
    [Floor((n+3)*(n+4)/10): n in [0..50] ]; // Vincenzo Librandi, Aug 21 2011
    
  • Maple
    A092202 := proc(n) op(1+(n mod 5),[0,1,0,-1,0]) ; end proc:
    A010891 := proc(n) op(1+(n mod 5),[1,-1,0,0,0]) ; end proc:
    A008732 := proc(n) (n+2)*(n+5)/10+(A010891(n-1)+2*A092202(n-1))/5 ; end proc:
  • Mathematica
    LinearRecurrence[{2, -1, 0, 0, 1, -2, 1}, {1, 2, 3, 4, 5, 7, 9}, 50] (* Jean-François Alcover, Jan 18 2018 *)
  • PARI
    a(n)=(n+3)*(n+4)\10 \\ Charles R Greathouse IV, Oct 07 2015
    
  • Sage
    [floor((n+3)*(n+4)/10) for n in (0..50)] # G. C. Greubel, Jul 30 2019
    

Formula

a(n) = floor( (n+3)*(n+4)/10 ) = (n+2)*(n+5)/10 + b(n)/5 where b(n) = A010891(n-2) + 2*A092202(n-1) = 0, 1, 1, 0, -2, ... with period length 5.
G.f.: 1/((1-x)^2*(1-x^5)).
a(n) = a(n-5) + n + 1. - Paul Barry, Jul 14 2004
From Mitch Harris, Sep 08 2008: (Start)
a(n) = Sum_{j=0..n+5} floor(j/5).
a(n-5) = (1/2)floor(n/5)*(2*n - 3 - 5*floor(n/5)). (End)
a(n) = A130520(n+5). - Philippe Deléham, Apr 05 2013
a(5n) = A000566(n+1), a(5n+1) = A005476(n+1), a(5n+2) = A005475(n+1), a(5n+3) = A147875(n+2), a(5n+4) = A028895(n+1); these formulas correspond to the 5 columns of the array shown in example. - Philippe Deléham, Apr 05 2013

A105384 Expansion of x/(1 + x + x^2 + x^3 + x^4).

Original entry on oeis.org

0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0
Offset: 0

Views

Author

Paul Barry, Apr 02 2005

Keywords

Comments

Inverse binomial transform of A103311. A transform of the Fibonacci numbers: apply the Chebyshev transform (1/(1+x^2), x/(1+x^2)) followed by the binomial involution (1/(1-x),-x/(1-x)) followed by the inverse binomial transform (1/(1+x), x/(1+x)) (expressed as Riordan arrays) to the -F(n); equivalently, apply (1/(1+x^2),-x/(1+x^2)) to -F(n). Periodic {0,1,-1,0,0}.
Essentially the same as A010891. - R. J. Mathar, Apr 07 2008

Formula

Euler transform of length 5 sequence [ -1, 0, 0, 0, 1].
G.f.: x(1-x)/(1-x^5);
a(n) = -sqrt(1/5 + 2*sqrt(5)/25)*cos(4*Pi*n/5 + Pi/10) + sqrt(5)*sin(4*Pi*n/5 + Pi/10)/5 + sqrt(1/5 - 2*sqrt(5)/25)*cos(2*Pi*n/5 + 3*Pi/10) + sqrt(5)*sin(2*Pi*n/5 + 3*Pi/10)/5.
a(n) = A010891(n-1). - R. J. Mathar, Apr 07 2008
a(n) + a(n-1) = A092202(n). - R. J. Mathar, Jun 23 2021

Extensions

Corrected by N. J. A. Sloane, Nov 05 2005

A105385 Expansion of (1-x^2)/(1-x^5).

Original entry on oeis.org

1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1, 0, -1, 0, 0, 1
Offset: 0

Views

Author

Paul Barry, Apr 02 2005

Keywords

Comments

Periodic {1,0,-1,0,0}.
Binomial transform is A103311(n+1). Consecutive pair sums of A105384.

Crossrefs

Cf. A092202 (essentially the same).
Cf. A198517 (absolute values).

Programs

  • Mathematica
    CoefficientList[Series[(1-x^2)/(1-x^5),{x,0,100}],x] (* or *) PadRight[{},100,{1,0,-1,0,0}] (* or *) LinearRecurrence[{-1,-1,-1,-1},{1,0,-1,0},100] (* Harvey P. Dale, Mar 10 2013 *)

Formula

G.f.: (1+x)/(1 + x + x^2 + x^3 + x^4);
a(n) = sqrt(1/5 - 2*sqrt(5)/25)*cos(4*Pi*n/5 + Pi/10) + sqrt(5)*sin(4*Pi*n/5 + Pi/10)/5 + sqrt(2*sqrt(5)/25 + 1/5)*cos(2*Pi*n/5 + 3*Pi/10) + sqrt(5)*sin(2*Pi*n/5 + 3*Pi/10)/5.
a(n) = A092202(n+1). - R. J. Mathar, Aug 28 2008
a(n) = a(n-1) - a(n-2) - a(n-3) - a(n-4); a(0)=1, a(1)=0, a(2)=-1, a(3)=0. - Harvey P. Dale, Mar 10 2013
Showing 1-4 of 4 results.