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.

A110551 Period 6: repeat [1, 3, 5, 5, 3, 1].

Original entry on oeis.org

1, 3, 5, 5, 3, 1, 1, 3, 5, 5, 3, 1, 1, 3, 5, 5, 3, 1, 1, 3, 5, 5, 3, 1, 1, 3, 5, 5, 3, 1, 1, 3, 5, 5, 3, 1, 1, 3, 5, 5, 3, 1, 1, 3, 5, 5, 3, 1, 1, 3, 5, 5, 3, 1, 1, 3, 5, 5, 3, 1, 1, 3, 5, 5, 3, 1, 1, 3, 5, 5, 3, 1, 1, 3, 5, 5, 3, 1, 1, 3, 5, 5, 3, 1, 1, 3, 5, 5, 3, 1
Offset: 0

Views

Author

Paul Barry, Jul 26 2005

Keywords

Comments

a(n) = A162699(n+1) (Modd 7) = A204453(A162699(n+1)), n>=0, where the nonnegative members of the seven residue classes Mod 7 (not to be confused with mod 7), called [m] for m=0..6, are given in the array A113807, if there the last row, starting with 7 is taken as class [0] after adding a 0 in front. Here only the classes [1], [3] and [5] are relevant. For Modd n residue classes see a comment on A203571. [Wolfdieter Lang, Feb 09 2012]
Continued fractions expansion of (8+sqrt(905))/29 = 1.3132144107925.. - R. J. Mathar, Mar 08 2012

Examples

			Modd 7 classes for positive odd numbers reduced mod 7: a(3)=5 because A162699(4)=9 (the fourth positive odd number not divisible by 7), and 9 is a member of the Modd 7 class [5] = {5,9,19,23,...}.
A162699: 1, 3, 5, 9, 11, 13, 15, 17, 19, 23, 25, 27,...
Modd 7:  1, 3, 5, 5,  3,  1,  1,  3,  5,  5,  3,  1,... [_Wolfdieter Lang_, Feb 09 2012]
		

Crossrefs

Programs

Formula

From R. J. Mathar, Oct 15 2014: (Start)
G.f.: ( 1+x+x^2 ) / ( (1-x)*(x^2-x+1) ).
a(n) = a(n-1) - a(n-2) + a(n-3) - a(n-4) + a(n-5) for n>4.
a(n) = 3 + 2*sin(Pi*n/3)/sqrt(3) - 2*cos(Pi*n/3).
a(n) = A001045(n+2) mod 6. (End)
From Wesley Ivan Hurt, Jun 29 2016: (Start)
a(n) = a(n-6) for n>5.
a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3) for n>2. (End)

A110569 Period 6: repeat [2, 1, 3, 3, 1, 2].

Original entry on oeis.org

2, 1, 3, 3, 1, 2, 2, 1, 3, 3, 1, 2, 2, 1, 3, 3, 1, 2, 2, 1, 3, 3, 1, 2, 2, 1, 3, 3, 1, 2, 2, 1, 3, 3, 1, 2, 2, 1, 3, 3, 1, 2, 2, 1, 3, 3, 1, 2, 2, 1, 3, 3, 1, 2, 2, 1, 3, 3, 1, 2, 2, 1, 3, 3, 1, 2, 2, 1, 3, 3, 1, 2, 2, 1, 3, 3, 1, 2, 2, 1, 3, 3, 1, 2, 2, 1, 3, 3, 1, 2
Offset: 0

Views

Author

Paul Barry, Jul 27 2005

Keywords

Comments

Permutation of {1, 2, 3}, followed by its reversal, repeated.

Crossrefs

Programs

Formula

a(n) = 1+(A078008(n) mod 3).
G.f.: (2-x+4*x^2-x^3+2*x^4) / (1-x+x^2-x^3+x^4-x^5).
a(n) = a(n-1) - a(n-2) + a(n-3) - a(n-4) + a(n-5) for n>4.
a(n) = 2 + cos(2*Pi*n/3)/2 - sqrt(3)*sin(2*Pi*n/3)/2 - cos(Pi*n/3)/2 + sqrt(3)*sin(Pi*n/3)/6.
a(n) = a(n-6) for n>5. - Wesley Ivan Hurt, Jun 27 2016

Extensions

Name changed by Wesley Ivan Hurt, Jun 27 2016

A110568 Period 6: repeat [1, 0, 2, 2, 0, 1].

Original entry on oeis.org

1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2
Offset: 0

Views

Author

Paul Barry, Jul 27 2005

Keywords

Comments

Permutation of {0, 1, 2}, followed by its reversal, repeated.

Crossrefs

Programs

  • Magma
    &cat [[1, 0, 2, 2, 0, 1]^^30]; // Wesley Ivan Hurt, Jun 28 2016
    
  • Maple
    A110568:=n->[1, 0, 2, 2, 0, 1][(n mod 6)+1]: seq(A110568(n), n=0..100); # Wesley Ivan Hurt, Jun 28 2016
  • Mathematica
    Mod[#,3]&/@CoefficientList[Series[(1-x)/(1-x-2x^2),{x,0,100}],x] (* Harvey P. Dale, Mar 30 2011 *)
    PadRight[{}, 100, {1, 0, 2, 2, 0, 1}] (* Wesley Ivan Hurt, Jun 28 2016 *)
    LinearRecurrence[{1,-1,1,-1,1},{1,0,2,2,0},100] (* Harvey P. Dale, Apr 03 2019 *)
  • PARI
    x='x+O('x^50); Vec((1-x+3*x^2-x^3+x^4)/(1-x+x^2-x^3+x^4-x^5)) \\ G. C. Greubel, Aug 31 2017

Formula

a(n) = A078008(n) mod 3.
G.f.: (1-x+3*x^2-x^3+x^4) / (1-x+x^2-x^3+x^4-x^5).
a(n) = a(n-1) - a(n-2) + a(n-3) - a(n-4) + a(n-5) for n>4.
a(n) = 1 + cos(2*Pi*n/3)/2 - sqrt(3)*sin(2*Pi*n/3)/2 - cos(Pi*n/3)/2 + sqrt(3)*sin(Pi*n/3)/6.
a(n) = a(n-6) for n > 5. - Wesley Ivan Hurt, Jun 28 2016
a(n) = ((n-1)*(-1)^(n-1) mod 3). - Wesley Ivan Hurt, Jan 07 2021

Extensions

Name changed by Wesley Ivan Hurt, Jun 28 2016

A111951 Period 8: repeat [0,3,1,2,2,1,3,0].

Original entry on oeis.org

0, 3, 1, 2, 2, 1, 3, 0, 0, 3, 1, 2, 2, 1, 3, 0, 0, 3, 1, 2, 2, 1, 3, 0, 0, 3, 1, 2, 2, 1, 3, 0, 0, 3, 1, 2, 2, 1, 3, 0, 0, 3, 1, 2, 2, 1, 3, 0, 0, 3, 1, 2, 2, 1, 3, 0, 0, 3, 1, 2, 2, 1, 3, 0, 0, 3, 1, 2, 2, 1, 3, 0, 0, 3, 1, 2, 2, 1, 3, 0, 0, 3, 1, 2, 2, 1, 3, 0, 0, 3, 1, 2, 2, 1, 3, 0
Offset: 0

Views

Author

Paul Barry, Aug 22 2005

Keywords

Comments

Permutation of {0,1,2,3} followed by its reversal, repeated.

Crossrefs

Programs

Formula

G.f.: (3x + x^2 + 2x^3 + 2x^4 + x^5 + 3x^6)/(1 - x^8);
a(n) = a(n-1) - a(n-2) + a(n-3) - a(n-4) + a(n-5) - a(n-6) + a(n-7);
a(n) = n(7n-1)/2 mod 4 = A022264(n) mod 4.
G.f.: -x*(3 - 2*x + 4*x^2 - 2*x^3 + 3*x^4) / ( (x-1)*(1+x^2)*(1+x^4) ). - R. J. Mathar, Feb 20 2015
a(n) = (3 + r/2 - s/2 + 2*cos(Pi*(1+2*n-r-s+t)/8) - 2*cos(Pi*(1-2*n+r-s+t)/8) - 2*sin(Pi*(1-2*n-r+s+t)/8))/2 where r = 2*sin(n*Pi/2), s = 2*cos(n*Pi/2) and t = cos(n*Pi). - Wesley Ivan Hurt, Oct 05 2018

Extensions

Name changed, the original name moved to comments. - Antti Karttunen, Aug 10 2017
Showing 1-4 of 4 results.