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 1-3 of 3 results.

A071408 a(n+1) - 2*a(n) + a(n-1) = (2/3)*(1 + w^(n+1) + w^(2*n+2)) with a(1)=0, a(2)=1, and where w is the imaginary cubic root of unity.

Original entry on oeis.org

0, 1, 4, 7, 10, 15, 20, 25, 32, 39, 46, 55, 64, 73, 84, 95, 106, 119, 132, 145, 160, 175, 190, 207, 224, 241, 260, 279, 298, 319, 340, 361, 384, 407, 430, 455, 480, 505, 532, 559, 586, 615, 644, 673, 704, 735, 766, 799, 832, 865, 900, 935, 970, 1007, 1044
Offset: 1

Views

Author

Robert G. Wilson v, Jun 24 2002

Keywords

Comments

w = exp(2*Pi*i/3)= (-1 - sqrt(-3))/2. Beginning with a(2) the first differences are 3,3,3,5,5,5,7,7,7,9,9,9,11, etc.

Links

Crossrefs

Cf. A071618.

Programs

  • Mathematica
    a[1] = 0; a[2] = 1; w = Exp[2Pi*I/3]; a[n_] := a[n] = Simplify[(2/3)(1 + w^n + w^(2n)) + 2a[n - 1] - a[n - 2]]; Table[ a[n], {n, 1, 60}]
Table[If[n<3,n-1,Floor[((n+1)^2-4)/3]],{n,1,100}] (*  Vladimir Joseph Stephan Orlovsky, Jan 30 2012 *)
LinearRecurrence[{2,-1,1,-2,1},{0,1,4,7,10},60] (* Harvey P. Dale, Jun 10 2016 *)
  • PARI
    a(n)=n*(n+2)\3 - 1 \\ Charles R Greathouse IV, Mar 02 2017

Formula

a(n) = A032765(n)-1.
a(n) = floor((n-1)*(n+1)*(n+3)/(3*n+3)). - Gary Detlefs, Jul 13 2010
a(n) = (n-1)^2 - A030511(n-1). - Wesley Ivan Hurt, Jun 19 2013
G.f.: x^2*(1+x)*(x^2-x-1) / ( (1+x+x^2)*(x-1)^3 ). - R. J. Mathar, Jun 23 2013
a(n) = n + floor(n*(n-1)/3) - 1. - Bruno Berselli, Mar 02 2017

A088166 Smallest integer divisible by Fibonacci(2n) such that the second partial quotient in the continued fraction expansion of a(n)/phi is 2 (phi is the golden ratio), n >= 2.

Original entry on oeis.org

12, 72, 504, 3410, 23184, 159094, 1089648, 7465176, 51170460, 350713222, 2403763488, 16475700746, 112925875764, 774004377960, 5305106018016, 36361732975514, 249227005939632, 1708227330997438, 11708364225400920
Offset: 2

Views

Author

Thomas Baruchel, Sep 21 2003

Keywords

Comments

Smallest integer divisible by Fibonacci(2n) such that the continued fraction expansion of a(n)/phi has period 2 or 4.

Crossrefs

Cf. A087954.

Formula

a(n) = Fibonacci(2n)*ceiling(Lucas(2n)/2).
Fibonacci(2n)*(A071618(n)+1). - Thomas Baruchel, Nov 26 2003
Empirical g.f.: -2*x^2*(x^7 - 7*x^6 - 17*x^4 + 131*x^3 - 6*x^2 + 6*x - 6) / ((x^2 - 7*x + 1)*(x^2 - 3*x + 1)*(x^4 + 3*x^3 + 8*x^2 + 3*x + 1)). - Colin Barker, Jan 11 2014

A072130 a(n+1) - 3*a(n) + a(n-1) = (2/3)*(1+w^(n+1)+w^(2*n+2)); a(1) = 0, a(2) = 1; where w is the cubic root of unity.

Original entry on oeis.org

0, 1, 5, 14, 37, 99, 260, 681, 1785, 4674, 12237, 32039, 83880, 219601, 574925, 1505174, 3940597, 10316619, 27009260, 70711161, 185124225, 484661514, 1268860317, 3321919439, 8696898000, 22768774561, 59609425685, 156059502494
Offset: 1

Views

Author

Robert G. Wilson v, Jun 24 2002

Keywords

Comments

w = exp(2*Pi*I/3) = (-1-sqrt(-3))/2.
The sequence (2/3)*(1+w^(n+1)+w^(2*n+2)) is "Period 3: repeat [0,2,0]."

Links

Crossrefs

Cf. A071618.

Programs

  • Mathematica
    a[1] = 0; a[2] = 1; w = Exp[2Pi*I/3]; a[n_] := (2/3)(1 + w^n + w^(2n)) + 3a[n - 1] - a[n - 2]; Table[ Simplify[ a[n]], {n, 1, 28}]
LinearRecurrence[{3,-1,1,-3,1},{0,1,5,14,37},30] (* Harvey P. Dale, Aug 19 2012 *)

Formula

G.f.: x^2*(1+x)*(1+x-x^2)/((1-x)*(1-3*x+x^2)*(1+x+x^2)). - Colin Barker, Jan 14 2012
a(n) = 3*a(n-1)- a(n-2)+ a(n-3)-3*a(n-4)+a(n-5). - Harvey P. Dale, Aug 19 2012
Showing 1-3 of 3 results.

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