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.

A131027 Period 6: repeat [4, 3, 1, 0, 1, 3].

Original entry on oeis.org

4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1, 0, 1, 3, 4, 3, 1
Offset: 1

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Author

Klaus Brockhaus, following a suggestion of Paul Curtz, Jun 10 2007

Keywords

Comments

Third column of triangular array T defined in A131022.
a(n) = abs(A078070(n+1)).
Determinants of the spiral knots S(3,k,(1,1)). a(k+4) = det(S(3,k,(1,1))). These knots are also the torus knots T(3,k). - Ryan Stees, Dec 13 2014

Examples

			For k=3, b(7)=sqrt(3)b(6)-b(5)=3-1=2, so det(S(3,3,(1,1)))=2^2=4.
		

Crossrefs

Cf. A087204, A131022, A078070. Other columns of T are in A088911, A131026, A131028, A131029, A131030.

Programs

  • Magma
    m:=105; [ [4, 3, 1, 0, 1, 3][(n-1) mod 6 + 1]: n in [1..m] ];
    
  • Maple
    A131027:=n->2+cos(n*Pi/3)+sqrt(3)*sin(n*Pi/3): seq(A131027(n), n=1..100); # Wesley Ivan Hurt, Sep 11 2014
  • Mathematica
    Table[2 + Cos[n*Pi/3] + Sqrt[3]*Sin[n*Pi/3], {n, 30}] (* Wesley Ivan Hurt, Sep 11 2014 *)
  • PARI
    {m=105; for(n=1, m, r=(n-1)%6; print1(if(r==0, 4, if(r==1||r==5, 3, if(r==3, 0, 1))), ","))}
    
  • Sage
    [(lucas_number2(n,2,1)-lucas_number2(n-1,1,1)) for n in range(4, 109)] # Zerinvary Lajos, Nov 10 2009

Formula

a(1) = 4, a(2) = a(6) = 3, a(3) = a(5) = 1, a(4) = 0, a(6) = 1; for n > 6, a(n) = a(n-6).
G.f.: (4-5*x+3*x^2)/((1-x)*(1-x+x^2)).
a(n) = 2+cos(n*Pi/3)+sqrt(3)*sin(n*Pi/3) = 2+(-1)^((n-1)/3)+(-1)^((1-n)/3). - Wesley Ivan Hurt, Sep 11 2014
a(k+4) = det(S(3,k,(1,1))) = (b(k+4))^2, where b(5)=1, b(6)=sqrt(3), b(k)=sqrt(3)*b(k-1) - b(k-2) = b(6)*b(k-1) - b(k-2). - Ryan Stees, Dec 13 2014
a(n) = 2 + 2*cos(Pi/3*(n-1)) = 2 + A087204(n-1) for n >= 1. - Werner Schulte, Jul 18 2017 and Peter Munn, Apr 28 2022