A131027 Period 6: repeat [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, 0, 1, 3, 4, 3, 1
Offset: 1
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.
Links
- A. Breiland, L. Oesper, and L. Taalman, p-Coloring classes of torus knots, Online Missouri J. Math. Sci., 21 (2009), 120-126.
- N. Brothers, S. Evans, L. Taalman, L. Van Wyk, D. Witczak, and C. Yarnall, Spiral knots, Missouri J. of Math. Sci., 22 (2010).
- M. DeLong, M. Russell, and J. Schrock, Colorability and determinants of T(m,n,r,s) twisted torus knots for n equiv. +/-1(mod m), Involve, Vol. 8 (2015), No. 3, 361-384.
- Seong Ju Kim, R. Stees, and L. Taalman, Sequences of Spiral Knot Determinants, Journal of Integer Sequences, Vol. 19 (2016), #16.1.4.
- Ryan Stees, Sequences of Spiral Knot Determinants, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
- Index entries for linear recurrences with constant coefficients, signature (2,-2,1).
Crossrefs
Programs
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Magma
m:=105; [ [4, 3, 1, 0, 1, 3][(n-1) mod 6 + 1]: n in [1..m] ];
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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
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Mathematica
Table[2 + Cos[n*Pi/3] + Sqrt[3]*Sin[n*Pi/3], {n, 30}] (* Wesley Ivan Hurt, Sep 11 2014 *)
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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))), ","))}
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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
Comments