|
%I #32 Sep 13 2018 11:26:30
%S 1,0,-2,-3,-1,3,5,2,-4,-7,-3,5,9,4,-6,-11,-5,7,13,6,-8,-15,-7,9,17,8,
%T -10,-19,-9,11,21,10,-12,-23,-11,13,25,12,-14,-27,-13,15,29,14,-16,
%U -31,-15,17,33,16,-18,-35,-17,19,37,18,-20,-39,-19,21,41,20,-22
%N A sequence generated from the Quadrifoil (flat knot).
%C a(3*n), n = 1,2,3... = 2*n + 1, unsigned. Odifreddi, p. 135 states: "Since the trefoil has polynomial x^2 - x + 1 and the quadrifoil (or flat knot) is the sum of two trefoils, its polynomial is (x^2 - x + 1) = x^4 - 2*x^3 + 3*x^2 - 2*x + 1."
%D P. Odifreddi, "The Mathematical Century; The 30 Greatest Problems of the Last 100 Years", Princeton University Press, page 135.
%H Alois P. Heinz, <a href="/A099471/b099471.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,-3,2,-1).
%F a(n) = M^n * [1 1 1 1], rightmost term; where M = the 4 X 4 companion matrix to the Quadrifoil polynomial x^4 - 2*x^3 + 3*x^2 - 2*x + 1: [0 1 0 0 / 0 0 1 0 / 0 0 0 1 / -1 2 -3 2].
%F G.f.: -(x^3-x^2+2*x-1) / (x^2-x+1)^2. - _Colin Barker_, May 25 2013
%F a(n+1) = 1 - sum(A101950(n-k+2, k+2), k=0..floor(n/2)) - _Johannes W. Meijer_, Aug 06 2013
%F From _A.H.M. Smeets_, Sep 13 2018 (Start)
%F a(3*k) = a(3*k-1) + a(3*k+1) for k > 0.
%F a(3*k) = (-1)^k*(2*k+1) for k >= 0.
%F a(3*k+1) = (-1)^k*k for k >= 0.
%F a(3*k+2) = (-1)^(k+1)*(k+2) for k >= 0. (End)
%e a(6) = 5 since M^6 * [1 1 1 1] = [ -3 -1 3 5].
%p a:= proc(n) local m, r; r:= 1+irem(n, 6, 'm');
%p [1, 0, -2, -3, -1, 3][r] +m*[4, 2, -2, -4, -2, 2][r]
%p end:
%p seq(a(n), n=0..80); # _Alois P. Heinz_, May 25 2013
%t Table[((9 + 6 n) Cos[Pi n/3] - 5 Sqrt[3] Sin[Pi n/3])/9, {n, 0, 20}] (* _Vladimir Reshetnikov_, Sep 09 2016 *)
%Y Cf. A099470.
%K sign,easy
%O 0,3
%A _Gary W. Adamson_, Oct 17 2004
|
