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%I #29 Sep 08 2022 08:45:15
%S 1,2,2,2,2,2,3,4,4,4,4,4,5,6,6,6,6,6,7,8,8,8,8,8,9,10,10,10,10,10,11,
%T 12,12,12,12,12,13,14,14,14,14,14,15,16,16,16,16,16,17,18,18,18,18,18,
%U 19,20,20,20,20,20,21,22,22,22,22,22,23,24,24,24,24,24,25,26,26,26,26,26
%N Count from 1, repeating 2n five times.
%C Could be called the Jones sequence of the knot 9_43, since the g.f. is the reciprocal of (a parameterization of) the Jones polynomial for 9_43.
%C Half the domination number of the knight's graph on a 2 X (n+1) chessboard. - _David Nacin_, May 28 2017
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2,-2,2,-2,2,-1).
%F G.f.: 1/((1-x+x^2)(1-x-x^3+x^4)) = 1/(1-2x+2x^2-2x^3+2x^4-2x^5+x^6);
%F a(n) = 2*a(n-1)-2*a(n-2)+2*a(n-3)-2*a(n-4)+2*a(n-5)-a(n-6), n>5;
%F a(n) = -cos(Pi*2n/3+Pi/3)/6+sqrt(3)*sin(Pi*2n/3+Pi/3)/18-sqrt(3)*cos(Pi*n/3+Pi/6)/6+sin(Pi*n/3+Pi/6)/2+(n+3)/3.
%F a(n) = Sum_{i=0..n+1} floor((i-1)/6) - floor((i-3)/6). - _Wesley Ivan Hurt_, Sep 08 2015
%F a(n) = A287393(n+1)/2. - _David Nacin_, May 28 2017
%t LinearRecurrence[{2, -2, 2, -2, 2, -1}, {1, 2, 2, 2, 2, 2}, 100] (* _Vincenzo Librandi_, Sep 09 2'15 *)
%t Table[If[EvenQ[n],{n,n,n,n,n},n],{n,30}]//Flatten (* _Harvey P. Dale_, Dec 15 2020 *)
%o (Magma) I:=[1,2,2,2,2,2]; [n le 6 select I[n] else 2*Self(n-1)-2*Self(n-2)+2*Self(n-3)-2*Self(n-4)+2*Self(n-5)-Self(n-6): n in [1..100]]; // _Vincenzo Librandi_, Sep 09 2015
%Y Cf. A099479, A287393.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Oct 18 2004
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