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%I #36 Mar 04 2024 02:53:29
%S 0,2,4,4,4,4,4,6,8,8,8,8,8,10,12,12,12,12,12,14,16,16,16,16,16,18,20,
%T 20,20,20,20,22,24,24,24,24,24,26,28,28,28,28,28,30,32,32,32,32,32,34,
%U 36,36,36,36,36,38,40,40,40,40,40,42,44,44,44,44,44,46
%N Domination number for knight graph on a 2 X n board.
%C Minimum number of knights required to dominate a 2 X n board.
%H Colin Barker, <a href="/A287393/b287393.txt">Table of n, a(n) for n = 0..1000</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Knight_(chess)">Knight (chess)</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2,-2,2,-2,2,-1).
%F a(n) = 2*(floor((n+4)/6) + floor((n+5)/6)).
%F From _Colin Barker_, May 26 2017: (Start)
%F G.f.: 2*x / ((1 - x)^2*(1 - x + x^2)*(1 + x + x^2)).
%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) for n>5.
%F (End)
%F a(n) = 2*A099480(n-1).
%e For n=3 we need a(3)=4 knights to dominate a 2 X 3 board.
%t Table[2*(Floor[(i+4)/6]+Floor[(i+5)/6]), {i, 0, 67}]
%t LinearRecurrence[{2,-2,2,-2,2,-1},{0,2,4,4,4,4},70] (* _Harvey P. Dale_, Jul 07 2020 *)
%o (Python) [2*((i+4)//6+(i+5)//6) for i in range(68)]
%o (PARI) concat(0, Vec(2*x / ((1 - x)^2*(1 - x + x^2)*(1 + x + x^2)) + O(x^100))) \\ _Colin Barker_, May 27 2017
%Y Cf. A099480, A287394.
%K nonn,easy
%O 0,2
%A _David Nacin_, May 24 2017
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