A287393 Domination number for knight graph on a 2 X n board.

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, 20, 20, 20, 20, 22, 24, 24, 24, 24, 24, 26, 28, 28, 28, 28, 28, 30, 32, 32, 32, 32, 32, 34, 36, 36, 36, 36, 36, 38, 40, 40, 40, 40, 40, 42, 44, 44, 44, 44, 44, 46

Offset: 0

Author

David Nacin, May 24 2017

Keywords

• nonn
• easy

Comments

Minimum number of knights required to dominate a 2 X n board.

Examples

			For n=3 we need a(3)=4 knights to dominate a 2 X 3 board.
		

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Wikipedia, Knight (chess)
• Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-2,2,-1).

Crossrefs

Cf. A099480, A287394.

Programs

Mathematica

Table[2*(Floor[(i+4)/6]+Floor[(i+5)/6]), {i, 0, 67}]
LinearRecurrence[{2,-2,2,-2,2,-1},{0,2,4,4,4,4},70] (* Harvey P. Dale, Jul 07 2020 *)

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

Python

[2*((i+4)//6+(i+5)//6) for i in range(68)]

Formula

a(n) = 2*(floor((n+4)/6) + floor((n+5)/6)).

From Colin Barker, May 26 2017: (Start)

G.f.: 2*x / ((1 - x)^2*(1 - x + x^2)*(1 + x + x^2)).

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.

(End)

a(n) = 2*A099480(n-1).