A274947 Irregular triangle read by rows: T(n,k) (n>=0, 0 <= k <= n^2) = least number of squares attacked by k queens on an n X n board.

0, 0, 1, 0, 4, 4, 4, 4, 0, 7, 8, 9, 9, 9, 9, 9, 9, 9, 0, 10, 13, 14, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 0, 13, 18, 20, 21, 22, 23, 23, 24, 24, 24, 24, 24, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 0, 16, 23, 27, 28, 30, 31, 32, 32, 33, 34, 34, 34, 34, 35, 35, 35, 35, 35, 35, 35, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36

Offset: 0

Author

N. J. A. Sloane, Jul 27 2016

Keywords

• nonn
• tabf

Comments

Place k queens on an n X n board so that the total number of squares attacked/occupied by the queens is minimized.

If enough terms were known, would provide an upper bound for A250000. For if A250000(n) = Q then T(n,Q) <= n^2 - Q, or equivalently A274948(n,Q) >= Q.

Values n^2 - T(n,n) are given in A001366.

Let X(n) be the smallest number so that no matter how you place X queens, they attack every square. That is, X is the minimal number such that T(n,k) = n^2 for all k >= X. Then X = n^2 - T(n,1) + 1 = A274948(n,1) + 1 = n^2 - 3*n + 3. More generally, T(n,k') <= n^2-k if and only if k' <= n^2-T(n,k). For example, we may place 2 queens on two squares of a 4 X 4 board and leave 4^2-T(4,2)=3 squares not attacked, so we may place 3 queens on these 3 squares instead and leave those two squares not attacked, ergo, T(4,3)=16-2. - Andrey Zabolotskiy, Jul 29 2016

Examples

			The triangle begins:
0
0, 1,
0, 4, 4, 4, 4,
0, 7, 8, 9, 9, 9, 9, 9, 9, 9,
0, 10, 13, 14, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16,
0, 13, 18, 20, 21, 22, 23, 23, 24, 24, 24, 24, 24, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25,
0, 16, 23, 27, 28, 30, 31, 32, 32, 33, 34, 34, 34, 34, 35, 35, 35, 35, 35, 35, 35, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36,
0, 19, 28, 33, 33, 38, 39, 42, 43, 43, 43, 44, 45, 45, 45, 45, 45, 47, 47, 47, 47, 47, 48, 48, 48, 48, 48, 48, 48, 48, 48, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49, 49,
0, 22, 33, 39, 40, 47, 49, 51, 53, 54, 55, 56, 57, 57, 58, 58, 59, 59, 60, 60, 60, 60, 60, 60, 60, 61, 62, 62, 62, 62, 62, 62, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64,
0, 25, 38, 45, 45, 54, 57, 61, 62, 63, 67, 68, 69, 70, 71, 72, 72, 72, 72, 73, 74, 75, 75, 75, 75, 76, 76, 76, 77, 77, 77, 77, 77, 77, 77, 77, 77, 79, 79, 79, 79, 79, 79, 79, 80, 80, 80, 80, 80, 80, 80, 80, 80, 80, 80, 80, 80, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81,
0, 28, 43, 51, 52, 63, 67, 70, 74, 76, 78, 81, 82, 84, 85, 86, 87, 88, 88, 89, 90, 90, 90, 91, 91, 92, 92, 93, 93, 93, 93, 94, 94, 94, 95, 95, 95, 95, 96, 96, 96, 96, 96, 96, 96, 96, 96, 96, 96, 97, 98, 98, 98, 98, 98, 98, 98, 98, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100,
...
(Rows 6 through 10 from _Rob Pratt_, Aug 02 2016)
The entry T(4,3) = 14 is achieved by
OXOX
OOOX
AOOO
OOAO
since the two squares marked A are not attacked by the three queens at X.
		

Links

• Bernard Lemaire and Pavel Vitushinkiy, Placing n non dominating queens on the n X n chessboard. Part I, French Federation of Mathematical Games.
• Bernard Lemaire and Pavel Vitushinkiy, Placing n non dominating queens on the n X n chessboard. Part II, French Federation of Mathematical Games.

Crossrefs

Cf. A001366, A274948, A250000.

Cf. A075458 (minimal number of queens needed to attack all the squares of an n X n board).

Row 8 subtracted from 64 is A342151.

Formula

T(n,1) = 3*n-2 for n >= 1.

Extensions

Corrections and more terms from Andrey Zabolotskiy, Jul 29 2016

More terms via integer linear programming from Rob Pratt, Aug 02 2016