A274947 - OEIS

%I #61 Jul 28 2023 01:59:54

%S 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,

%T 16,16,16,16,16,16,0,13,18,20,21,22,23,23,24,24,24,24,24,25,25,25,25,

%U 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

%N 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.

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

%C 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.

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

%C 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

%H Bernard Lemaire and Pavel Vitushinkiy, <a href="https://cs.nyu.edu/~gottlieb/tr/overflow/1996-may-jun-1.pdf">Placing n non dominating queens on the n X n chessboard. Part I</a>, French Federation of Mathematical Games.

%H Bernard Lemaire and Pavel Vitushinkiy, <a href="https://web.archive.org/web/20220217203847/https://www.ffjm.org/upload/fichiers/THE_PROBLEM_OF_N_NON_DOMINATING_part_II.pdf">Placing n non dominating queens on the n X n chessboard. Part II</a>, French Federation of Mathematical Games.

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

%e The triangle begins:

%e 0

%e 0, 1,

%e 0, 4, 4, 4, 4,

%e 0, 7, 8, 9, 9, 9, 9, 9, 9, 9,

%e 0, 10, 13, 14, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16,

%e 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,

%e 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,

%e 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,

%e 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,

%e 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,

%e 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,

%e ...

%e (Rows 6 through 10 from _Rob Pratt_, Aug 02 2016)

%e The entry T(4,3) = 14 is achieved by

%e OXOX

%e OOOX

%e AOOO

%e OOAO

%e since the two squares marked A are not attacked by the three queens at X.

%Y Cf. A001366, A274948, A250000.

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

%Y Row 8 subtracted from 64 is A342151.

%K nonn,tabf

%O 0,5

%A _N. J. A. Sloane_, Jul 27 2016

%E Corrections and more terms from _Andrey Zabolotskiy_, Jul 29 2016

%E More terms via integer linear programming from _Rob Pratt_, Aug 02 2016