A306954 - OEIS

%I #9 Mar 17 2019 12:04:08

%S 1,4,0,9,1,0,16,2,1,1,25,4,1,1,1,36,5,2,1,1,1,49,7

%N Triangle read by rows: T(n, k), for 1 <= k <= n, is the maximum integer q such that k non-attacking armies of q queens can be placed on an n X n chessboard.

%C One has T(n, k) = 0 exactly when (n, k) = (2, 2) or (3, 3).

%C One has T(n, n) = 1 except when n = 2 or 3 (that is, when A000170(n) = 0).

%C For a fixed k, the sequence T(-, k) is nondecreasing.

%C For a fixed n, the sequence T(n, -) is nonincreasing.

%C For a fixed nonzero p, the sequence (T(k + p, k))_k is nonincreasing.  Indeed, given a configuration with k+1 armies on a k+p+1 chessboard, remove the row and column containing a given queen; this row and column can contain only queens of one army, so this yields a configuration of k armies on a k+p chessboard.

%C One has T(n, 1) = n^2 and T(n, 2) = A250000(n).

%C For a fixed k, one has, asymptotically in n, that 1/2.(n/k)^2 <= T(n, k) <= (n/k)^2, which can be proved as follows.

%C For the upper bound, let R(n, k) be defined as T(n, k) with rooks instead of queens.  Then, T(n, k) <= R(n, k) ~ (n/k)^2. Indeed, for 1 <= i <= k, say the i^th army controls a_i rows and b_i columns. The sum of the a_i's, as well as the sum of the b_i's, is at most n.  The size of the i^th army is at most a_i b_i. Therefore, one wants to maximize min(a_i b_i, i = 1..k). Ignoring rounding, the maximum is reached when all the a_i's and b_i's are equal.

%C For the lower bound, consider the n X n chessboard as a k X k grid with cells of size (n/k) X (n/k). Consider a configuration of k non-attacking queens on the k X k chessboard as in A000170(k). Place each of the k armies inside one cell occupied in that configuration. The precise placement is as follows: the army occupies the square whose vertices are the midpoints of the sides of the cell, hence has size 1/2.(n/k)^2. These armies are non-attacking.

%H Arthur O'Dwyer, <a href="https://quuxplusone.github.io/blog/2019/01/21/peaceful-encampments-round-2/">Peaceful Encampments, round 2</a> (blog post) for the "continuous case", that is, n = +oo.

%e Triangle begins:

%e    1

%e    4  0

%e    9  1  0

%e   16  2  1  1

%e   25  4  1  1  1

%e   36  5  2  1  1  1

%e   49  7  .  .  1  1  1

%e   64  9  .  .  .  1  1  1

%e   81 12  .  .  .  .  1  1  1

%Y Column 1 gives A000290, n >= 1.

%Y Cf. A000170 (non-attacking queens).

%Y Column 2 gives A250000 (case of two armies).

%K nonn,tabl,more

%O 1,2

%A _Benoit Jubin_, Mar 17 2019