OFFSET
1,2
COMMENTS
One has T(n, k) = 0 exactly when (n, k) = (2, 2) or (3, 3).
One has T(n, n) = 1 except when n = 2 or 3 (that is, when A000170(n) = 0).
For a fixed k, the sequence T(-, k) is nondecreasing.
For a fixed n, the sequence T(n, -) is nonincreasing.
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.
One has T(n, 1) = n^2 and T(n, 2) = A250000(n).
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.
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.
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.
LINKS
Arthur O'Dwyer, Peaceful Encampments, round 2 (blog post) for the "continuous case", that is, n = +oo.
EXAMPLE
Triangle begins:
1
4 0
9 1 0
16 2 1 1
25 4 1 1 1
36 5 2 1 1 1
49 7 . . 1 1 1
64 9 . . . 1 1 1
81 12 . . . . 1 1 1
CROSSREFS
KEYWORD
AUTHOR
Benoit Jubin, Mar 17 2019
STATUS
approved