Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.
%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