cp's OEIS Frontend

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.

Length of n-th row: 1 + (n-1)n/2 (for a configuration for T(n,(n-1)n/2), consider n circles of radius 1 and centers at (k/n,0) for 1<=k<=n).

The generating function down the column k=1 is 1+z^2 *C^2(z) *[C^2(z)+C(z^2)]/ (2*[1-z*C(z)]) = 1+ z^2 +4*z^3 +15*z^4+ 50*z^5+...where C(z) = 1+z+2*z^2+4*z^3+... is the g.f. of A000081 divided by z; eq. (78) in arXiv:1603.00077. - R. J. Mathar, Mar 05 2016

a(n) = Theta(2^n), and more precisely, for n >= 4, (13/16)*(3/16)2^n <= a(n) <= (3/16)*2^n.

Indeed, from the formula, one gets a(n) <= (3/16)*2^n, and injecting this in the formula, one gets a(n) >= 2*a(n - 1) - (3/32)*n. Then by induction, and using the formula sum(k*2^k, k = 1 .. n) = (n - 1)*2^(n + 1) + 2, one obtains a(n) >= (13/16)*(3/16)2^n + (3/32)*n + (3/8).

Sequence motivated by the study of certain replicate tilings, where each tile can be replaced by a square number of tiles.

Adding multiples of 3=2^2-1 to the numbers 1=1^2, 9=3^2 and 17=3^2+(3^2-1), one obtains all the integers not in the sequence.

Initializing at 0, 1 or 2 gives constant sequences (except for the first term when initializing at 0).

Initializing at 3 gives 3,6,720,720!,...

If the initial term is 2 the sequence is 2, 4, 256, 256^256, ...

The term a(n) is either n+1 or a square. All the squares appear and they appear in increasing order. Every other term is a square, except when the index is a square, in which case, the corresponding term is also a square (which shifts the pattern). See FORMULA for a more precise statement.

Initializing at 0, 1, 2 or 4 gives A014221.