

A357723


Number of ways to place a nonattacking black king and white king on an n X n board, up to rotation and reflection.


2



0, 0, 0, 5, 21, 63, 135, 270, 462, 770, 1170, 1755, 2475, 3465, 4641, 6188, 7980, 10260, 12852, 16065, 19665, 24035, 28875, 34650, 40986, 48438, 56550, 65975, 76167, 87885, 100485, 114840, 130200, 147560, 166056, 186813, 208845, 233415, 259407, 288230, 318630
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OFFSET

0,4


COMMENTS

Rotations and reflections of placements are not counted. (If they were then see A035286.)
a(8)=462 is the number of states in the KvK endgame in an eightfoldreducing chess tablebase on 8 X 8 boards.
When kings are unlabeled, see A279111. The ratio a(n)/A279111(n) is bounded in the interval [1, 2] and converges to 2, because the number of placements in which the kings' positions can be swapped by an automorphism is O(n^2), while the sequence itself is O(n^4).
When there are pawns on the board and the position is only equivalent under reflection in the x axis, see A357740.


LINKS



FORMULA

a(n) = n^4/8  (5/8)*n^2 + 1/2 if n is odd, else n^4/8  (7/8)*n^2 + (3/4)*n.
a(n) = 2*a(n1) + 2*a(n2)  6*a(n3) + 6*a(n5)  2*a(n6)  2*a(n7) + a(n8).
a(n) = n^4/8  (3/4)*n^2 + (3/8)*n + 1/4 + ((1/8)*n^2 + (3/8)*n  1/4)*(1)^n.
a(n) = (n^4 + (2*(n mod 2)7)*n^2 + 6*(1(n mod 2))*n + (n mod 2)*4)/8.
a(n) = (n2)*(n1)*(n^2 + 3*n + 2*(n mod 2))/8.
G.f.: x^3*(3*x^3  11*x^2  11*x  5)/((x+1)^3*(x1)^5).
E.g.f.: (x*(x^3 + 6*x^2  4)*cosh(x) + (x^4 + 6*x^3 + 2*x^2 + 4)*sinh(x))/8.  Stefano Spezia, Jan 28 2023


EXAMPLE

For n=3, the a(3) = 5 solutions are
... ... ..b b.. .b.
... ..b ... ... ...
w.b w.. w.. .w. .w.


PROG

(Python)
a=(lambda n: ((n2)*(n1)*(n**2+3*n+n%2*2)//8))


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



