cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-1 of 1 results.

A357723 Number of ways to place a non-attacking black king and white king on an n X n board, up to rotation and reflection.

Original entry on oeis.org

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
Offset: 0

Views

Author

Natalia L. Skirrow, Oct 10 2022

Keywords

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 eightfold-reducing 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.
A quasipolynomial of degree 4 and period 2. - Charles R Greathouse IV, Feb 02 2023

Examples

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

Links

Crossrefs

Cf. A035286, A279111, A357740.

Programs

  • PARI
    a(n)=(n-2)*(n-1)*(n^2+3*n+n%2*2)\8 \\ Charles R Greathouse IV, Feb 02 2023
  • Python
    a=(lambda n: ((n-2)*(n-1)*(n**2+3*n+n%2*2)//8))

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(n-1) + 2*a(n-2) - 6*a(n-3) + 6*a(n-5) - 2*a(n-6) - 2*a(n-7) + a(n-8).
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) = (n-2)*(n-1)*(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*(x-1)^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
Showing 1-1 of 1 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.