A357740 Number of non-equivalent ways under symmetry in one axis that 2 non-attacking kings of different colors can be placed on an n X n board.

0, 0, 17, 78, 234, 520, 1035, 1806, 2996, 4608, 6885, 9790, 13662, 18408, 24479, 31710, 40680, 51136, 63801, 78318, 95570, 115080, 137907, 163438, 192924, 225600, 262925, 303966, 350406, 401128, 458055, 519870, 588752, 663168, 745569, 834190, 931770, 1036296, 1150811, 1273038

Offset: 1

Author

Natalia L. Skirrow, Oct 11 2022

Keywords

• nonn
• easy

Comments

The number of king positions over which you iterate when making tablebases of positions containing pawns, wherein it is only equivalent under reflection in the x axis.

Examples

			For n=3, the a(3)=17 solutions are
   |   |  K|   |   |  K|k  |k  |k K| K |K  |K  | K |k  |k  | k | k |
   |  K|   |k  |k K|k  |   |  K|   |   |   |   |   |   |   |   |   |
k K|k  |k  |  K|   |   |  K|   |   |k  |k  | k | k | K |K  |K  | K |
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (2,2,-6,0,6,-2,-2,1).

Crossrefs

Cf. A035286 (no symmetry), A357723 (8-fold symmetry).

Programs

PARI

a(n) = {(n-2)*(n-1)*((n+3)*n - 2 + (n % 2))/2} \\ Andrew Howroyd, Dec 31 2022

Python

a=(lambda n: (n-2)*(n-1)*((n+3)*n-2+n%2)//2)

Formula

a(n) = n^4/2 - 4*n^2 + (9/2)*n - 1 if n is odd else n^4/2 - (9/2)*n^2 + 6*n - 2;

a(n) = n^4/2 - (17/4)*n^2 + (21/4)*n - 3/2 + (-1)^n*(-(1/4)*n^2 + (3/4)*n - 1/2);

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-2)*(n-1)*((n+3)*n - 2 + (n mod 2))/2.

G.f.: x^3*(17 + 44*x + 44*x^2 - 2*x^3 - 5*x^4 - 2*x^5)/((1 - x)^5*(1 + x)^3). - Andrew Howroyd, Dec 31 2022

E.g.f.: 2+(e^x*(2*x^4 + 12*x^3 - 3*x^2 + 6*x - 6) - e^(-x)*(x^2 + 2*x + 2))/4 = (cosh(x)*(x^4 + 6*x^3 - 2*x^2 + 2*x - 4) + sinh(x)*(x^4 + 6*x^3 - x^2 + 4*x - 2))/2 + 2.