A292080 Number of nonequivalent ways to place n non-attacking rooks on an n X n board with no rook on 2 main diagonals up to rotations and reflections of the board.

1, 0, 0, 0, 2, 2, 14, 84, 630, 6096, 55336, 672160, 7409300, 104999520, 1366363752, 22068387264, 331233939624, 6005919062528, 102144359744192, 2054811316442112, 39053339674065360, 863259240785840640, 18132529836143846560, 436899062862222484480

Offset: 0

Author

Andrew Howroyd, Sep 12 2017

Keywords

• nonn

Comments

For odd n, there are no symmetrical configurations of non-attacking rooks without a rook in the main diagonal, so a(2n+1) = A003471(2n+1) / 8. For even n, configurations with rotational and diagonal symmetry are possible.

Examples

			Case n=4: The 2 nonequivalent solutions are:
   _ x _ _     _ x _ _
   x _ _ _     _ _ _ x
   _ _ _ x     x _ _ _
   _ _ x _     _ _ x _
Case n=5: The 2 nonequivalent solutions are:
   _ x _ _ _   _ x _ _ _
   x _ _ _ _   _ _ _ _ x
   _ _ _ x _   x _ _ _ _
   _ _ _ _ x   _ _ x _ _
   _ _ x _ _   _ _ _ x _
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..100
• Andrew Howroyd, Nonequivalent and Symmetric Solutions

Crossrefs

Cf. A000166, A000903, A003471, A037224, A053871, A064280.

Programs

Mathematica

sf[n_] := n! * SeriesCoefficient[Exp[-x ] / (1 - x), {x, 0, n}];
F[n_] := (Clear[v]; v[_] = 0; For[m = 4, m <= n, m++, v[m] = (m - 1)*v[m - 1] + 2*If[OddQ[m], (m - 1)*v[m - 2], (m - 2)*If[m == 4, 1, v[m - 4]]]]; v[n]);
d[n_] := Sum[(-1)^(n-k)*Binomial[n, k]*(2k)!/(2^k*k!), {k, 0, n}];
R[n_] := If[OddQ[n], 0, (n - 1)!*2/(n/2 - 1)!];
a[0] = 1; a[n_] := (F[n] + If[OddQ[n], 0, m = n/2; 2^m * sf[m] + 2*R[m] + 2*d[m]])/8;
Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Dec 28 2017, after Andrew Howroyd *)

PARI

\\ here sf is A000166, F is A003471, D is A053871, R(n) is A037224(2n).
sf(n) = {n! * polcoeff( exp(-x + x * O(x^n)) / (1 - x), n)}
F(n) = {my(v = vector(n)); for(n=4,length(v),v[n]=(n-1)*v[n-1]+2*if(n%2==1,(n-1)*v[n-2],(n-2)*if(n==4,1,v[n-4]))); v[n]}
D(n) = {sum(k=0, n, (-1)^(n-k) * binomial(n,k) * (2*k)!/(2^k*k!))}
R(n) = {if(n%2==1, 0, (n-1)!*2/(n/2-1)!)}
a(n) = {(F(n) + if(n%2==1, 0, my(m=n/2); 2^m * sf(m) + 2*R(m) + 2*D(m)))/8}

Formula

a(2n+1) = A003471(2n+1) / 8, a(2n) = (A003471(2n) + 2^n * A000166(n) + 2*A037224(2*n) + 2*A053871(n)) / 8.