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%I #28 Jan 14 2018 03:23:26
%S 1,0,0,0,2,2,14,84,630,6096,55336,672160,7409300,104999520,1366363752,
%T 22068387264,331233939624,6005919062528,102144359744192,
%U 2054811316442112,39053339674065360,863259240785840640,18132529836143846560,436899062862222484480
%N 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.
%C 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.
%H Andrew Howroyd, <a href="/A292080/b292080.txt">Table of n, a(n) for n = 0..100</a>
%H Andrew Howroyd, <a href="/A292080/a292080.txt">Nonequivalent and Symmetric Solutions</a>
%F a(2n+1) = A003471(2n+1) / 8, a(2n) = (A003471(2n) + 2^n * A000166(n) + 2*A037224(2*n) + 2*A053871(n)) / 8.
%e Case n=4: The 2 nonequivalent solutions are:
%e _ x _ _ _ x _ _
%e x _ _ _ _ _ _ x
%e _ _ _ x x _ _ _
%e _ _ x _ _ _ x _
%e Case n=5: The 2 nonequivalent solutions are:
%e _ x _ _ _ _ x _ _ _
%e x _ _ _ _ _ _ _ _ x
%e _ _ _ x _ x _ _ _ _
%e _ _ _ _ x _ _ x _ _
%e _ _ x _ _ _ _ _ x _
%t sf[n_] := n! * SeriesCoefficient[Exp[-x ] / (1 - x), {x, 0, n}];
%t 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]);
%t d[n_] := Sum[(-1)^(n-k)*Binomial[n, k]*(2k)!/(2^k*k!), {k, 0, n}];
%t R[n_] := If[OddQ[n], 0, (n - 1)!*2/(n/2 - 1)!];
%t 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;
%t Table[a[n], {n, 0, 23}] (* _Jean-François Alcover_, Dec 28 2017, after _Andrew Howroyd_ *)
%o (PARI) \\ here sf is A000166, F is A003471, D is A053871, R(n) is A037224(2n).
%o sf(n) = {n! * polcoeff( exp(-x + x * O(x^n)) / (1 - x), n)}
%o 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]}
%o D(n) = {sum(k=0, n, (-1)^(n-k) * binomial(n,k) * (2*k)!/(2^k*k!))}
%o R(n) = {if(n%2==1, 0, (n-1)!*2/(n/2-1)!)}
%o a(n) = {(F(n) + if(n%2==1, 0, my(m=n/2); 2^m * sf(m) + 2*R(m) + 2*D(m)))/8}
%Y Cf. A000166, A000903, A003471, A037224, A053871, A064280.
%K nonn
%O 0,5
%A _Andrew Howroyd_, Sep 12 2017
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