%I #23 Jun 14 2016 12:50:49

%S 1,2,1,8,14,4,1,18,98,184,100,8,1,32,356,1704,3532,2816,632,16,1,50,

%T 940,8480,38932,89256,93800,37600,3856,32,1,72,2050,29900,242292,

%U 1109184,2800016,3653280,2180656,474368,23264,64

%N Triangle read by rows: T(n,k) = number of configurations of k non-attacking bishops on the white squares of an n X n chessboard (for n even, 0 <= k < n).

%D R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 1, 1997; see section 2.4.

%H Vincenzo Librandi, <a href="/A088960/b088960.txt">Table of n, a(n) for n = 2..1000</a>

%H Irving Kaplansky and John Riordan, <a href="http://projecteuclid.org/euclid.dmj/1077473616">The problem of the rooks and its applications</a>, Duke Mathematical Journal 13.2 (1946): 259-268. See Section 9.

%H Irving Kaplansky and John Riordan, <a href="/A274105/a274105.pdf">The problem of the rooks and its applications</a>, in Combinatorics, Duke Mathematical Journal, 13.2 (1946): 259-268. See Section 9. [Annotated scanned copy]

%H S.-M. Ma, T. Mansour, M. Schork. <a href="http://arxiv.org/abs/1308.0169">Normal ordering problem and the extensions of the Stirling grammar</a>, arXiv preprint arXiv:1308.0169, 2013

%F Generating function for fixed n = rook polynomial of Ferrers board with shape (2, 2, 4, 4, 6, 6, 8, 8, ..., (n-2), (n-2), n)

%e T(4,1) = 8 because there are 8 white squares on the 4 X 4 board to put one bishop; T(4,3) = 4 because we must place one bishop on each of three principal diagonal lines, which can be accomplished in 2*1*2=4 ways.

%e Triangle begins:

%e 1, 2

%e 1, 8, 14, 4

%e 1, 18, 98, 184, 100, 8

%e 1, 32, 356, 1704, 3532, 2816, 632, 16

%t T[n_, k_] := (Sum[(-1)^j*Binomial[n - k - 1, j]/(n - k - 1)!*(n - k + 1 - j)^(n/2)*(n - k - j)^(n/2 - 1), {j, 0, n - k - 1}]); Flatten[Table[T[n, k], {n, 2, 12, 2}, {k, 0, n - 1}]] (* _Vaclav Kotesovec_, Mar 24 2011 *)

%K nonn,tabf

%O 2,2

%A Brant Jones (brant(AT)math.washington.edu), Oct 28 2003