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A088960
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).
3
1, 2, 1, 8, 14, 4, 1, 18, 98, 184, 100, 8, 1, 32, 356, 1704, 3532, 2816, 632, 16, 1, 50, 940, 8480, 38932, 89256, 93800, 37600, 3856, 32, 1, 72, 2050, 29900, 242292, 1109184, 2800016, 3653280, 2180656, 474368, 23264, 64
OFFSET
2,2
REFERENCES
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 1, 1997; see section 2.4.
LINKS
Irving Kaplansky and John Riordan, The problem of the rooks and its applications, Duke Mathematical Journal 13.2 (1946): 259-268. See Section 9.
Irving Kaplansky and John Riordan, The problem of the rooks and its applications, in Combinatorics, Duke Mathematical Journal, 13.2 (1946): 259-268. See Section 9. [Annotated scanned copy]
S.-M. Ma, T. Mansour, M. Schork. Normal ordering problem and the extensions of the Stirling grammar, arXiv preprint arXiv:1308.0169, 2013
FORMULA
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)
EXAMPLE
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.
Triangle begins:
1, 2
1, 8, 14, 4
1, 18, 98, 184, 100, 8
1, 32, 356, 1704, 3532, 2816, 632, 16
MATHEMATICA
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 *)
CROSSREFS
Sequence in context: A360608 A151501 A102735 * A263528 A121360 A199928
KEYWORD
nonn,tabf
AUTHOR
Brant Jones (brant(AT)math.washington.edu), Oct 28 2003
STATUS
approved