

A088960


Triangle read by rows: T(n,k) = number of configurations of k nonattacking 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
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OFFSET

2,2


REFERENCES

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


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 2..1000
Irving Kaplansky and John Riordan, The problem of the rooks and its applications, Duke Mathematical Journal 13.2 (1946): 259268. See Section 9.
Irving Kaplansky and John Riordan, The problem of the rooks and its applications, in Combinatorics, Duke Mathematical Journal, 13.2 (1946): 259268. 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, ..., (n2), (n2), 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: A214279 A151501 A102735 * A263528 A121360 A199928
Adjacent sequences: A088957 A088958 A088959 * A088961 A088962 A088963


KEYWORD

nonn,tabf


AUTHOR

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


STATUS

approved



