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A054500
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Indicator sequence for classification of nonattacking queens on n X n toroidal board.
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4
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1, 5, 7, 11, 13, 13, 13, 13, 17, 17, 17, 17, 17, 19, 19, 19, 23, 23, 23, 25, 25, 25, 25, 25, 25, 25, 25, 29, 29, 29, 29, 29
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OFFSET
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1,2
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COMMENTS
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The three sequences A054500/A054501/A054502 are used to classify solutions to the problem of "Nonattacking queens on a 2n+1 X 2n+1 toroidal board" by their symmetry; solutions are considered equivalent iff they differ only by rotation, reflection or torus shift.
For brevity, let i(n) = A054500(n) (indicator sequence), m(n) = A054501(n) (multiplicity) and c(n) = A054502(n) (count).
i(n) = k means that there are solutions for the k X k board and that m(n) and c(n) refer to it. There are c(n) inequivalent solutions which may be extended to m(n) different representations each (i.e., m(n) permutations).
This gives two formulas: A007705(n) = Sum (c(k) * m(k)), A053994(n) = Sum (c(k)), where the sum is taken over all k for which i(k) = 2n+1, for both formulas. Note that m(n) is always a divisor of 8 * i(n)^2.
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REFERENCES
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A. P. Street and R. Day, Sequential binary arrays II: Further results on the square grid, pp. 392-418 of Combinatorial Mathematics IX. Proc. Ninth Australian Conference (Brisbane, August 1981). Ed. E. J. Billington, S. Oates-Williams and A. P. Street. Lecture Notes Math., 952. Springer-Verlag, 1982 (for getting equivalence classes).
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LINKS
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I. Rivin, I. Vardi and P. Zimmermann, The n-queens problem, Amer. Math.Monthly, 101 (1994), 629-639 (for finding the solutions).
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EXAMPLE
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For a 19 X 19 toroidal board, you have three entries in the indicator sequence A054500; their count terms (A054502) give 354 = 4 + 132 + 218 inequivalent solutions; together with their multiplicity (A054501) they add up to 4*76 + 132*1444 + 218*2888 = 820496 solutions at all.
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CROSSREFS
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KEYWORD
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nonn,nice,hard
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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