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A208529
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Number of permutations of n > 1 having exactly 2 points on the boundary of their bounding square.
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11
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2, 2, 4, 12, 48, 240, 1440, 10080, 80640, 725760, 7257600, 79833600, 958003200, 12454041600, 174356582400, 2615348736000, 41845579776000, 711374856192000, 12804747411456000, 243290200817664000, 4865804016353280000, 102181884343418880000
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OFFSET
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2,1
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COMMENTS
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A bounding square for a permutation of n is the square with sides parallel to the coordinate axis containing (1,1) and (n,n), and the set of points P of a permutation p is the set {(k,p(k)) for 0 < k < n+1}.
a(n) is the number of permutations of n symbols that commute with a transposition: a permutation p of {1,...,n} has exactly two points on the boundary of their bounding square if and only if p commutes with transposition (1, n). - Luis Manuel Rivera Martínez, Feb 27 2014
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LINKS
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FORMULA
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a(n) = 2(n-2)!.
G.f.: G(0), where G(k)= 1 + 1/(1 - x*(k+1)/(x*(k+1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
G.f.: 2 + 2*x/Q(0), where Q(k) = 1 - 2*x*(2*k+1) - x^2*(2*k+1)*(2*k+2)/( 1 - 2*x*(2*k+2) - x^2*(2*k+2)*(2*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Sep 23 2013
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EXAMPLE
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a(2) = 2 because {(1,1),(2,2)} and {(1,2),(2,1)} each have two points on the bounding square.
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MAPLE
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MATHEMATICA
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Table[2(n-2)!, {n, 2, 10}]
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PROG
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(Python)
import math
def a(n):
return 2*math.factorial(n-2)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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