

A098916


Permanent of the n X n (0,1)matrices with ijth entry equal to zero iff (i=1,j=1),(i=1,j=n),(i=n,j=1) and (i=n,j=n).


4



0, 4, 36, 288, 2400, 21600, 211680, 2257920, 26127360, 326592000, 4390848000, 63228211200, 971415244800, 15866448998400, 274611617280000, 5021469573120000, 96746980442112000, 1959126353952768000
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OFFSET

3,2


COMMENTS

The number of all possible ways to permute n distinct aligned balls, one is blue, 2 are red and the remaining are green, such that no red ball occurs by the side of the blue ball. It may generalized to r red balls: a(n,r) = (nr1)(nr)(n2)!.  Alessandro Nicolosi (xxalenicxx(AT)hotmail.com), Jul 12 2006
A formula for the permanents of these n X n matrices(A) can be easily derived by minor expansion along the first row: a(n)=per(A)=(n2)*per(B), where B is the n1 X n1 (0,1)matrix with bij=0 iff (i=n,j=1) and (i=n,j=n). A new minor expansion along the last row of B yields: per(B)=(n3)*per(C)=(n3)*(n2)! since C is the n2 X n2 1matrix. Hence: a(n)=(n2)*(n3)*(n2)!.  Herman Jamke (hermanjamke(AT)fastmail.fm), May 13 2007
Number of permutations of n1 having exactly 4 points P on the boundary of their bounding square. (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}).  David Nacin, Feb 27 2012
a(n) is also the number of permutations of n symbols that 4commute with a transposition (see A233440 for definition): a permutation p of {1,...,n} has exactly four points on the boundary of their bounding square if and only if p 4commutes with transposition (1, n).  Luis Manuel Rivera MartÃnez, Feb 27 2014


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 3..200
E. Deutsch, Permutations and their bounding squares, Math Magazine, 85(1) (2012), p. 63.
Luis Manuel Rivera, Integer sequences and kcommuting permutations, arXiv preprint arXiv:1406.3081, 2014


FORMULA

a(n) = (n2)*(n3)*(n2)!.  Herman Jamke (hermanjamke(AT)fastmail.fm), May 13 2007


EXAMPLE

a(3) = 0 because no configuration is allowed, the 2 red balls always occurs by the side of the blue ball. a(4) = 4 because we can have 4 possible permutations: b,g1,r1,r2 b,g1,r2,r1 r1,r2,g1,b r2,r1,g1,b.


MAPLE

a:= n>(n2)*(n3)*(n2)!: seq(a(n), n=3..20); # Zerinvary Lajos, Jul 01 2007


MATHEMATICA

a[n_, r_] := (nr1)(nr)(n2)! (* Alessandro Nicolosi (xxalenicxx(AT)hotmail.com), Jul 12 2006 *)
Table[(n2)*(n3)*(n2)!, {n, 3, 30}] (* Vincenzo Librandi, Feb 27 2012 *)


PROG

(PARI) permRWNb(a)=n=matsize(a)[1]; if(n==1, return(a[1, 1])); sg=1; in=vectorv(n); x=in; x=a[, n]sum(j=1, n, a[, j])/2; p=prod(i=1, n, x[i]); for(k=1, 2^(n1)1, sg=sg; j=valuation(k, 2)+1; z=12*in[j]; in[j]+=z; x+=z*a[, j]; p+=prod(i=1, n, x[i], sg)); return(2*(2*(n%2)1)*p) for(n=3, 24, a=matrix(n, n, i, j, 1); a[1, 1]=0; a[1, n]=0; a[n, 1]=0; a[n, n]=0; print1(permRWNb(a)", ")) for(n=3, 24, print1((n2)*(n3)*(n2)!", ")) \\ Herman Jamke (hermanjamke(AT)fastmail.fm), May 13 2007
(Python)
import math
def a(n):
.return (n2)*(n3)*math.factorial(n2) # David Nacin, Feb 27 2012


CROSSREFS

Cf. A208528, A208529.
Sequence in context: A176097 A173429 A172134 * A316297 A180170 A277174
Adjacent sequences: A098913 A098914 A098915 * A098917 A098918 A098919


KEYWORD

nonn


AUTHOR

Simone Severini, Oct 17 2004


EXTENSIONS

More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), May 13 2007


STATUS

approved



