3,2

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) = (n-r-1)(n-r)(n-2)! - 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)=(n-2)*per(B), where B is the n-1 X n-1 (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)=(n-3)*per(C)=(n-3)*(n-2)! since C is the n-2 X n-2 1-matrix. Hence: a(n)=(n-2)*(n-3)*(n-2)! - Herman Jamke (hermanjamke(AT)fastmail.fm), May 13 2007

a(n)=(n-2)*(n-3)*(n-2)! - Herman Jamke (hermanjamke(AT)fastmail.fm), May 13 2007

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

a:=n->((n+1)!-n!)*(n-1): seq(a(n), n=1..18); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 01 2007

a[n_, r_] := (n-r-1)(n-r)(n-2)! - Alessandro Nicolosi (xxalenicxx(AT)hotmail.com), Jul 12 2006

(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^(n-1)-1, sg=-sg; j=valuation(k, 2)+1; z=1-2*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((n-2)*(n-3)*(n-2)!", ")) - Herman Jamke (hermanjamke(AT)fastmail.fm), May 13 2007

Sequence in context: A075144 A043024 A144889 this_sequence A059416 A108019 A093186

Adjacent sequences: A098913 A098914 A098915 this_sequence A098917 A098918 A098919

nonn

Simone Severini (ss54(AT)york.ac.uk), Oct 17 2004

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