A110711 Number of linear arrangements of n blue, n red and n green items such that first and last elements have the same color but there are no adjacent items of the same color.

0, 6, 42, 288, 1992, 13980, 99432, 715344, 5196336, 38056284, 280658100, 2082218160, 15528409920, 116331315360, 874985339760, 6604555554720, 50010373864416, 379760762209692, 2891169309592548, 22062102167330592

Offset: 1

Author

Max Alekseyev, Aug 04 2005

Keywords

• nonn

Comments

The number of linear arrangements is given by A110706 (first and last elements are not adjacent) and A110707 (first and last elements are adjacent) and the number of circular arrangements (counted up to rotations) is given by A110710.

Crossrefs

Cf. A110706, A110707, A110710.

Programs

Maple

ogf := 6*((x-2)*hypergeom([1/3,1/3],[1], 27*x^2/((8*x-1)*(x+1)^2)) + 2*hypergeom([1/3,1/3],[2], 27*x^2/((8*x-1)*(x+1)^2))) / ((1-2* x)*(1+x)^(2/3)*(1-8*x)^(1/3));
series(ogf, x=0, 30); # Mark van Hoeij, Jan 22 2013

PARI

a(n) = 6 * sum(k=0,n\2, binomial(n-1,k) * ( binomial(n-1,k)*binomial(2*n-1-2*k,n+1) + binomial(n-1,k+1)*binomial(2*n-2*k-2,n+1) ))

Formula

a(n) = 6 * Sum_{k=0..floor(n/2)} binomial(n-1, k) * ( binomial(n-1, k)*binomial(2n-1-2k, n+1) + binomial(n-1, k+1)*binomial(2n-2k-2, n+1) ).

a(n) = A110706(n) - A110707(n).

a(n) = ((n-3)*A000172(n-1) + n*A000172(n))/(n+1). - Mark van Hoeij, Jul 14 2010

Conjecture: -(n+1)*(n-2)*a(n) + (7*n^2 - 13*n + 4)*a(n-1) + 8*(n-2)^2*a(n-2) = 0. - R. J. Mathar, Nov 01 2015