A110707 Number of linear arrangements of n blue, n red and n green items such that there are no adjacent items of the same color (first and last elements considered as adjacent).

6, 24, 132, 804, 5196, 34872, 240288, 1688244, 12040188, 86892384, 633162360, 4650680640, 34390540320, 255773538240, 1911730760832, 14350853162676, 108139250403804, 817629606524112, 6200696697358344, 47152195812692664

Offset: 1

Author

Max Alekseyev, Aug 04 2005

Keywords

• nonn

Comments

The number of linear arrangements is given by A110706 and the number of circular arrangements counted up to rotations is given by A110710.

Crossrefs

Cf. A110706, A110710, A110711.

Programs

Mathematica

b = Binomial; a[n_] := 2*Sum[b[n-1, k]*(b[n-1, k]*(b[2*n+1-2*k, n+1] - 3* b[2*n-1-2*k, n+1]) + b[n-1, k+1]*(b[2*n-2*k, n+1] - 3*b[2*n-2*k-2, n+1]) ), {k, 0, n/2}]; Array[a, 20] (* Jean-François Alcover, Dec 04 2015, adapted from PARI *)

PARI

a(n) = 2 * sum(k=0,n\2, binomial(n-1,k) * ( binomial(n-1,k)*(binomial(2*n+1-2*k,n+1)-3*binomial(2*n-1-2*k,n+1)) + binomial(n-1,k+1)*(binomial(2*n-2*k,n+1)-3*binomial(2*n-2*k-2,n+1)) ))

Formula

a(n) = 2 * Sum[k=0..[n/2]] binomial(n-1, k) * ( binomial(n-1, k)*(binomial(2n+1-2k, n+1)-3*binomial(2n-1-2k, n+1)) + binomial(n-1, k+1)*(binomial(2n-2k, n+1)-3*binomial(2n-2k-2, n+1)) )

a(n) = A110706(n) - A110711(n)

a(n) = 2*A000172(n-1)+2*A000172(n) - Mark van Hoeij, Jul 14 2010

Conjecture: n^2*a(n) -3*n*(2*n-1)*a(n-1) -3*(n-1)*(5*n-12)*a(n-2) -8*(n-3)^2*a(n-3)=0. - R. J. Mathar, Jul 26 2014

a(n) ~ 3^(3/2) * 2^(3*n - 1) / (Pi*n). - Vaclav Kotesovec, Nov 09 2024