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

1,1

Comments

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

Links

Table of n, a(n) for n=1..20.

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

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 *)

Prog

(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)) ))

Crossrefs

Cf. A110706, A110710, A110711.

Sequence in context: A219622 A052170 A027224 * A047712 A188330 A126267

Adjacent sequences: A110704 A110705 A110706 * A110708 A110709 A110710

Keyword

nonn

Author

Max Alekseyev, Aug 04 2005

Status

approved