A110711 - OEIS

%I #20 Jan 12 2019 20:44:08

%S 0,6,42,288,1992,13980,99432,715344,5196336,38056284,280658100,

%T 2082218160,15528409920,116331315360,874985339760,6604555554720,

%U 50010373864416,379760762209692,2891169309592548,22062102167330592

%N 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.

%C 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.

%F 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) ).

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

%F a(n) = ((n-3)*A000172(n-1) + n*A000172(n))/(n+1). - _Mark van Hoeij_, Jul 14 2010

%F 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

%p 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));

%p series(ogf, x=0, 30); # _Mark van Hoeij_, Jan 22 2013

%o (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) ))

%Y Cf. A110706, A110707, A110710.

%K nonn

%O 1,2

%A _Max Alekseyev_, Aug 04 2005