cp's OEIS Frontend

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).

%I A110707 #15 Nov 09 2024 08:31:42
%S A110707 6,24,132,804,5196,34872,240288,1688244,12040188,86892384,633162360,
%T A110707 4650680640,34390540320,255773538240,1911730760832,14350853162676,
%U A110707 108139250403804,817629606524112,6200696697358344,47152195812692664
%N 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).
%C A110707 The number of linear arrangements is given by A110706 and the number of circular arrangements counted up to rotations is given by A110710.
%F A110707 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)) )
%F A110707 a(n) = A110706(n) - A110711(n)
%F A110707 a(n) = 2*A000172(n-1)+2*A000172(n) - _Mark van Hoeij_, Jul 14 2010
%F A110707 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
%F A110707 a(n) ~ 3^(3/2) * 2^(3*n - 1) / (Pi*n). - _Vaclav Kotesovec_, Nov 09 2024
%t A110707 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 *)
%o A110707 (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)) ))
%Y A110707 Cf. A110706, A110710, A110711.
%K A110707 nonn
%O A110707 1,1
%A A110707 _Max Alekseyev_, Aug 04 2005