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A110707
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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).
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8
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6, 24, 132, 804, 5196, 34872, 240288, 1688244, 12040188, 86892384, 633162360, 4650680640, 34390540320, 255773538240, 1911730760832, 14350853162676, 108139250403804, 817629606524112, 6200696697358344, 47152195812692664
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OFFSET
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1,1
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COMMENTS
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The number of linear arrangements is given by A110706 and the number of circular arrangements counted up to rotations is given by A110710.
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LINKS
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FORMULA
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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)) )
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
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MATHEMATICA
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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 *)
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PROG
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(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)) ))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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