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FORMULA
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a(n) = 9*sqrt(3)*GAMMA(n+5/3)*GAMMA(n+4/3)*27^n/(Pi*GAMMA(2*n+3)).
a(n) = binomial(2*n+n-1, n);
Let denote P(n) = the number of integer partitions of n,
p(i) = the number of parts of the i-th partition of n,
d(i) = the number of different parts of the i-th partition of n,
m(i,j) = multiplicity of the j-th part of the i-th partition of n.
Furthermore let sum_{i=1}^{P(n)} be a sum over i and
prod_{j=1}^{d(i)} a product over j.
Then one has:
a(n)=sum_{i=1}^{P(n)} (2*n)!/((2*n-p(i))!*(prod_{j=1}^{d(i)} m(i,j)!)).
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EXAMPLE
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Let [1,1,1], [1,2] and [3] be the integer partitions of n=3.
Then [0,0,0,1,1,1],[0,0,0,0,1,2] and [0,0,0,0,0,3] are the corresponding partitions occupying 2*n = 6 positions.
We have to take into account the multiplicities of the parts including the multiplicities of the zeros.
Then
[0,0,0,1,1,1] --> 6!/(3!*3!) = 20
[0,0,0,0,1,2] --> 6!/(4!*1!*1!) = 30
[0,0,0,0,0,3] --> 6!/(5!*1!) = 6
and thus a(n=3)=20+30+6=56.
a(n=2)=10, since we have 10 ordered partitions of n=2
where the parts are distributed over 2*n=4 boxes:
[0, 0, 0, 2]
[0, 0, 1, 1]
[0, 0, 2, 0]
[0, 1, 0, 1]
[0, 1, 1, 0]
[0, 2, 0, 0]
[1, 0, 0, 1]
[1, 0, 1, 0]
[1, 1, 0, 0]
[2, 0, 0, 0].
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