OFFSET
2,1
COMMENTS
Contribution from George Beck, Jan 08 2011: (Start)
The number of multiset partitions of the n-multiset M={0,0,...,0,1,2} (with n-2 zeros) is sum_{k=0..(n-2)}( (n-k) * p(k) ) where p(k) is the number of partitions of k.
Proof:
For each k = 0, 1, ..., n-2, partition k zeros and add the remaining n-k-2 zeros to the block {1, 2}, to give p(k) partitions.
For each k, partition k zeros and add the remaining n-k-2 zeros to the two blocks {1} and {2} in all possible 1 + n-k-2 ways, which gives (1 + n-k-2) * p(k) partitions.
Together, the number of partitions of M is sum_{k=0..n-2}( (n-k) * p(k) ). (End)
FORMULA
a(n) ~ sqrt(3) * exp(Pi*sqrt(2*n/3)) / (2*Pi^2). - Vaclav Kotesovec, Jun 23 2015
EXAMPLE
a(7) = 64 because (7,5,3,2,1,1) dot (2,3,4,5,6,7) = 14+15+12+10+6+7= 64.
MATHEMATICA
f[n_] := Sum[(n - k) PartitionsP[k], {k, 0, n - 2}]; Array[f, 39, 2]
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Alford Arnold, May 22 2003
EXTENSIONS
More terms from Ray Chandler, Oct 11 2003
STATUS
approved