Identifier
Values
[] => 1
[1] => 1
[2] => 1
[1,1] => 1
[3] => 1
[2,1] => 3
[1,1,1] => 1
[4] => 1
[3,1] => 4
[2,2] => 2
[2,1,1] => 6
[1,1,1,1] => 1
[5] => 1
[4,1] => 5
[3,2] => 5
[3,1,1] => 10
[2,2,1] => 10
[2,1,1,1] => 10
[1,1,1,1,1] => 1
[6] => 1
[5,1] => 6
[4,2] => 6
[4,1,1] => 15
[3,3] => 3
[3,2,1] => 30
[3,1,1,1] => 20
[2,2,2] => 5
[2,2,1,1] => 30
[2,1,1,1,1] => 15
[1,1,1,1,1,1] => 1
[7] => 1
[6,1] => 7
[5,2] => 7
[5,1,1] => 21
[4,3] => 7
[4,2,1] => 42
[4,1,1,1] => 35
[3,3,1] => 21
[3,2,2] => 21
[3,2,1,1] => 105
[3,1,1,1,1] => 35
[2,2,2,1] => 35
[2,2,1,1,1] => 70
[2,1,1,1,1,1] => 21
[1,1,1,1,1,1,1] => 1
[8] => 1
[7,1] => 8
[6,2] => 8
[6,1,1] => 28
[5,3] => 8
[5,2,1] => 56
[5,1,1,1] => 56
[4,4] => 4
[4,3,1] => 56
[4,2,2] => 28
[4,2,1,1] => 168
[4,1,1,1,1] => 70
[3,3,2] => 28
[3,3,1,1] => 84
[3,2,2,1] => 168
[3,2,1,1,1] => 280
[3,1,1,1,1,1] => 56
[2,2,2,2] => 14
[2,2,2,1,1] => 140
[2,2,1,1,1,1] => 140
[2,1,1,1,1,1,1] => 28
[1,1,1,1,1,1,1,1] => 1
[9] => 1
[8,1] => 9
[7,2] => 9
[7,1,1] => 36
[6,3] => 9
[6,2,1] => 72
[6,1,1,1] => 84
[5,4] => 9
[5,3,1] => 72
[5,2,2] => 36
[5,2,1,1] => 252
[5,1,1,1,1] => 126
[4,4,1] => 36
[4,3,2] => 72
[4,3,1,1] => 252
[4,2,2,1] => 252
[4,2,1,1,1] => 504
[4,1,1,1,1,1] => 126
[3,3,3] => 12
[3,3,2,1] => 252
[3,3,1,1,1] => 252
[3,2,2,2] => 84
[3,2,2,1,1] => 756
[3,2,1,1,1,1] => 630
[3,1,1,1,1,1,1] => 84
[2,2,2,2,1] => 126
[2,2,2,1,1,1] => 420
[2,2,1,1,1,1,1] => 252
[2,1,1,1,1,1,1,1] => 36
[1,1,1,1,1,1,1,1,1] => 1
[10] => 1
[9,1] => 10
[8,2] => 10
[8,1,1] => 45
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Description
The Kreweras number of an integer partition.
This is defined for $\lambda \vdash n$ with $k$ parts as
$$\frac{1}{n+1}\binom{n+1}{n+1-k,\mu_1(\lambda),\ldots,\mu_n(\lambda)}$$
where $\mu_j(\lambda)$ denotes the number of parts of $\lambda$ equal to $j$, see [1]. This formula indeed counts the number of noncrossing set partitions where the ordered block sizes are the partition $\lambda$.
These numbers refine the Narayana numbers $N(n,k) = \frac{1}{k}\binom{n-1}{k-1}\binom{n}{k-1}$ and thus sum up to the Catalan numbers $\frac{1}{n+1}\binom{2n}{n}$.
This is defined for $\lambda \vdash n$ with $k$ parts as
$$\frac{1}{n+1}\binom{n+1}{n+1-k,\mu_1(\lambda),\ldots,\mu_n(\lambda)}$$
where $\mu_j(\lambda)$ denotes the number of parts of $\lambda$ equal to $j$, see [1]. This formula indeed counts the number of noncrossing set partitions where the ordered block sizes are the partition $\lambda$.
These numbers refine the Narayana numbers $N(n,k) = \frac{1}{k}\binom{n-1}{k-1}\binom{n}{k-1}$ and thus sum up to the Catalan numbers $\frac{1}{n+1}\binom{2n}{n}$.
References
[1] Reiner, V., Sommers, E. Weyl group $q$-Kreweras numbers and cyclic sieving arXiv:1605.09172
Code
def statistic(la):
la = list(la)
n = sum(la)
k = len(la)
multi = [n+1-k]+[ la.count(j) for j in [1..n] ]
return multinomial(multi)/(n+1)
Created
May 31, 2016 at 14:57 by Christian Stump
Updated
Jun 19, 2023 at 10:40 by Martin Rubey
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