Your data matches 1 statistic following compositions of up to 3 maps.
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Matching statistic: St001387
Mapping
St001387: Integer partitions ⟶ ℤResult quality: 100% ●values known / values provided: 100%●distinct values known / distinct values provided: 100%
Values
[1]
 => 1
[2]
 => 1
[1,1]
 => 2
[3]
 => 1
[2,1]
 => 6
[1,1,1]
 => 3
[4]
 => 1
[3,1]
 => 12
[2,2]
 => 6
[2,1,1]
 => 12
[1,1,1,1]
 => 4
[5]
 => 1
[4,1]
 => 20
[3,2]
 => 30
[3,1,1]
 => 30
[2,2,1]
 => 20
[2,1,1,1]
 => 20
[1,1,1,1,1]
 => 5
[6]
 => 1
[5,1]
 => 30
[4,2]
 => 90
[4,1,1]
 => 60
[3,3]
 => 20
[3,2,1]
 => 120
[3,1,1,1]
 => 60
[2,2,2]
 => 15
[2,2,1,1]
 => 30
[2,1,1,1,1]
 => 30
[1,1,1,1,1,1]
 => 6
[7]
 => 1
[6,1]
 => 42
[5,2]
 => 210
[5,1,1]
 => 105
[4,3]
 => 140
[4,2,1]
 => 420
[4,1,1,1]
 => 140
[3,3,1]
 => 105
[3,2,2]
 => 105
[3,2,1,1]
 => 210
[3,1,1,1,1]
 => 105
[2,2,2,1]
 => 42
[2,2,1,1,1]
 => 42
[2,1,1,1,1,1]
 => 42
[1,1,1,1,1,1,1]
 => 7
[8]
 => 1
[7,1]
 => 56
[6,2]
 => 420
[6,1,1]
 => 168
[5,3]
 => 560
[5,2,1]
 => 1120
Description
Number of standard Young tableaux of the skew shape tracing the border of the given partition. Let $\lambda \vdash n$ be a diagram with the given partition as shape. Add $n$ additional boxes, one in each column $1,\dotsc,n$, and let this be $\mu$. The statistic is the number of standard Young tableaux of skew shape $\mu/\lambda$, which is equal to $\frac{n!}{\prod_{i} (\mu_i - \lambda_i)!}$. For example, $\lambda=[2,1,1]$ gives $\mu = [4,2,1,1]$. The first row in the skew shape $\mu/\lambda$ has two boxes, so the number of SYT of shape $\mu/\lambda$ is then $4!/2 = 12$. This statistic shows up in the study of skew specialized Macdonald polynomials, where a type of charge statistic give rise to a $q$-analogue of the above formula.