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Identifier

St001785: Integer partitions ⟶ ℤ

Values

[1] => 1

[2] => 0

[1,1] => 2

[3] => 0

[2,1] => 1

[1,1,1] => 2

[4] => 0

[3,1] => 0

[2,2] => 0

[2,1,1] => 3

[1,1,1,1] => 2

[5] => 0

[4,1] => 0

[3,2] => 0

[3,1,1] => 0

[2,2,1] => 3

[2,1,1,1] => 2

[1,1,1,1,1] => 2

[6] => 0

[5,1] => 0

[4,2] => 0

[4,1,1] => 0

[3,3] => 0

[3,2,1] => 1

[3,1,1,1] => 0

[2,2,2] => 0

[2,2,1,1] => 6

[2,1,1,1,1] => 2

[1,1,1,1,1,1] => 2

[7] => 0

[6,1] => 0

[5,2] => 0

[5,1,1] => 0

[4,3] => 0

[4,2,1] => 0

[4,1,1,1] => 0

[3,3,1] => 0

[3,2,2] => 0

[3,2,1,1] => 4

[3,1,1,1,1] => 0

[2,2,2,1] => 3

[2,2,1,1,1] => 4

[2,1,1,1,1,1] => 2

[1,1,1,1,1,1,1] => 2

[8] => 0

[7,1] => 0

[6,2] => 0

[6,1,1] => 0

[5,3] => 0

[5,2,1] => 0

[5,1,1,1] => 0

[4,4] => 0

[4,3,1] => 0

[4,2,2] => 0

[4,2,1,1] => 0

[4,1,1,1,1] => 0

[3,3,2] => 0

[3,3,1,1] => 0

[3,2,2,1] => 6

[3,2,1,1,1] => 2

[3,1,1,1,1,1] => 0

[2,2,2,2] => 0

[2,2,2,1,1] => 6

[2,2,1,1,1,1] => 4

[2,1,1,1,1,1,1] => 2

[1,1,1,1,1,1,1,1] => 2

[9] => 0

[8,1] => 0

[7,2] => 0

[7,1,1] => 0

[6,3] => 0

[6,2,1] => 0

[6,1,1,1] => 0

[5,4] => 0

[5,3,1] => 0

[5,2,2] => 0

[5,2,1,1] => 0

[5,1,1,1,1] => 0

[4,4,1] => 0

[4,3,2] => 0

[4,3,1,1] => 0

[4,2,2,1] => 0

[4,2,1,1,1] => 0

[4,1,1,1,1,1] => 0

[3,3,3] => 0

[3,3,2,1] => 4

[3,3,1,1,1] => 0

[3,2,2,2] => 0

[3,2,2,1,1] => 9

[3,2,1,1,1,1] => 2

[3,1,1,1,1,1,1] => 0

[2,2,2,2,1] => 3

[2,2,2,1,1,1] => 4

[2,2,1,1,1,1,1] => 4

[2,1,1,1,1,1,1,1] => 2

[1,1,1,1,1,1,1,1,1] => 2

[10] => 0

[9,1] => 0

[8,2] => 0

[8,1,1] => 0

[7,3] => 0

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Description

The number of ways to obtain a partition as the multiset of antidiagonal lengths of the Ferrers diagram of a partition.
Given a partition $\lambda\vdash n$, let $\alpha(\lambda)$ be the partition given by the lengths of the antidiagonals of the Ferrers diagram of $\lambda$. Then, the value of the statistic on $\mu$ is the number of times $\mu$ appears in the multiset $\{\{\alpha(\lambda)\mid \lambda\vdash n\}\}$.

References

[1] Amdeberhan, T. Agreement between two sets of data on partitions MathOverflow:420033

Code

def alpha(la):
    la = Partition(la)
    if not la:
        return la    
    result = [0]*(2*max(len(la), la[0])-1)
    for i, j in la.cells():
        result[i+j] += 1
    return Partition(sorted(result, reverse=True))

@cached_function
def frequencies(l):
    return {d: l.count(d) for d in set(l)}

def statistic(la):
    F = frequencies(tuple(alpha(mu) for mu in Partitions(la.size())))
    return F.get(la, 0)

Created

Apr 14, 2022 at 15:39 by Martin Rubey

Updated

Apr 14, 2022 at 15:39 by Martin Rubey