- FindStat

Identifier

St000749: Integer partitions ⟶ ℤ

Values

[1] => 0

[2] => 0

[1,1] => 0

[3] => 0

[2,1] => 1

[1,1,1] => 0

[4] => 0

[3,1] => 0

[2,2] => 2

[2,1,1] => 0

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

[5] => 0

[4,1] => 1

[3,2] => 0

[3,1,1] => 1

[2,2,1] => 0

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

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

[6] => 0

[5,1] => 0

[4,2] => 0

[4,1,1] => 2

[3,3] => 0

[3,2,1] => 1

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

[2,2,2] => 0

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

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

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

[7] => 0

[6,1] => 1

[5,2] => 1

[5,1,1] => 0

[4,3] => 1

[4,2,1] => 0

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

[3,3,1] => 0

[3,2,2] => 0

[3,2,1,1] => 0

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

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

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

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

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

[8] => 0

[7,1] => 0

[6,2] => 2

[6,1,1] => 0

[5,3] => 2

[5,2,1] => 1

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

[4,4] => 2

[4,3,1] => 1

[4,2,2] => 1

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

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

[3,3,2] => 1

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

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

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

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

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

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

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

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

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

[9] => 0

[8,1] => 1

[7,2] => 0

[7,1,1] => 1

[6,3] => 3

[6,2,1] => 0

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

[5,4] => 3

[5,3,1] => 2

[5,2,2] => 2

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

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

[4,4,1] => 2

[4,3,2] => 2

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

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

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

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

[3,3,3] => 2

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

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

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

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

[3,2,1,1,1,1] => 0

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

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

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

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

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

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

[10] => 0

[9,1] => 0

[8,2] => 0

[8,1,1] => 2

[7,3] => 0

>>> Load all 287 entries. <<<

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$F_{1} = 1$

$F_{2} = 2$

$F_{3} = 2 + q$

$F_{4} = 4 + q^{2}$

$F_{5} = 4 + 3\ q$

$F_{6} = 8 + q + 2\ q^{2}$

$F_{7} = 8 + 6\ q + q^{3}$

$F_{8} = 8 + 8\ q + 6\ q^{2}$

$F_{9} = 8 + 7\ q + 11\ q^{2} + 4\ q^{3}$

$F_{10} = 16 + 5\ q + 6\ q^{2} + 11\ q^{3} + 4\ q^{4}$

$F_{11} = 16 + 18\ q + 4\ q^{2} + 5\ q^{3} + 11\ q^{4} + 2\ q^{5}$

$F_{12} = 32 + 10\ q + 17\ q^{2} + 3\ q^{3} + 4\ q^{4} + 9\ q^{5} + 2\ q^{6}$

Description

The smallest integer d such that the restriction of the representation corresponding to a partition of n to the symmetric group on n-d letters has a constituent of odd degree.
For example, restricting

$S_{(6,3)}$ to

$\mathfrak S_8$ yields

$S_{(5,3)}\oplus S_{(6,2)}$ of degrees (number of standard Young tableaux) 28 and 20, none of which are odd. Restricting to

$\mathfrak S_7$ yields

$S_{(4,3)}\oplus 2S_{(5,2)}\oplus S_{(6,1)}$ of degrees 14, 14 and 6. However, restricting to

$\mathfrak S_6$ yields

$S_{(3,3)}\oplus 3S_{(4,2)}\oplus 3S_{(5,1)}\oplus S_6$ of degrees 5,9,5 and 1. Therefore, the statistic on the partition

$(6,3)$ gives 3.
This is related to

$2$ -saturations of Welter's game, see [1, Corollary 1.2].

References

[1] Irie, Y. p-Saturations of Welter's Game and the Irreducible Representations of Symmetric Groups arXiv:1604.07214

Code

def branching_symmetric_group(la, p):
    """
    Return a dictionary from partitions to multiplicities.
    """
    la = Partition(la)
    l = {la: 1}
    for i in range(la.size()-p):
        l_new = dict()
        for mu in l:
            for r, _ in mu.removable_cells():
                nu = mu.remove_cell(r)
                l_new[nu] = l_new.get(nu, 0) + l[mu]
        l = l_new
    return l

def statistic(la):
    """Return the largest number such that the restriction of the
    irreducible representation corresponding to la has a component
    relative prime to 2.
    """
    la = Partition(la)    
    for m in range(la.size(), 0, -1):
        if any(gcd(StandardTableaux(mu).cardinality(), 2) == 1
               for mu in branching_symmetric_group(la, m)):
            return la.size()-m

Created

Apr 05, 2017 at 11:28 by Martin Rubey

Updated

Sep 14, 2018 at 18:56 by Martin Rubey