- FindStat

Identifier

St001711: Integer partitions ⟶ ℤ

Values

=>
                            Cc0002;cc-rep
                            
                            
                            

                        []=>1
[1]=>1
[2]=>1
[1,1]=>1
[3]=>1
[2,1]=>3
[1,1,1]=>1
[4]=>1
[3,1]=>1
[2,2]=>1
[2,1,1]=>5
[1,1,1,1]=>1
[5]=>1
[4,1]=>5
[3,2]=>1
[3,1,1]=>1
[2,2,1]=>5
[2,1,1,1]=>7
[1,1,1,1,1]=>1
[6]=>3
[5,1]=>1
[4,2]=>1
[4,1,1]=>9
[3,3]=>1
[3,2,1]=>3
[3,1,1,1]=>1
[2,2,2]=>9
[2,2,1,1]=>17
[2,1,1,1,1]=>9
[1,1,1,1,1,1]=>1
[7]=>1
[6,1]=>3
[5,2]=>1
[5,1,1]=>1
[4,3]=>1
[4,2,1]=>7
[4,1,1,1]=>13
[3,3,1]=>19
[3,2,2]=>1
[3,2,1,1]=>5
[3,1,1,1,1]=>1
[2,2,2,1]=>15
[2,2,1,1,1]=>37
[2,1,1,1,1,1]=>11
[1,1,1,1,1,1,1]=>1
[8]=>1
[7,1]=>1
[6,2]=>3
[6,1,1]=>3
[5,3]=>1
[5,2,1]=>3
[5,1,1,1]=>1
[4,4]=>1
[4,3,1]=>5
[4,2,2]=>1
[4,2,1,1]=>29
[4,1,1,1,1]=>17
[3,3,2]=>1
[3,3,1,1]=>37
[3,2,2,1]=>5
[3,2,1,1,1]=>7
[3,1,1,1,1,1]=>1
[2,2,2,2]=>33
[2,2,2,1,1]=>45
[2,2,1,1,1,1]=>65
[2,1,1,1,1,1,1]=>13
[1,1,1,1,1,1,1,1]=>1
                    

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Description

The number of permutations such that conjugation with a permutation of given cycle type yields the squared permutation.
Let $\alpha$ be any permutation of cycle type $\lambda$. This statistic is the number of permutations $\pi$ such that
$$ \alpha\pi\alpha^{-1} = \pi^2.$$

References

[1] Homolya, S., Szigeti, Jenő Solving equations in the symmetric group arXiv:2104.03593

Code

def statistic(la):
    total = 0
    cycles = []
    for p in la:
        cycles.append(tuple(range(total+1, total+p+1)))
        total += p
    a = Permutation(cycles)
    return sum(1 for pi in Permutations(len(a)) if a*pi == pi^2*a)

Created

Apr 09, 2021 at 10:57 by Martin Rubey

Updated

Apr 09, 2021 at 10:57 by Martin Rubey