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Identifier
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