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Identifier

St001461: Permutations ⟶ ℤ

Values

[1] => 1

[1,2] => 2

[2,1] => 1

[1,2,3] => 3

[1,3,2] => 2

[2,1,3] => 2

[2,3,1] => 1

[3,1,2] => 1

[3,2,1] => 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

[1,2,3,4,5] => 5

[1,2,3,5,4] => 4

[1,2,4,3,5] => 4

[1,2,4,5,3] => 3

[1,2,5,3,4] => 3

[1,2,5,4,3] => 4

[1,3,2,4,5] => 4

[1,3,2,5,4] => 3

[1,3,4,2,5] => 3

[1,3,4,5,2] => 2

[1,3,5,2,4] => 2

[1,3,5,4,2] => 3

[1,4,2,3,5] => 3

[1,4,2,5,3] => 2

[1,4,3,2,5] => 4

[1,4,3,5,2] => 3

[1,4,5,2,3] => 2

[1,4,5,3,2] => 2

[1,5,2,3,4] => 2

[1,5,2,4,3] => 3

[1,5,3,2,4] => 3

[1,5,3,4,2] => 4

[1,5,4,2,3] => 2

[1,5,4,3,2] => 3

[2,1,3,4,5] => 4

[2,1,3,5,4] => 3

[2,1,4,3,5] => 3

[2,1,4,5,3] => 2

[2,1,5,3,4] => 2

[2,1,5,4,3] => 3

[2,3,1,4,5] => 3

[2,3,1,5,4] => 2

[2,3,4,1,5] => 2

[2,3,4,5,1] => 1

[2,3,5,1,4] => 1

[2,3,5,4,1] => 2

[2,4,1,3,5] => 2

[2,4,1,5,3] => 1

[2,4,3,1,5] => 3

[2,4,3,5,1] => 2

[2,4,5,1,3] => 1

[2,4,5,3,1] => 1

[2,5,1,3,4] => 1

[2,5,1,4,3] => 2

[2,5,3,1,4] => 2

[2,5,3,4,1] => 3

[2,5,4,1,3] => 1

[2,5,4,3,1] => 2

[3,1,2,4,5] => 3

[3,1,2,5,4] => 2

[3,1,4,2,5] => 2

[3,1,4,5,2] => 1

[3,1,5,2,4] => 1

[3,1,5,4,2] => 2

[3,2,1,4,5] => 4

[3,2,1,5,4] => 3

[3,2,4,1,5] => 3

[3,2,4,5,1] => 2

[3,2,5,1,4] => 2

[3,2,5,4,1] => 3

[3,4,1,2,5] => 2

[3,4,1,5,2] => 1

[3,4,2,1,5] => 2

[3,4,2,5,1] => 1

[3,4,5,1,2] => 1

[3,4,5,2,1] => 1

[3,5,1,2,4] => 1

[3,5,1,4,2] => 2

>>> Load all 1200 entries. <<<

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$F_{1} = q$

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

$F_{3} = 2\ q + 3\ q^{2} + q^{3}$

$F_{4} = 7\ q + 10\ q^{2} + 6\ q^{3} + q^{4}$

$F_{5} = 34\ q + 45\ q^{2} + 30\ q^{3} + 10\ q^{4} + q^{5}$

$F_{6} = 206\ q + 258\ q^{2} + 170\ q^{3} + 70\ q^{4} + 15\ q^{5} + q^{6}$

Description

The number of topologically connected components of the chord diagram of a permutation.
The chord diagram of a permutation

$\pi\in\mathfrak S_n$ is obtained by placing labels

$1,\dots,n$ in cyclic order on a cycle and drawing a (straight) arc from

$i$ to

$\pi(i)$ for every label

$i$ .
This statistic records the number of topologically connected components in the chord diagram. In particular, if two arcs cross, all four labels connected by the two arcs are in the same component.
The permutation

$\pi\in\mathfrak S_n$ stabilizes an interval

$I=\{a,a+1,\dots,b\}$ if

$\pi(I)=I$ . It is stabilized-interval-free, if the only interval

$\pi$ stablizes is

$\{1,\dots,n\}$ . Thus, this statistic is

$1$ if

$\pi$ is stabilized-interval-free.

References

[1] Callan, D. Counting Stabilized-Interval-Free Permutations arXiv:math/0310157
[2] Coefficients of power series A(x) such that n-th term of A(x)^n = n! x^(n-1) for n > 0. OEIS:A075834

Code

def factor(pi):
    """
    Return smallest i such that pi([1,...,i]) equals [1,...,i]
    """
    pi = list(pi)
    for i in range(1, len(pi)+1):
        A = pi[:i]
        if max(A) == i:
            return Permutation(A), Permutation([e-i for e in pi[i:]])

def rotate(pi):
    r = len(pi)
    pi1 = [x % r + 1 for x in pi]
    return Permutation(pi1[-1:] + pi1[:-1])

def orbit(pi):
    o = [pi]
    new_pi = rotate(pi)
    while pi != new_pi:
        o += [new_pi]
        new_pi = rotate(new_pi)
    return o

@cached_function
def statistic(pi):
    o = orbit(pi)
    for pi in o:
        A, B = factor(pi)
        if len(B) > 0:
            return statistic(A) + statistic(B)
    return 1

Created

Aug 09, 2019 at 21:30 by Martin Rubey

Updated

Aug 09, 2019 at 21:30 by Martin Rubey