Identifier
Values
[1] => -1
[2] => 2
[1,1] => -2
[3] => 3
[2,1] => 0
[1,1,1] => -3
[4] => 4
[3,1] => 0
[2,2] => 2
[2,1,1] => -2
[1,1,1,1] => -4
[5] => 5
[4,1] => 0
[3,2] => 3
[3,1,1] => -1
[2,2,1] => 1
[2,1,1,1] => -3
[1,1,1,1,1] => -5
[6] => 6
[5,1] => 0
[4,2] => 4
[4,1,1] => -1
[3,3] => 3
[3,2,1] => 1
[3,1,1,1] => -3
[2,2,2] => 2
[2,2,1,1] => -2
[2,1,1,1,1] => -4
[1,1,1,1,1,1] => -6
[7] => 7
[6,1] => 0
[5,2] => 5
[5,1,1] => -1
[4,3] => 4
[4,2,1] => 1
[4,1,1,1] => -2
[3,3,1] => 1
[3,2,2] => 3
[3,2,1,1] => -1
[3,1,1,1,1] => -4
[2,2,2,1] => 2
[2,2,1,1,1] => -3
[2,1,1,1,1,1] => -5
[1,1,1,1,1,1,1] => -7
[8] => 8
[7,1] => 0
[6,2] => 6
[6,1,1] => -1
[5,3] => 5
[5,2,1] => 1
[5,1,1,1] => -2
[4,4] => 4
[4,3,1] => 1
[4,2,2] => 4
[4,2,1,1] => -1
[4,1,1,1,1] => -4
[3,3,2] => 3
[3,3,1,1] => 0
[3,2,2,1] => 2
[3,2,1,1,1] => -3
[3,1,1,1,1,1] => -5
[2,2,2,2] => 2
[2,2,2,1,1] => -2
[2,2,1,1,1,1] => -4
[2,1,1,1,1,1,1] => -6
[1,1,1,1,1,1,1,1] => -8
[9] => 9
[8,1] => 0
[7,2] => 7
[7,1,1] => -1
[6,3] => 6
[6,2,1] => 1
[6,1,1,1] => -2
[5,4] => 5
[5,3,1] => 1
[5,2,2] => 5
[5,2,1,1] => -1
[5,1,1,1,1] => -3
[4,4,1] => 1
[4,3,2] => 4
[4,3,1,1] => 0
[4,2,2,1] => 2
[4,2,1,1,1] => -2
[4,1,1,1,1,1] => -5
[3,3,3] => 3
[3,3,2,1] => 2
[3,3,1,1,1] => -3
[3,2,2,2] => 3
[3,2,2,1,1] => -1
[3,2,1,1,1,1] => -4
[3,1,1,1,1,1,1] => -6
[2,2,2,2,1] => 3
[2,2,2,1,1,1] => -3
[2,2,1,1,1,1,1] => -5
[2,1,1,1,1,1,1,1] => -7
[1,1,1,1,1,1,1,1,1] => -9
[10] => 10
[9,1] => 0
[8,2] => 8
[8,1,1] => -1
[7,3] => 7
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Description
Dyson's crank of a partition.
Let $\lambda$ be a partition and let $o(\lambda)$ be the number of parts that are equal to 1 (St000475The number of parts equal to 1 in a partition.), and let $\mu(\lambda)$ be the number of parts that are strictly larger than $o(\lambda)$ (St000473The number of parts of a partition that are strictly bigger than the number of ones.). Dyson's crank is then defined as
$$crank(\lambda) = \begin{cases} \text{ largest part of }\lambda & o(\lambda) = 0\\ \mu(\lambda) - o(\lambda) & o(\lambda) > 0. \end{cases}$$
Let $\lambda$ be a partition and let $o(\lambda)$ be the number of parts that are equal to 1 (St000475The number of parts equal to 1 in a partition.), and let $\mu(\lambda)$ be the number of parts that are strictly larger than $o(\lambda)$ (St000473The number of parts of a partition that are strictly bigger than the number of ones.). Dyson's crank is then defined as
$$crank(\lambda) = \begin{cases} \text{ largest part of }\lambda & o(\lambda) = 0\\ \mu(\lambda) - o(\lambda) & o(\lambda) > 0. \end{cases}$$
References
[1] Andrews, G. E., Garvan, F. G. Dyson's crank of a partition MathSciNet:0929094
Code
def statistic(L):
ones = list(L).count(1)
if ones == 0:
return L[0]
else:
return sum(1 for part in L if part > ones) - ones
Created
Apr 19, 2016 at 10:47 by Christian Stump
Updated
Oct 29, 2017 at 21:22 by Martin Rubey
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