A338001 - OEIS

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A338001

Irregular triangle read by rows, a refinement of A271708.

1, 0, 1, 0, 2, 2, 0, 6, 2, 3, 0, 24, 8, 4, 3, 4, 0, 120, 8, 12, 6, 6, 4, 5, 0, 720, 48, 16, 48, 18, 6, 18, 8, 8, 5, 6, 0, 5040, 48, 48, 240, 18, 24, 12, 72, 12, 8, 24, 10, 10, 6, 7, 0, 40320, 384, 96, 192, 1440, 36, 36, 24, 36, 360, 32, 12, 32, 16, 96, 15, 10, 30, 12, 12, 7, 8

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OFFSET

0,5

COMMENTS

Row n of the triangle gives the sizes of the centralizers of any permutation of cycle type given by the partitions of n with max. part k.

T(n, k) divides n! if k > 0 and in this case the n!/T(n, k) give, up to order, the rows of A036039.

LINKS

Table of n, a(n) for n=0..74.

S. W. Golomb and P. Gaal, On the number of permutations of n objects with greatest cycle length k, Adv. in Appl. Math., 20(1), 1998, 98-107.

EXAMPLE

Triangle rows start:

0: [1];

1: [0], [1];

2: [0], [2], [2];

3: [0], [6], [2], [3];

4: [0], [24], [8, 4], [3], [4];

5: [0], [120], [8, 12], [6, 6], [4], [5];

6: [0], [720], [48, 16, 48], [18, 6, 18], [8, 8], [5], [6];

7: [0], [5040], [48, 48, 240], [18, 24, 12, 72], [12, 8, 24], [10, 10], [6], [7];

For n = 4 the partition of 4 with cycle type [2, 2] has centralizer size 8, and the partition [2, 1, 1] has centralizer size 4. Therefore in column 2 in the above triangle the pair [8, 4] appears.

PROG

(SageMath)

def A338001(n):

R = []

for k in (0..n):

P = Partitions(n, max_part=k, inner=[k])

q = [p.aut() for p in P]

R.append(q if q != [] else [0])

return flatten(R)

for n in (0..7): print(A338001(n))

CROSSREFS

Cf. A271708, A110143 (row sums), A052810 (row length), A126074, A036039.