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

Offset: 0

Author

Peter Luschny, Nov 13 2020

Keywords

• nonn
• tabf

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.

Examples

			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.
		

Links

• 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.

Crossrefs

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

Programs

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))