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A338001
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Irregular triangle read by rows, a refinement of A271708.
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1
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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
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0,5
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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];
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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.
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PROG
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(SageMath)
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)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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