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例子
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T(4,1)=1,因为p=3412是{1,2,3,4}与属1的唯一对合。这很容易从{1,2,…,n}的置换p有亏格0这一事实得出结论,当且仅当p的循环分解给出{1,2、…,n{的非交叉分区,并且p的每个循环都在增加(参见Dulucq-Simion参考的引理2.1)。[此外,对于p=3412=(13)(24),我们得到cp'=2341*3412=4123=(1432),因此g(p)=(1/2)(4+1-2-1)=1。]
三角形开始:
[ 1] 1,
[ 2] 2, 0,
[ 3] 4, 0, 0,
[ 4] 9, 1, 0, 0,
[5]21,5,0,0,0,
[6]51,25,0,0,0,0,
[ 7] 127, 105, 0, 0, 0, 0, 0,
[ 8] 323, 420, 21, 0, 0, 0, 0, 0,
[ 9] 835, 1596, 189, 0, 0, 0, 0, 0, 0,
[10] 2188, 5880, 1428, 0, 0, 0, 0, 0, 0, 0,
[11] 5798, 21120, 8778, 0, 0, 0, 0, 0, 0, 0, 0,
[12] 15511, 74415, 48741, 1485, 0, 0, 0, 0, 0, 0, 0, 0,
[13] 41835, 258115, 249249, 19305, 0, 0, 0, 0, 0, 0, 0, 0, 0,
[14] 113634, 883883, 1201200, 191763, 0, 0, 0, 0, 0, 0, 0, ...,
[15] 310572, 2994355, 5519514, 1525095, 0, 0, 0, 0, 0, 0, 0, ...,
[16] 853467, 10051860, 24408384, 10667800, 225225, 0, 0, 0, ...,
[17] 2356779, 33479460, 104552448, 67581800, 3828825, 0, 0, ...,
...
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