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A329358型 |
| 二元展开式具有等长Lyndon和co-Lyndon因式分解的数。 |
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2
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1, 3, 5, 7, 9, 15, 17, 21, 27, 31, 33, 45, 51, 63, 65, 73, 74, 83, 85, 86, 89, 93, 99, 107, 119, 127, 129, 138, 150, 153, 163, 165, 174, 177, 185, 189, 195, 203, 205, 219, 231, 255, 257, 266, 273, 274, 278, 291, 294, 297, 302, 305, 310, 313, 323, 325, 333, 341
(列表;图表;参考文献;听;历史;文本;内部格式)
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抵消
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1,2
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评论
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我们将两个或多个有限序列的Lyndon乘积定义为通过将序列混洗在一起而获得的字典最大序列。例如,(231)与(213)的Lyndon乘积为(232131),(221)与(213)的乘积为(222131),(122)与(2121)的乘积为(2122121)。Lyndon词是相对于Lyndon乘积为素数的有限序列。等价地,Lyndon单词是严格小于其所有循环旋转的有限序列。每个有限序列对Lyndon单词都有一个唯一的(无序)因子分解,如果这些因子按字典序递减排列,那么它们的串联等于它们的Lyndon乘积。例如,(1001)对Lyndon因式分解(001)(1)进行了排序。
类似地,co-Lyndon乘积是通过将序列混在一起可以获得的词典编纂最小序列,co-Lindon单词是相对于co-Lyndon乘积为素数的有限序列,或者等价地,是词典编纂严格大于其所有循环旋转的有限序列。例如,(1001)对co-Lyndon因子分解(1)(100)进行了排序。
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链接
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配方奶粉
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例子
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初始项的二元展开及其Lyndon和co-Lyndon因式分解:
1: (1) = (1) = (1)
3: (11) = (1)(1) = (1)(1)
5: (101) = (1)(01) = (10)(1)
7: (111) = (1)(1)(1) = (1)(1)(1)
9: (1001) = (1)(001) = (100)(1)
15: (1111) = (1)(1)(1)(1) = (1)(1)(1)(1)
17: (10001) = (1)(0001) = (1000)(1)
21: (10101) = (1)(01)(01) = (10)(10)(1)
27: (11011) = (1)(1)(011) = (110)(1)(1)
31: (11111) = (1)(1)(1)(1)(1) = (1)(1)(1)(1)(1)
33: (100001) = (1)(00001) = (10000)(1)
45: (101101) = (1)(011)(01) = (10)(110)(1)
51: (110011) = (1)(1)(0011) = (1100)(1)(1)
63: (111111) = (1)(1)(1)(1)(1)(1) = (1)(1)(1)(1)(1)(1)
65: (1000001) = (1)(000001) = (100000)(1)
73: (1001001) = (1)(001)(001) = (100)(100)(1)
74: (1001010) = (1)(00101)(0) = (100)(10)(10)
83: (1010011) = (1)(01)(0011) = (10100)(1)(1)
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数学
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lynQ[q_]:=数组[Union[{q,RotateRight[q,#]}]=={q,旋转右[q,#]}&,长度[q]-1,1,And];
lynfac[q_]:=如果[Length[q]==0,{},函数[i,前缀[lynfac[Drop[q,i]],Take[q,i]][Last[Select[Range[Length[q]],lynQ[Take[q,#]]&]]];
colynQ[q_]:=数组[Union[{RotateRight[q,#],q}]=={Rotate Right[q,#],q}&,Length[q]-1,1,And];
colynfac[q_]:=如果[Length[q]==0,{},函数[i,前缀[colynfac[Drop[q,i]],Take[q,i]]@Last[Select[Range[Length[q]],colynQ[Take[q,#]]&]]];
选择[Range[100],Length[lynfac[IntegerDigits[#,2]]==长度[colynfac[Integer Digits[#,2]]&]
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交叉参考
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关键字
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非n
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作者
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状态
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经核准的
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