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例子
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G.f.=1+2*x+4*x^2+7*x^3+12*x^4+19*x^5+30*x^6+45*x^7+67*x^8+。。。
对于n=5,考虑n+1的分区:
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.编号
第6部分,共1部分
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6 .......................... 0
3 + 3 ...................... 0
4 + 2 ...................... 0
2 + 2 + 2 .................. 0
5 + 1 ...................... 1
3 + 2 + 1 .................. 1
4 + 1 + 1 .................. 2
2 + 2 + 1 + 1 .............. 2
3 + 1 + 1 + 1 .............. 三
2 + 1 + 1 + 1 + 1 .......... 4
1 + 1 + 1 + 1 + 1 + 1 ...... 6
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35-16=19
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6的分区集合的第一列的和和和第二列的和之间的差是35-16=19,并且等于6的所有分区中的1的数量,因此该序列的第六项是a(5)=19。
(结束)
在偏移量为1的情况下,a(1)=1到a(6)=19个2*n的分区,其最大部分>n:
(2) (4)(6)(8)(A)(C)
(31) (42) (53) (64) (75)
(51) (62) (73) (84)
(411) (71) (82) (93)
(521)(91)(A2)
(611)(622)(B1)
(5111) (631) (732)
(721) (741)
(811) (822)
(6211) (831)
(7111) (921)
(61111)(A11)
(7221)
(7311)
(8211)
(9111)
(72111)
(81111)
(711111)
在偏移量为1的情况下,a(1)=1到a(6)=19个2*n的分区,其部分数>n:
(11) (211) (2211) (22211) (222211) (2222211)
(1111) (3111) (32111) (322111) (3222111)
(21111) (41111) (331111) (3321111)
(111111) (221111) (421111) (4221111)
(311111) (511111) (4311111)
(2111111) (2221111) (5211111)
(11111111) (3211111) (6111111)
(4111111) (22221111)
(22111111) (32211111)
(31111111) (33111111)
(211111111) (42111111)
(1111111111) (51111111)
(222111111)
(321111111)
(411111111)
(2211111111)
(3111111111)
(21111111111)
(111111111111)
(结束)
多集{1^5,2^1}的a(5)=19个多集分区为:
1: {{1, 1, 1, 1, 1, 2}}
2: {{1, 1, 1, 1, 1}, {2}}
3: {{1, 1, 1, 1, 2}, {1}}
4: {{1, 1, 1, 1}, {1, 2}}
5: {{1, 1, 1, 1}, {1}, {2}}
6: {{1, 1, 1, 2}, {1, 1}}
7: {{1, 1, 1, 2}, {1}, {1}}
8: {{1, 1, 1}, {1, 1, 2}}
9: {{1, 1, 1}, {1, 1}, {2}}
10: {{1, 1, 1}, {1, 2}, {1}}
11: {{1, 1, 1}, {1}, {1}, {2}}
12: {{1, 1, 2}, {1, 1}, {1}}
13: {{1, 1, 2}, {1}, {1}, {1}}
14: {{1, 1}, {1, 1}, {1, 2}}
15: {{1, 1}, {1, 1}, {1}, {2}}
16: {{1, 1}, {1, 2}, {1}, {1}}
17: {{1, 1}, {1}, {1}, {1}, {2}}
18: {{1, 2}, {1}, {1}, {1}, {1}}
19: {{1}, {1}, {1}, {1}, {1}, {2}}
(结束)
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