借助软件,利用非常充分的数据验证了一下,得到三元,四元,五元,六元的均衡样本个数的递推公式分别如下:
(注释:K为奇数时,偶数n的样本个数为0,所以下面的3,5元反映的是去掉所有的偶数项的0之后的新序列)
三元: an=2an-1-2an-3+an-4
初始值{0, 1, 2, 5}
四元:an=2an-1-an-3-an-4+2an-6-an-7
初始值{0, 0, 0, 1, 1, 3, 5}
五元:an=2an-1-an-3-2an-5+2an-6+an-8-2an-10+an-11
初始值{0, 0, 1, 3, 12, 32, 73, 141, 252, 414, 649}
六元:an=an-1+2an-2-an-3-an-4-an-5-an-6+2an-8+2an-9-an-11-an-12-an-13-an-14+2an-15+an-16-an-17
初始值{0, 0, 0, 0, 0, 1, 1, 4, 8, 18, 32, 58, 94, 151, 227, 338, 480}
比如,六元均衡样本,用mathematica进行迭代:
LinearRecurrence[{1, 2, -1, -1, -1, -1, 0, 2, 2, 0, -1, -1, -1, -1, 2,1, -1}, {0, 0, 0, 0, 0, 1, 1, 4, 8, 18, 32, 58, 94, 151, 227, 338, 480}, 100]
{0, 0, 0, 0, 0, 1, 1, 4, 8, 18, 32, 58, 94, 151, 227, 338, 480, 676, 920, 1242, 1636, 2137, 2739, 3486, 4370, 5444, 6698, 8196, 9926, 11963, 14293, 17002, 20076, 23612, 27594, 32134, 37212, 42955, 49341, 56512, 64444, 73294, 83036, 93844, 105690, 118765, 133037, 148718, 165772, 184430, 204654, 226694, 250510, 276373, 304239, 334402, 366814, 401792, 439284, 479632, 522780, 569093, 618513, 671430, 727782, 787986, 851974, 920192, 992568, 1069575, 1151137, 1237756, 1329352, 1426456, 1528984, 1637498, 1751908, 1872809, 2000105, 2134424, 2275666, 2424490, 2580792, 2745266, 2917802, 3099129,3289131, 3488574, 3697336, 3916220, 4145098, 4384810, 4635224, 4897217, 5170651, 5456444, 5754450, 6065628, 6389826, 6728044} |