主帖换个马甲——正整数n的无约束划分。
a(1)=01, {1},
a(2)=02, {2}, {11},
a(3)=03, {3}, {12}, {111},
a(4)=05, {4}, {13}, {22}, {112}, {1111},
a(5)=07, {5}, {14}, {23}, {113}, {122}, {1112}, {11111},
a(6)=11, {6}, {15}, {24}, {33}, {114}, {123}, {222}, {1113}, {1122}, {11112}, {111111},
a(7)=15, {7}, {16}, {25}, {34}, {115}, {124}, {133}, {223}, {1114}, {1123}, {12222}, {11113}, {11122}, {111112}, {1111111},
a(8)=22, {8}, {17}, {26}, {35}, {44}, {116}, {125}, {134}, {224}, {233}, {1115}, {1124}, {1133}, {1223}, {2222}, ......
a(9)=30, {9}, {18}, {27}, {36}, {45}, {117}, {126}, {135}, {144}, {225}, {234}, {333}, {1116}, {1125}, {1134}, ......
得到这样一串数——1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718,
4565, 5604, 6842, 8349, 10143, 12310, 14883, 17977, 21637, 26015, 31185, 37338, 44583, 53174, 63261, 75175, 89134, 105558, 124754, ...... ——A000041
- Table[PartitionsP[n], {n, 45}]
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详见《将正整数 n 拆分成 3 个不全相等的正整数相加,有几种不同的拆分法?》
在电影《知无涯者》中看到麦克马洪手算了前200个整数的拆分数,在观影时我注意到这个细节,自己编程算出了p(200) = 3972999029388,我才意识到每个时代的数学家都有着自己的伟大之处。
谢谢 Nicolas2050 !谢谢 Nicolas2050 !谢谢 Nicolas2050 ! |