换句话说,a(n)是最小的数m,使得m有n个不同的除数d_1。。。,d_n使d_1++d_n=m.(d_i不需要是m的所有除数)例如,a(6)=m=24,因为24的除数是1、2、3、4、6、8、12、24和1+2+3+4+6+8=24。
在下面的三角形中,第n行给出了n个除数a(1)。。。,a(7);a(n)=第n行的总和:
1
- -
1 2 3
1 2 3 6
1 2 3 6 12
1 2 3 4 6 8
1 2 3 6 8 12 16
然而,对于a(n)=m的给定值,可能有多种方法来选择d_1。。。,d_n使得d_1++d_n=米。
例如,对于n=10,a(10)=120,有以下同样有效的解:
[1, 2, 3, 4, 5, 6, 15, 20, 24, 40]
[1, 2, 3, 4, 5, 8, 10, 12, 15, 60]
[1, 2, 3, 4, 5, 8, 12, 15, 30, 40]
[1, 2, 3, 4, 6, 8, 12, 20, 24, 40]
[1, 2, 3, 5, 6, 8, 10, 15, 30, 40]
[1, 2, 3, 5, 8, 10, 12, 15, 24, 40]
[1, 2, 3, 5, 8, 12, 15, 20, 24, 30]
[1, 2, 4, 5, 6, 8, 10, 20, 24, 40]
[1, 2, 4, 6, 8, 10, 15, 20, 24, 30]
[1, 3, 4, 5, 6, 10, 12, 15, 24, 40]
[1, 3, 4, 5, 6, 12, 15, 20, 24, 30]
[1, 3, 4, 5, 8, 10, 15, 20, 24, 30]
[1, 3, 5, 6, 8, 10, 12, 15, 20, 40]
[1, 4, 5, 6, 8, 10, 12, 20, 24, 30]
[2, 3, 4, 5, 6, 8, 10, 12, 30, 40]
[2, 3, 4, 6, 8, 10, 12, 15, 20, 40]
[2, 3, 5, 6, 8, 10, 12, 20, 24, 30]
词典学上最早的解决方案是:
..编号…m:d_1 d_2。。。dn(数字)
-------------------------
..1....1: 1
..2....0: - -
..3....6: 1, 2, 3
..4...12: 1, 2, 3, 6
..5...24: 1, 2, 3, 6, 12
..6...24: 1, 2, 3, 4, 6, 8
..7...48: 1, 2, 3, 4, 6, 8, 24
..8...60: 1, 2, 3, 4, 5, 10, 15, 20
..9...84: 1, 2, 3, 4, 6, 7, 12, 21, 28
.10..120: 1, 2, 3, 4, 5, 6, 15, 20, 24, 40
...