将 2,3,4,…,n 排成数列 {a(k)：k=1,2,3,…,n-1},使 a(k) 都是 k 的倍数,有几种排法？

王守恩

将2,3,4,5,6,…,n排成数列{a(k):k=1,2,3,4,5,…,n-1},使a(k)都是k的倍数,有几种排法?
a(00)=1,{1},   
a(01)=1,{2},   
a(02)=1,{3},   
a(03)=2,{2,4},{4},  
a(04)=1,{5},
a(05)=3,{2,6}{3,6}{6},
a(06)=1,{7},
a(07)=4,{2,4,8},{2,8},{4,8},{8},   {2,4,8}表示:a(1)=2,a(2)=4,a(4)=8,......
a(08)=2,{3,9},{9},
a(09)=3,{2,10},{5,10},{10},
a(10)=1,{11},
a(11)=8,{2,4,12},{2,6,12},{2,12},{3,6,12},{3,12},{4,12},{6,12},{12},
a(12)=1,{13},
a(13)=3,{2,14},{7,14},{14},
a(14)=3,{3,15},{5,15},{15},
a(15)=8,{2,4,8,16},{2,4,16},{2,8,16},{2,16},{4,8,16},{4,16},{8,16},{16},
a(16)=1,{17},
a(17)=8,{2,6,18},{2,18},{3,6,18},{3,9,18},{3,18},{6,18},{9,18},{18},
a(18)=1,{19},
a(19)=8,{2,4,20},{2,10,20},{2,20},{4,20},{5,10,20},{5,20},{10,20,}{20},
{1, 1, 1, 2, 1, 3, 1, 4, 2, 3, 1, 8, 1, 3, 3, 8, 1, 8, 1, 8, 3, 3, 1, 20, 2, 3, 4, 8, 1, 13, 1, 16, 3, 3,
3, 26, 1, 3, 3, 20, 1, 13, 1, 8, 8, 3, 1, 48, 2, 8, 3, 8, 1, 20, 3, 20, 3, 3, 1, 44, 1, 3, 8, 32, 3, 13,
1, 8, 3, 13, 1, 76, 1, 3, 8, 8, 3, 13, 1, 48, 8, 3, 1, 44, 3, 3, 3, 20, 1, 44, 3, 8, 3, 3, 3, 112}

1. a[0]=1;a[1]=1;a[n_]=a[n]=a/@Most[Divisors[n]]//Total;a/@Range[96]

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可有好的通项公式？谢谢各位！

王守恩

将2,3,4,5,6,…,n排成数列{a(k):k=1,2,3,4,5,…,n-1},使a(k)都是k的倍数,有几种排法?

将3,4,5,6,…,n排成数列{a(k):k=1,2,3,4,5,…,n-2},使a(k)都是k的倍数,有几种排法?

将4,5,6,…,n排成数列{a(k):k=1,2,3,4,5,…,n-3},使a(k)都是k的倍数,有几种排法?

lihp2020

为啥 一定要 求 通项公式？？

lihp2020

这个 明显和质数 和分解因式有关
你们写出 质数的  方程？
也就是 a[n] ={2,3,5,7,11}  an表示第n个质数？？

将2,3,4,5,6,…,n排成数列{a(k):k=1,2,3,4,5,…,n-1},使a(k)都是k的倍数,有几种排法?
如果n是质数   就1个排发  n是两个不同的质数相乘 就  是3  ？？

王守恩

lihp2020 发表于 2023-10-25 06:08
这个 明显和质数 和分解因式有关
你们写出 质数的  方程？
也就是 a[n] ={2,3,5,7,11}  an表示第n个质数 ...

Table[DivisorSigma[0, n], {n, 1, 30}]
{1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6, 2, 4, 4, 5, 2, 6, 2, 6, 4, 4, 2, 8, 3, 4, 4, 6, 2, 8}
Table[Divisors[n], {n, 1, 30}]
{{1}, {1, 2}, {1, 3}, {1, 2, 4}, {1, 5}, {1, 2, 3, 6}, {1, 7}, {1, 2, 4, 8}, {1, 3, 9}, {1, 2, 5, 10}, {1, 11},
{1, 2, 3, 4, 6, 12}, {1, 13}, {1, 2, 7, 14}, {1, 3, 5, 15}, {1, 2, 4, 8, 16}, {1, 17}, {1, 2, 3, 6, 9, 18},
{1, 19}, {1, 2, 4, 5, 10, 20}, {1, 3, 7, 21}, {1, 2, 11, 22}, {1, 23}, {1, 2, 3, 4, 6, 8, 12, 24},
{1, 5, 25}, {1, 2, 13, 26}, {1, 3, 9, 27}, {1, 2, 4, 7, 14, 28}, {1, 29}, {1, 2, 3, 5, 6, 10, 15, 30}}
Table[FactorInteger[n], {n, 1, 30}]
{{{1, 1}}, {{2, 1}}, {{3, 1}}, {{2, 2}}, {{5, 1}}, {{2, 1}, {3, 1}}, {{7, 1}}, {{2, 3}}, {{3, 2}}, {{2, 1}, {5, 1}},
{{11, 1}}, {{2, 2}, {3, 1}}, {{13, 1}}, {{2, 1}, {7, 1}}, {{3, 1}, {5, 1}}, {{2, 4}}, {{17, 1}}, {{2, 1}, {3, 2}},
{{19, 1}}, {{2, 2}, {5,1}}, {{3, 1}, {7, 1}}, {{2, 1}, {11, 1}}, {{23, 1}}, {{2, 3}, {3, 1}}, {{5, 2}},
{{2, 1}, {13, 1}}, {{3, 3}}, {{2, 2}, {7, 1}}, {{29,1}}, {{2, 1}, {3, 1}, {5, 1}}}

王守恩

将4,5,6,…,2023排成数列{a(k):k=1,2,3,4,5,…,2020},使a(k)都是k的倍数,有几种排法?

王守恩

将4,5,6,…,2024排成数列{a(k):k=1,2,3,4,5,…,2021},使a(k)都是k的倍数,有几种排法?
答: 有8*44*3=1056种排法。
a(1)=8,
01: 7,119,2023,
02: 7,2023,
03: 17,119,2023,
04: 17,289,2023,
05: 17,2023,
06: 119,2023,
07: 289,2023,
08: 2023,
a(2)=44,
01: 4,8,88,2024,
02: 4,8,184,2024,
03: 4,8,2024,
04: 4,44,88,2024,
05: 4,44,1012,2024,
06: 4,44,2024,
07: 4,88,2024,
08: 4,92,184,2024,
09: 4,92,1012,2024,
10: 4,92,2024,
11: 4,184,2024,
12: 4,1012,2024,
13: 4,2024,
14: 8,88,2024,
15: 8,184,2024,
16: 8,2024,
17: 22,44,88,2024,
18: 22,44,1012,2024,
19: 22,44,2024,
20: 22,88,2024,
21: 22,506,1012,2024,
22: 22,506,2024,
23: 22,1012,2024,
24: 22,2024,
25: 44,88,2024,
26: 44,1012,2024,
27: 44,2024,
28: 46,92,184,1012,2024,
29: 46,92,1012,2024,
30: 46,92,2024,
31: 46,184,2024,
32: 46,506,1012,2024,
33: 46,506,2024,
34: 46,1012,2024,
35: 46,2024,
36: 88,2024,
37: 92,184,2024,
38: 92,1012,2024,
39: 92,2024,
40: 184,2024,
41: 506,1012,2024,
42: 506,2024,
43: 1012,2024,
44: 2024,
a(3)=3,
01: 6,2022,
02: 1011,2022,
03: 2022,

王守恩

将10,11,12,…,56排成数列{a(k):k=1,2,3,4,5,…,47},使a(k)都是k的倍数,有几种排法?
答: 有1*2*1*1*1*4*1*1*3=24种排法。
a(1)=1,
1: 53,
a(2)=2,
1: 10,50,
2: 50,
a(3)=1,
1: 51,
a(4)=1,
1: 52,
a(5)=1,
1: 55,
a(6)=4,
1: 12,24,48,
2: 12,48,
3: 24,48,
4: 48,
a(7)=1,
1: 49,
a(8)=1,
1: 56,
a(9)=3,
1: 18,54,
2: 27,54,
3: 54,

王守恩

lihp2020 发表于 2023-10-25 02:51
为啥 一定要 求 通项公式？？

通项公式是可以有的！
将3,4,5,6,…,n排成数列{a(k):k=1,2,3,4,5,…,n-2},使a(k)都是k的倍数,有几种排法?
Table[b[n_] :=If[n<2,1,Total[b/@Most@Divisors[n]]];{b[k]*b[2k-1],b[k]*b[2k+1]},{k,21}]//Flatten
{1, 1, 1, 1, 1, 1, 2, 4, 2, 1, 3, 3, 1, 3, 12, 4, 2, 2, 3, 9, 3, 1, 8, 16, 2, 4, 12, 3, 3, 3, 8, 24, 3, 3,
24, 8, 1, 3, 24, 8, 3, 3, 3, 24, 8, 1, 20, 40, 4, 6, 9, 3, 4, 12, 24, 24, 3, 1, 13, 13, 1, 8, 128, 48,
9, 3, 3, 9, 9, 3, 26, 26, 1, 8, 24, 9, 9, 3, 20, 160, 8, 1, 13, 39, 3, 3, 24, 8, 8, 24, 9, 9, 3, 3, 144,
48, 2, 16, 64, 8, 3, 3, 8, 104, 13, 1, 20, 20, 3, 9, 60, 20, 3, 9, 9, 24, 8, 3, 132, 88}

lihp2020

我也看不懂 这个代码写的通项公式 可否用数学语言解释一下