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

这个"通项公式"应该可以看懂。谢谢northwolves(数学研发论坛)!
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?
a(03)=b(01)*b(03)=1,{3},
a(04)=b(03)*b(02)=1,{3,4},
a(05)=b(02)*b(05)=1,{5,4,3},
a(06)=b(05)*b(03)=1,{5,6,3,4},
a(07)=b(03)*b(07)=1,{7,6,3,4,5},
a(08)=b(07)*b(04)=2,{7,4,3,8,5,6},{7,8,3,4,5,6},
a(09)=b(04)*b(09)=4,{3,4,8,9,5,6,7},{3,8,9,4,5,6,7},{9,4,3,8,5,6,7}{9,8,3,4,5,6,7},
a(10)=b(09)*b(05)=2,{3,10,9,4,5,6,7,8},{9,10,3,4,5,6,7,8},
a(11)=b(05)*b(11)=1,{11,10,3,4,5,6,7,8,9},
a(12)=b(11)*b(06)=3,{11,4,3,12,5,6,7,8,9,10},{11,6,3,4,5,12,7,8,9,10},{11,12,3,4,5,6,7,8,9,10},
a(13)=b(06)*b(13)=3,{13,4,3,12,5,6,7,8,9,10,11},{13,6,3,4,5,12,7,8,9,10,11},{13,12,3,4,5,6,7,8,9,10,11},
a(14)=b(13)*b(07)=1,{13,14,3,4,5,6,7,8,9,10,11,12},
......
得到一串数(OEIS是没有的):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,...
Table[Total[1/Divisors[n]]; {b[Floor[n/2]]*b[2 Floor[n/2] - Cos[n*Pi]]}, {n,3, 210}]
{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, 2, 3, 9, 12, 32, 8, 32, 96,
9, 3, 13, 39, 3, 20,160, 8, 3, 3, 13, 39, 3, 3, 228, 228, 3, 8, 24, 3, 8, 8, 8, 64, 24, 9, 39, 13, 1, 3,
144, 144, 24, 8, 3, 39, 13, 1, 44, 88, 6, 24, 24, 3, 3, 24, 160, 60, 3, 1, 44, 44, 3, 9, 24, 24, 9, 9,
9, 60, 60, 3, 112, 112, 1, 13, 104, 8, 8, 8, 26, 78, 3, 3, 39, 39, 3, 8, 160, 60, 39}

王守恩

类似的"通项公式"。谢谢northwolves(数学研发论坛)!
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?

Table[Total[1/Divisors[n]];{0, b[6 k + 1]*b[3 k + 1] {b[2 k], b[2 k + 1]},
((b[6k+4]-2b[3k+2])b[3k+1]+b[6k+2]*b[3k+2])b[2k+1],b[6k+5]*b[3k+2]{b[2k+1],b[2k+2]}},{k,0,35}]

{0, 1, 1, 1, 1, 1, 0, 2, 2, 6, 1, 2, 0, 2, 1, 12, 4, 12, 0, 9, 3, 11, 1, 4, 0, 8, 4, 22, 6, 9, 0, 24, 8, 24, 3, 24, 0, 8,
1, 28, 8, 24, 0, 9, 9, 33, 3, 8, 0, 32, 4, 28, 3, 24, 0, 192, 24, 28, 1, 8, 0, 8, 3, 144, 144, 144, 0, 9, 3, 45, 3,
60, 0, 20, 2, 22, 18, 27, 0, 60, 80, 272, 4, 8, 0, 24, 3, 28, 8, 104, 0, 117, 9, 11, 3, 48, 0, 32, 6, 252, 24,24,
0, 24, 24, 84, 3, 26, 0, 78, 3, 284, 20, 60, 0, 27, 27, 33, 9, 60, 0, 40, 2, 11, 12, 156, 0, 416,32, 416, 3, 24,
0, 24, 24, 224, 64, 24, 0, 39, 13, 57, 3, 144, 0, 144, 6, 22, 6, 24, 0, 64, 24, 348, 27, 72, 0, 8, 1, 160, 144,
2880, 0, 60, 9, 33, 3, 20, 0, 120, 18, 135, 9, 9, 0, 480, 160, 68, 1, 44, 0, 132, 3, 116, 24, 72,0,27,72,360,
24, 96, 0, 32, 3, 102, 24, 104, 0, 338, 26, 102, 3, 24, 0, 24, 9, 204, 180, 780, 0, 39, 3, 11, 3, 228,......}

cgl_74

王守恩 发表于 2023-11-10 01:33
这个"通项公式"应该可以看懂。谢谢northwolves(数学研发论坛)!
将3,4,5,6,…,n排成数列{a(k):k=1,2,3,4,5, ...

这个递推公式是啥呢？要写出来。
并举个例子：比如n=3*5*5*7*7*7*11*11时，有多少种排法呢？

王守恩

类似的"通项公式"。谢谢northwolves(数学研发论坛)!
将5,6,7,…,n排成数列{a(k):k=1,2,3,4,5,…,n-4},使a(k)都是k的倍数,有几种排法?

Table[Total[1/Divisors[n]];{b[12 k + 1]*b[6 k + 1]*b[4 k + 1] {b[3 k + 0], b[3 k + 1]},
b[12 k + 5]*b[4 k + 1]*b[3 k + 1] {b[6 k + 1], b[6 k + 3]}, 0, 0, b[12k+7]*b[4k+3]*b[3k+2]{b[6k+3],
b[6 k + 5]}, b[12 k + 11]*b[6 k + 5]*b[4 k + 3] {b[3 k + 2], b[3 k + 3]}, 0,  0}, {k, 0, 18}]

{1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 2, 2, 4, 0, 0, 2, 1, 1, 3, 0, 0, 12, 4, 2, 6, 0, 0, 12, 4, 12, 6, 0, 0, 2, 3,
3, 9, 0, 0, 9, 3, 3, 24, 0, 0, 32, 4, 2, 4, 0, 0, 36, 9, 3, 3, 0, 0, 9, 24, 72, 216, 0, 0, 3, 3, 3, 24, 0, 0, 16,
2, 6, 18, 0, 0, 96, 32, 32, 12, 0, 0, 9, 9, 3, 24, 0, 0, 24, 3, 3, 60, 0, 0, 120, 12, 12, 18, 0, 0, 27, 9, 9,
12, 0, 0, 12, 24, 24, 24, 0, 0, 27, 9, 9, 117, 0, 0, 26, 2, 4, 32, 0, 0, 128, 48, 48, 9, 0, 0, 72, 72, 24, 72,
0, 0, 9, 3, 9, 78, 0, 0, 156, 6, 2, 16, 0, 0, 72, 27, 81, 81, 0, 0, 3, 20, 60, 480, 0, 0, 24, 3, 3, 39, 0, 0, 234,
18, 9, 9, 0, 0, 192, 64, 8, 8, 0, 0, 24, 9, 27, 27, 0, 0, 72, 72, 24, 1152, 0, 0, 144, 6, 6, 48, 0, 0, 64, 8, 24,
9, 0, 0, 27, 72, 72, 936, 0, 0, 13, 1, 3, 60, 0, 0, 60, 9, 9, 27, 0, 0, 480, 160, 160, 24, 0, 0}

王守恩

  挑战一下？

将6,7,8,9,…,28排成数列{a(k):k=1,2,3,4,5,…,23},使a(k)都是k的倍数,有几种排法?

王守恩

将6,7,8,9,…,28排成数列{a(k):k=1,2,3,4,5,…,23},使a(k)都是k的倍数,有206种排法。