数字串的通项公式

王守恩

northwolves 发表于 2023-10-25 14:42
{{1,1,2,1,3,1,4,2,3},{1,1,1,1,1,2,4,2,1},{1,1,1,0,2,2,6,1,2}}

a(01)=1,
a(02)=1,
a(03)=1,
a(04)=2,
a(05)=1,
a(06)=3,
a(07)=1,
a(08)=4,
a(09)=2,
......
{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}
a[0]=1;a[1]=1;a[n_]:=a[n]=a/@Most[Divisors[n]]//Total;a/@Range[36]
如果我们只是想把某个数单独拉出来, 譬如: a(36)=26,应该怎样编排？谢谢！

northwolves

王守恩 发表于 2023-11-7 17:34
a(01)=1,
a(02)=1,
a(03)=1,

1. b[n_] := If[n < 2, 1, Sum[b[i], {i, Most@Divisors[n]}]];
   
2. {b[36], b[360], b[3600]}

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{26, 604, 11008}

northwolves

或者

1. b[n_] := If[n < 2, 1, b /@ Most@Divisors[n] // Total];
   
2. {b[36], b[360], b[3600]}

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王守恩

northwolves 发表于 2023-11-7 17:57
或者

第2问: 如果我们想把若干个数单独拉出来且相乘, 譬如:  a(2023)*a(2024/2)*a(2022/3)*=8*44*3=1056, 详见418#。
譬如:  a(53)*a(50/2)*a(51/3)*a(52/4)*a(55/5)*a(48/6)*a(49/7)*a(56/8)*a(54/9)=1*2*1*1*1*4*1*1*3=24, 详见420#。

northwolves

1. b[n_]:=If[n<2,1,b/@Most@Divisors[n]//Total];
   
2. s={b[2023],b[2024/2],b[2022/3]};
   
3. {s,Times@@s}

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{{8, 44, 3}, 1056}

northwolves

1. b[n_] := If[n < 2, 1, b /@ Most@Divisors[n] // Total];
   
2. s = {53, 50/2, 51/3, 52/4, 55/5, 48/6, 49/7, 56/8, 54/9};
   
3. {s, b /@ s, Times @@ b /@ s}

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{{53,25,17,13,11,8,7,7,6},{1,2,1,1,1,4,1,1,3},24}

王守恩

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(01)=b(03)=1,{3},
a(02)=b(03)*b(04)=1,{3,4},
a(03)=b(04)*b(05)=1,{5,4,3},
a(04)=b(05)*b(06)=1,{5,6,3,4},
a(05)=b(06)*b(07)=1,{7,6,3,4,5},
a(06)=b(07)*b(08)=2,{7,4,3,8,5,6},{7,8,3,4,5,6},
a(07)=b(08)*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(08)=b(09)*b(10)=2,{3,10,9,4,5,6,7,8},{9,10,3,4,5,6,7,8},
a(09)=b(10)*b(11)=1,{11,10,3,4,5,6,7,8,9},
a(10)=b(11)*b(12)=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(11)=b(12)*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(12)=b(13)*b(14)=1,{13,14,3,4,5,6,7,8,9,10,11,12},
a(13)=b(14)*b(15)=3,
得到这样一串数: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,...
把这串数分成A,B,
A=1,1,1,2,2,3,1,12,2,3,3,08,2,12,3,08,...
B=1,1,1,4,1,3,3,04,2,9,1,16,4,03,3,24,...

1. A:Table[b[n_] := If[n < 2, 1, b /@ Most@Divisors[n] // Total]; s = {b[k], b[2 k - 1]}; {s, Times @@ s}, {k, 1, 19}]

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{{{1, 1}, 1}, {{1, 1}, 1}, {{1, 1}, 1}, {{2, 1}, 2}, {{1, 2},  2}, {{3, 1}, 3}, {{1, 1}, 1}, {{4, 3}, 12}, {{2, 1}, 2}, {{3, 1}, 3}, {{1, 3}, 3}, {{8, 1}, 08}, {{1, 2}, 2}, {{3, 4}, 12}, {{3, 1}, 3}, {{8, 1}, 08}, {{1, 3}, 3}, {{8, 3}, 24}, {{1, 1}, 1}}

1. B:Table[b[n_] := If[n < 2, 1, b /@ Most@Divisors[n] // Total]; s = {b[k], b[2 k + 1]}; {s, Times @@ s}, {k, 1, 19}]

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{{{1, 1}, 1}, {{1, 1}, 1}, {{1, 1}, 1}, {{2, 2}, 4}, {{1, 1},  1}, {{3, 1}, 3}, {{1, 3}, 3}, {{4, 1}, 04}, {{2, 1}, 2}, {{3, 3}, 9}, {{1, 1}, 1}, {{8, 2}, 16}, {{1, 4}, 4}, {{3, 1}, 03}, {{3, 1}, 3}, {{8, 3}, 24}, {{1, 3}, 3}, {{8, 1}, 08}, {{1, 3}, 3}}
A,B可以化简吗？A,B可以合并吗？A,B合并后可以只出现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,...？谢谢各位！

northwolves

1. Flatten@Table[b[n_]:=If[n<2,1,Total[b/@Most@Divisors[n]]];{b[k]*b[2 k-1],b[k]*b[2 k+1]},{k,18}]

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{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}

northwolves

1. b[n_]:=If[n<2,1,Total[b/@Most@Divisors[n]]];Flatten@Array[b[#]*{b[2 #-1],b[2#+1]}&,18]

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王守恩

northwolves 发表于 2023-11-8 11:06

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?

把这串数分成6k+1, 6k+2, 6k+3, 6k+4, 6k+4, 6k+5, 6k+6  7部分。

6k+1: Table[b[n_] := If[n < 2, 1, b /@ Most@Divisors[n] // Total]; s = {b[(6 k - 1)/1], b[(6 k + 0)/3], b[(6 k + 1)/2]}; {s, Times @@ s}, {k, 1, 12}]
6k+2: Table[b[n_] := If[n < 2, 1, b /@ Most@Divisors[n] // Total]; s = {b[(6 k+ 0) /3], b[(6 k + 1)/1], b[(6 k + 2)/2]}; {s, Times @@ s}, {k, 1, 12}]
{{{1, 1, 2}, 2}, {{2, 1, 1}, 02}, {{3, 1, 3}, 9}, {{4, 2, 1}, 08}, {{3, 1, 8}, 24}, {{8, 1, 1}, 08}, {{3, 1, 3}, 09}, {{8, 2, 2}, 32}, {{8, 3, 8}, 192}, {{8, 1, 1}, 8}, {{3, 1, 3}, 9}, {{20, 1, 1}, 20}}
6k+3: Table[b[n_] := If[n < 2, 1, b /@ Most@Divisors[n] // Total]; s = {b[(6 k + 1)/1], b[(6 k + 2)/2], b[(6 k + 3)/3]}; {s, Times @@ s}, {k, 1, 12}]
{{{1, 2, 1}, 2}, {{1, 1, 1}, 01}, {{1, 3, 1}, 3}, {{2, 1, 2},  04}, {{1, 8, 1}, 08}, {{1, 1, 1}, 01}, {{1, 3, 3}, 09}, {{2, 2, 1}, 04}, {{3, 8, 1}, 24}, {{1, 1, 3}, 3}, {{1, 3, 1}, 3}, {{1, 1, 2}, 2}}
6k+4: Table[b[n_] := If[n < 2, 1, b /@ Most@Divisors[n] // Total]; s = {b[(6 k + 2)/1], b[(6 k + 3)/3], b[(6 k + 4)/2]}; {s, Times @@ s}, {k, 1, 12}]
{{{4, 1, 1}, 4}, {{3, 1, 4}, 12}, {{8, 1, 1}, 8}, {{3, 2, 3}, 18}, {{16, 1, 1}, 16}, {{3, 1, 8}, 24}, {{8, 3, 1}, 24}, {{8, 1, 3},24}, {{20, 1, 1}, 20}, {{3, 3, 16}, 144}, {{8, 1, 3}, 24}, {{3, 2, 3}, 18}}
6k+4: Table[b[n_] := If[n < 2, 1, b /@ Most@Divisors[n] // Total]; s = {b[(6 k + 2)/2], b[(6 k + 3)/3], b[(6 k + 4)/1]}; {s, Times @@ s}, {k, 1, 12}]
{{{2, 1, 3}, 6}, {{1, 1, 8}, 08}, {{3, 1, 3}, 9}, {{1, 2, 8}, 16}, {{8, 1, 3}, 24}, {{1, 1, 20}, 20}, {{3, 3, 3}, 27}, {{2, 1, 8}, 16}, {{8, 1, 3}, 24}, {{1, 3, 32}, 96}, {{3, 1, 13}, 39}, {{1, 2, 8}, 16}}
6k+5: Table[b[n_] := If[n < 2, 1, b /@ Most@Divisors[n] // Total]; s = {b[(6 k + 3)/3], b[(6 k + 4)/2], b[(6 k + 5)/1]}; {s, Times @@ s}, {k, 1, 12}]
{{{1, 1, 1}, 1}, {{1, 4, 1}, 04}, {{1, 1, 1}, 1}, {{2, 3, 1}, 06}, {{1, 1, 3}, 03}, {{1, 8, 1}, 08}, {{3, 1, 1}, 03}, {{1, 3, 1}, 03}, {{1, 1, 1}, 01}, {{3, 16, 3}, 144}, {{1, 3, 1}, 3}, {{2, 3, 3}, 18}}
6k+6: Table[b[n_] := If[n < 2, 1, b /@ Most@Divisors[n] // Total]; s = {b[(6 k + 4)/2], b[(6 k + 5)/1], b[(6 k + 6)/3]}; {s, Times @@ s}, {k, 1, 12}]
{{{1, 1, 2}, 2}, {{4, 1, 3}, 12}, {{1, 1, 4}, 4}, {{3, 1, 3}, 09}, {{1, 3, 8}, 24}, {{8, 1, 3}, 24}, {{1, 1, 8}, 08}, {{3, 1, 8}, 24}, {{1, 1, 8}, 08}, {{16, 3, 3}, 144}, {{3, 1, 20}, 60}, {{3, 3, 3}, 27}}

得到这样一串数(2个6k+4需合并):
0,2, 2, (4+6), 1, 2, 0, 2, 1, (12+8), 4, 12, 0, 9, 3, (8+9), 1, 4, 0, 8, 4, (18+16), 6, 9, 0, 24, 8, (16+24), 3, 24, 0, 8, 1, (24+20), 8, 24, 0, 9, 9, (24+27), 3, 8, 0, 32, 4, (24+16), 3, 24, 0, 192, 24, (20+24), 1, 8, ...

我还是找不到通项公式。