求通项公式

wayne

已知$f(n)$的定义域和值域都是正整数，且$f(n)$严格单调递增，且任取正整数$n$,存在$f(f(n))=3n$,求$f(n)$的通项公式。

northwolves

$f(1)=1\rightarrow f(f(1))=1!=3\rightarrow f(1)!=1$
$f(1)=3\rightarrow f(3)=3\rightarrow f(1)=1,f(2)=2$
$f(1)=2,f(2)=3$
$(3)=f(f(2))=6$
$f(6)=f(f(3))=9$
$f(3^n)=3f(3^{n-1})=\ldots=3^n* f(1)=2*3^n$
$ f(3^n + k) = 2\cdot3^n + k ,k \in [0,3^n]$

northwolves

$ f(3^n + k) = 2\cdot3^n + k $
$f(2\cdot3^n+k)=3*(3^n+k)$
$k\in[0,3^n]$

northwolves

1. f[n_]:=Block[{r=3^Floor[Log[3,n]]},If[n<=2r,n+r,3(n-r)]];Table[f[n],{n,100}]
   
2. f /@{1992, 2024, 10^20, 10^100}

复制代码

{2,3,6,7,8,9,12,15,18,19,20,21,22,23,24,25,26,27,30,33,36,39,42,45,48,51,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,84,87,90,93,96,99,102,105,108,111,114,117,120,123,126,129,132,135,138,141,144,147,150,153,156,159,162,163,164,165,166,167,168,169,170,171,172,173,174,175,176,177,178,179,180,181}

{3789,3885,190581010868487640791,15228080143043843084895232761630250394879802048576763864267558971910557498410330867878474031283071683}

mathe

A003605

northwolves

王守恩 发表于 2024-4-6 07:36
正整数中:  使得十进制表示的非零子序列不能被 3 整除。

1, 2, 4, 5, 7, 8, 10, 11, 14, 17, 20, 22, 25,  ...

1. a=With[{k=3},Select[Range@8000,NoneTrue[DeleteCases[FromDigits/@Rest@Subsequences[IntegerDigits@#],0],Mod[#,k]==0&]&]];s=Total@(1/a);{Length@a,a,s,N[s,10]}

复制代码

{123,{1,2,4,5,7,8,10,11,14,17,20,22,25,28,40,41,44,47,50,52,55,58,70,71,74,77,80,82,85,88,100,101,104,107,110,140,170,200,202,205,208,220,250,280,400,401,404,407,410,440,470,500,502,505,508,520,550,580,700,701,704,707,710,740,770,800,802,805,808,820,850,880,1000,1001,1004,1007,1010,1040,1070,1100,1400,1700,2000,2002,2005,2008,2020,2050,2080,2200,2500,2800,4000,4001,4004,4007,4010,4040,4070,4100,4400,4700,5000,5002,5005,5008,5020,5050,5080,5200,5500,5800,7000,7001,7004,7007,7010,7040,7070,7100,7400,7700,8000},2175919945339027979739835924147121496956377212405817719/692584026612285902036909113880759445107872284941840000,3.141741452}

wayne

修改一下题目，升级讨论：  已知$f(n)$的定义域和值域都是正整数，且$f(n)$严格单调递增，且任取正整数$n$,存在$f(f(n))=(2f(1)-1)n$, 求$f(n)$的通项公式。

northwolves

wayne 发表于 2024-4-6 15:12
修改一下题目，升级讨论：  已知$f(n)$的定义域和值域都是正整数，且$f(n)$严格单调递增，且任取正整数$n$, ...

1. 
   
2. f[f1_,n_]:=Block[{r=If[f1==1,0,(2f1-1)^Floor[Log[2f1-1,n]]]},f1*n-(f1-1)^2*r+(f1-1)*Abs[n-f1*r]];
   
3. Table[{f1,Table[f[f1,n],{n,30}]},{f1,10}]
   
4. //MatrixForm

复制代码

1        {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30}
2        {2,3,6,7,8,9,12,15,18,19,20,21,22,23,24,25,26,27,30,33,36,39,42,45,48,51,54,55,56,57}
3        {3,4,5,10,15,16,17,18,19,20,21,22,23,24,25,30,35,40,45,50,55,60,65,70,75,76,77,78,79,80}
4        {4,5,6,7,14,21,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,56,63}
5        {5,6,7,8,9,18,27,36,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66}
6        {6,7,8,9,10,11,22,33,44,55,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85}
7        {7,8,9,10,11,12,13,26,39,52,65,78,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108}
8        {8,9,10,11,12,13,14,15,30,45,60,75,90,105,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135}
9        {9,10,11,12,13,14,15,16,17,34,51,68,85,102,119,136,153,154,155,156,157,158,159,160,161,162,163,164,165,166}
10        {10,11,12,13,14,15,16,17,18,19,38,57,76,95,114,133,152,171,190,191,192,193,194,195,196,197,198,199,200,201}

northwolves

或者 $r=(2f_1-1)^{\lfloorlog_{2f_1-1}n\rfloor},f(n)=n+(f_1-1)(r+(n-f_1*r)+|n-f_1*r|)$