数字串的通项公式

王守恩

dlpg070 发表于 2020-1-31 09:40
谢谢你的有趣数列
显然，你是有答案的，不然，怎么能给出数列的那么多项
我也找到了通项公式，等我消化 ...

有见过这样的通项公式吗？

a(n)=24, 39, 58, 81, 108, 139, 174, 213, 256, 303, 354, 409, 468,
  531, 598, 669, 744, 823, 906, 993, 1084, 1179, 1278, 1381, 1488,
1599, 1714, 1833, 1956, 2083, 2214, 2349, 2488, 2631, 2778, 2929,
3084, 3243, 3406, 3573, 3744, 3919, 4098, 4281,............

\(\D a(n)=(2n-1)\lim_{\theta\to 0}\frac{\prod_{k=2}^{n}\tan(k\theta)-\prod_{k=2}^{n}\sin(k\theta)}{\prod_{k=1}^{n-1}\tan(k\theta)-\prod_{k=1}^{n-1}\sin(k\theta)}\)

王守恩

王守恩 发表于 2020-2-1 13:30
有见过这样的通项公式吗？

a(n)=24, 39, 58, 81, 108, 139, 174, 213, 256, 303, 354, 409, 468,

这样想，会简单些。

1+2+3+4+5+6+7+8+9+10+11+12+13
=1×2/2+2×2/2+3×2/2+4×2/2+5×2/2
=(1×2/2+2×2/2)+3×2/2+4×2/2+5×2/2
=(2×3)/2+3×4/2+4×2/2+5×2/2
=(2×3/2+3×2/2)+4×2/2+5×2/2
=(3×4)/2+4×2/2+5×2/2
=(3×4/2+4×2/2)+5×2/2
=(4×5)/2+5×2/2+6×2/2
=(4×5/2+5×2/2)+6×2/2
=(5×6)/2+6×2/2+7×2/2
=(5×6/2+6×2/2)+7×2/2
=(6×7)/2+7×2/2+8×2/2
=(6×7/2+7×2/2)+8×2/2
=(7×8)/2+8×2/2+9×2/2

1×2+2×3+3×4+4×5+5×6+6×7+7×8+8×9
=1×2×3/3+2×3×3/3+3×4×3/3+4×5×3/3
=(1×2×3/3+2×3×3/3)+3×4×3/3+4×5×3/3
=(2×3×4)/3+3×4×3/3+4×5×3/3+5×6×3/3
=(2×3×4/3+3×4×3/3)+4×5×3/3+5×6×3/3
=(3×4×5)/3+4×5×3/3+5×6×3/3
=(3×4×5/3+4×5×3/3)+5×6×3/3
=(4×5×6)/3+5×6×3/3+6×7×3/3
=(4×5×6/3+5×6×3/3)+6×7×3/3
=(5×6×7)/3+6×7×3/3+7×8×3/3
=(5×6×7/3+6×7×3/3)+7×8×3/3
=(6×7×8)/3+7×8×3/3+8×9×3/3

1×2×3+2×3×4+3×4×5+4×5×6+5×6×7+6×7×8+7×8×9
=1×2×3×4/4+2×3×4×4/4+3×4×5×4/4+4×5×6×4/4
=(1×2×3×4/4+2×3×4×4/4)+3×4×5×4/4+4×5×6×4/4
=(2×3×4×5)/4+3×4×5×4/4+4×5×6×4/4+5×6×7×4/4
=(2×3×4×5/4+3×4×5×4/4)+4×5×6×4/4+5×6×7×4/4
=(3×4×5×6)/4+4×5×6×4/4+5×6×7×4/4
=(3×4×5×6/4+4×5×6×4/4)+5×6×7×4/4
=(4×5×6×7)/4+5×6×7×4/4+6×7×8×4/4
=(4×5×6×7/4+5×6×7×4/4)+6×7×8×4/4
=(5×6×7×8)/4+6×7×8×4/4+7×8×9×4/4

1×2×3×4+2×3×4×5+3×4×5×6+4×5×6×7+5×6×7×8+6×7×8×9
=1×2×3×4×5/5+2×3×4×5×5/5+3×4×5×6×5/5+4×5×6×7×5/5
=(1×2×3×4×5/5+2×3×4×5×5/5)+3×4×5×6×5/5+4×5×6×7×5/5
=(2×3×4×5×6)/5+3×4×5×6×5/5+4×5×6×7×5/5+5×6×7×8×5/5
=(2×3×4×5×6/5+3×4×5×6×5/5)+4×5×6×7×5/5+5×6×7×8×5/5
=(3×4×5×6×7)/5+4×5×6×7×5/5+5×6×7×8×5/5
=(3×4×5×6×7/5+4×5×6×7×5/5)+5×6×7×8×5/5
=(4×5×6×7×8)/5+5×6×7×8×5/5+6×7×8×9×5/5

王守恩

王守恩 发表于 2020-2-1 13:30
有见过这样的通项公式吗？

a(n)=24, 39, 58, 81, 108, 139, 174, 213, 256, 303, 354, 409, 468,

真有点  “小题大作” 。

\(\D A(n)=\lim_{x\to\infty}\bigg(x^2\ln\frac{x-n}{x-n-1}-\frac{2x+1}{2}\bigg)\)

   n={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, 31, 32, 33, 34, 35, 36,
37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, ........}

A(n)={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, 31, 32, 33, 34, 35, 36,
37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, ........}

dlpg070

王守恩 发表于 2020-3-7 18:28
真有点  “小题大作” 。

\(\D A(n)=\lim_{x\to\infty}\bigg(x^2\ln\frac{x-n}{x-n-1}-\frac{2x+1}{2 ...

实际上是一道极限计算题
求 a(n)=n
虽然简单,但OEIS中有:
A000027                The positive integers. Also called the natural numbers, the whole numbers or the counting numbers, but these terms are ambiguous.

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,

王守恩

dlpg070 发表于 2020-3-8 10:14
实际上是一道极限计算题
求 a(n)=n
虽然简单,但OEIS中有:

这个也有吗？

    当 n = 0, 1, 2,  3,   4,   5,    6,    7,    8,   9, ......
    A(n)=0, 1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, ......


\(\D A(n)=\lim_{x\to\infty}\cos\big((n-1)\pi\big)\bigg(x^n\ln\frac{x+1}{x}+\sum_{k=1}^{n-1}\frac{x^{n-k}}{\cos(k\pi)*k}\bigg)\)

看得懂么？我不知道怎么发。删来删去把不该删的删了？

Table[Limit[Cos[(n - 1)*\[Pi]] (x^n Log[(x + 1)/x] + [\[Sum]\, \(k = 1\), \(n - 1\)]
\Frac[\(x\)^\(n - k\), \(Cos[k*\[Pi]]*k\)]\)),  x -> \[Infinity]], {n, 0, 35}]

{0, 1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10, 1/11, 1/12,
1/13, 1/14, 1/15, 1/16, 1/17, 1/18, 1/19, 1/20, 1/21, 1/22, 1/23,
1/24, 1/25, 1/26, 1/27, 1/28, 1/29, 1/30, 1/31, 1/32, 1/33, 1/34, 1/35}

dlpg070

王守恩 发表于 2020-3-8 13:30
这个也有吗？

    当 n = 0, 1, 2,  3,   4,   5,    6,    7,    8,   9, ......

公式和代码都正确,你的Mathematica不错嘛
下面是我的代码:

1. Clear["Global`*"];
   
2. 
   
3. a[n_]:=a[n]=Limit[Cos[(n-1)*Pi]*((Log[(x+1)/(x)])*x^n+(Sum[x^(n-k)/(k*Cos[k*Pi]),{k,1,n-1}])),x->\[Infinity]];
   
4. Table[a[n],{n,1,30}]

复制代码

输出:
{1,1/2,1/3,1/4,1/5,1/6,1/7,1/8,1/9,1/10,1/11,1/12,1/13,1/14,1/15,1/16,1/17,1/18,1/19,1/20,1/21,1/22,1/23,1/24,1/25,1/26,1/27,1/28,1/29,1/30}

dlpg070

dlpg070 发表于 2020-3-9 08:22
公式和代码都正确,你的Mathematica不错嘛
下面是我的代码:

你的代码很好,
下面是我模仿你的代码:
预先定义函数稍微好一点

1. Table[Limit[
   
2.   Cos[(n - 1)*
   
3.      Pi]*((Log[(x + 1)/(x)])*
   
4.       x^n + (Sum[x^(n - k)/(k*Cos[k*Pi]), {k, 1, n - 1}])),
   
5.   x -> \[Infinity]], {n, 1, 30}]

复制代码

王守恩

dlpg070 发表于 2020-3-9 17:42
你的代码很好,
下面是我模仿你的代码:
预先定义函数稍微好一点

《有趣的回文》的那个 “东东” 我还是不会用。譬如：

1, 11, 112, 1123, 11235, 112358, 11235813, ......

dlpg070

王守恩 发表于 2020-3-20 08:15
《有趣的回文》的那个 “东东” 我还是不会用。譬如：

1, 11, 112, 1123, 11235, 112358, 11235813, . ...

不知道 你的 :1, 11, 112, 1123, 11235, 112358, 11235813, ......
是哪里来的?
不是我的" 东东"吧.

王守恩

dlpg070 发表于 2020-3-21 09:03
不知道 你的 :1, 11, 112, 1123, 11235, 112358, 11235813, ......
是哪里来的?
不是我的" 东东"吧.

这样的通项也是挺好的。

\(\D a(n)=\sum_{k=1}^{\infty}\frac{k^n}{2^k}\)

1, 2, 6, 26, 150, 1082, 9366, 94586, 1091670, 14174522, 204495126, 3245265146,
56183135190, 1053716696762, 21282685940886, 460566381955706, 10631309363962710,
260741534058271802, 6771069326513690646, 185603174638656822266, .........