数字串的通项公式

王守恩

这些数字串应该如何来描述？

1. Table[CoefficientList[Series[x^2/((1-xk)(1-(k-1)x(x+1))),{x,0,16}],x],{k,2,7}]

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{0, 0, 1, 3, 8, 19, 43, 94, 201, 423, 880, 1815, 3719, 7582, 15397, 31171, 62952, 126891, 255379,
{0, 0, 1, 5, 21, 79, 281, 963, 3217, 10547, 34089, 108955, 345137, 1085331, 3392377, 10549739,
{0, 0, 1, 7, 40, 205, 991, 4612, 20905, 92935, 407056, 1762117, 7556095, 32148940, 135892321,
{0, 0, 1, 9, 65, 421, 2569, 15085,  86241,  483429,  2669305,  14564061,  78699089,  421880725,
{0, 0, 1, 11, 96, 751, 5531, 39186, 270241,1827071,12166176,80043931,521516711,3370600266,
{0, 0, 1, 13, 133, 1219, 10513,  87199,  703921,  5570263,  43409905,  334234615,  2548342369,
{0, 0, 1, 15, 176, 1849, 18271, 173608, 1605297, 14549487,129860704,1145089065,9998390207,

northwolves

$a(n,k)=(2k-1)a(n,k-1)-(k-1)^2a(n,k-2)-k(k-1)a(n,k-3),a(1)=0,a(2)=1,a(3)=2k-1$

northwolves

$a(n,k)=\text{Round}\left[k^n-\frac{1}{2}(1+\frac{k+1}{\sqrt{(k-1) (k+3)}}) (\frac{k-1+\sqrt{(k-1) (k+3)}}{2})^n\right]$

王守恩

(1)+(2)=k^n,(1)=k^n-(2)怎么编排也出不来。

1. (1)Table[CoefficientList[Series[x^2/((1 - x (k + 1)) (1 - k x ( x + 1))), {x, 0, 21}], x], {k, 1, 7}]

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{0, 0, 1, 3, 8, 19, 43, 94, 201, 423, 880, 1815, 3719, 7582, 15397, 31171, 62952, 126891, 255379,
{0, 0, 1, 5, 21, 79, 281, 963, 3217, 10547, 34089, 108955, 345137, 1085331, 3392377, 10549739,
{0, 0, 1, 7, 40, 205, 991, 4612, 20905, 92935, 407056, 1762117, 7556095, 32148940, 135892321,
{0, 0, 1, 9, 65, 421, 2569, 15085,  86241,  483429,  2669305,  14564061,  78699089,  421880725,
{0, 0, 1, 11, 96, 751, 5531, 39186, 270241,1827071,12166176,80043931,521516711,3370600266,

1. (2)Table[LinearRecurrence[{k, k}, {1, k + 1}, 23], {k, 1, 8}]

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{1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711,
{1, 3, 8, 22, 60, 164, 448, 1224, 3344, 9136,  24960, 68192, 186304, 508992, 1390592, 3799168,
{1, 4, 15, 57, 216, 819, 3105,  11772,  44631,  169209,  641520,  2432187,  9221121,  34959924,
{1, 5, 24, 116, 560, 2704,  13056,  63040,  304384,  1469696,  7096320,  34264064,  165441536,
{1, 6, 35, 205, 1200, 7025, 41125, 240750, 1409375,8250625,48300000,282753125,1655265625,

  

nyy

王守恩 发表于 2023-9-4 12:09
(1)+(2)=k^n,(1)=k^n-(2)怎么编排也出不来。

{0, 0, 1, 3, 8, 19, 43, 94, 201, 423, 880, 1815, 3719, 75 ...

我给你找下通项公式：

1. Clear["Global`*"];(*清除所有变量*)
   
2. xx={0,0,1,3,8,19,43,94,201,423,880,1815,3719,7582,15397,31171,62952,126891,255379}
   
3. aaa=FindSequenceFunction[xx,n]//FullSimplify
   

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0,0,1,3,8,19,43,94,201,423,880,1815,3719,7582,15397,31171,62952,126891,255379的通项公式
通项公式结果表达式
\[\frac{1}{10} \left(5\times2^n+\left(\frac{1}{2} \left(1-\sqrt{5}\right)\right)^n \left(-5+\sqrt{5}\right)-\left(\frac{1}{2} \left(1+\sqrt{5}\right)\right)^n \left(5+\sqrt{5}\right)\right)\]

nyy

王守恩 发表于 2023-9-4 12:09
(1)+(2)=k^n,(1)=k^n-(2)怎么编排也出不来。

{0, 0, 1, 3, 8, 19, 43, 94, 201, 423, 880, 1815, 3719, 75 ...

{0,0,1,5,21,79,281,963,3217,10547,34089,108955,345137,1085331,3392377,10549739}

我都有现成代码，你也不算一下。自己修改一下就行了！

1. Clear["Global`*"];(*清除所有变量*)
   
2. xx={0,0,1,5,21,79,281,963,3217,10547,34089,108955,345137,1085331,3392377,10549739}
   
3. aaa=FindSequenceFunction[xx,n]//FullSimplify
   

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通项公式
\[\frac{1}{12} \left(4\times3^n+\left(1-\sqrt{3}\right)^n \left(-3+\sqrt{3}\right)-\left(1+\sqrt{3}\right)^n \left(3+\sqrt{3}\right)\right)\]

northwolves

nyy 发表于 2023-9-4 15:15
{0,0,1,5,21,79,281,963,3217,10547,34089,108955,345137,1085331,3392377,10549739}

我都有现成代码， ...

1. Table[Round[3^n-(Sqrt[3]+1)^(n+2)/(4Sqrt[3])],{n,0,10}]

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northwolves

1. Table[{k,Table[Round[k^n-((1/2) (1+(k+1)/Sqrt[(k-1)(k+3)])) ((k-1+Sqrt[(k-1)(k+3)])/2)^n],{n,0,10}]},{k,2,9}]//MatrixForm

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\begin{array}{cc}
2 & \{0,0,1,3,8,19,43,94,201,423,880\} \\
3 & \{0,0,1,5,21,79,281,963,3217,10547,34089\} \\
4 & \{0,0,1,7,40,205,991,4612,20905,92935,407056\} \\
5 & \{0,0,1,9,65,421,2569,15085,86241,483429,2669305\} \\
6 & \{0,0,1,11,96,751,5531,39186,270241,1827071,12166176\} \\
7 & \{0,0,1,13,133,1219,10513,87199,703921,5570263,43409905\} \\
8 & \{0,0,1,15,176,1849,18271,173608,1605297,14549487,129860704\} \\
9 & \{0,0,1,17,225,2665,29681,317817,3311425,33816905,340073361\} \\
\end{array}

王守恩

本来是这样一道题: f(n)表示是由 0,1,2,3 组成的长度为 n 的所有序列中连续两个 0 出现的次数总和。
f(n)=0, 1, 8, 48, 256, 1280, 6144, 28672, ......

1. Table[CoefficientList[Series[x/(1 - k x - x)^2, {x, 0, 21}], x], {k, 1, 7}]

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{0, 1, 4, 12, 32, 80, 192, 448, 1024, 2304, 5120, 11264, 24576, 53248, 114688,  245760,  524288,  
{0, 1, 6, 27, 108, 405, 1458, 5103, 17496, 59049, 196830, 649539, 2125764, 6908733, 22320522,
{0, 1, 8, 48, 256, 1280, 6144, 28672, 131072, 589824, 2621440, 11534336,50331648,218103808,
{0, 1, 10, 75, 500, 3125, 18750, 109375, 625000,  3515625,  19531250,  107421875,  585937500,
{0, 1, 12, 108, 864, 6480, 46656,326592,2239488,15116544,100776960,665127936,4353564672,
用你们的方法来表示这些数字串，我还是不会(一下子领悟不了)。
当然，OEIS是不可能有这些共性方法的。

northwolves

王守恩 发表于 2023-9-4 16:31
本来是这样一道题: f(n)表示是由 0,1,2,3 组成的长度为 n 的所有序列中连续两个 0 出现的次数总和。
f(n)=0 ...

f(n)=0, 1, 8, 48, 256, 1280, 6144, 28672, ......

$f_n=(n - 1)*4^{n-2}$