数字串的通项公式

dlpg070

dlpg070 发表于 2019-5-19 11:23
转贴一段代码

为一起解除你的疑惑,
试着分析一下第二句代码的运行过程,供你理解代码时参考
nn=8
            2^i +2^j+2^k  {i, 2, nn}, {j, 1, i - 1}, {k, 0, j - 1}
a1 = 7,     2^2 +2^1 +2^0 (i=2,j=1,k=0)

a2 = 11,    2^3 +2^1 +2^0 (i=3,j=1,k=0)
a3 = 13,    2^3 +2^2 +2^0 (i=3,j=2,k=0)
a4 = 14,    2^3 +2^2 +2^1 (i=3,j=2,k=1)

a5 = 19,    2^4 +2^1 +2^0 (i=4,j=1,k=0)
a6 = 21,    2^4 +2^2 +2^0 (i=4,j=2,k=0)
a7 = 22,    2^4 +2^2 +2^1 (i=4,j=2,k=1)
a8 = 25,    2^4 +2^3 +2^0 (i=4,j=3,k=0)
a9 = 26,    2^4 +2^3 +2^1 (i=4,j=3,k=1)
a10= 28,    2^4 +2^3 +2^2 (i=4,j=3,k=2)
---
依此循环直到指定的i最大值nn

dlpg070

dlpg070 发表于 2019-5-23 17:23
为一起解除你的疑惑,
试着分析一下第二句代码的运行过程,供你理解代码时参考
nn=8

关于a(n)的算式:
多数数列没有解析式,,显函数
此数列a(n)没有解析式,显函数
但 A014311给出2个a(n)的关系式,隐函数
1 a(n)的关系式,隐函数: A000120(a(n)) = 3. [Reinhard Zumkeller, May 03 2012]
2 a(n)的递推关系式:a(n+1) = A057168(a(n)). - M. F. Hasler, Aug 27 2014

下面给出利用A000120求A014311的例子: A000120=3 对应 A014311
A000120         = {0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2,   3,   2,   3,   3,   4,   1,  2,    2,   3,   2,  3,   3,   4,   2,   3,   3,   4,   3,   4,   4,   5,   1,   2,   2]
A000120序号n= {0, 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]
A014311         = {                            7,              11,        13, 14,                          19,      21, 22,             25, 26,       28,                       ]
A014311序号n= {                            1,               2,          3,  4,                              5,       6,  7,                8,  9,          10,                       ]

下面代码可以求任意第n项的数值 a(n)

1. (* 求 an*)
   
2. Clear["Global`*"]
   
3. f[n_] := f[n] = Count[IntegerDigits[n, 2], 1];
   
4. id = 0;
   
5. idmax = 10;
   
6. idmax = Input["求第n项的an,输入项数n"];
   
7. x = 1;
   
8. out = 0;
   
9. While[id < idmax,
   
10. If[f[x] == 3, id = id + 1; out = x];
    
11. x += 1
    
12. ]
    
13. Print["n=", idmax, " an=", out]

复制代码

下面代码可以求数列的前n项 a(n)

1. (* 求 前n项 *)
   
2. Clear["Global`*"]
   
3. f[n_] := f[n] = Count[IntegerDigits[n, 2], 1];
   
4. an = {};
   
5. id = 0;
   
6. idmax = 10;
   
7. idmax = Input["求前n项,输入项数n:"];
   
8. x = 1;
   
9. out = 0;
   
10. While[id < idmax,
    
11. If[f[x] == 3, id = id + 1; out = x; AppendTo[an, x]];
    
12. x += 1
    
13. ]
    
14. Print["n=", idmax, " an=", out, "\n清单:"]
    
15. an

复制代码

王守恩

dlpg070 发表于 2019-5-24 22:09
关于a(n)的算式:
多数数列没有解析式,,显函数
此数列a(n)没有解析式,显函数

甲乙分别从 1～(n+1) 与 1～n 中取 3 个不同的整数，a(n) 为甲数之和大于乙数之和的取法种数。
3, 27, 126, 423, 1151, 2705, 5704, 11063, 20074, 34499, 56671, 89606, 137125, 203986,
296025, 420309, 585297, 801012, 1079223, 1433637, 1880100, 2436810, 3124538, 3966860,
4990399, 6225077, 7704376, 9465611, 11550211, 14004011, 16877554, 20226403, 24111462,
28599309, 33762537, 39680106, 46437705, 54128124, 62851635, 72716385, 83838797, 96343982,
110366161, 126049097, 143546536, 163022660, 184652548, 208622648,.............

\(\D a(n)=\sum_{k=6}^{3n }\bigg(\sum_{i=\lceil\frac{k+3}{3}\rceil}^{\min(n+1,k-3)}\sum_{j=\lceil\frac{k-i+1}{2}\rceil}^{\min(i-1,k-i-1)}1\bigg)
               \sum_{k=6}^{k-1}\bigg(\sum_{i=\lceil\frac{k+3}{3}\rceil}^{\min(n,k-3)}\sum_{j=\lceil\frac{k-i+1}{2}\rceil}^{\min(i-1,k-i-1)}1\bigg)\)

王守恩

dlpg070 发表于 2019-5-24 22:09
关于a(n)的算式:
多数数列没有解析式,,显函数
此数列a(n)没有解析式,显函数

用\(\D f_{k}(n)\ \)表示\(\ \D x_{1}+2x_{2}+3x_{3}+\cdots\cdots+kx_{k}=n\ \)的非负整数解的个数。
\(\D f_{1}(n)=\coefficientlist\big[\series\big[\prod_{i=1}^1\ \frac{1}{1-x^i},\ (x,\  0,\  n)\big],\ x\big]\)
\(\D f_{1}(n)=1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1\)
\(\D f_{2}(n)=\coefficientlist\big[\series\big[\prod_{i=1}^2\ \frac{1}{1-x^i},\ (x,\  0,\  n)\big],\ x\big]\)
\(\D f_{2}(n)=1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15\)
\(\D f_{3}(n)=\coefficientlist\big[\series\big[\prod_{i=1}^3\ \frac{1}{1-x^i},\ (x,\  0,\  n)\big],\ x\big]\)
\(\D f_{3}(n)=1, 1, 2, 3, 4, 5, 7, 8, 10, 12, 14, 16, 19, 21, 24, 27, 30, 33, 37, 40, 44, 48, 52, 56, 61, 65, 70\)
\(\D f_{4}(n)=\coefficientlist\big[\series\big[\prod_{i=1}^4\ \frac{1}{1-x^i},\ (x,\  0,\  n)\big],\ x\big]\)
\(\D f_{4}(n)=1, 1, 2, 3, 5, 6, 9, 11, 15, 18, 23, 27, 34, 39, 47, 54, 64, 72, 84, 94, 108, 120, 136, 150, 169\)
\(\D f_{5}(n)=\coefficientlist\big[\series\big[\prod_{i=1}^5\ \frac{1}{1-x^i},\ (x,\  0,\  n)\big],\ x\big]\)
\(\D f_{5}(n)=1, 1, 2, 3, 5, 7, 10, 13, 18, 23, 30, 37, 47, 57, 70, 84, 101, 119, 141, 164, 192, 221, 255, 291\)
\(\D f_{6}(n)=\coefficientlist\big[\series\big[\prod_{i=1}^6\ \frac{1}{1-x^i},\ (x,\  0,\  n)\big],\ x\big]\)
\(\D f_{6}(n)=1, 1, 2, 3, 5, 7, 11, 14, 20, 26, 35, 44, 58, 71, 90, 110, 136, 163, 199, 235, 282, 331, 391, 454\)
\(\D f_{7}(n)=\coefficientlist\big[\series\big[\prod_{i=1}^7\ \frac{1}{1-x^i},\ (x,\  0,\  n)\big],\ x\big]\)
\(\D f_{7}(n)=1, 1, 2, 3, 5, 7, 11, 15, 21, 28, 38, 49, 65, 82, 105, 131, 164, 201, 248, 300, 364, 436, 522, 618\)
\(\D f_{8}(n)=\coefficientlist\big[\series\big[\prod_{i=1}^8\ \frac{1}{1-x^i},\ (x,\  0,\  n)\big],\ x\big]\)
\(\D f_{8}(n)=1, 1, 2, 3, 5, 7, 11, 15, 22, 29, 40, 52, 70, 89, 116, 146, 186, 230, 288, 352, 434, 525, 638, 764\)
\(\D f_{9}(n)=\coefficientlist\big[\series\big[\prod_{i=1}^9\ \frac{1}{1-x^i},\ (x,\  0,\  n)\big],\ x\big]\)
\(\D f_{9}(n)=1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 41, 54, 73, 94, 123, 157, 201, 252, 318, 393, 488, 598, 732, 887\)

dlpg070

王守恩 发表于 2019-9-19 06:59
甲乙分别从 1～(n+1) 与 1～n 中取 3 个不同的整数，a(n) 为甲数之和大于乙数之和的取法种数。
3, 27 ...

请复查公式是否有手误,我无法验算
a(5)= 126?实际排一下

王守恩

dlpg070 发表于 2019-9-21 12:26
请复查公式是否有手误,我无法验算
a(5)= 126?实际排一下

数列：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, 51, 52, 53, 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, 82, 83, 84, 85, 86, 87, 88, 89,.........

\(通项公式：\D\bigg\lfloor\frac{e/2}{ e - (1 + 1/n)^n}\bigg\rfloor\)

dlpg070

王守恩 发表于 2019-9-19 11:34
用\(\D f_{k}(n)\ \)表示\(\ \D x_{1}+2x_{2}+3x_{3}+\cdots\cdots+kx_{k}=n\ \)的非负整数解的个数。
...

114#的9个数列OEIS都有,整理如下(前引导0可能不同)
f1(n)= A000012                The simplest sequence of positive numbers: the all 1's sequence.
f2(n)= A004526                Nonnegative integers repeated, floor(n/2).       
f3(n)= A001399                a(n) = number of partitions of n into at most 3 parts; also partitions of n+3 in which the greatest part is 3; also number of unlabeled multigraphs with 3 nodes and n edges.
f4(n)= A026810                Number of partitions of n in which the greatest part is 4.       
f5(n)= A001401                Number of partitions of n into at most 5 parts.
f6(n)= A026812                Number of partitions of n in which the greatest part is 6.               
f7(n)= A026813                Number of partitions of n in which the greatest part is 7.               
       A008636                Number of partitions of n into at most 7 parts.       
f8(n)= A026814                Number of partitions of n in which the greatest part is 8.
f9(n)= A026815                Number of partitions of n in which the greatest part is 9.

dlpg070

王守恩 发表于 2019-9-21 14:19
数列：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, ...

116# 这是哪来的数列?说明什么?

王守恩

dlpg070 发表于 2019-9-21 12:26
请复查公式是否有手误,我无法验算
a(5)= 126?实际排一下

\D a(n)=\sum_{k=6}^{3n }\bigg(\sum_{i=\lceil\frac{k+3}{3}\rceil}^{\min(n+1,k-3)}\sum_{j=\lceil\frac{k-i+1}{2}\rceil}^{\min(i-1,k-i-1)}1\bigg)
             \sum_{k=6}^{k-1}\bigg(\sum_{i=\lceil\frac{k+3}{3}\rceil}^{\min(  n,  k-3)}\sum_{j=\lceil\frac{k-i+1}{2}\rceil}^{\min(i-1,k-i-1)}1\bigg)

a(n)=\sum_{k=6}^{3n }(\sum_{i=\lceil\frac{k+3}{3}\rceil}^{\min(n+1,k-3)}\sum_{j=\lceil\frac{k-i+1}{2}\rceil}^{\min(i-1,k-i-1)}1)
         \sum_{k=6}^{k-1}(\sum_{i=\lceil\frac{k+3}{3}\rceil}^{\min(  n,  k-3)}\sum_{j=\lceil\frac{k-i+1}{2}\rceil}^{\min(i-1,k-i-1)}1)

Table[\!\(\*UnderoverscriptBox[\(\[Sum]\), \(k = 6\), \(3  n\)]
\(\((\*UnderoverscriptBox[\(\[Sum]\), \(i = Ceiling[\*FractionBox[\(k + 3\), \(3\)]]\), \(Min[n + 1, k - 3]\)]
\(\*UnderoverscriptBox[\(\[Sum]\), \(j = Ceiling[\*FractionBox[\(k - i + 1\), \(2\)]]\), \(Min[i - 1,  k - i - 1]\)]1\))\)
\(\*UnderoverscriptBox[\(\[Sum]\), \(k = 6\), \(k - 1\)]
\((\*UnderoverscriptBox[\(\[Sum]\), \(i = Ceiling[\*FractionBox[\(k + 3\), \(3\)]]\), \(Min[n, k - 3]\)]
\(\*UnderoverscriptBox[\(\[Sum]\), \(j = Ceiling[\*FractionBox[\(k - i + 1\), \(2\)]]\), \(Min[i - 1, k - i - 1]\)]1\))\)\)\)\), {n, 3, 50}]

dlpg070

王守恩 发表于 2019-9-19 06:59
甲乙分别从 1～(n+1) 与 1～n 中取 3 个不同的整数，a(n) 为甲数之和大于乙数之和的取法种数。
3, 27 ...

116# 的数列和公式很新颖,OEIS没有录入
我看不懂,提出2个疑点,认为是你的手误
经过几天讨论,问题没有解决,
再深入谈一下
疑点1 a(x)是否正确? 你只要给出你的公式原始10项结果就清楚了 a(1)--- a(10) ,是 3,27,126,---吗
疑点2 按题意,数列是 3,27,126,---吗,希望不按公式,按题意具体排一下,特别是 第3项是126吗
       这是你的强项, 你没有回应
现在给出我的初步分析计算结果:
a(1)=0
a(2)=1
a(3)=3
a(4)=27
a(5)=127(不是 126, 似乎以后数列都不对了,不只是手误而已)
---
但愿我错了,请认真算一算
必要时我可以给出 127项的清单,请你审查