悬赏挑战：求2025的两两之差乘积最大的堆磊数组

hujunhua

目标式为\[
\prod_{i=1}^nx_i·\prod_{1≤i<j≤n}(x_j-x_i)\tag1
\] 共`C_{n+1}^2` 项因子，我们给配上 正系数 后为\[
\prod_{i=1}^na_ix_i·\prod_{1≤i<j≤n}a_{ji}(x_j-x_i)\tag2
\]所配 正系数 使得\[
a_ix_i=a_jx_j=a_{ji}(x_j-x_i)=\frac1{C_n^2},\text{for 1}≤i<j≤n\tag3,\]\[
\sum_{i=1}^na_ix_i+\sum_{1≤i<j≤n}a_{ji}(x_j-x_i)=\sum_{i=1}^nx_i=1\tag4
\]由平均不等式得(1)的极大值等于   \[\frac{{C_n^2}^{-C_n^2}}{\D\prod_{i=1}^na_i·\prod_{1≤i<j≤n}a_{ji}}\]
将(4)式左边展开归集各项 `x_i`的系数，左右对比得\[
\sum_{j=1}^na_{ij}=1,\text{for 1}≤i≤n \tag5\\\text{there } a_{ij}=-a_{ji}, a_{ii}=a_i
\]即 \[\{a_{ij}\}_{n×n}.(1,1,...,1)^τ=(1,1,...,1)^τ\tag6\](待续......）

wayne

我来个临门一脚，锦上添花。
定义：$\text{HypergeometricU}(a,b,z)=\frac{1}{\Gamma (a)}\int_0^{\infty } t^{a-1} e^{-zt} (t+1)^{b-a-1} dt$
https://mathworld.wolfram.com/Co ... ftheSecondKind.html

1. Table[{n,  Expand[(n (1 + n))^-n HypergeometricU[-n, 2, n (1 + n) x]]}, {n,   20}]

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$\begin{array}{l}
\{1,x-1\} \\
\left\{2,x^2-x+\frac{1}{6}\right\} \\
\left\{3,x^3-x^2+\frac{x}{4}-\frac{1}{72}\right\} \\
\left\{4,x^4-x^3+\frac{3 x^2}{10}-\frac{3 x}{100}+\frac{3}{4000}\right\} \\
\left\{5,x^5-x^4+\frac{x^3}{3}-\frac{2 x^2}{45}+\frac{x}{450}-\frac{1}{33750}\right\} \\
\left\{6,x^6-x^5+\frac{5 x^4}{14}-\frac{25 x^3}{441}+\frac{25 x^2}{6174}-\frac{5 x}{43218}+\frac{5}{5445468}\right\} \\
\left\{7,x^7-x^6+\frac{3 x^5}{8}-\frac{15 x^4}{224}+\frac{75 x^3}{12544}-\frac{45 x^2}{175616}+\frac{45 x}{9834496}-\frac{45}{1927561216}\right\} \\
\left\{8,x^8-x^7+\frac{7 x^6}{18}-\frac{49 x^5}{648}+\frac{245 x^4}{31104}-\frac{245 x^3}{559872}+\frac{245 x^2}{20155392}-\frac{35 x}{241864704}+\frac{35}{69657034752}\right\} \\
\left\{9,x^9-x^8+\frac{2 x^7}{5}-\frac{56 x^6}{675}+\frac{98 x^5}{10125}-\frac{98 x^4}{151875}+\frac{98 x^3}{4100625}-\frac{28 x^2}{61509375}+\frac{7 x}{1845281250}-\frac{7}{747338906250}\right\} \\
\left\{10,x^{10}-x^9+\frac{9 x^8}{22}-\frac{54 x^7}{605}+\frac{378 x^6}{33275}-\frac{7938 x^5}{9150625}+\frac{3969 x^4}{100656875}-\frac{1134 x^3}{1107225625}+\frac{1701 x^2}{121794818750}-\frac{567 x}{6698715031250}+\frac{567}{3684293267187500}\right\} \\
\end{array}$

yuange1975

今天难得大牛们这么有兴趣呀。

这个方程竟然有这么好的表达式。

mathe

wayne 发表于 2025-3-18 16:12
我来个临门一脚，锦上添花。
定义：$\text{HypergeometricU}(a,b,z)=\frac{1}{\Gamma (a)}\int_0^{\infty } ...

对于本题来说，除了计算出这些多项式以外，我们还需要有效计算出f(x)所有根之间差值的乘积。
由于我们需要计算\(c_n=\prod_{0\le s \lt t\le n}(x_t-x_s)\), 而我们知道\(f'(x_t)=\prod_{s\neq t}(x_t-x_s)\), 所以\(\prod_{h=0}^n f'(x_h)=c_n^2\)
对于整系数函数f(x)的所有根\(x_0,x_1,...,x_n\),如果整系数u(x)次数小于f(x)的次数，而且g(x)如果以\(u(x_0),u(x_1),...,u(x_n)\)为根
那么必然f(x)|g(u(x)), 所以我们只需要计算\(1,u(x),u(x)^2,...,u(x)^n \)关于f(x)的余式，然后找出让它们的线性组合为0的数，就得到了g(x)的系数。由此可以知道g(x)的系数也必然式有理数。
所以可以知道\(c_n^2\)是有理数。
比如我们以\(f(x)=x(x^4-x^3+3\frac{x^2}{10}-\frac{3x}{100}+\frac{3}{4000})\)为例子，
其导函数\(f'(x)=5x^4 - 4x^3 + \frac{9x^2}{10} - \frac{3x}{50} + \frac{3}{4000}\)
计算其平方，三次方，四次方, 五次方分别得到
\(f'(x)^2  = \frac{23x^4}{800} - \frac{51x^3}{4000} + \frac{63x^2}{40000} - \frac{21x}{400000} + \frac9{16000000} \pmod {f(x)}\)

3029/3200000*x^4 - 6897/16000000*x^3 + 8019/160000000*x^2 - 2127/1600000000*x + 27/64000000000

366527/12800000000*x^4 - 828507/64000000000*x^3 + 957393/640000000000*x^2 - 250833/6400000000000*x + 81/256000000000000

8981881/10240000000000*x^4 - 20313789/51200000000000*x^3 + 23478039/512000000000000*x^2 - 6150327/5120000000000000*x + 243/1024000000000000000

然后计算可以得到
-729+3240000000f'(x)^2 -1920000000000f'(x)^3 -920000000000000f'(x)^4+32000000000000000f'(x)^5=0 mod f(x)

由此我们得到g(x)=-729+3240000000x^2 -1920000000000x^3 -920000000000000x^4+32000000000000000x^5

所以得出\(c_5^2 =\frac{729}{32000000000000000} \)

然后用计算机计算可以得到\(c_n^2\)从n=2开始前若干项为
1,1/108,1/2985984,729/32000000000000000,1/591156752014160156250000，9765625/109557337770936610818023422262773996059220992，3783403212890625/1619858932012200054170409405890921220815687456856366256032153761480704， 6623272195791015625/294560779218133654515180236763892077549977268256656953492166117844174864184344920047952592896，
678223072849/11096305323949011816861066510597411011570633698060804377804944160211021719764090676108025945723056793212890625，
39438493315563029337860799974920421481/1061148425924421578260900012285364200845300304297028581803271221532799110908103465676662309221282797274781413377972572043006493913708254694938659667968750000000000，
631644458369351923465728759765625/153956337797082801312602155132712527728419594801950662056170258285109651302959752178425709020550957668607054925323137651844148105032708573883446988584465575377522022147098035857822689787904，
51417595546226495302701057798528605140745639801025390625/755573185563836669348934647683428862478268474128686950564848866904165670839702535558649669116053363648131386269441357509416861330849908144626185171615083059582306115466534841065483021413743025661462136242550349454054130435648464042726855203094528

mathe

数值计算经验对于和为S的情况，最佳n大概在0.4S,
比如S=10, n=4(5个数）可以达到最优结果， S=60, n=24(25个数）达到最优结果。S=100, n=39 (40个数）达到最优结果。
但是对于更大的和，我这边pari/gp计算效率已经不行了。

wayne

mathe 发表于 2025-3-19 11:00
数值计算经验对于和为S的情况，最佳n大概在0.4S,
比如S=10, n=4(5个数）可以达到最优结果， S=60, n=24(25 ...

粗略拼凑了下，跟mathe的若干项数据都对上了，那应该就是 $c_n^2=\frac{(n+1)^{n+1}}{(n (n+1))^{n (n+1)}} (\prod _{i=1}^n i^i)^2$, 需要补充证明过程。

1. Table[(1 + n)^(1 + n) (n (1 + n))^(-n (1 + n))Hyperfactorial[n]^2, {n, 100}]

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又出现了一个特殊函数，Hyperfactorial，https://mathworld.wolfram.com/Hyperfactorial.html

yuange1975

Plot[n(n+1)*ln2025+(n+1)*ln(n+1)-n(n+1)*ln(n(n+1))+2*ln(Hyperfactorial(n))
,{n,748,749}]

应该是n=748取到最大值。

yuange1975

如果没有算错，那么最大值就是

2025^280126*749^(749/2)/560252^280126*Hyperfactorial(748)=9.76*10^61327

wayne

wayne 发表于 2025-3-19 12:36
粗略拼凑了下，跟mathe的若干项数据都对上了，那应该就是 $c_n^2=\frac{(n+1)^{n+1}}{(n (n+1))^{n (n+1)} ...

计算了一下，确实如同yigo的猜想，接近于$\frac{1}{e}=0.36787944117144232160$的比例。
其实，是可以证明的，令导函数为0。得$log S =\frac{-2 \text{LogGamma}(n+1)-\log (n+1)+2 n \log (n (n+1))+\log (n (n+1))-1+\log (2 \pi )}{2 n+1}$
再设$S=kn$,代入得求极限，$logk = lim_{n->+\infty}\frac{-2 \text{logGamma}(n+1)-\log (n+1)+2 n \log (n (n+1))+\log (n (n+1))-1+\log (2 \pi )}{2 n+1}-\log (n) =1$
其实这里，应该是令$n=kS$，对S取极限更恰当。但涉及到表达式的逆解，无法解析表达。
LogGamma函数的定义： https://mathworld.wolfram.com/LogGammaFunction.html
这里，给出数据{{S,n},1/e-n/S}

1. {{10^1,4},-0.03212055883}
   
2. {{10^2,39},-0.02212055883}
   
3. {{10^3,371},-0.003120558829}
   
4. {{10^4,3683},-0.0004205588286}
   
5. {{10^5,36793},-0.00005055882856}
   
6. {{10^6,367886},-6.558828558*10^-6}
   
7. {{10^7,3678802},-7.588285577*10^-7}
   
8. {{10^8,36787953},-8.882855768*10^-8}
   
9. {{10^9,367879451},-9.828557678*10^-9}
   
10. {{10^10,3678794423},-1.128557678*10^-9}
    
11. {{10^11,36787944129},-1.185576784*10^-10}
    
12. {{10^12,367879441185},-1.355767840*10^-11}
    
13. {{10^13,3678794411729},-1.457678404*10^-12}
    
14. {{10^14,36787944117160},-1.576784045*10^-13}
    
15. {{10^15,367879441171459},-1.667840448*10^-14}
    
16. {{10^16,3678794411714441},-1.778404476*10^-15}
    
17. {{10^17,36787944117144251},-1.884044762*10^-16}
    
18. {{10^18,367879441171442342},-2.040447623*10^-17}
    
19. {{10^19,3678794411714423237},-2.104476230*10^-18}
    
20. {{10^20,36787944117144232182},-2.244762298*10^-19}
    
21. {{10^21,367879441171442321619},-2.347622984*10^-20}
    
22. {{10^22,3678794411714423215980},-2.476229839*10^-21}
    
23. {{10^23,36787944117144232159578},-2.562298385*10^-22}
    
24. {{10^24,367879441171442321595551},-2.722983854*10^-23}
    
25. {{10^25,3678794411714423215955266},-2.829838539*10^-24}
    
26. {{10^26,36787944117144232159552406},-2.898385391*10^-25}
    
27. {{10^27,367879441171442321595523801},-3.083853913*10^-26}
    
28. {{10^28,3678794411714423215955237733},-3.138539133*10^-27}
    
29. {{10^29,36787944117144232159552377049},-3.285391326*10^-28}
    
30. {{10^30,367879441171442321595523770195},-3.353913255*10^-29}
    
31. {{10^31,3678794411714423215955237701650},-3.539132554*10^-30}
    
32. {{10^32,36787944117144232159552377016182},-3.591325542*10^-31}
    
33. {{10^33,367879441171442321595523770161498},-3.713255419*10^-32}
    
34. {{10^34,3678794411714423215955237701614647},-3.832554189*10^-33}
    
35. {{10^35,36787944117144232159552377016146127},-4.025541889*10^-34}
    
36. {{10^36,367879441171442321595523770161460908},-4.055418887*10^-35}
    
37. {{10^37,3678794411714423215955237701614608717},-4.254188869*10^-36}
    
38. {{10^38,36787944117144232159552377016146086788},-4.341888690*10^-37}
    
39. {{10^39,367879441171442321595523770161460867490},-4.418886897*10^-38}
    
40. {{10^40,3678794411714423215955237701614608674504},-4.588868968*10^-39}
    
41. {{10^41,36787944117144232159552377016146086744628},-4.688689682*10^-40}
    
42. {{10^42,367879441171442321595523770161460867445859},-4.786896823*10^-41}
    
43. {{10^43,3678794411714423215955237701614608674458160},-4.868968232*10^-42}
    
44. {{10^44,36787944117144232159552377016146086744581163},-4.989682322*10^-43}
    
45. {{10^45,367879441171442321595523770161460867445811182},-5.096823217*10^-44}
    
46. {{10^46,3678794411714423215955237701614608674458111363},-5.268232165*10^-45}
    
47. {{10^47,36787944117144232159552377016146086744581113157},-5.382321655*10^-46}
    
48. {{10^48,367879441171442321595523770161460867445811131087},-5.523216549*10^-47}
    
49. {{10^49,3678794411714423215955237701614608674458111310374},-5.632165492*10^-48}
    
50. {{10^50,36787944117144232159552377016146086744581113103234},-5.721654922*10^-49}
    

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S=10^100的时候，n=3678794411714423215955237701614608674458111274356742252026138246134833618107261680792454171386183680, $|\frac{1}{e}-\frac{n}{S}| = 3.596093609*10^-45$

mathe

和为S时，我们需要查看\(c_n^2 S^{n(n+1)}\)什么时候达到最大，我们只需要查看\(\frac{c_{n+1}^2 S^{(n+1)(n+2)}}{c_n^2 S^{n(n+1)}}\)什么时候小于1即可。
消去重复表达式可以表示为
\(\left(\frac{S^{1+\frac1n}}{(n+2)(1+\frac2n)^{\frac n2}}\right)^{2n}\)
我们需要找到第一个让上述表达式不超过1的n即可。
很显然，对于充分大的n,S逼近(n+2)e,或者说逼近ne