数字串的通项公式

northwolves

王守恩 发表于 2023-8-3 17:51
我们这串数: 0, 4, 48, 1440, 65280, ..., 通项公式(A189849/2^n)与 A189849不大一样。
如果用这串数， ...

n对夫妻入住某酒店的n个标间，夫妻不得同屋

王守恩

(1),

1. Table[Minimize[{1/Sin[a] + n/Cos[a], \[Pi]/2 > a > 0}, {a}], {n, 1, 9}] // ToRadicals // FullSimplify

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   {2 Sqrt[2],   Sqrt[5 + 6 2^(1/3) + 3 2^(2/3)],    Sqrt[10 + 9 3^(1/3) + 3 3^(2/3)],    Sqrt[17 + 6 2^(1/3) + 12 2^(2/3)],
   Sqrt[26 + 15 5^(1/3) + 3 5^(2/3)],  Sqrt[37 + 18 6^(1/3) + 3 6^(2/3)],  Sqrt[50 + 21 7^(1/3) + 3 7^(2/3)],  5 Sqrt[5],   Sqrt[82 + 9 3^(1/3) + 27 3^(2/3)]}
(1)可以有通项公式:

1. Table[Solve[{s == Sqrt[n^2 + 1 + 3 n Power[n, (3)^-1] + 3 Power[n^2, (3)^-1]]}, {s}], {n, 1, 9}] // ToRadicals // FullSimplify

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s -> 2 Sqrt[2], s -> Sqrt[5 + 6 2^(1/3) + 3 2^(2/3)], s -> Sqrt[10 + 9 3^(1/3) + 3 3^(2/3)], s -> Sqrt[17 + 6 2^(1/3) + 12 2^(2/3)],
s -> Sqrt[26 + 15 5^(1/3) + 3 5^(2/3)], s -> Sqrt[37 + 18 6^(1/3) + 3 6^(2/3)], s -> Sqrt[50 + 21 7^(1/3) + 3 7^(2/3)], s -> 5 Sqrt[5], s -> Sqrt[82 + 9 3^(1/3) + 27 3^(2/3)]}
(2),

1. Table[Minimize[{2/Sin[a] + n/Cos[a], \[Pi]/2 > a > 0}, {a}], {n, 1, 9}] // ToRadicals // FullSimplify

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   {Sqrt[5 + 6 2^(1/3) + 3 2^(2/3)],  4 Sqrt[2],  Root[-2197 - 465 #^2 - 39 #^4 + #^6&,2,0],  2 Sqrt[5 + 6 2^(1/3) + 3 2^(2/3)],  {Root[-24389 - 177 #^2 - 87 #^4 + #^6&,2,0],
   2 Sqrt[10 + 9 3^(1/3) + 3 3^(2/3)],  Sqrt[53 + 21 2^(2/3) 7^(1/3) + 6 2^(1/3) 7^(2/3)],  2 Sqrt[17 + 6 2^(1/3) + 12 2^(2/3)],  Sqrt[85 + 18 6^(1/3) + 27 6^(2/3)]}
(3),

1. Table[Minimize[{3/Sin[a] + n/Cos[a], \[Pi]/2 > a > 0}, {a}], {n, 1, 9}] // ToRadicals // FullSimplify

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   {Sqrt[10 + 9 3^(1/3) + 3 3^(2/3)],  Root[-2197 - 465 #^2 - 39 #^4 + #^6&,2,0],  6 Sqrt[2],   Sqrt[25 + 18 6^(1/3) + 12 6^(2/3)],  Sqrt[34 + 15 3^(2/3) 5^(1/3) + 9 3^(1/3) 5^(2/3)],
   3 Sqrt[5 + 6 2^(1/3) + 3 2^(2/3)],   Sqrt[58 + 21 3^(2/3) 7^(1/3) + 9 3^(1/3) 7^(2/3)],   Sqrt[73 + 36 3^(1/3) + 48 3^(2/3)],   3 Sqrt[10 + 9 3^(1/3) + 3 3^(2/3)]]}
......
(1)可以有通项公式;   (2),(3),(4)...可以有通项公式吗？

王守恩

(1),(2),(3),(4)...可以有统一的通项公式。

若  \(\pi/2 > a > 0,\frac{m}{\sin(a)}+\frac{n}{\cos(a)}\)  的最小值=\(\D(m^{2/3}+n^{2/3})^{3/2}\)

northwolves

$f(a)=k/Sin[a] + n/Cos[a]$
$f'(a)=\frac{kTan[a]}{Cos[a]}- \frac{n}{Tan[a]Sin[a]}$
$f'(a)=0->Tan[a]=(\frac{k}{n})^{\frac{1}{3}}$
此时 $k/Sin[a] + n/Cos[a]=(k+nTan[a])\sqrt{1+\frac{1}{Tan^2[a])}=(k^{\frac{2}{3}}+n^{\frac{2}{3}})^{\frac{3}{2}}$

王守恩

有这样一串数:4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128, 144, 169, 196, 216, 225,...A001597
1,每个数-1，取倒数相加，和是多少？1/3+1/7+1/8+1/15+1/24+1/26+1/31+1/35+1/48+1/63+1/80+1/99+1/120+...
2,每个数-指数(较小者)，取倒数相加，和是多少？1/2+1/5+1/7+1/14+1/23+24+1/27+1/34+1/47+1/62+1/79+1/98+...       

northwolves

$\sum_{k=2}^{∞}\frac{1}{n^k}=\frac{1}{n-1}-\frac{1}{n}$
$1/3+1/7+1/8+1/15+1/24+1/26+1/31+1/35+1/48+1/63+1/80+1/99+1/120+...=1-\frac{1}{2}+\frac1 2-\frac1 3+...=1$

northwolves

1. N[Total[1/Union@Select[Flatten@Table[n^k-k,{n,2,10000},{k,2,Floor[Log[10^8]/Log[n]]}],#<10^8&]],20]

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1.4085485959754331397

northwolves

第2题猜测极限值$\sqrt2$

王守恩

一个数列的通项式，据说难倒英雄汉！
每一行的和都是"兔子数列"。每一列都是"杨辉三角"中的列。

1. Cn:Table[(n+k)!/((2k)!(n-k)!),{n,0,9},{k,0,n}]//TableForm

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{1},
{1, 1},
{1, 3, 1},
{1, 6, 5, 1},
{1, 10, 15, 7, 1},
{1, 15, 35, 28, 9, 1},
{1, 21, 70, 84, 45, 11, 1},
{1, 28, 126, 210, 165, 66, 13, 1},
{1, 36, 210, 462, 495, 286, 91, 15, 1},
{1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1}

1. An:Table[(n+k)!/((2k)!(n-k)!),{n,1,9},{k,0,n}]//TableForm

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{1, 1},
{1, 3, 1},
{1, 6, 5, 1},
{1, 10, 15, 7, 1},
{1, 15, 35, 28, 9, 1},
{1, 21, 70, 84, 45, 11, 1},
{1, 28, 126, 210, 165, 66, 13, 1},
{1, 36, 210, 462, 495, 286, 91, 15, 1},
{1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1}}]

1. Dn:Table[(n+k+1)!/((2k1)!(n-k)!),{n,0,9},{k,0,n}]//TableForm

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{1},
{2, 1},
{3, 4, 1},
{4, 10, 6, 1},
{5, 20, 21, 8, 1},
{6, 35, 56, 36, 10, 1},
{7, 56, 126, 120, 55, 12, 1},
{8, 84, 252, 330, 220, 78, 14, 1},
{9, 120, 462, 792, 715, 364, 105, 16, 1},
{10,165, 792, 1716, 2002, 1365, 560, 136, 18, 1}}]

1. Bn:Table[(n+k+1)!/((2k+1)!(n-k)!),{n,1,9},{k,0,n}]//TableForm

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{2, 1},
{3, 4, 1},
{4, 10, 6, 1},
{5, 20, 21, 8, 1},
{6, 35, 56, 36, 10, 1},
{7, 56, 126, 120, 55, 12, 1},
{8, 84, 252, 330, 220, 78, 14, 1},
{9, 120, 462, 792, 715, 364, 105, 16, 1},
{10,165, 792, 1716, 2002, 1365, 560, 136, 18, 1}

王守恩

用足够多的a,b,c凑出n,  一共有多少种方法？

1. CoefficientList[Series[1/((1-x^a)(1-x^b)(1-x^c)),{x,0,n]],x]

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