Wilson prime and 组合数学 (Fall 2017)/Problem Set 1: Difference between pages

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A '''Wilson prime''' is a special kind of [[prime number]]. A prime number ''p'' is a Wilson prime if (and only if [ [[iff]] ])
*每道题目的解答都要有<font color="red" size=4>完整的解题过程</font>。中英文不限。


<math>\frac{\left(p-1\right)! + 1}{p^2} = n\,\!</math>  
== Problem 1 ==
#有<math>k</math>种不同的明信片,每种明信片有无限多张,寄给<math>n</math>个人,每人一张,有多少种方法?
#有<math>k</math>种不同的明信片,每种明信片有无限多张,寄给<math>n</math>个人,每人一张,每个人必须收到不同种类的明信片,有多少种方法?
#有<math>k</math>种不同的明信片,每种明信片有无限多张,寄给<math>n</math>个人,每人收到<math>r</math>张不同的明信片(但不同的人可以收到相同的明信片),有多少种方法?
#只有一种明信片,共有<math>m</math>张,寄给<math>n</math>个人,全部寄完,每个人可以收多张明信片或者不收明信片,有多少种方法?
#有<math>k</math>种不同的明信片,其中第<math>i</math>种明信片有<math>m_i</math>张,寄给<math>n</math>个人,全部寄完,每个人可以收多张明信片或者不收明信片,有多少种方法?


where ''n'' is a [[positive number|positive]] [[integer]] (sometimes called [[natural number]]). Wilson primes were first described by [[Emma Lehmer]].<ref>[http://www.google.de/url?sa=t&source=web&cd=10&ved=0CGgQFjAJ&url=http%3A%2F%2Fgradelle.educanet2.ch%2Fchristian.aebi%2F.ws_gen%2F14%2FEmma_Lehmer_1938.pdf&rct=j&q=lehmer%20on%20congruences%20involving%20bernoulli%20numbers%20and%20the%20quotients%20of%20fermat%20and%20wilson&ei=QVHcTK2iAZDysgae9o2iBA&usg=AFQjCNGuZ93z06kDmxXutGU8S_ADA6FgZw&cad=rja On congruences involving Bernoulli numbers and the quotients of Fermat and Wilson], Ann. of. Math. '''39'''(1938), 350-360.</ref>
== Problem 2 ==
Find the number of ways to select <math>2n</math> balls from <math>n</math> identical blue balls, <math>n</math> identical red balls and <math>n</math> identical green balls.
* Give a combinatorial proof for the problem.
* Give an algebraic proof for the problem.


The only known Wilson primes are [[5 (number)|5]], [[13 (number)|13]], and 563 {{OEIS|id=A007540}}; if any others exist, they must be greater than 5{{e|8}}.<ref>[http://www.loria.fr/~zimmerma/records/Wieferich.status Status of the search for Wilson primes], also see Crandall ''et al.'' 1997</ref> It has been [[conjecture]]d<ref>[http://primes.utm.edu/glossary/page.php?sort=WilsonPrime The Prime Glossary: Wilson prime]</ref> that  there are an infinite number of Wilson primes, and that the number of Wilson primes in an interval {{math|[''x'' , ''y'']}} is about
== Problem 3 ==
*一个长度为<math>n</math>的“山峦”是如下由<math>n</math>个"/"和<math>n</math>个"\"组成的,从坐标<math>(0,0)</math><math>(0,2n)</math>的折线,但任何时候都不允许低于<math>x</math>轴。例如下图:


<math>\frac{\log \left ( \log y \right )}{\log x}</math>.
    /\
  /  \/\/\    /\/\
  /        \/\/    \/\/\
  ----------------------
:长度为<math>n</math>的“山峦”有多少?


Compare this with [[Wilson's theorem]], which states that every prime ''p'' divides (''p'' − 1)! + 1.
*一个长度为<math>n</math>的“地貌”是由<math>n</math>个"/"和<math>n</math>个"\"组成的,从坐标<math>(0,0)</math>到<math>(0,2n)</math>的折线,允许低于<math>x</math>轴。长度为<math>n</math>的“地貌”有多少?


== Related pages ==
== Problem 4==
* [[Wieferich prime]]
李雷和韩梅梅竞选学生会主席,韩梅梅获得选票 <math>p</math> 张,李雷获得选票 <math>q</math> 张,<math>p>q</math>。我们将总共的 <math>p+q</math> 张选票一张一张的点数,有多少种选票的排序方式使得在整个点票过程中,韩梅梅的票数一直高于李雷的票数?等价地,假设选票均匀分布的随机排列,以多大概率在整个点票过程中,韩梅梅的票数一直高于李雷的票数。
* [[Wall-Sun-Sun prime]]
* [[Wolstenholme prime]]


==Notes==
==Problem 5==
<references/>
A <math>2\times n</math> rectangle is to be paved with <math>1\times 2</math> identical blocks and <math>2\times 2</math> identical blocks. Let <math>f(n)</math> denote the number of ways that can be done. Find a recurrence relation for <math>f(n)</math>, solve the recurrence relation.


==References==
== Problem 6 ==
* {{cite journal | author=N. G. W. H. Beeger | title=Quelques remarques sur les congruences r<sup>p-1</sup> ≡ 1 (mod p<sup>2</sup>) et (p-1!) ≡ -1 (mod p<sup>2</sup>) | journal=[[Messenger of Mathematics|The Messenger of Mathematics]] | volume=43 | pages=72-84 | year=1913-1914}}
* <math>s_n</math>表示长度为<math>n</math>,没有2个连续的1的二进制串的数量,即
* {{cite journal | author=Karl Goldberg | title=A table of Wilson quotients and the third Wilson prime | journal=J. Lond. Math. Soc. | volume=28 | pages=252–256 | year=1953 | doi=10.1112/jlms/s1-28.2.252 }}
*:<math>s_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-1, x_ix_{i+1}\neq 11\}|</math>
* {{cite book | title=The new book of prime number records | author=Paulo Ribenboim | authorlink=Paulo Ribenboim | publisher=[[Springer-Verlag]] | year=1996 | isbn=0-387-94457-5 | pages=346 }}
:求 <math>s_n</math>
* {{cite journal | author=Richard E. Crandall | coauthors=Karl Dilcher, Carl Pomerance | title=A search for Wieferich and Wilson primes | journal=Math. Comput. | volume=66 | issue=217 | pages=433–449 | year=1997 | doi=10.1090/S0025-5718-97-00791-6 }}
* {{cite book | title=Prime Numbers: A Computational Perspective | author=Richard E. Crandall | coauthors=Carl Pomerance | publisher=Springer-Verlag | year=2001 | page=29 | isbn=0-387-94777-9 }}
* {{cite journal | author=Takashi Agoh | coauthors=Karl Dilcher, Ladislav Skula | title=Wilson quotients for composite moduli | journal=Math. Comput. | volume=67 | issue=222 | pages=843-861 | year=1998 | url=http://en.wikipedia.org/w/index.php?title=Wilson_prime&action=edit&section=4}}
* {{cite journal | author=Erna H. Pearson | title=On the Congruences (p-1)! ≡ -1 and 2<sup>p-1</sup> ≡ 1 (mod p<sup>2</sup>) | journal=Math. Comput. | volume=17 | pages=194-195 | year=1963 | url=http://www.ams.org/journals/mcom/1963-17-082/S0025-5718-1963-0159780-0/S0025-5718-1963-0159780-0.pdf}}


==Other websites==
*令<math>t_n</math>表示长度为<math>n</math>,没有3个连续的1的二进制串的数量,即
*:<math>t_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-2, x_ix_{i+1}x_{i+2}\neq 111\}|</math>。
*#给出计算<math>t_n</math>的递归式,并给出足够的初始值。
*#计算<math>t_n</math>的生成函数<math>T(x)=\sum_{n\ge 0}t_n x^n</math>,给出生成函数<math>T(x)</math>的闭合形式。


* [http://primes.utm.edu/glossary/page.php?sort=WilsonPrime The Prime Glossary: Wilson prime]
注意:只需解生成函数的闭合形式,无需展开。
* {{MathWorld|urlname=WilsonPrime|title=Wilson prime}}
* [http://www.loria.fr/~zimmerma/records/Wieferich.status Status of the search for Wilson primes]
* [http://www.google.de/url?sa=t&source=web&cd=3&ved=0CC4QFjAC&url=http%3A%2F%2Fciteseerx.ist.psu.edu%2Fviewdoc%2Fdownload%3Fdoi%3D10.1.1.102.6544%26rep%3Drep1%26type%3Dpdf&rct=j&q=a%20table%20of%20wilson%20quotients%20and%20the%20third%20wilson%20prime&ei=nk_cTPjcJMv4sgaLn4yiBA&usg=AFQjCNHrvGLzlN26fFSUfC1sh5keBqvpWA&cad=rja Wilson Quotients for composite moduli]
* [http://www.google.de/url?sa=t&source=web&cd=10&ved=0CGgQFjAJ&url=http%3A%2F%2Fgradelle.educanet2.ch%2Fchristian.aebi%2F.ws_gen%2F14%2FEmma_Lehmer_1938.pdf&rct=j&q=lehmer%20on%20congruences%20involving%20bernoulli%20numbers%20and%20the%20quotients%20of%20fermat%20and%20wilson&ei=QVHcTK2iAZDysgae9o2iBA&usg=AFQjCNGuZ93z06kDmxXutGU8S_ADA6FgZw&cad=rja On congruences involving Bernoulli numbers and the quotients of Fermat and Wilson]


 
== Problem 7 ==
{{math-stub}}
Let <math>a_n</math> be a sequence of numbers satisfying the recurrence relation:
 
:<math>p a_n+q a_{n-1}+r a_{n-2}=0</math>
[[Category:Number theory]]
with initial condition <math>a_0=s</math> and <math>a_1=t</math>, where <math>p,q,r,s,t</math> are constants such that <math>{p}+q+r=0</math>, <math>p\neq 0</math> and <math>s\neq t</math>. Solve the recurrence relation.

Revision as of 13:14, 17 September 2017

  • 每道题目的解答都要有完整的解题过程。中英文不限。

Problem 1

  1. [math]\displaystyle{ k }[/math]种不同的明信片,每种明信片有无限多张,寄给[math]\displaystyle{ n }[/math]个人,每人一张,有多少种方法?
  2. [math]\displaystyle{ k }[/math]种不同的明信片,每种明信片有无限多张,寄给[math]\displaystyle{ n }[/math]个人,每人一张,每个人必须收到不同种类的明信片,有多少种方法?
  3. [math]\displaystyle{ k }[/math]种不同的明信片,每种明信片有无限多张,寄给[math]\displaystyle{ n }[/math]个人,每人收到[math]\displaystyle{ r }[/math]张不同的明信片(但不同的人可以收到相同的明信片),有多少种方法?
  4. 只有一种明信片,共有[math]\displaystyle{ m }[/math]张,寄给[math]\displaystyle{ n }[/math]个人,全部寄完,每个人可以收多张明信片或者不收明信片,有多少种方法?
  5. [math]\displaystyle{ k }[/math]种不同的明信片,其中第[math]\displaystyle{ i }[/math]种明信片有[math]\displaystyle{ m_i }[/math]张,寄给[math]\displaystyle{ n }[/math]个人,全部寄完,每个人可以收多张明信片或者不收明信片,有多少种方法?

Problem 2

Find the number of ways to select [math]\displaystyle{ 2n }[/math] balls from [math]\displaystyle{ n }[/math] identical blue balls, [math]\displaystyle{ n }[/math] identical red balls and [math]\displaystyle{ n }[/math] identical green balls.

  • Give a combinatorial proof for the problem.
  • Give an algebraic proof for the problem.

Problem 3

  • 一个长度为[math]\displaystyle{ n }[/math]的“山峦”是如下由[math]\displaystyle{ n }[/math]个"/"和[math]\displaystyle{ n }[/math]个"\"组成的,从坐标[math]\displaystyle{ (0,0) }[/math][math]\displaystyle{ (0,2n) }[/math]的折线,但任何时候都不允许低于[math]\displaystyle{ x }[/math]轴。例如下图:
   /\
  /  \/\/\    /\/\
 /        \/\/    \/\/\
 ----------------------
长度为[math]\displaystyle{ n }[/math]的“山峦”有多少?
  • 一个长度为[math]\displaystyle{ n }[/math]的“地貌”是由[math]\displaystyle{ n }[/math]个"/"和[math]\displaystyle{ n }[/math]个"\"组成的,从坐标[math]\displaystyle{ (0,0) }[/math][math]\displaystyle{ (0,2n) }[/math]的折线,允许低于[math]\displaystyle{ x }[/math]轴。长度为[math]\displaystyle{ n }[/math]的“地貌”有多少?

Problem 4

李雷和韩梅梅竞选学生会主席,韩梅梅获得选票 [math]\displaystyle{ p }[/math] 张,李雷获得选票 [math]\displaystyle{ q }[/math] 张,[math]\displaystyle{ p\gt q }[/math]。我们将总共的 [math]\displaystyle{ p+q }[/math] 张选票一张一张的点数,有多少种选票的排序方式使得在整个点票过程中,韩梅梅的票数一直高于李雷的票数?等价地,假设选票均匀分布的随机排列,以多大概率在整个点票过程中,韩梅梅的票数一直高于李雷的票数。

Problem 5

A [math]\displaystyle{ 2\times n }[/math] rectangle is to be paved with [math]\displaystyle{ 1\times 2 }[/math] identical blocks and [math]\displaystyle{ 2\times 2 }[/math] identical blocks. Let [math]\displaystyle{ f(n) }[/math] denote the number of ways that can be done. Find a recurrence relation for [math]\displaystyle{ f(n) }[/math], solve the recurrence relation.

Problem 6

  • [math]\displaystyle{ s_n }[/math]表示长度为[math]\displaystyle{ n }[/math],没有2个连续的1的二进制串的数量,即
    [math]\displaystyle{ s_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-1, x_ix_{i+1}\neq 11\}| }[/math]
[math]\displaystyle{ s_n }[/math]
  • [math]\displaystyle{ t_n }[/math]表示长度为[math]\displaystyle{ n }[/math],没有3个连续的1的二进制串的数量,即
    [math]\displaystyle{ t_n=|\{x\in\{0,1\}^n\mid \forall 1\le i\le n-2, x_ix_{i+1}x_{i+2}\neq 111\}| }[/math]
    1. 给出计算[math]\displaystyle{ t_n }[/math]的递归式,并给出足够的初始值。
    2. 计算[math]\displaystyle{ t_n }[/math]的生成函数[math]\displaystyle{ T(x)=\sum_{n\ge 0}t_n x^n }[/math],给出生成函数[math]\displaystyle{ T(x) }[/math]的闭合形式。

注意:只需解生成函数的闭合形式,无需展开。

Problem 7

Let [math]\displaystyle{ a_n }[/math] be a sequence of numbers satisfying the recurrence relation:

[math]\displaystyle{ p a_n+q a_{n-1}+r a_{n-2}=0 }[/math]

with initial condition [math]\displaystyle{ a_0=s }[/math] and [math]\displaystyle{ a_1=t }[/math], where [math]\displaystyle{ p,q,r,s,t }[/math] are constants such that [math]\displaystyle{ {p}+q+r=0 }[/math], [math]\displaystyle{ p\neq 0 }[/math] and [math]\displaystyle{ s\neq t }[/math]. Solve the recurrence relation.