具体数学第二版第四章习题（5)

61 假设$frac{hat{m}}{hat{n}}$是$frac{m^{'}}{n^{'}}$，现在证明$frac{hat{m}}{hat{n}}=frac{m^{''}}{n^{''}}$

$hat{m}perp hat{n},frac{hat{m}}{hat{n}}>frac{m^{'}}{n^{'}},N=frac{n+N}{n^{'}}n^{'}-ngeq hat{n}>(frac{n+N}{n^{'}}-1)n^{'}-n=N-n^{'}>0 ightarrow frac{hat{m}}{hat{n}}geq frac{m^{''}}{n^{''}}$(这里为什么可以推出来这个结论)。

然后如果等于号不满足，那么$n^{''}=(hat{m}n^{'}-m^{'}hat{n})n^{''}=n^{'}(hat{m}n^{''}-m^{''}hat{n})+hat{n}(m^{''}n^{'}-m^{'}n^{''})geq n^{'}+hat{n}>N$出现矛盾

62 $e=2^{-1}+(2^{-2}+2^{-3}-2^{-6}-2^{-7})+(2^{-12}+2^{-13}-2^{-20}-2^{-21})+..=2^{-1}+sum_{kgeq 0}(2^{-(2k^{2}+6k+3)}-2^{-(4k^{2}+10k+7)})$

63(1) 首先，假设$n=4$时，不存在$a,b,c$满足$a^{4}+b^{4} eq c^{4}$

如果$n>4$且$n$不是素数使得$a^{n}+b^{n}=c^{d}$，由于存在一个因子$d$使得$n=kd$，那么$a^{n}+b^{n}=c^{n} ightarrow  (a^{frac{n}{d}})^{d}+(b^{frac{n}{d}})^{d}=(c^{frac{n}{d}})^{d}$。所以最小的一定是素数。

(2) 设$a+b=x ightarrow left (frac{c^{p}}{x}=frac{a^{p}+b^{p}}{x}=frac{a^{p}+(x-a)^{p}}{x}  ight )equiv pa^{p-1}(mod(x))$.。另外最小的反例满足$aperp x$.

如果$p$不能整除$x$,那么$Gcd(x,frac{c^{p}}{x})=Gcd(x,pa^{p-1})=1$，所以一定存在$m$使得$x=m^{p}$

如果$p$可以整除$x$，那么$frac{c^{p}}{x}$能被$p$整除，但是不能被$p^{2}$整除，所以有$x=p^{p-1}m^{p}$

64 生成数列的代码

#include <iostream>
#include <sstream>
#include <string>
#include <vector>

// The max number of sign of Pn
constexpr int kMaxSignNumber = 300;

// The max N to generate
constexpr int kMaxN = 20;

class Node {
 public:
  void Add(const std::string &s) {
    if (Size() + 1 <= kMaxSignNumber) {
      ops.emplace_back(s);
    }
  }

  void Add(int x, int y) { Add(std::to_string(x) + "/" + std::to_string(y)); }

  std::string Get(int x) {
    --x;
    if (0 <= x && x < Size()) {
      return ops[x];
    }
    return "";
  }

  std::string ToString() const {
    std::stringstream ss;
    for (size_t i = 0; i < ops.size(); ++i) {
      if (i % 2 == 0) {
        ss << Transform(ops[i]);
      } else {
        ss << ops[i];
      }
    }
    return ss.str();
  }

 private:
  std::string Transform(const std::string &num) const {
    auto p = num.find_first_of('/');
    return "\frac{" + num.substr(0, p) + "}{" + num.substr(p + 1) + "}";
  }

  int Size() const { return static_cast<int>(ops.size()); }

  std::vector<std::string> ops;
};

Node a[kMaxN + 1];

int main() {
  for (int i = 0; i < kMaxSignNumber / 2; ++i) {
    a[1].Add(i, 1);
    a[1].Add("<");
  }
  for (int N = 1; N < kMaxN; ++N) {
    for (int k = 1; k <= kMaxSignNumber; ++k) {
      for (int j = (k - 1) * N + 1; j < k * N; ++j) {
        a[N + 1].Add(a[N].Get(j));
      }
      if (k * N % 2 == 1) {
        a[N + 1].Add(k - 1, N + 1);
        a[N + 1].Add("=");
      } else {
        a[N + 1].Add(a[N].Get(k * N));
        a[N + 1].Add(k - 1, N + 1);
      }
      a[N + 1].Add(a[N].Get(k * N));
    }
  }
  for (int i = 1; i <= kMaxN; ++i) {
    std::cout << i << " : " << a[i].ToString() << "
";
  }
  return 0;
}

1 : $frac{0}{1}<frac{1}{1}<frac{2}{1}<frac{3}{1}<frac{4}{1}<frac{5}{1}<frac{6}{1}<frac{7}{1}<frac{8}{1}<frac{9}{1}<frac{10}{1}<frac{11}{1}<frac{12}{1}<frac{13}{1}<frac{14}{1}<frac{15}{1}<frac{16}{1}<frac{17}{1}<frac{18}{1}<frac{19}{1}<frac{20}{1}<frac{21}{1}<frac{22}{1}<frac{23}{1}<frac{24}{1}<frac{25}{1}<frac{26}{1}<frac{27}{1}<frac{28}{1}<frac{29}{1}<frac{30}{1}<frac{31}{1}<frac{32}{1}<frac{33}{1}<frac{34}{1}<frac{35}{1}<frac{36}{1}<frac{37}{1}<frac{38}{1}<frac{39}{1}$
2 : $frac{0}{2}=frac{0}{1}<frac{1}{2}<frac{2}{2}=frac{1}{1}<frac{3}{2}<frac{4}{2}=frac{2}{1}<frac{5}{2}<frac{6}{2}=frac{3}{1}<frac{7}{2}<frac{8}{2}=frac{4}{1}<frac{9}{2}<frac{10}{2}=frac{5}{1}<frac{11}{2}<frac{12}{2}=frac{6}{1}<frac{13}{2}<frac{14}{2}=frac{7}{1}<frac{15}{2}<frac{16}{2}=frac{8}{1}<frac{17}{2}<frac{18}{2}=frac{9}{1}<frac{19}{2}<frac{20}{2}=frac{10}{1}<frac{21}{2}<frac{22}{2}=frac{11}{1}<frac{23}{2}<frac{24}{2}=frac{12}{1}<frac{25}{2}<frac{26}{2}$
3 : $frac{0}{2}=frac{0}{3}=frac{0}{1}<frac{1}{3}<frac{1}{2}<frac{2}{3}<frac{2}{2}=frac{3}{3}=frac{1}{1}<frac{4}{3}<frac{3}{2}<frac{5}{3}<frac{4}{2}=frac{6}{3}=frac{2}{1}<frac{7}{3}<frac{5}{2}<frac{8}{3}<frac{6}{2}=frac{9}{3}=frac{3}{1}<frac{10}{3}<frac{7}{2}<frac{11}{3}<frac{8}{2}=frac{12}{3}=frac{4}{1}<frac{13}{3}<frac{9}{2}<frac{14}{3}<frac{10}{2}=frac{15}{3}=frac{5}{1}<frac{16}{3}<frac{11}{2}<frac{17}{3}<frac{12}{2}=frac{18}{3}=frac{6}{1}<frac{19}{3}$
4 : $frac{0}{2}=frac{0}{4}=frac{0}{3}=frac{0}{1}<frac{1}{4}<frac{1}{3}<frac{2}{4}=frac{1}{2}<frac{2}{3}<frac{3}{4}<frac{2}{2}=frac{4}{4}=frac{3}{3}=frac{1}{1}<frac{5}{4}<frac{4}{3}<frac{6}{4}=frac{3}{2}<frac{5}{3}<frac{7}{4}<frac{4}{2}=frac{8}{4}=frac{6}{3}=frac{2}{1}<frac{9}{4}<frac{7}{3}<frac{10}{4}=frac{5}{2}<frac{8}{3}<frac{11}{4}<frac{6}{2}=frac{12}{4}=frac{9}{3}=frac{3}{1}<frac{13}{4}<frac{10}{3}<frac{14}{4}=frac{7}{2}<frac{11}{3}<frac{15}{4}$
5 : $frac{0}{2}=frac{0}{4}=frac{0}{5}=frac{0}{3}=frac{0}{1}<frac{1}{5}<frac{1}{4}<frac{1}{3}<frac{2}{5}<frac{2}{4}=frac{1}{2}<frac{3}{5}<frac{2}{3}<frac{3}{4}<frac{4}{5}<frac{2}{2}=frac{4}{4}=frac{5}{5}=frac{3}{3}=frac{1}{1}<frac{6}{5}<frac{5}{4}<frac{4}{3}<frac{7}{5}<frac{6}{4}=frac{3}{2}<frac{8}{5}<frac{5}{3}<frac{7}{4}<frac{9}{5}<frac{4}{2}=frac{8}{4}=frac{10}{5}=frac{6}{3}=frac{2}{1}<frac{11}{5}<frac{9}{4}<frac{7}{3}<frac{12}{5}<frac{10}{4}$
6 : $frac{0}{2}=frac{0}{4}=frac{0}{6}=frac{0}{5}=frac{0}{3}=frac{0}{1}<frac{1}{6}<frac{1}{5}<frac{1}{4}<frac{2}{6}=frac{1}{3}<frac{2}{5}<frac{2}{4}=frac{3}{6}=frac{1}{2}<frac{3}{5}<frac{4}{6}=frac{2}{3}<frac{3}{4}<frac{4}{5}<frac{5}{6}<frac{2}{2}=frac{4}{4}=frac{6}{6}=frac{5}{5}=frac{3}{3}=frac{1}{1}<frac{7}{6}<frac{6}{5}<frac{5}{4}<frac{8}{6}=frac{4}{3}<frac{7}{5}<frac{6}{4}=frac{9}{6}=frac{3}{2}<frac{8}{5}<frac{10}{6}=frac{5}{3}<frac{7}{4}$
7 : $frac{0}{2}=frac{0}{4}=frac{0}{6}=frac{0}{7}=frac{0}{5}=frac{0}{3}=frac{0}{1}<frac{1}{7}<frac{1}{6}<frac{1}{5}<frac{1}{4}<frac{2}{7}<frac{2}{6}=frac{1}{3}<frac{2}{5}<frac{3}{7}<frac{2}{4}=frac{3}{6}=frac{1}{2}<frac{4}{7}<frac{3}{5}<frac{4}{6}=frac{2}{3}<frac{5}{7}<frac{3}{4}<frac{4}{5}<frac{5}{6}<frac{6}{7}<frac{2}{2}=frac{4}{4}=frac{6}{6}=frac{7}{7}=frac{5}{5}=frac{3}{3}=frac{1}{1}<frac{8}{7}<frac{7}{6}<frac{6}{5}<frac{5}{4}<frac{9}{7}$
8 : $frac{0}{2}=frac{0}{4}=frac{0}{6}=frac{0}{8}=frac{0}{7}=frac{0}{5}=frac{0}{3}=frac{0}{1}<frac{1}{8}<frac{1}{7}<frac{1}{6}<frac{1}{5}<frac{2}{8}=frac{1}{4}<frac{2}{7}<frac{2}{6}=frac{1}{3}<frac{3}{8}<frac{2}{5}<frac{3}{7}<frac{2}{4}=frac{4}{8}=frac{3}{6}=frac{1}{2}<frac{4}{7}<frac{3}{5}<frac{5}{8}<frac{4}{6}=frac{2}{3}<frac{5}{7}<frac{6}{8}=frac{3}{4}<frac{4}{5}<frac{5}{6}<frac{6}{7}<frac{7}{8}<frac{2}{2}=frac{4}{4}=frac{6}{6}=frac{8}{8}$
9 : $frac{0}{2}=frac{0}{4}=frac{0}{6}=frac{0}{8}=frac{0}{9}=frac{0}{7}=frac{0}{5}=frac{0}{3}=frac{0}{1}<frac{1}{9}<frac{1}{8}<frac{1}{7}<frac{1}{6}<frac{1}{5}<frac{2}{9}<frac{2}{8}=frac{1}{4}<frac{2}{7}<frac{2}{6}=frac{3}{9}=frac{1}{3}<frac{3}{8}<frac{2}{5}<frac{3}{7}<frac{4}{9}<frac{2}{4}=frac{4}{8}=frac{3}{6}=frac{1}{2}<frac{5}{9}<frac{4}{7}<frac{3}{5}<frac{5}{8}<frac{4}{6}=frac{6}{9}=frac{2}{3}<frac{5}{7}<frac{6}{8}=frac{3}{4}<frac{7}{9}$

10 :$ frac{0}{2}=frac{0}{4}=frac{0}{6}=frac{0}{8}=frac{0}{10}=frac{0}{9}=frac{0}{7}=frac{0}{5}=frac{0}{3}=frac{0}{1}<frac{1}{10}<frac{1}{9}<frac{1}{8}<frac{1}{7}<frac{1}{6}<frac{2}{10}=frac{1}{5}<frac{2}{9}<frac{2}{8}=frac{1}{4}<frac{2}{7}<frac{3}{10}<frac{2}{6}=frac{3}{9}=frac{1}{3}<frac{3}{8}<frac{4}{10}=frac{2}{5}<frac{3}{7}<frac{4}{9}<frac{2}{4}=frac{4}{8}=frac{5}{10}=frac{3}{6}=frac{1}{2}<frac{5}{9}<frac{4}{7}<frac{6}{10}=frac{3}{5}<frac{5}{8}$
11 : $frac{0}{2}=frac{0}{4}=frac{0}{6}=frac{0}{8}=frac{0}{10}=frac{0}{11}=frac{0}{9}=frac{0}{7}=frac{0}{5}=frac{0}{3}=frac{0}{1}<frac{1}{11}<frac{1}{10}<frac{1}{9}<frac{1}{8}<frac{1}{7}<frac{1}{6}<frac{2}{11}<frac{2}{10}=frac{1}{5}<frac{2}{9}<frac{2}{8}=frac{1}{4}<frac{3}{11}<frac{2}{7}<frac{3}{10}<frac{2}{6}=frac{3}{9}=frac{1}{3}<frac{4}{11}<frac{3}{8}<frac{4}{10}=frac{2}{5}<frac{3}{7}<frac{4}{9}<frac{5}{11}<frac{2}{4}=frac{4}{8}=frac{5}{10}=frac{3}{6}=frac{1}{2}<frac{6}{11}<frac{5}{9}<frac{4}{7}<frac{6}{10}=frac{3}{5}<frac{5}{8}<frac{7}{11}<frac{4}{6}=frac{6}{9}=frac{2}{3}<frac{7}{10}<frac{5}{7}<frac{8}{11}<frac{6}{8}=frac{3}{4}<frac{7}{9}<frac{8}{10}=frac{4}{5}<frac{9}{11}<frac{5}{6}<frac{6}{7}<frac{7}{8}<frac{8}{9}<frac{9}{10}<frac{10}{11}<frac{2}{2}=frac{4}{4}=frac{6}{6}=frac{8}{8}=frac{10}{10}=frac{11}{11}=frac{9}{9}=frac{7}{7}=frac{5}{5}$
12 : $frac{0}{2}=frac{0}{4}=frac{0}{6}=frac{0}{8}=frac{0}{10}=frac{0}{12}=frac{0}{11}=frac{0}{9}=frac{0}{7}=frac{0}{5}=frac{0}{3}=frac{0}{1}<frac{1}{12}<frac{1}{11}<frac{1}{10}<frac{1}{9}<frac{1}{8}<frac{1}{7}<frac{2}{12}=frac{1}{6}<frac{2}{11}<frac{2}{10}=frac{1}{5}<frac{2}{9}<frac{2}{8}=frac{3}{12}=frac{1}{4}<frac{3}{11}<frac{2}{7}<frac{3}{10}<frac{2}{6}=frac{4}{12}=frac{3}{9}=frac{1}{3}<frac{4}{11}<frac{3}{8}<frac{4}{10}=frac{2}{5}<frac{5}{12}<frac{3}{7}$

首先可以发现，相等的数字出现的规律为$frac{2m}{2n},frac{4m}{4n},frac{6m}{6n},..,frac{rm}{rn},...frac{5m}{5n},frac{3m}{3n},frac{m}{n}$

假设$P_{N}$是正确的，来证明$P_{N+1}$是正确的。证明两部分：

（1）如果$kN$是奇数，那么$frac{k-1}{N+1}=P_{N,kN}$

（2）如果$kN$是偶数，那么$P_{N,kN-1}P_{N,kN}frac{k-1}{N+1}P_{N,kN}P_{N,kN+1}$，其中$P_{N,kN}$为比较符号。

首先计算$P_{N}$中小于$frac{k-1}{N+1}$的数字个数。

$sum_{n=1}^{N}sum_{m}[0leq frac{m}{n}<frac{k-1}{N+1}]=sum_{n=1}^{N}left lceil frac{(k-1)n}{N+1} ight ceil=sum_{n=1}^{N}left lfloor frac{(k-1)n+N}{N+1} ight floor=frac{(k-2)N}{2}+frac{d-1}{2}+dleft lfloor frac{N}{d} ight floor$

最后一步由第三章的公式3.32得到。其中$d=Gcd(k-1,n+1) ightarrow Nequiv d-1(mod(d)) ightarrow dleft lfloor frac{N}{d} ight floor=N-(d-1)$，所以总的个数为$frac{kN-d+1}{2}$

 其中，$P_{N}$中与$frac{k-1}{N+1}$相等且在$P_{N+1}$中在$frac{k-1}{N+1}$之前的个数为$frac{1}{2}(d-1-[d=2q])$

(1)如果$kN$是奇数，那么$d$为偶数，那么$frac{k-1}{N+1}$在$P_{N}$中位于$frac{kN-d+1}{2}+frac{1}{2}(d-1-1)=frac{kN-1}{2}$个数字之后，所以$frac{k-1}{N+1}$前面有$kN-1$个字符，所以$frac{k-1}{N+1}=P_{N,kN}$

(2)如果$kN$是偶数，则$d$为奇数，则$frac{k-1}{N+1}$在$P_{N}$中位于$frac{kN}{2}$个数字之后。如果$d=1$那么$P_{N}$中没有数字等于$frac{k-1}{N+1}$,此时$P_{N,kN}$是小于号。如果$d$为偶数，那么$frac{k-1}{N+1}$在两个相等的元素中间，此时$P_{N,kN}$为等于号。

65 可能是

66 应该是

67 总的趋势是如果第一个数和第$n$个数得到的$f(1,n)=frac{a_{n}}{Gcd(a_{1},a_{n})}$不是最大的话，说明他们中间的位置很少，那么中间的某两个元素$i,j$得到的$Gcd(a_{i},a_{j})$会很小，从而使得$f(i,j)geq n$

68  可能存在

69 应该成立

70 由这里第24题的结论有，等式成立当且仅当$upsilon _{3}(n)=upsilon _{2}(n)$.$upsilon _{p}(n)$表示$n$的$p$进制中各位数字之和。1和6满足，$upsilon _{3}(1)=upsilon _{2}(1)=1,upsilon _{3}(6)=upsilon _{2}(6)=2$.所以应该有很多

71 看起来并不多

72 $a=-1$时所有素数都满足。任意的$a$应该都有无穷个$n$满足

73 应该是

74 大约是$p(1-e^{-1})$