罗瑟定理

在数学中，Rosser定理指出，对于所有大于1的n，第n个素数大于n与n的自然对数的乘积。
数学上
对于n> = 1，如果p _n是第n个素数，则
p _n > n *(ln n)

Illustrative Examples:
      
       For n = 1, nth prime number = 2
    2 > 1 * ln(1)
Explanations, the above relation inherently holds true and it verifies the 
statement of Rosser's Theroem clearly.
Some more examples are, 
       For n = 2, nth prime number = 3
    3 > 2 * ln(2)
       For n = 3, nth prime number = 5
    5 > 3 * ln(3)
       For n = 4, nth prime number = 7
    7 > 4 * ln(4)
       For n = 5, nth prime number = 11
    11 > 5 * ln(5)
       For n = 6, nth prime number = 13
    13 > 6 * ln(6)

代码方法：
使用质数筛有效生成质数，并分别检查每个质数的Rosser定理的条件。

C++

// CPP code to verify Rosser's Theorem
#include 
using namespace std;
  
#define show(x) cout << #x << " = " << x << "\n";
  
vector prime;
  
// Sieve of Eratosthenes
void sieve()
{
    // prime sieve to generate prime numbers efficiently
    int n = 1e5;
    vector isprime(n + 2, true);
    isprime[0] = isprime[1] = false;
    for (long long i = 2; i <= n; i++) {
        if (isprime[i]) {
            for (long long j = i * i; j <= n; j += i)
                isprime[j] = false;
        }
    }
    // store primes in prime[] vector
    for (int i = 0; i <= n; i++)
        if (isprime[i])
            prime.push_back(i);
}
  
// Verifies ROSSER'S THEOREM for all numbers 
// smaller than n.
void verifyRosser(int n)
{
    cout << "ROSSER'S THEOREM: nth prime "
            "number > n * (ln n)\n";
    for (int i = 0; i < n; i++)
        if (prime[i] > (i + 1) * log(i + 1)) {
            cout << "For n = " << i + 1
                 << ", nth prime number = "
                 << prime[i] << "\n\t"
                 << prime[i] << " > " << i + 1 
                 << " * ln(" << i + 1 << ")\n";
        }
}
  
  
int main()
{
    sieve();
    verifyRosser(20);
    return 0;
}

Java

// Java code to verify Rosser's Theorem
  
import java.util.*;
class GFG
{
    static Vector prime=new Vector(); 
    // Sieve of Eratosthenes
    static void sieve()
    {
        // prime sieve to generate prime numbers efficiently
        int n = 10000;
        boolean []isprime=new boolean[n+2];
        for(int i=0;i n * (ln n)");
                  
        for (int i = 0; i < n; i++)
            if (prime.get(i) > (i + 1) * Math.log(i + 1)) {
                System.out.println( "For n = " + (i+1) 
                    + ", nth prime number = "
                    + prime.get(i) + "\n\t"
                    + prime.get(i) + " > " + (i + 1)
                    + " * ln(" + (i + 1) + ")");
            }
    }
      
    // Driver code
    public static void main(String [] args)
    {
        sieve();
        verifyRosser(20);
          
    }
}
  
// This code is contributed by ihritik

Python3

# Python3 code to verify Rosser's Theorem
import math
prime = [];
  
# Sieve of Eratosthenes
def sieve():
      
    # prime sieve to generate 
    # prime numbers efficiently
    n = 100001;
    isprime = [True] * (n + 2);
    isprime[0] = False;
    isprime[1] = False;
    for i in range(2, n + 1):
        if(isprime[i]):
            j = i * i;
            while (j <= n):
                isprime[j] = False;
                j += i;
      
    # store primes in 
    # prime[] vector
    for i in range(n + 1):
        if (isprime[i]):
            prime.append(i);
  
# Verifies ROSSER'S THEOREM 
# for all numbers smaller than n.
def verifyRosser(n):
      
    print("ROSSER'S THEOREM: nth", 
          "prime number > n * (ln n)");
    for i in range(n):
        if (prime[i] > int((i + 1) * math.log(i + 1))):
            print("For n =", (i + 1), ", nth prime number =", 
                   prime[i], "\n\t", prime[i], " >", (i + 1), 
                                      "* ln(", (i + 1), ")");
  
# Driver Code
sieve();
verifyRosser(20);
  
# This code is contributed 
# by mits

C#

// C# code to verify Rosser's Theorem
using System;
using System.Collections.Generic;
      
class GFG
{
    static List prime = new List(); 
      
    // Sieve of Eratosthenes
    static void sieve()
    {
        // prime sieve to generate 
        // prime numbers efficiently
        int n = 10000;
        bool []isprime = new bool[n + 2];
        for(int i = 0; i < n; i++)
        isprime[i] = true;
          
        isprime[0] = false;
        isprime[1] = false;
        for (int i = 2; i <= n; i++) 
        {
            if (isprime[i])
            {
                for (int j = i * i; 
                         j <= n; j += i)
                    isprime[j] = false;
            }
        }
          
        // store primes in prime[] vector
        for (int i = 0; i <= n; i++)
            if (isprime[i])
                prime.Add(i);
    }
      
    // Verifies ROSSER'S THEOREM for 
    // all numbers smaller than n.
    static void verifyRosser(int n)
    {
        Console.WriteLine("ROSSER'S THEOREM: " +    
                          "nth prime number > n * (ln n)");
                  
        for (int i = 0; i < n; i++)
            if (prime[i] > (i + 1) * Math.Log(i + 1)) 
            {
                Console.WriteLine( "For n = " + (i + 1) + 
                                ", nth prime number = " + 
                           prime[i] + "\n\t" + prime[i] + 
                             " > " + (i + 1) + " * ln(" + 
                                          (i + 1) + ")");
            }
    }
      
    // Driver code
    public static void Main(String [] args)
    {
        sieve();
        verifyRosser(20);
    }
}
  
// This code is contributed by PrinciRaj1992

PHP

 n * (ln n)\n";
    for ($i = 0; $i < $n; $i++)
        if ($prime[$i] > (int)(($i + 1) * log($i + 1)))
        {
            echo "For n = " . ($i + 1).
                 ", nth prime number = " . 
                  $prime[$i] . "\n\t" . 
                  $prime[$i] . " > " . 
                  ($i + 1) . " * ln(" . 
                  ($i + 1) . ")\n";
        }
}
  
// Driver Code
sieve();
verifyRosser(20);
  
// This code is contributed 
// by mits
?>

输出：

ROSSER'S THEOREM: nth prime number > n * (ln n)
For n = 1, nth prime number = 2
    2 > 1 * ln(1)
For n = 2, nth prime number = 3
    3 > 2 * ln(2)
For n = 3, nth prime number = 5
    5 > 3 * ln(3)
For n = 4, nth prime number = 7
    7 > 4 * ln(4)
For n = 5, nth prime number = 11
    11 > 5 * ln(5)
For n = 6, nth prime number = 13
    13 > 6 * ln(6)
For n = 7, nth prime number = 17
    17 > 7 * ln(7)
For n = 8, nth prime number = 19
    19 > 8 * ln(8)
For n = 9, nth prime number = 23
    23 > 9 * ln(9)
For n = 10, nth prime number = 29
    29 > 10 * ln(10)
For n = 11, nth prime number = 31
    31 > 11 * ln(11)
For n = 12, nth prime number = 37
    37 > 12 * ln(12)
For n = 13, nth prime number = 41
    41 > 13 * ln(13)
For n = 14, nth prime number = 43
    43 > 14 * ln(14)
For n = 15, nth prime number = 47
    47 > 15 * ln(15)
For n = 16, nth prime number = 53
    53 > 16 * ln(16)
For n = 17, nth prime number = 59
    59 > 17 * ln(17)
For n = 18, nth prime number = 61
    61 > 18 * ln(18)
For n = 19, nth prime number = 67
    67 > 19 * ln(19)
For n = 20, nth prime number = 71
    71 > 20 * ln(20)