初等检验的Vantieghems定理

Vantieghems定理是数成为质数的必要和充分条件。它指出，对于自然数n为素数， $2^i - 1$在哪里$0 < i < n$ ，与$n~(mod~(2^n - 1))$ 。
换句话说，当且仅当数字n为质数。
${\displaystyle \prod _{1\leq i\leq n-1}\left(2^{i}-1\right)\equiv n\mod \left(2^{n}-1\right).}$
例子：

• 对于n = 3，最终乘积为(2 ¹ – 1)*(2 ² – 1)= 1 * 3 =3。3与3 mod 7一致。我们从表达式3 *(mod(2 ³ – 1))，因此3是质数。
• 对于n = 5，最终乘积为1 * 3 * 7 * 15 =315。315与5(mod 31)相等，因此5为质数。
• 对于n = 7，最终乘积为1 * 3 * 7 * 15 * 31 * 63 =615195。615195与7(mod 127)一致，因此7为质数。
• 对于n = 4，最终乘积1 * 3 * 7 =21。21不等于4(mod 15)，因此4是合成的。

高于定理的另一种陈述方式是$(2^n - 1)$分界$\prod _{1\leq i\leq n-1}\left(2^{i}-1\right) - n$ ，则n为质数。

C++

// C++ code to verify Vantieghem's Theorem
#include 
using namespace std;
 
void checkVantieghemsTheorem(int limit)
{
    long long unsigned prod = 1;
    for (long long unsigned n = 2; n < limit; n++) {
 
        // Check if above condition is satisfied
        if (((prod - n) % ((1LL << n) - 1)) == 0)
            cout << n << " is prime\n";
 
        // product of previous powers of 2
        prod *= ((1LL << n) - 1);
    }
}
 
// Driver code
int main()
{
    checkVantieghemsTheorem(10);
    return 0;
}

Java

// Java code to verify Vantieghem's Theorem
import java.util.*;
class GFG
{
 
static void checkVantieghemsTheorem(int limit)
{
    long prod = 1;
    for (long n = 2; n < limit; n++)
    {
 
        // Check if above condition is satisfied
        if (((prod - n < 0 ? 0 : prod - n) % ((1 << n) - 1)) == 0)
            System.out.print(n + " is prime\n");
 
        // product of previous powers of 2
        prod *= ((1 << n) - 1);
    }
}
 
// Driver code
public static void main(String []args)
{
    checkVantieghemsTheorem(10);
}
}
 
// This code is contributed by rutvik_56.

Python3

# Python3 code to verify Vantieghem's Theorem
def checkVantieghemsTheorem(limit):
     
    prod = 1
    for n in range(2, limit):
         
        # Check if above condition is satisfied
        if n == 2:
            print(2, "is prime")
        if (((prod - n) % ((1 << n) - 1)) == 0):
            print(n, "is prime")
             
        # Product of previous powers of 2
        prod *= ((1 << n) - 1)
     
# Driver code
checkVantieghemsTheorem(10)
 
# This code is contributed by shubhamsingh10

C#

// C# code to verify Vantieghem's Theorem
using System;
class GFG
{
  static void checkVantieghemsTheorem(int limit)
  {
    long prod = 1;
    for (long n = 2; n < limit; n++)
    {
 
      // Check if above condition is satisfied
      if (((prod - n < 0 ? 0 : prod - n) % ((1 << (int)n) - 1)) == 0)
        Console.Write(n + " is prime\n");
 
      // product of previous powers of 2
      prod *= ((1 << (int)n) - 1);
    }
  }
 
  // Driver code
  public static void Main()
  {
    checkVantieghemsTheorem(10);
  }
}
 
// This code is contributed by pratham76.

Javascript

 
// Javascript code to verify Vantieghem's Theorem
 
function checkVantieghemsTheorem( limit)
{
    let prod = 1;
    for (let n = 2; n < limit; n++) {
 
        // Check if above condition is satisfied
        if (n == 2)
            document.write(2 + " is prime" + "</br>");
        if (((prod - n) % ((1 << n) - 1)) == 0)
            document.write( n + " is prime" + "</br>");
 
        // product of previous powers of 2
        prod *= ((1 << n) - 1);
    }
}
 
// Driver Code
checkVantieghemsTheorem(10);
 
// This code is contributed by jana_sayantan.

输出：

2 is prime
3 is prime
5 is prime
7 is prime

上面的代码不适用于大于11的n值。它会导致prod评估中的溢出。

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