Euclid–Mullin序列

给定整数N ，任务是打印Euclid-Mullin序列的前N个元素。
Euclid-Mullin序列是素数序列，其中每个元素是一个的最小素数因子加上所有较早元素的乘积。
该序列以古希腊数学家Euclid命名。

例子：

Input: N = 14
Output: 2 3 7 43 13 53 5 6221671 38709183810571 139 2801 11 17 5471

方法： Euclid-Mullin序列是质数序列，其中序列的第n个数是：
$a(n) = Least prime factor of ( 1 + \sum_{x=1}^{n-1} a(x)) \\$
因此，我们将运行一个从1到N的循环，并采用一个初始为1的变量乘积，其中将包含所有先前元素的乘积。然后，我们将在O(sqrt(n))时间中找到(1 +乘积)的最小素因数，并打印数字。请注意，由于乘积太大并且找到最小的素因数需要花费大量时间，因此代码无法在第14个元素后打印数字。

下面是上述方法的实现：

Java

// Java implementation of the approach
import java.math.BigInteger;
class GFG {
  
    // Function to return the smallest prime factor of n
    static BigInteger smallestPrimeFactor(BigInteger n)
    {
  
        // Initialize i = 2
        BigInteger i = BigInteger.valueOf(2);
  
        // While i <= sqrt(n)
        while ((i.multiply(i)).compareTo(n) <= 0) {
  
            // If n is divisible by i
            if (n.mod(i).compareTo(BigInteger.ZERO) == 0)
                return i;
  
            // Increment i
            i = i.add(BigInteger.ONE);
        }
        return n;
    }
  
    // Function to print the first n
    // terms of the required sequence
    static void solve(BigInteger n)
    {
        // To store the product of the previous terms
        BigInteger product = BigInteger.ONE;
  
        // Traverse the prime numbers
        BigInteger i = BigInteger.ZERO;
        while (i.compareTo(n) < 0) {
  
            // Current term will be smallest prime
            // factor of (1 + product of all previous terms)
            BigInteger num = smallestPrimeFactor(product.add(BigInteger.ONE));
  
            // Print the current term
            System.out.print(num + " ");
  
            // Update the product
            product = product.multiply(num);
            i = i.add(BigInteger.ONE);
        }
    }
  
    // Driver code
    public static void main(String[] args)
    {
  
        // Find the first 14 terms of the sequence
        BigInteger b = BigInteger.valueOf(14);
        solve(b);
    }
}

Python3

# Python3 implementation of the approach 
  
# Function to return the smallest prime factor of n 
def smallestPrimeFactor(n): 
  
    # Initialize i = 2 
    i = 2
  
    # While i <= sqrt(n) 
    while (i * i) <= n : 
  
        # If n is divisible by i 
        if n % i == 0: 
            return i 
  
        # Increment i 
        i += 1
    return n 
  
# Function to print the first n 
# terms of the required sequence 
def solve(n): 
  
    # To store the product of the previous terms 
    product = 1
  
    # Traverse the prime numbers 
    i = 0
    while i < n: 
  
        # Current term will be smallest prime 
        # factor of (1 + product of all previous terms)
        num = smallestPrimeFactor(product + 1) 
  
        # Print the current term 
        print(num, end=' ') 
  
        # Update the product 
        product = product * num 
        i += 1
  
# Driver code 
# Find the first 14 terms of the sequence 
b = 14
solve(b)
  
# This code is contributed by divyamohan123

输出：

2 3 7 43 13 53 5 6221671 38709183810571 139 2801 11 17 5471