Woodall Primes

Woodall素数是素数，也是Woodall数。

查找小于N的Woodall素数

给定数字N ，打印所有小于或等于N的Woodall素数。

例如：

Input: N = 10
Output: 7

Input: N = 500
Output: 7, 23, 383

方法：这个想法是使用Eratosthenes筛子来检查数字是质数还是效率不高。然后，对从1到N的整数进行迭代，并对每个数字检查是否为质数，是否为Woodall数。如果数字是素数也是Woodall数，则它是Woodall素数。

下面是上述算法的实现：

Python3

# Python3 implementation to print all Woodall  
# primes smaller than or equal to n.  
     
# Function to check if a number 
# N is Woodall   
def isWoodall(x) : 
        
    # If number is even, return false. 
    if (x % 2 == 0) : 
        return False
     
    # If x is 1, return true. 
    if (x == 1) : 
        return True
         
    x = x + 1  # Add 1 to make x even 
     
    # While x is divisible by 2 
    p = 0
    while (x % 2 == 0) : 
            
        # Divide x by 2 
        x = x / 2
     
        # Count the power 
        p = p + 1
     
        # If at any point power and  
        # x became equal, return true. 
        if (p == x) : 
            return True
            
    return False
        
# Function to generate all primes and checking  
# whether number is Woodall or not  
def printWoodallPrimesLessThanN(n): 
        
    # Create a boolean array "prime[0..n]" and  
    # initialize all entries it as true. A value  
    # in prime[i] will finally be false if i is  
    # Not a prime, else true.  
    prime = [True] * (n + 1);  
    p = 2; 
    while (p * p <= n): 
            
        # If prime[p] is not changed,  
        # then it is a prime  
        if (prime[p]):  
                
            # Update all multiples of p  
            for i in range(p * 2, n + 1, p):  
                prime[i] = False; 
        p += 1; 
            
    # Print all Woodall prime numbers  
    for p in range(2, n + 1):  
            
        # checking whether the given number  
        # is prime Woodall or not  
        if (prime[p] and isWoodall(p)):  
            print(p, end = " ");  
        
# Driver Code  
n = 1000; 
printWoodallPrimesLessThanN(n)

输出：

7 23 383