Pierpont Prime in C++

In this problem, we are given a number n. Our task is to print all Pierpont prime numbers less than n.

Pierpont Prime number is a special type of prime number that is of the form,

p= 2ⁱ . 3^k + 1.

Where p is a prime number, and i and k are some integers.

Let’s take an example to understand the problem,

Input − n = 50

Output − 2, 3, 5, 7, 13, 17, 19, 37

To solve this problem, we have to find all the prime numbers that follow the condition. For this, we will find a number with factors of powers of 2 and 3. And find all prime numbers. And print those numbers that are both, a prime number that follows the condition.

Example

Program to show an implementation of our solution,

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void printPierpontPrimes(int n){
   bool arr[n+1];
   memset(arr, false, sizeof arr);
   int two = 1, three = 1;
   while (two + 1 < n) {
      arr[two] = true;
      while (two * three + 1 < n) {
         arr[three] = true;
         arr[two * three] = true;
         three *= 3;
      }
      three = 1;
      two *= 2;
   }
   vector<int> primes;
   for (int i = 0; i < n; i++)
   if (arr[i])
      primes.push_back(i + 1);
   memset(arr, false, sizeof arr);
   for (int p = 2; p * p < n; p++) {
      if (arr[p] == false)
         for (int i = p * 2; i< n; i += p)
            arr[i] = true;
   }
   for (int i = 0; i < primes.size(); i++)
      if (!arr[primes[i]])
      cout<<primes[i]<<"\t";
}
int main(){
   int n = 50;
   cout<<"All Pierpont Prime Numbers less than "<<n<<" are :\n";
   printPierpontPrimes(n);
   return 0;
}

Output

All Pierpont Prime Numbers less than 50 are :
2    3    5    7    13    17    19    37