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