Prime Triplet (original) (raw)

Last Updated : 31 Aug, 2022

Prime Triplet is a set of three prime numbers of the form (p, p+2, p+6) or (p, p+4, p+6). This is the closest possible grouping of three prime numbers, since one of every three sequential odd numbers is a multiple of three, and hence not prime (except for 3 itself) except (2, 3, 5) and (3, 5, 7).

Examples :

Input : n = 15 Output : 5 7 11 7 11 13

Input : n = 25 Output : 5 7 11 7 11 13 11 13 17 13 17 19 17 19 23

A simple solution is to traverse through all numbers from 1 to n-6. For every number, i check if i, i+2, i+6, or i, i+4, i+6 are primes. If yes, print triplets.

An efficient solution is Sieve of Eratosthenes to first find all prime numbers so that we can quickly check if a number is prime or not.
Below is the implementation of the approach.

C++ `

// C++ program to find prime triplets smaller // than or equal to n. #include <bits/stdc++.h> using namespace std;

// function to detect prime number // here we have used sieve method // https://www.geeksforgeeks.org/dsa/sieve-of-eratosthenes/ // to detect prime number void sieve(int n, bool prime[]) { for (int p = 2; p * p <= n; p++) {

    // If prime[p] is not changed, then it is a prime
    if (prime[p] == true) {

        // Update all multiples of p
        for (int i = p * 2; i <= n; i += p)
            prime[i] = false;
    }
}

}

// function to print prime triplets void printPrimeTriplets(int n) { // Finding all primes from 1 to n bool prime[n + 1]; memset(prime, true, sizeof(prime)); sieve(n, prime);

cout << "The prime triplets from 1 to " 
      << n << "are :" << endl;
for (int i = 2; i <= n-6; ++i) {

    // triplets of form (p, p+2, p+6)
    if (prime[i] && prime[i + 2] && prime[i + 6])
        cout << i << " " << (i + 2) << " " << (i + 6) << endl;

    // triplets of form (p, p+4, p+6)
    else if (prime[i] && prime[i + 4] && prime[i + 6])
        cout << i << " " << (i + 4) << " " << (i + 6) << endl;
}

}

int main() { int n = 25; printPrimeTriplets(n); return 0; }

Java

// Java program to find prime triplets // smaller than or equal to n. import java.io.; import java.util.;

class GFG {

// function to detect prime number // here we have used sieve method // https://www.geeksforgeeks.org/dsa/sieve-of-eratosthenes/ // to detect prime number static void sieve(int n, boolean prime[]) { for (int p = 2; p * p <= n; p++) {

        // If prime[p] is not changed,
        //then it is a prime
        if (prime[p] == true) {

            // Update all multiples of p
            for (int i = p * 2; i <= n; i += p)
                prime[i] = false;
        }
    }
}

// function to print prime triplets
static void printPrimeTriplets(int n)
{
    // Finding all primes from 1 to n
    boolean prime[]=new boolean[n + 1];
    Arrays.fill(prime,true);
    sieve(n, prime);
    
    System.out.println("The prime triplets"+
                       " from 1 to " + n + "are :");
    
    for (int i = 2; i <= n-6; ++i) {

        // triplets of form (p, p+2, p+6)
        if (prime[i] && prime[i + 2] && prime[i + 6])
            System.out.println( i + " " + (i + 2) + 
                                " " + (i + 6));

        // triplets of form (p, p+4, p+6)
        else if (prime[i] && prime[i + 4] && 
                 prime[i + 6])
            
            System.out.println(i + " " + (i + 4) +
                               " " + (i + 6));
    }
}

public static void main(String args[])
{
    int n = 25;
    printPrimeTriplets(n);
}

}

/This code is contributed by Nikita Tiwari./

Python3

Python 3 program to find

prime triplets smaller

than or equal to n.

function to detect prime number

using sieve method

https://www.geeksforgeeks.org/dsa/sieve-of-eratosthenes/

to detect prime number

def sieve(n, prime) :

p = 2

while (p * p <= n ) :
    
    # If prime[p] is not changed
    # , then it is a prime
    if (prime[p] == True) :
        
        # Update all multiples of p
        i = p * 2
    
        while ( i <= n ) :
            prime[i] = False
            i = i + p
    
    p = p + 1
    

function to print

prime triplets

def printPrimeTriplets(n) :

# Finding all primes 
# from 1 to n
prime = [True] * (n + 1)
sieve(n, prime)

print( "The prime triplets from 1 to ",
                           n , "are :")

for i in range(2, n - 6 + 1) :
    
    # triplets of form (p, p+2, p+6)
    if (prime[i] and prime[i + 2] and
                        prime[i + 6]) :
        print( i , (i + 2) , (i + 6))
        
    # triplets of form (p, p+4, p+6)
    elif (prime[i] and prime[i + 4] and
                        prime[i + 6]) :
        print(i , (i + 4) , (i + 6))
        

Driver code

n = 25 printPrimeTriplets(n)

This code is contributed by Nikita Tiwari.

C#

// C# program to find prime // triplets smaller than or // equal to n. using System;

class GFG {

// function to detect // prime number static void sieve(int n, bool[] prime) { for (int p = 2; p * p <= n; p++) {

    // If prime[p] is not changed,
    // then it is a prime
    if (prime[p] == false) 
    {

        // Update all multiples of p
        for (int i = p * 2; 
                 i <= n; i += p)
            prime[i] = true;
    }
}

}

// function to print // prime triplets static void printPrimeTriplets(int n) { // Finding all primes // from 1 to n bool[] prime = new bool[n + 1]; sieve(n, prime);

Console.WriteLine("The prime triplets " + 
                           "from 1 to " + 
                           n + " are :");

for (int i = 2; i <= n - 6; ++i) 
{

    // triplets of form (p, p+2, p+6)
    if (!prime[i] && 
        !prime[i + 2] && 
        !prime[i + 6])
        Console.WriteLine(i + " " + (i + 2) + 
                              " " + (i + 6));

    // triplets of form (p, p+4, p+6)
    else if (!prime[i] && 
             !prime[i + 4] && 
             !prime[i + 6])
        Console.WriteLine(i + " " + (i + 4) + 
                              " " + (i + 6));
}

}

// Driver Code public static void Main() { int n = 25; printPrimeTriplets(n); } }

// This code is contributed by mits

PHP

JavaScript

`

Output :

The prime triplets from 1 to 25 are : 5 7 11 7 11 13 11 13 17 13 17 19 17 19 23

Time Complexity: O(n*log(log(n)))

Auxiliary Space: O(n)