Data Modelling & AI Data Structure & Algorithm

Prime Triplet

21 June 2025

2

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/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/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/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

<?php
// PHP program to find prime 
// triplets smaller than or
// equal to n.
 
// function to print 
// prime triplets
function printPrimeTriplets($n)
{
    // Finding all primes
    // from 1 to n
    $prime = array_fill(0, $n + 1, true);
     
    // to detect prime number
    for ($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 ($i = $p * 2; $i <= $n; $i += $p)
                $prime[$i] = false;
        }
    }
     
    echo "The prime triplets from 1 to " . 
                          $n . " are :\n";
    for ($i = 2; $i <= $n-6; ++$i) 
    {
 
        // triplets of form (p, p+2, p+6)
        if ($prime[$i] && $prime[$i + 2] && 
                          $prime[$i + 6])
            echo $i. " " . ($i + 2) . 
                     " " . ($i + 6) . "\n";
 
        // triplets of form (p, p+4, p+6)
        else if ($prime[$i] && $prime[$i + 4] && 
                               $prime[$i + 6])
            echo $i. " " . ($i + 4) . 
                     " " . ($i + 6) . "\n";
    } 
}
 
// Driver Code
$n = 25;
printPrimeTriplets($n);
 
// This code is contributed by mits.
?>

Javascript

<script>
// Javascript program to find prime
// triplets smaller than or
// equal to n.
 
// function to print
// prime triplets
function printPrimeTriplets(n)
{
    // Finding all primes
    // from 1 to n
    let prime = new Array(n + 1).fill(true);
     
    // to detect prime number
    for (let 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 (let i = p * 2; i <= n; i += p)
                prime[i] = false;
        }
    }
     
    document.write("The prime triplets from 1 to " + n + " are :<br>");
    for (let i = 2; i <= n-6; ++i)
    {
 
        // triplets of form (p, p+2, p+6)
        if (prime[i] && prime[i + 2] &&
                        prime[i + 6])
            document.write(i + " " + (i + 2) +
                    " " + (i + 6) + "<br>");
 
        // triplets of form (p, p+4, p+6)
        else if (prime[i] && prime[i + 4] &&
                            prime[i + 6])
            document.write(i + " " + (i + 4) +
                    " " + (i + 6) + "<br>");
    }
}
 
// Driver Code
let n = 25;
printPrimeTriplets(n);
 
// This code is contributed by gfgking
</script>

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)

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Prime Triplet

C++

Java

Python3

C#

PHP

Javascript

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