Count distinct prime triplets up to N such that sum of two primes is equal to the third prime

Given an integer N, the task is to count the number of distinct prime triplets (a, b, c) from the range [1, N] such that a < b < c ≤ N and a + b = c.
Note: Two prime tuples are distinct if at least one of the primes present in them are different.

Examples:

Input: N = 6
Output: 1
Explanation: Among numbers in the range [1, 6], the only prime triplet is (2, 3, 5) (Since 2 + 3 = 5).

Input: N = 10
Output: 2
Explanation: The distinct prime triplets satisfying the condition are (2, 3, 5), (2, 5, 7).

Approach: The problem can be solved based on the observation stated below:

Observation:

For every prime number p from 1 to N, it is a part of a triplet if and only if it can be represented as a sum of two prime numbers. 
Since a prime number is an odd number, it must be equal to the sum of an even number and an odd number. 
Hence the only even prime is 2. Therefore, for a prime number p to constitute a unique tuple (2, p-2, p), the number p – 2 must be a prime number.

Follow the steps below to solve the problem:

• Initialize a variable, say count = 0, to store the number of prime triplets.
• Iterate from 1 to N and for each number p, check if this number p and p – 2 are prime or not.
• If they are prime, then increment count by 1.
• Print the value of count.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if a
// number is a prime or not
bool isPrime(int N)
{
    if (N <= 1)
        return false;

    for (int i = 2; i <= sqrt(N); i++) {
        if (N % i == 0)
            return false;
    }

    return true;
}

// Function to count the number
// of valid prime triplets
void countPrimeTuples(int N)
{
    // Stores the count
    // of prime triplets
    int count = 0;

    // Iterate from 2 to N and check for each
    // p, whether p & (p - 2) are prime or not
    for (int i = 2; i <= N; i++) {

        if (isPrime(i) && isPrime(i - 2))
            count++;
    }

    // Print the count obtained
    cout << count;
}

// Driver Code
int main()
{
    int N = 6;
    countPrimeTuples(N);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG
{

  // Function to check if a
  // number is a prime or not
  static boolean isPrime(int N)
  {
      if (N <= 1)
          return false;
      for (int i = 2; i <= Math.sqrt(N); i++)
      {
          if (N % i == 0)
              return false;
      }
      return true;
  }

  // Function to count the number
  // of valid prime triplets
  static void countPrimeTuples(int N)
  {

      // Stores the count
      // of prime triplets
      int count = 0;

      // Iterate from 2 to N and check for each
      // p, whether p & (p - 2) are prime or not
      for (int i = 2; i <= N; i++)
      {
          if (isPrime(i) && isPrime(i - 2))
              count++;
      }

      // Print the count obtained
      System.out.println(count);
  }

  // Driver Code
  public static void main (String[] args)
  {
    int N = 6;
    countPrimeTuples(N);
  }
}

// This code is contributed by Dharanendra L V.

Python3

# Python3 program for the above approach
import math

# Function to check if a
# number is a prime or not
def isPrime(N) :
    if (N <= 1) :
        return False
    for i in range(2, int(math.sqrt(N) + 1)):
        if (N % i == 0) :
            return False
    return True

# Function to count the number
# of valid prime triplets
def countPrimeTuples(N) :

    # Stores the count
    # of prime triplets
    count = 0

    # Iterate from 2 to N and check for each
    # p, whether p & (p - 2) are prime or not
    for i in range(2, N + 1):

        if (isPrime(i) and isPrime(i - 2)) :
            count += 1

    # Print count obtained
    print(count)

# Driver Code
N = 6
countPrimeTuples(N)

# This code is contributed by susmitakundugoaldanga.

C#

// C# program for the above approach
using System;
public class GFG
{

  // Function to check if a
  // number is a prime or not
  static bool isPrime(int N)
  {
      if (N <= 1)
          return false;
      for (int i = 2; i <= Math.Sqrt(N); i++)
      {
          if (N % i == 0)
              return false;
      }
      return true;
  }

  // Function to count the number
  // of valid prime triplets
  static void countPrimeTuples(int N)
  {

      // Stores the count
      // of prime triplets
      int count = 0;

      // Iterate from 2 to N and check for each
      // p, whether p & (p - 2) are prime or not
      for (int i = 2; i <= N; i++)
      {
          if (isPrime(i) && isPrime(i - 2))
              count++;
      }

      // Print the count obtained
      Console.WriteLine(count);
  }

  // Driver Code
  static public void Main ()
  {
    int N = 6;
    countPrimeTuples(N);
  }
}

// This code is contributed by Dharanendra L V.

Javascript

<script>

// Javascript program for the above approach

// Function to check if a
// number is a prime or not
function isPrime(N)
{
    if (N <= 1)
        return false;

    for (let i = 2; i <= Math.sqrt(N); i++) {
        if (N % i == 0)
            return false;
    }

    return true;
}

// Function to count the number
// of valid prime triplets
function countPrimeTuples(N)
{
    // Stores the count
    // of prime triplets
    let count = 0;

    // Iterate from 2 to N and check for each
    // p, whether p & (p - 2) are prime or not
    for (let i = 2; i <= N; i++) {

        if (isPrime(i) && isPrime(i - 2))
            count++;
    }

    // Print the count obtained
    document.write(count);
}

// Driver Code

let N = 6;
countPrimeTuples(N);

// This code is contributed by _saurabh_jaiswal

</script>

Output:

1

Time Complexity: O(N^3/2)
Auxiliary Space: O(1)