Find a sequence of N prime numbers whose sum is a composite number

Given an integer N and the task is to find a sequence of N prime numbers whose sum is a composite number.

Examples:

Input: N = 5
Output: 2 3 5 7 11
2 + 3 + 5 + 7 + 11 = 28 which is composite.



Input: N = 6
Output: 3 5 7 11 13 17

Approach: The sum of two prime numbers is always even which is composite as they are odd numbers except 2. There are two cases now,

  1. When N is even then we can print any N prime numbers except 2 and their sum will always be even i.e. odd numbers when added even number of times will give an even sum.
  2. When N is odd then we can print 2 and any other N – 1 primes to make sure that the sum is even. Since, N – 1 is even so the sum will be even for any primes except 2 then we add 2 as the Nth number to make sure that the sum remains even.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define MAXN 100000
  
// To store prime numbers
vector<int> v;
  
// Function to find and store
// all the primes <= n
void SieveOfEratosthenes(int n)
{
    // Create a boolean array "prime[0..n]" and initialize
    // all entries it as true. A value in prime[i] will
    // finally be false if i is Not a prime, else true.
    bool prime[n + 1];
    memset(prime, true, sizeof(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 greater than or
            // equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked.
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
  
    // Store all the prime numbers
    for (int p = 2; p <= n; p++)
        if (prime[p])
            v.push_back(p);
}
  
// Function to print the required sequence
void GetSequence(int n)
{
  
    // If n is even then we do not include 2
    // and start the sequence from the 2nd
    // smallest prime else we include 2
    int i = (n % 2 == 0) ? 1 : 0;
  
    int cnt = 0;
    // Print the sequence
    while (cnt < n) {
        cout << v[i] << " ";
        i++;
        cnt++;
    }
}
  
// Driver code
int main()
{
    int n = 6;
    SieveOfEratosthenes(MAXN);
  
    GetSequence(n);
  
    return 0;
}

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Java

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// Java implementation of the above approach
import java.util.*;
  
class GFG
{
      
static int MAXN = 100000;
  
// To store prime numbers
static Vector<Integer> v = new Vector<Integer>();
  
// Function to find and store
// all the primes <= n
static void SieveOfEratosthenes(int n)
{
    // Create a boolean array "prime[0..n]" and initialize
    // all entries it as true. A value in prime[i] will
    // finally be false if i is Not a prime, else true.
    boolean[] prime = new boolean[n + 1];
    Arrays.fill(prime,true);
  
    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 greater than or
            // equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked.
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
  
    // Store all the prime numbers
    for (int p = 2; p <= n; p++)
        if (prime[p])
            v.add(p);
}
  
// Function to print the required sequence
static void GetSequence(int n)
{
  
    // If n is even then we do not include 2
    // and start the sequence from the 2nd
    // smallest prime else we include 2
    int i = (n % 2 == 0) ? 1 : 0;
  
    int cnt = 0;
      
    // Print the sequence
    while (cnt < n)
    {
        System.out.print(v.get(i) + " ");
        i++;
        cnt++;
    }
}
  
// Driver code
public static void main(String[] args)
{
    int n = 6;
    SieveOfEratosthenes(MAXN);
  
    GetSequence(n);
}
}
  
// This code is contributed by Princi Singh

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python

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# Python3 implementation of the approach
  
MAXN=100000
  
# To store prime numbers
v=[]
  
# Function to find and store
# all the primes <= n
def SieveOfEratosthenes(n):
  
    # Create a boolean array "prime[0..n]" and initialize
    # all entries it as true. A value in prime[i] will
    # finally be false if i is Not a prime, else true.
    prime=[True for i in range(n + 1)]
  
    for p in range(2,n+1):
  
        # If prime[p] is not changed, then it is a prime
        if (prime[p] == True):
  
            # Update all multiples of p greater than or
            # equal to the square of it
            # numbers which are multiple of p and are
            # less than p^2 are already been marked.
            for i in range(2 * p, n + 1, p):
                prime[i] = False
  
    # Store all the prime numbers
    for p in range(2, n + 1):
        if (prime[p]):
            v.append(p)
  
# Function to print the required sequence
def GetSequence(n):
  
    # If n is even then we do not include 2
    # and start the sequence from the 2nd
    # smallest prime else we include 2
    if n % 2 == 0:
        i = 1
    else:
        i = 0
  
    cnt = 0
    # Print the sequence
    while (cnt < n):
        print(v[i],end=" ")
        i += 1
        cnt += 1
  
  
# Driver code
n = 6
SieveOfEratosthenes(MAXN)
  
GetSequence(n)
  
# This code is contributed by mohit kumar 29

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C#

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// C# implementation of the approach
using System;
using System.Collections.Generic;
  
class GFG 
      
static int MAXN = 100000; 
  
// To store prime numbers 
static List<int> v = new List<int>(); 
  
// Function to find and store 
// all the primes <= n 
static void SieveOfEratosthenes(int n) 
    // Create a boolean array "prime[0..n]" and initialize 
    // all entries it as true. A value in prime[i] will 
    // finally be false if i is Not a prime, else true. 
    Boolean[] prime = new Boolean[n + 1]; 
    for(int i = 0; i < n + 1; i++)
        prime[i] = true;
  
    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 greater than or 
            // equal to the square of it 
            // numbers which are multiple of p and are 
            // less than p^2 are already been marked. 
            for (int i = p * p; i <= n; i += p) 
                prime[i] = false
        
    
  
    // Store all the prime numbers 
    for (int p = 2; p <= n; p++) 
        if (prime[p]) 
            v.Add(p); 
  
// Function to print the required sequence 
static void GetSequence(int n) 
  
    // If n is even then we do not include 2 
    // and start the sequence from the 2nd 
    // smallest prime else we include 2 
    int i = (n % 2 == 0) ? 1 : 0; 
  
    int cnt = 0; 
      
    // Print the sequence 
    while (cnt < n) 
    
        Console.Write(v[i] + " "); 
        i++; 
        cnt++; 
    
  
// Driver code 
public static void Main(String[] args) 
    int n = 6; 
    SieveOfEratosthenes(MAXN); 
  
    GetSequence(n); 
  
/* This code is contributed by PrinciRaj1992 */

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Output:

3 5 7 11 13 17


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