Lemoine’s Conjecture

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Any odd integer greater than 5 can be expressed as a sum of an odd prime (all primes other than 2 are odd) and an even semiprime. A semiprime number is a product of two prime numbers. This is called Lemoine’s conjecture.
Examples :

7 = 3 + (2 × 2),
where 3 is a prime number (other than 2) and 4 (= 2 × 2) is a semiprime number.
11 = 5 + (2 × 3)
where 5 is a prime number and 6(= 2 × 3) is a semiprime number.
9 = 3 + (2 × 3) or 9 = 5 + (2 × 2)
47 = 13 + 2 × 17 = 37 + 2 × 5 = 41 + 2 × 3 = 43 + 2 × 2

Below is the implementation of Lemoine’s Conjecture:

C++

// C++ code to verify Lemoine's Conjecture
// for any odd number >= 7
#include<bits/stdc++.h>
using namespace std;
 
// Function to check if a number is
// prime or not
bool isPrime(int n)
{
    if (n < 2)
        return false;
         
    for (int i = 2; i <= sqrt(n); i++) {
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// Representing n as p + (2 * q) to satisfy
// lemoine's conjecture
void lemoine(int n)
{
    // Declaring a map to hold pairs (p, q)
    map<int, int> pr;
     
    // Declaring an iterator for map
    map<int, int>::iterator it;
    it = pr.begin();
     
    // Finding various values of p for each q
    // to satisfy n = p + (2 * q)
    for (int q = 1; q <= n / 2; q++) 
    { 
        int p = n - 2 * q;
         
        // After finding a pair that satisfies the
        // equation, check if both p and q are
        // prime or not
        if (isPrime(p) && isPrime(q))
         
            // If both p and q are prime, store
            // them in the map
            pr.insert(it, pair<int, int>(p, q));
    }
     
    // Displaying all pairs (p, q) that satisfy
    // lemoine's conjecture for the number 'n'
    for (it = pr.begin(); it != pr.end(); ++it)
        cout << n << " = " << it->first
             << " + (2 * " << it->second << ")\n";
}
 
// Driver Function
int main()
{
    int n = 39;
    cout << n << " can be expressed as " << endl;
     
    // Function calling
    lemoine(n);
     
    return 0;
}
 
// This code is contributed by Saagnik Adhikary

Java

// Java code to verify Lemoine's Conjecture
// for any odd number >= 7
 
import java.util.*;
 
class GFG {
    static class pair {
        int first, second;
 
        public pair(int first, int second) {
            this.first = first;
            this.second = second;
        }
    }
 
    // Function to check if a number is
    // prime or not
    static boolean isPrime(int n) {
        if (n < 2)
            return false;
 
        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (n % i == 0)
                return false;
        }
        return true;
    }
 
    // Representing n as p + (2 * q) to satisfy
    // lemoine's conjecture
    static void lemoine(int n) {
        // Declaring a map to hold pairs (p, q)
        HashMap<Integer, pair> pr = new HashMap<>();
 
        // Declaring an iterator for map
        // HashMap<Integer,Integer>::iterator it;
        // it = pr.begin();
        int i = 0;
        // Finding various values of p for each q
        // to satisfy n = p + (2 * q)
        for (int q = 1; q <= n / 2; q++) {
            int p = n - 2 * q;
 
            // After finding a pair that satisfies the
            // equation, check if both p and q are
            // prime or not
            if (isPrime(p) && isPrime(q))
 
                // If both p and q are prime, store
                // them in the map
                pr.put(i, new pair(p, q));
            i++;
        }
 
        // Displaying all pairs (p, q) that satisfy
        // lemoine's conjecture for the number 'n'
        for (Map.Entry<Integer, pair> it : pr.entrySet())
            System.out.print(n + " = " + it.getValue().first + " + (2 * " + it.getValue().second + ")\n");
    }
 
    // Driver Function
    public static void main(String[] args) {
        int n = 39;
        System.out.print(n + " can be expressed as " + "\n");
 
        // Function calling
        lemoine(n);
 
    }
}
 
// This code contributed by aashish1995

Python3

# Python code to verify Lemoine's Conjecture
# for any odd number >= 7
from math import sqrt
 
# Function to check if a number is
# prime or not
def isPrime(n: int) -> bool:
    if n < 2:
        return False
 
    for i in range(2, int(sqrt(n)) + 1):
        if n % i == 0:
            return False
 
    return True
 
# Representing n as p + (2 * q) to satisfy
# lemoine's conjecture
def lemoine(n: int) -> None:
 
    # Declaring a map to hold pairs (p, q)
    pr = dict()
 
    # Finding various values of p for each q
    # to satisfy n = p + (2 * q)
    for q in range(1, n // 2 + 1):
        p = n - 2 * q
 
        # After finding a pair that satisfies the
        # equation, check if both p and q are
        # prime or not
        if isPrime(p) and isPrime(q):
 
            # If both p and q are prime, store
            # them in the map
            if p not in pr:
                pr[p] = q
 
    # Displaying all pairs (p, q) that satisfy
    # lemoine's conjecture for the number 'n'
    for it in pr:
        print("%d = %d + (2 * %d)" % (n, it, pr[it]))
 
# Driver Code
if __name__ == "__main__":
    n = 39
    print(n, "can be expressed as ")
 
    # Function calling
    lemoine(n)
 
# This code is contributed by
# sanjeev2552

C#

// C# code to verify Lemoine's Conjecture
// for any odd number >= 7
using System;
using System.Collections.Generic;
 
class GFG 
{
    public
   class pair 
   {
         public
           int first, second;
 
          public pair(int first, int second) 
          {
              this.first = first;
              this.second = second;
          }
      }
 
      // Function to check if a number is
      // prime or not
      static bool isPrime(int n)
      {
          if (n < 2)
              return false;
 
          for (int i = 2; i <= Math.Sqrt(n); i++)
          {
              if (n % i == 0)
                  return false;
          }
          return true;
      }
 
      // Representing n as p + (2 * q) to satisfy
      // lemoine's conjecture
      static void lemoine(int n)
      {
 
          // Declaring a map to hold pairs (p, q)
          Dictionary<int, pair> pr = new Dictionary<int, pair>();
 
          // Declaring an iterator for map
          // Dictionary<int,int>::iterator it;
          // it = pr.begin();
          int i = 0;
 
          // Finding various values of p for each q
          // to satisfy n = p + (2 * q)
          for (int q = 1; q <= n / 2; q++) 
          {
              int p = n - 2 * q;
 
              // After finding a pair that satisfies the
              // equation, check if both p and q are
              // prime or not
              if (isPrime(p) && isPrime(q))
 
                  // If both p and q are prime, store
                  // them in the map
                  pr.Add(i, new pair(p, q));
              i++;
          }
 
          // Displaying all pairs (p, q) that satisfy
          // lemoine's conjecture for the number 'n'
          foreach (KeyValuePair<int, pair> it in pr)
              Console.Write(n + " = " + it.Value.first + 
                            " + (2 * " + it.Value.second + ")\n");
      }
 
      // Driver code
      public static void Main(String[] args)
      {
          int n = 39;
          Console.Write(n + " can be expressed as " + "\n");
 
          // Function calling
          lemoine(n);
      }
  }
 
  // This code is contributed by aashish1995 

Javascript

<script>
// JavaScript code to verify Lemoine's Conjecture
// for any odd number >= 7
 
// Function to check if a number is
// prime or not
function isPrime(n)
{
    if (n < 2)
        return false;
         
    for (let i = 2; i <= Math.sqrt(n); i++) {
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// Representing n as p + (2 * q) to satisfy
// lemoine's conjecture
function lemoine(n)
{
    // Declaring a map to hold pairs (p, q)
    let pr = new Map();
     
    // Finding various values of p for each q
    // to satisfy n = p + (2 * q)
    for (let q = 1; q <= Math.floor(n / 2); q++) 
    { 
        let p = n - 2 * q;
         
        // After finding a pair that satisfies the
        // equation, check if both p and q are
        // prime or not
        if (isPrime(p) && isPrime(q))
         
            // If both p and q are prime, store
            // them in the map
            if(!pr.has(p)){
                pr.set(p, q);
            }
    }
     
    // Displaying all pairs (p, q) that satisfy
    // lemoine's conjecture for the number 'n'
    for (const [key, value] of pr)
        console.log(n, "=", key, " + (2 *", value, ")");
}
 
// Driver Function
let n = 39;
console.log(n, "can be expressed as ");
 
// Function calling
lemoine(n);
 
// This code is contributed by Gautam goel (gautamgoel962)
</script>

Output :

39 can be expressed as :
39 = 5 + (2 * 17)
39 = 13 + (2 * 13)
39 = 17 + (2 * 11)
39 = 29 + (2 * 5)

Time Complexity: O(n*(sqrt(n)+log(n)))
Auxiliary Space: O(n)

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