Lemoine's Conjecture (original) (raw)
Last Updated : 20 Sep, 2022
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 TrueRepresenting 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
`
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)