Find n-th Fortunate Number

A Fortunate number is the smallest integer m > 1 such that, for a given positive integer n, pn + m is a prime number. Here pn is the product of the first n prime numbers, i.e prime factorials (or primorials) of order n.

For example :

p3 = 2 × 3 × 5 = 30
p4 = 2 × 3 × 5 × 7 = 210
p5 = 2 × 3 × 5 × 7 × 11 = 2310

Now, the smallest difference m between the prime factorial pn and the first prime number greater than pn for which (m > 1), is a prime number.

Examples :

Input : n = 3
Output : 7
Explanation : 7 must be added to the product
of first n prime numbers to make the product 
prime. 2 x 3 x 5 = 30, need to add 7 to make 
it 37, which is a prime

Input : n = 5
Output : 23

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach : To find the nth Fortunate number, calculate the product of the first n prime numbers (primorial). Let this product be p. Then we find prime number greater than p and return the difference between the found prime number and p.

p4 + 13 = 223, where m = 13, a fortunate number
p5 + 23 = 2333, where m = 23, a fortunate number
p6 + 17 = 30047, where m = 17, a fortunate number

C++

// C++ program to find n-th Fortunate number 
#include <bits/stdc++.h> 
using namespace std; 
  
bool isPrime(int n) 
{ 
    // Corner cases 
    if (n <= 1)  return false; 
    if (n <= 3)  return true; 
   
    // This is checked so that we can skip  
    // middle five numbers in below loop 
    if (n%2 == 0 || n%3 == 0) return false; 
   
    for (int i=5; i*i<=n; i=i+6) 
        if (n%i == 0 || n%(i+2) == 0) 
           return false; 
   
    return true; 
} 
  
// Function to Find primorial of order n  
// (product of first n prime numbers). 
long long int primorial(long long int n) 
{ 
    long long int p = 2; 
    n--; 
    for (int i = 3; n != 0; i++) { 
        if (isPrime(i)) { 
            p = p * i; 
            n--; 
        } 
        i++; 
    } 
    return p; 
} 
  
// Function to find next prime number greater 
// than n 
long long int findNextPrime(long long int n) 
{ 
    // Note that difference (or m) should be 
    // greater than 1. 
    long long int nextPrime = n + 2; 
  
    // loop continuously until isPrime 
    // returns true for a number above n 
    while (true) { 
  
        // Ignoring the prime number that  
        // is 1 greater than n 
        if (isPrime(nextPrime)) 
            break; 
  
        nextPrime++; 
    } 
  
    return nextPrime; 
} 
  
// Returns n-th Fortunate number 
long long int fortunateNumber(int n) 
{ 
   long long int p = primorial(n); 
   return findNextPrime(p) - p; 
} 
  
// Driver function 
int main() 
{ 
    long long int n = 5; 
    cout << fortunateNumber(n) << "\n"; 
    return 0; 
} 

Java

// Java program to find n-th Fortunate number 
import java.lang.*; 
import java.util.*; 
  
class GFG 
{ 
      
    public static boolean isPrime(int n) 
    { 
        // Corner cases 
        if (n <= 1) return false; 
        if (n <= 3) return true; 
      
        // This is checked so that we can skip  
        // middle five numbers in below loop 
        if (n % 2 == 0 || n % 3 == 0) return false; 
      
        for (int i = 5; i * i <= n; i = i + 6) 
            if (n % i == 0 || n % (i + 2) == 0) 
            return false; 
      
        return true; 
    } 
      
    // Function to Find primorial of order n  
    // (product of first n prime numbers). 
    public static int primorial(int n) 
    { 
        int p = 2; 
        n--; 
        for (int i = 3; n != 0; i++) { 
            if (isPrime(i) == true) { 
                p = p * i; 
                n--; 
            } 
            i++; 
        } 
        return p; 
    } 
  
    // Function to find next prime number greater 
    // than n 
    public static int findNextPrime(int n) 
    { 
        // Note that difference (or m) should be 
        // greater than 1. 
        int nextPrime = n + 2; 
      
        // loop continuously until isPrime 
        // returns true for a number above n 
        while (true) { 
      
            // Ignoring the prime number that  
            // is 1 greater than n 
            if (isPrime(nextPrime) == true) 
                break; 
      
            nextPrime++; 
       } 
  
    return nextPrime; 
    } 
      
    // Returns n-th Fortunate number 
    public static int fortunateNumber(int n) 
    { 
        int p = primorial(n); 
        return findNextPrime(p)-p; 
    } 
      
    //Driver function 
    public static void main (String[] args) { 
        int n = 5; 
        System.out.println(fortunateNumber(n)); 
    } 
} 
  
/*This code is contributed by Akash Singh*/

Python3

# Python3 program to find 
# n-th Fortunate number 
  
def isPrime(n): 
  
    # Corner cases 
    if (n <= 1): return False
    if (n <= 3): return True
  
    # This is checked so that we can skip  
    # middle five numbers in below loop 
    if (n % 2 == 0 or n % 3 == 0): 
        return False
      
    i = 5
    while(i * i <= n): 
        if (n % i == 0 or
            n % (i + 2) == 0): 
            return False
        i += 6
          
    return True
  
  
# Function to Find primorial of order n  
# (product of first n prime numbers). 
def primorial(n): 
  
    p = 2; n -= 1; i = 3
    while(n != 0): 
        if (isPrime(i)):  
            p = p * i 
            n -= 1
          
        i += 1
      
    return p 
  
  
# Function to find next prime  
# number greater than n 
def findNextPrime(n): 
  
    # Note that difference (or m)  
    # should be greater than 1. 
    nextPrime = n + 2
  
    # loop continuously until isPrime 
    # returns true for a number above n 
    while (True): 
  
        # Ignoring the prime number that  
        # is 1 greater than n 
        if (isPrime(nextPrime)): 
            break
  
        nextPrime += 1
      
    return nextPrime 
  
# Returns n-th Fortunate number 
def fortunateNumber(n): 
    p = primorial(n) 
    return findNextPrime(p) - p 
  
# Driver Code 
n = 5
print(fortunateNumber(n)) 
  
# This code is contributed by Anant Agarwal. 

C#

// C# program to find  
// n-th Fortunate number 
using System; 
  
class GFG 
{ 
    public static bool isPrime(int n) 
    { 
        // Corner cases 
        if (n <= 1) return false; 
        if (n <= 3) return true; 
      
        // This is checked so that  
        // we can skip middle five  
        // numbers in below loop 
        if (n % 2 == 0 || n % 3 == 0)  
            return false; 
      
        for (int i = 5;  
                 i * i <= n; i = i + 6) 
            if (n % i == 0 ||  
                n % (i + 2) == 0) 
            return false; 
      
        return true; 
    } 
      
    // Function to Find primorial  
    // of order n (product of first 
    // n prime numbers). 
    public static int primorial(int n) 
    { 
        int p = 2; 
        n--; 
        for (int i = 3; n != 0; i++)  
        { 
            if (isPrime(i) == true)  
            { 
                p = p * i; 
                n--; 
            } 
            i++; 
        } 
        return p; 
    } 
  
    // Function to find next  
    // prime number greater than n 
    public static int findNextPrime(int n) 
    { 
        // Note that difference (or m)  
        // should be greater than 1. 
        int nextPrime = n + 2; 
      
        // loop continuously until  
        // isPrime returns true  
        // for a number above n 
        while (true) 
        { 
      
            // Ignoring the prime number  
            // that is 1 greater than n 
            if (isPrime(nextPrime) == true) 
                break; 
      
            nextPrime++; 
    } 
  
    return nextPrime; 
    } 
      
    // Returns n-th  
    // Fortunate number 
    public static int fortunateNumber(int n) 
    { 
        int p = primorial(n); 
        return findNextPrime(p) - p; 
    } 
      
    // Driver Code 
    public static void Main () 
    { 
        int n = 5; 
        Console.WriteLine(fortunateNumber(n)); 
    } 
} 
  
// This code is contributed 
// by anuj_67. 

PHP

<?php 
// PHP program to find n-th 
// Fortunate number 
  
function isPrime($n) 
{ 
      
    // Corner cases 
    if ($n <= 1)  
        return false; 
    if ($n <= 3)  
        return true; 
  
    // This is checked so that we can skip  
    // middle five numbers in below loop 
    if ($n % 2 == 0 || $n % 3 == 0) 
        return false; 
  
    for($i = 5; $i * $i <= $n; 
                  $i = $i + 6) 
        if ($n % $i == 0 || $n % ($i + 2) == 0) 
        return false; 
  
    return true; 
} 
  
// Function to Find primorial of order n  
// (product of first n prime numbers). 
function primorial($n) 
{ 
    $p = 2; 
    $n--; 
    for ($i = 3; $n != 0; $i++)  
    { 
        if (isPrime($i)) 
        { 
            $p = $p * $i; 
            $n--; 
        } 
        $i++; 
    } 
    return $p; 
} 
  
// Function to find next prime  
// number greater than n 
function findNextPrime($n) 
{ 
      
    // Note that difference (or m) 
    // should be greater than 1. 
    $nextPrime = $n + 2; 
  
    // loop continuously until isPrime 
    // returns true for a number above n 
    while (true)  
    { 
  
        // Ignoring the prime number that  
        // is 1 greater than n 
        if (isPrime($nextPrime)) 
            break; 
  
        $nextPrime++; 
    } 
  
    return $nextPrime; 
} 
  
// Returns n-th Fortunate number 
function fortunateNumber($n) 
{ 
    $p = primorial($n); 
    return findNextPrime($p) - $p; 
} 
  
    // Driver Code 
    $n = 5; 
    echo fortunateNumber($n) , "\n"; 
  
// This code is contributed by ajit 
?> 

Output:

Optimization : The above solution can be optimized using Sieve of Eratosthenes.

My Personal Notes

Suggest a Topic

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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