# Generate a list of n consecutive composite numbers (An interesting method)

Given a number n, generate a list of n composite numbers.

Examples:

Input : 5
Output : 122, 123, 124, 125

Input : 10
Output : 3628802, 3628803, 3628804, 3628805, 3628806,
3628807, 3628808, 3628809, 3628810


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

The idea here is using the properties of . Since , then numbers , all divide . Therefore is divisible by 2, is divisible by 3 ….. is divisible by n. And by above pattern they are consecutive composites.

We find (n+1)!, then we print numbers (n+1)! + 2, (n+1)! + 3, …. (n+1)! + (n + 1).

Below is the implementation of above approach:

## C++

 // CPP program to print n consecutive composite  // numbers.  #include  using namespace std;     // function to find factorial of given   // number  unsigned long long int factorial(unsigned int n)  {          unsigned long long int res = 1;      for (int i=2; i<=n; i++)          res *= i;      return res;  }     // Prints n consecutive numbers.   void printNComposite(int n)  {      unsigned long long int fact = factorial(n+1);      for (int i = 2; i <= n+1; ++i)           cout << fact + i << " ";   }     // Driver program to test above function  int main()  {      int n = 4;      printNComposite(n);      return 0;  }

## Java

 // Java program to print n consecutive composite   // numbers     class GFG {     // function to find factorial of given   // number       static long factorial(int n) {          long res = 1;          for (int i = 2; i <= n; i++) {              res *= i;          }          return res;      }     // Prints n consecutive numbers.       static void printNComposite(int n) {          long fact = factorial(n + 1);          for (int i = 2; i <= n + 1; ++i) {              System.out.print(fact + i + " ");          }      }     // Driver program to test above function       public static void main(String[] args) {          int n = 4;          printNComposite(n);         }  }

## Python3

 # Python3 program to print n consecutive  # composite numbers.     # function to find factorial   # of given number  def factorial( n):         res = 1;      for i in range(2, n + 1):          res *= i;      return res;     # Prints n consecutive numbers.   def printNComposite(n):      fact = factorial(n + 1);      for i in range(2, n + 2):           print(fact + i, end = " ");      # Driver Code  n = 4;  printNComposite(n);         # This code is contributed by mits

## C#

 // C# program to print n consecutive composite   // numbers  using System;                         public class Program{      // function to find factorial of given   // number       static long factorial(int n) {          long res = 1;          for (int i = 2; i <= n; i++) {              res *= i;          }          return res;      }      // Prints n consecutive numbers.       static void printNComposite(int n) {          long fact = factorial(n + 1);          for (int i = 2; i <= n + 1; ++i) {              Console.Write(fact + i + " ");          }      }      // Driver program to test above function       public static void Main() {          int n = 4;          printNComposite(n);          }  }     // This code is contributed by Rajput-Ji

## PHP

 

Output:

122 123 124 125


The above solution causes overflow very soon (for small values of n). We can use technique to find factorial of large number to avoid overflow.

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Improved By : vt_m, Rajput-Ji, Mithun Kumar

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