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

Last Updated : 11 Jul, 2025

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


 


The idea here is using the properties of n!   . Since n! = 1\times2\times3....(n-1)\times n   , then numbers 1, 2, 3..... (n-1), n   , all divide n!   . Therefore n!+2   is divisible by 2, n!+3   is divisible by 3 ..... n!+n   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 <iostream>
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
<?php
// PHP program to print n consecutive
// composite numbers.

// function to find factorial of given 
// number
function factorial( $n)
{ 
    $res = 1;
    for ($i = 2; $i <= $n; $i++)
        $res *= $i;
    return $res;
}

// Prints n consecutive numbers. 
function printNComposite(int $n)
{
    $fact = factorial($n + 1);
    for($i = 2; $i <= $n + 1; ++$i) 
        echo $fact + $i ," "; 
}

    // Driver Code
    $n = 4;
    printNComposite($n);
    
// This code is contributed by anuj_67.
?>
JavaScript
<script>

// JavaScript program to print n consecutive composite 
// numbers

// function to find factorial of given 
// number 
    function factorial(n) {
        let res = 1;
        for (let i = 2; i <= n; i++) {
            res *= i;
        }
        return res;
    }
  
// Prints n consecutive numbers. 
    function printNComposite(n) {
        let fact = factorial(n + 1);
        for (let i = 2; i <= n + 1; ++i) {
            document.write(fact + i + " ");
        }
    }

// Driver code
    
        let n = 4;
        printNComposite(n);
    
    // This code is contributed by code_hunt.
</script>

Output: 
122 123 124 125

 

Time Complexity: O(n)
Auxiliary Space: O(1)

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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