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

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 <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.