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

Last Updated : 06 Apr, 2023

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 using namespace std; // function to find factorial of given // numberunsigned 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 functionint 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 numberdef 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 Coden = 4;printNComposite(n);     # This code is contributed by mits

## C#

 // C# program to print n consecutive composite // numbersusing 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

 

## Javascript

 

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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