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 // 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 // numbersclass 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
<?php// PHP program to print n consecutive// composite numbers.// function to find factorial of given // numberfunction 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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