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

05 Apr, 2021
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++
Java
Python3
C#
PHP
Javascript
// 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 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 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# 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 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.?>
<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>
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. 
vt_m
Rajput-Ji
Mithun Kumar
code_hunt
factorial
Mathematical
Mathematical
factorial
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