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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 n!. Since n! = 1	imes2	imes3....(n-1)	imes 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.
?>

Output:

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.



This article is attributed to GeeksforGeeks.org

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