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\times2\times3....(n-1)\times 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++

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// 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;
}

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Java

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// 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);
  
    }
}

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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#

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// 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

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PHP

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<?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.
?>

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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.



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