Print first N Mosaic numbers


Given an integer N, the task is to print first N Mosaic numbers. A Mosaic number can be expressed as follows:

If N = p1a1p2a2…pkak in the prime facrorization of N
where p1 ,p2 … pk are prime numbers.
Then the Nth Mosaic number is equal to ((p1)*(a1))*((p2)*(a2))*…*((pk)*(ak))

Examples:



Input : N=10
Output : 1 2 3 4 5 6 7 6 6 10
For N = 4, N = 22 .
4th Mosaic number = 2*2 = 4
For N=8 , N= 2 3
8th Mosaic number = 2*3 = 6
Similarly print first N Mosaic numbers

Input : N=5
Output : 1 2 3 4 5

Approach:
Run a loop from 1 to N and for every i we have to find all the prime factors and also the powers of the factors in the number by dividing the number by the factor until the factor divides the number. The ith Mosaic number will then be the product of the found prime factors and their powers.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the nth mosaic number
int mosaic(int n)
{
    int i, ans = 1;
  
    // Iterate from 2 to the number
    for (i = 2; i <= n; i++) {
  
        // If i is the factor of n
        if (n % i == 0 && n > 0) {
            int count = 0;
  
            // Find the count where i^count
            // is a factor of n
            while (n % i == 0) {
  
                // Divide the number by i
                n /= i;
  
                // Increase the count
                count++;
            }
  
            // Multiply the answer with
            // count and i
            ans *= count * i;
        }
    }
  
    // Return the answer
    return ans;
}
  
// Function to print first N Mosaic numbers
void nMosaicNumbers(int n)
{
    for (int i = 1; i <= n; i++)
        cout << mosaic(i) << " ";
}
  
// Driver code
int main()
{
    int n = 10;
    nMosaicNumbers(n);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG
{
  
    // Function to return the nth mosaic number
    static int mosaic(int n)
    {
        int i, ans = 1;
  
        // Iterate from 2 to the number
        for (i = 2; i <= n; i++)
        {
  
            // If i is the factor of n
            if (n % i == 0 && n > 0)
            {
                int count = 0;
  
                // Find the count where i^count
                // is a factor of n
                while (n % i == 0
                {
  
                    // Divide the number by i
                    n /= i;
  
                    // Increase the count
                    count++;
                }
  
                // Multiply the answer with
                // count and i
                ans *= count * i;
            }
        }
  
        // Return the answer
        return ans;
    }
  
    // Function to print first N Mosaic numbers
    static void nMosaicNumbers(int n)
    {
        for (int i = 1; i <= n; i++)
            System.out.print( mosaic(i)+ " ");
    }
  
    // Driver code
    public static void main(String[] args) 
    {
  
        int n = 10;
        nMosaicNumbers(n);
    }
}
  
// This code contributed by Rajput-Ji

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C#

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// C# implementation of the approach 
using System;
  
class GFG 
  
    // Function to return the nth mosaic number 
    static int mosaic(int n) 
    
        int i, ans = 1; 
  
        // Iterate from 2 to the number 
        for (i = 2; i <= n; i++) 
        
  
            // If i is the factor of n 
            if (n % i == 0 && n > 0) 
            
                int count = 0; 
  
                // Find the count where i^count 
                // is a factor of n 
                while (n % i == 0) 
                
  
                    // Divide the number by i 
                    n /= i; 
  
                    // Increase the count 
                    count++; 
                
  
                // Multiply the answer with 
                // count and i 
                ans *= count * i; 
            
        
  
        // Return the answer 
        return ans; 
    
  
    // Function to print first N Mosaic numbers 
    static void nMosaicNumbers(int n) 
    
        for (int i = 1; i <= n; i++) 
            Console.Write( mosaic(i)+ " "); 
    
  
    // Driver code 
    public static void Main() 
    
  
        int n = 10; 
        nMosaicNumbers(n); 
    
  
// This code is contributed by AnkitRai01

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Python

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# Python implementation of the approach
  
# Function to return the nth mosaic number
def mosaic( n):
    ans = 1
  
    # Iterate from 2 to the number
    for i in range(2,n+1):
  
        # If i is the factor of n
        if (n % i == 0 and n > 0):
            count = 0;
  
            # Find the count where i^count
            # is a factor of n
            while (n % i == 0): 
  
                # Divide the number by i
                n = n// i
  
                # Increase the count
                count+=1;
              
  
            # Multiply the answer with
            # count and i
            ans *= count * i;
      
  
    # Return the answer
    return ans;
  
  
# Function to print first N Mosaic numbers
def nMosaicNumbers(n):
    for i in range(1,n+1):
        print mosaic(i), 
  
  
# Driver code
n = 10;
nMosaicNumbers(n);
  
# This code is coributed by CrazyPro

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Output:

1 2 3 4 5 6 7 6 6 10


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Improved By : Rajput-Ji, AnkitRai01, CrazyPro