Print first N Mosaic numbers

• Last Updated : 05 Oct, 2021

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 factorization 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++

 // C++ implementation of the approach#include using namespace std; // Function to return the nth mosaic numberint 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 numbersvoid nMosaicNumbers(int n){    for (int i = 1; i <= n; i++)        cout << mosaic(i) << " ";} // Driver codeint main(){    int n = 10;    nMosaicNumbers(n);     return 0;}

Java

 // Java implementation of the approachclass 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

C#

 // C# implementation of the approachusing 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

Python

 # Python implementation of the approach # Function to return the nth mosaic numberdef 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 numbersdef nMosaicNumbers(n):    for i in range(1,n+1):        print mosaic(i),  # Driver coden = 10;nMosaicNumbers(n); # This code is contributed by CrazyPro

Javascript


Output:
1 2 3 4 5 6 7 6 6 10

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