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# Print first N Mosaic numbers

• Last Updated : 31 May, 2022

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 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;` `}`

## Java

 `// 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`

## C#

 `// 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`

## Python

 `# 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 contributed by CrazyPro`

## Javascript

 ``

Output:

`1 2 3 4 5 6 7 6 6 10`

Time Complexity: O(logn)

Auxiliary Space: O(1)

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