Number formed after K times repeated addition of smallest divisor of N

Given two integers N and K, the task is to generate the final outcome of performing K operations which involves adding the smallest divisor, other than 1, of the current value of N to it at every step.
Example:

Input: N = 9, K = 4 
Output: 18 
Explanation:
Divisors of 9 are {1, 3, 9} 
1st Operation: N = 9 + 3 = 12 
2nd Operation: N = 12 + 2 = 14 
3rd Operation: N = 14 + 2 = 16 
4th Operation: N = 16 + 2 = 18
Input: N = 16, K = 3 
Output: 22 

Naive Approach: The brute force approach for this problem is to perform the operation K times and then print the final number.
Efficient Approach: The trick here is that if the given N is even, smallest divisor will be always 2 for all the K operations. Hence, the required K^th number will be simply 

required number = N + K * 2

Also if N is odd, the smallest divisor will be odd. Hence adding them will result in an even value (odd + odd = even). Therefore now the above trick can be applied for (K-1) operations, i.e., 

required number = N + smallest divisor of N + (K – 1) * 2

Below is the implementation of the above approach: 

C++

// C++ program to find the Kth number
// formed after repeated addition of
// smallest divisor of N
#include<bits/stdc++.h>
#include <cmath>
using namespace std;

void FindValue(int N, int K)
{

    // If N is even
    if( N % 2 == 0 )
    {
        N = N + 2 * K;
    }

    // If N is odd
    else
    {
        int i;

        for(i = 2; i < sqrt(N) + 1; i++)
        {
           if(N % i == 0)
              break;
        }

        // Add smallest divisor to N
        N = N + i;

        // Updated N is even
        N = N + 2 * ( K - 1 );
    }
    cout << N << endl;
}

// Driver code
int main()
{
    int N = 9;
    int K = 4;

    FindValue( N, K );
}

// This code is contributed by Surendra_Gangwar

Java

// Java program to find the Kth number
// formed after repeated addition of
// smallest divisor of N
import java.util.*;
class GFG{

static void FindValue(int N, int K)
{

    // If N is even
    if( N % 2 == 0 )
    {
        N = N + 2 * K;
    }

    // If N is odd
    else
    {
        int i;

        for(i = 2; i < Math.sqrt(N) + 1; i++)
        {
            if(N % i == 0)
                break;
        }

        // Add smallest divisor to N
        N = N + i;

        // Updated N is even
        N = N + 2 * ( K - 1 );
    }
    System.out.print(N);
}

// Driver code
public static void main(String args[])
{
    int N = 9;
    int K = 4;

    FindValue( N, K );
}
}

// This code is contributed by Nidhi_biet

Python3

# Python3 program to find the
# Kth number formed after
# repeated addition of
# smallest divisor of N

import math

def FindValue(N, K):

    # If N is even
    if( N % 2 == 0 ):
        N = N + 2 * K

    # If N is odd
    else:

        # Find the smallest divisor
        for i in range( 2, (int)(math.sqrt(N))+1 ):
            if( N % i == 0):
                break

        # Add smallest divisor to N
        N = N + i

        # Updated N is even
        N = N + 2 * ( K - 1 )

    print(N)

# Driver code
if __name__ == "__main__":
    N = 9
    K = 4
    FindValue( N, K )

C#

// C# program to find the Kth number
// formed after repeated addition of
// smallest divisor of N
using System;
class GFG{

static void FindValue(int N, int K)
{

    // If N is even
    if( N % 2 == 0 )
    {
        N = N + 2 * K;
    }

    // If N is odd
    else
    {
        int i;

        for(i = 2; i < Math.Sqrt(N) + 1; i++)
        {
            if(N % i == 0)
                break;
        }

        // Add smallest divisor to N
        N = N + i;

        // Updated N is even
        N = N + 2 * ( K - 1 );
    }
    Console.WriteLine(N);
}

// Driver code
public static void Main()
{
    int N = 9;
    int K = 4;

    FindValue( N, K );
}
}

// This code is contributed by Code_Mech

Javascript

<script>
// JavaScript program to find the Kth number
// formed after repeated addition of
// smallest divisor of N
function FindValue(N, K)
{

    // If N is even
    if( N % 2 == 0 )
    {
        N = N + 2 * K;
    }

    // If N is odd
    else
    {
        let i;

        for(i = 2; i < Math.sqrt(N) + 1; i++)
        {
        if(N % i == 0)
            break;
        }

        // Add smallest divisor to N
        N = N + i;

        // Updated N is even
        N = N + 2 * ( K - 1 );
    }
    document.write(N + "<br>");
}

// Driver code
let N = 9;
let K = 4;

FindValue( N, K );

// This code is contributed by Surbhi Tyagi.
</script>

Output:

18

Time complexity: O(?N )

Space Complexity: O(1)