Perfect totient number

A Perfect totient number is an integer that is equal to the sum of its iterated totients. Perfect totient number is denoted by .
For example:

, Now, , and 9 == 6 + 2 + 1, Therefore, 9 is a perfect totient number.

Check if N is a Perfect totient number

Given an integer N, the task is to check N is a Perfect totient number.
Examples:

Input: N = 9
Output: Yes
Input: N = 10
Output: No

Approach: The idea is to find the Euler Totient Value of the given number, suppose we get the Euler Totient Value of N as V, then we will again find the Euler Totient Value of V until the new Euler Totient Value V becomes 1. We will also keep the sum of all the Euler Totient Value values V we got till now and check if the sum is equal to N or not. If equal then the given number is a perfect Euler Totient number.
Below is the implementation of the above approach:

C++

// C++ implementation to find 
// the number of digits in
// a Nth fibonacci number
 
#include <bits/stdc++.h>

using namespace std;
 
// Function to find the Totient 
// number of the given value

int phi(int n) 
{ 

    // Initialize result as n

    int result = n; 
 
    // Consider all prime factors 

    // of n and subtract their 

    // multiples from result 

    for (int p = 2; p * p <= n; ++p) { 

        // Check if p is a prime factor. 

        if (n % p == 0) { 

            // If yes, then update N 

            // and result 

            while (n % p == 0) 

                n /= p; 

            result -= result / p; 

        } 

    } 
 
    // If n has a prime factor 

    // greater than sqrt(n) 

    // (There can be at-most one

    // such prime factor) 

    if (n > 1) 

        result -= result / n; 

    return result; 
} 
 
// Function to check if the number 
// is a perfect totient number

int isPerfectTotientNum(int n)
{

    // store original value of n

    int temp = n;

    int sum = 0;

    // loop to calculate sum

    // of iterated totients

    while(n > 1){

        sum = sum + phi(n);

        n = phi(n);

    }

    // condition for Perfect

    // Totient Number

    if(sum == temp)

    return true;

    return false;
}
 
// Driver Code 

int main()
{

    int n = 9; 

    if(isPerfectTotientNum(n))

        cout << "Yes";

    else

        cout << "No";

    return 0;
}

Java

// Java implementation to find the  
// number of digits in a Nth 
// fibonacci number

class GFG{
 
// Function to find the Totient 
// number of the given value

static int phi(int n) 
{ 

    // Initialize result as n

    int result = n; 
 
    // Consider all prime factors 

    // of n and subtract their 

    // multiples from result 

    for(int p = 2; p * p <= n; ++p)

    {

       // Check if p is a prime factor

       if (n % p == 0) 

       { 

           // If yes, then update N 

           // and result 

           while (n % p == 0)

           { 

               n /= p;

           } 

           result -= result / p; 

       } 

    } 
 
    // If n has a prime factor 

    // greater than sqrt(n) 

    // (There can be at-most one

    // such prime factor) 

    if (n > 1) 

        result -= result / n; 

    return result; 
} 
 
// Function to check if the number 
// is a perfect totient number

static boolean isPerfectTotientNum(int n)
{

    // Store original value of n

    int temp = n;

    int sum = 0;

    // Loop to calculate sum

    // of iterated totients

    while(n > 1)

    {

        sum = sum + phi(n);

        n = phi(n);

    }

    // Condition for Perfect

    // Totient Number

    if(sum == temp)

       return true;

    return false;
}
 
// Driver Code 

public static void main(String[] args)
{

    int n = 9; 

    if(isPerfectTotientNum(n))

    {

        System.out.println("Yes");

    }

    else

    {

        System.out.println("No");

    }
}
}
 
// This code is contributed by Ritik Bansal

Python3

# Python3 implementation to find
# the number of digits in
# a Nth fibonacci number
 
# Function to find the Totient
# number of the given value

def phi(n):

    # Initialize result as n

    result = n
 
    # Consider all prime factors

    # of n and subtract their

    # multiples from result

    for p in range(2, n):

        if p * p > n:

            break
 
        # Check if p is a prime factor.

        if (n % p == 0):
 
            # If yes, then update N

            # and result

            while (n % p == 0):

                n //= p

            result -= result // p
 
    # If n has a prime factor

    # greater than sqrt(n)

    # (There can be at-most one

    # such prime factor)

    if (n > 1):

        result -= result // n

    return result
 
# Function to check if the number
# is a perfect totient number

def isPerfectTotientNum(n):

    # Store original value of n

    temp = n

    sum = 0
 
    # Loop to calculate sum

    # of iterated totients

    while (n > 1):

        sum = sum + phi(n)

        n = phi(n)

    # Condition for Perfect

    # Totient Number

    if (sum == temp):

        return True
 
    return False
 
# Driver Code

if __name__ == '__main__':

    n = 9

    if (isPerfectTotientNum(n)):

        print("Yes")

    else:

        print("No")
 
# This code is contributed by mohit kumar 29

// C# implementation to find the 
// number of digits in a Nth 
// fibonacci number

using System;

class GFG{
 
// Function to find the Totient 
// number of the given value

static int phi(int n) 
{ 

    // Initialize result as n

    int result = n; 
 
    // Consider all prime factors 

    // of n and subtract their 

    // multiples from result 

    for(int p = 2; p * p <= n; ++p)

    {

       // Check if p is a prime factor

       if (n % p == 0) 

       { 

           // If yes, then update N 

           // and result 

           while (n % p == 0)

           { 

               n /= p;

           } 

           result -= result / p; 

       } 

    } 
 
    // If n has a prime factor 

    // greater than sqrt(n) 

    // (There can be at-most one

    // such prime factor) 

    if (n > 1) 

        result -= result / n; 

    return result; 
} 
 
// Function to check if the number 
// is a perfect totient number

static bool isPerfectTotientNum(int n)
{

    // Store original value of n

    int temp = n;

    int sum = 0;

    // Loop to calculate sum

    // of iterated totients

    while(n > 1)

    {

        sum = sum + phi(n);

        n = phi(n);

    }

    // Condition for Perfect

    // Totient Number

    if(sum == temp)

    return true;

    return false;
}
 
// Driver Code 

public static void Main()
{

    int n = 9; 

    if(isPerfectTotientNum(n))

    {

        Console.Write("Yes");

    }

    else

    {

        Console.Write("No");

    }
}
}
 
// This code is contributed by Code_Mech

Javascript

<script>
// Javascript implementation to find the  
// number of digits in a Nth 
// fibonacci number
 
    // Function to find the Totient

    // number of the given value

    function phi( n)

    {
 
        // Initialize result as n

        let result = n;
 
        // Consider all prime factors

        // of n and subtract their

        // multiples from result

        for ( let p = 2; p * p <= n; ++p) 

        {
 
            // Check if p is a prime factor

            if (n % p == 0)

            {
 
                // If yes, then update N

                // and result

                while (n % p == 0)

                {

                    n = parseInt(n/p);

                }

                result -= parseInt(result / p);

            }

        }
 
        // If n has a prime factor

        // greater than sqrt(n)

        // (There can be at-most one

        // such prime factor)

        if (n > 1)

            result -= parseInt(result / n);

        return result;

    }
 
    // Function to check if the number

    // is a perfect totient number

    function isPerfectTotientNum( n) {
 
        // Store original value of n

        let temp = n;

        let sum = 0;
 
        // Loop to calculate sum

        // of iterated totients

        while (n > 1) {

            sum = sum + phi(n);

            n = phi(n);

        }
 
        // Condition for Perfect

        // Totient Number

        if (sum == temp)

            return true;
 
        return false;

    }
 
    // Driver Code

    let n = 9;
 
    if (isPerfectTotientNum(n)) {

        document.write("Yes");

    } else {

        document.write("No");

    }
// This code is contributed by aashish1995 
</script>

Output:

Yes

Time Complexity: O(n^1/2)

Auxiliary Space: O(1)

References: https://en.wikipedia.org/wiki/Perfect_totient_number

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