Perfect totient number - GeeksforGeeks

A Perfect totient number is an integer that is equal to the sum of its iterated totients. Perfect totient number is denoted by $\varphi^i(n)$ .
For example:

$\varphi^i (9) = 6$ , Now, $\varphi^i (6) = 2$ , $\varphi^i (2) = 1$ 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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#

// 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 

Output:

Yes

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

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My Personal Notes

Check if N is a Perfect totient number

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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