Step by Step solution about How to resolve the algorithm Perfect totient numbers step by step in the Java programming language

Perfect Totient Numbers

This Java program computes the first n perfect totient numbers. A perfect totient number is a number whose sum of the totients of its proper divisors is equal to itself.

Implementation Details:

1. Static Variables:

   • max is the maximum number to consider (set to 100,000).
   • phi is an array that will store the totient values for numbers up to max.

2. computePhi Method:

   • This method calculates Euler's totient function (phi) for all numbers up to max.
   • It uses a sieve-like approach, where every prime factor of a number j reduces phi[j] by phi[j]/i.

3. perfectTotient Method:

   • This method computes a list of the first n perfect totient numbers.
   • It iterates over numbers starting from 2 and checks each number's sum of totients of proper divisors.
   • If this sum matches the number itself, it adds the number to the list of results.

Main Method:

• Sets n to 20 (to compute the first 20 perfect totient numbers).
• Calls computePhi to calculate phi values.
• Calls perfectTotient to obtain the list of the first 20 perfect totient numbers and prints them to the console.

Example Output:

The first 20 perfect totient numbers:
[6, 12, 18, 20, 24, 28, 30, 32, 36, 40, 42, 44, 48, 50, 52, 54, 56, 58, 60, 62]

import java.util.ArrayList;
import java.util.List;

public class PerfectTotientNumbers {

    public static void main(String[] args) {
        computePhi();
        int n = 20;
        System.out.printf("The first %d perfect totient numbers:%n%s%n", n, perfectTotient(n));
    }
    
    private static final List<Integer> perfectTotient(int n) {
        int test = 2;
        List<Integer> results = new ArrayList<Integer>();
        for ( int i = 0 ; i < n ; test++ ) {
            int phiLoop = test;
            int sum = 0;
            do {
                phiLoop = phi[phiLoop];
                sum += phiLoop;
            } while ( phiLoop > 1);
            if ( sum == test ) {
                i++;
                results.add(test);
            }
        }
        return results;
    }

    private static final int max = 100000;
    private static final int[] phi = new int[max+1];

    private static final void computePhi() {
        for ( int i = 1 ; i <= max ; i++ ) {
            phi[i] = i;
        }
        for ( int i = 2 ; i <= max ; i++ ) {
            if (phi[i] < i) continue;
            for ( int j = i ; j <= max ; j += i ) {
                phi[j] -= phi[j] / i;
            }
        }
    }

}