Assignment #76 Collatz Sequence

The counter for the number of steps may be 1 higher than the number. When I made it, I wanted to count the starting number as the first step.

Code

    /// Name: Koosha Kimelman
    /// Period: 7
    /// Program Name: Collatz Sequence
    /// File Name: Collatz.java
    /// Date Completed: 11/20/15
    
    import java.util.Scanner;
    
    public class Collatz {
    
        public static void main(String[] args) throws Exception {
        
            Scanner keyboard = new Scanner(System.in);
            
            int n, x;
            x = 1;
            String na = "N/A";
            
            System.out.print("Starting number: ");
            n = keyboard.nextInt();
            
            System.out.println(n);
            
            if (n != 1 && n > 0) {    // This is in case  someone chooses 1 (or something lower than it) as their starting number
                while (n != 1) {
                    Thread.sleep(80);
                    if (n % 2 == 0)
                        n = n/2;
                    else if (n % 2 == 1)
                        n = ((n*3)+1);
                    System.out.println(n);
                    x++;
                }
                System.out.println("It took " + x + " steps to get to 1.");
            }
            else if (n == 1)
                System.out.println("You were supposed to choose a number OTHER than 1.");
            else if (n < 1)
                System.out.println("You were supposed to choose a number greater than 1.");
        }
    }
    

Picture of the output