Question

Hailstones, also known as the Collatz sequence, are a mathematical curiosity. For any number in the sequence, the next number in the sequence is determined by two simple rules: If the current number n is odd, the next number in the sequence is equal to 3 * n + 1 If the current number n is even instead, the next number in the sequence is equal to one hald of n (i.e., n divided by 2) We repeat this process until it produces 1 (which is where the "mathematical curiosity" comes into play; so far, mathematicians have been unable to prove that this process always (eventually) produces the value 1, but no one has been able to find a counterexample that doesn't eventually reach 1). A Hailstone sequence can begin with any positive integer. For example, the integer 6 produces the hailstone sequence 6, 3, 10, 5, 16, 8, 4, 2, 1 In the code cell below, use the input() command to read in an integer from the user (assume that this integer is always positive and greater than 1). Then use a loop (we recommend a while loop) and any extra variables that you may need to follow the rules above to generate a Hailstone sequence from your starting number. DO NOT print out the Hailstone numbers that you generate; instead, your code should print exactly one number when it finishes: the total number of values in the Hailstone sequence that you generated (including both the starting value and the final 1 that ends the sequence).

Examples

The starting value 2 produces the output 2 (internally, it generates the sequence 2, 1).

Answer 1

Please look at my code and in case of indentation issues check the screenshots.

------------main.py----------------

def main():
   n = int(input("Enter the integer to begin(greater than 1): "))
   count = 0               #to count the number of values in the sequence
   while n!= 1:           #if number is not equal to 1
       count = count + 1   #increase the count by 1
       if n%2 == 1:       #if the number is odd
           n = 3*n+1       #next number is 3n+1
       else:               #if number is even
           n = n/2           #next number is n/2
   count = count + 1       #to count the last number in sequence which is 1
   print(count)           #print the count

main()

--------------Screenshots-------------------

----------------------Output------------------------------------

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