def seq3np1(n):

""" Print the 3n+1 sequence from n, terminating when it reaches 1."""

while n != 1:

print(n)

if n % 2 == 0: # n is even

n = n // 2

else: # n is odd

n = n * 3 + 1

print(n) # the last print is 1

seq3np1(3)

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code it in python

Activity 2: Repeat the call to seq3npl using a range of values, up to and including an upper bound. Now that we have a function that can return the number of iterations required to get to 1, we can use it to check a wide range of starting values. In fact, instead of just doing one value at a time, we can call the function iteratively, each time passing in a new value. Create a simple for loop using a loop variable called start that provides values from 1 up to 50. Call the seq3mpl function once for each value of start. Modify the print statement to also print the value of start. copy and paste your code here