recursion value being repeated in python

Question

def sequence(n):
    while n!=1:
        print(n)
        if n%2==0:
            sequence(n/2)
        else:
            sequence(n*3+1)


sequence(53)

This loop is not terminating. I don't understand why.

You never change n inside your function so the while loop never terminates. Perhaps you meant if n != 1:? — quamrana
– quamrana, Commented Jun 30, 2020 at 15:51

Arkajyoti Banerjee · Accepted Answer · 2020-06-30 15:59:44Z

1

Remove the while loop. You don't need that as the recursive function is doing its job.

Here is the corrected version of your code:

def sequence(n):
        print(n)
        
        if n == 1:
            return
        elif n%2==0:
            sequence(n//2)
        else:
            sequence(n*3+1)


sequence(53)

Change the n/2 to n//2 so that it doesn't become a float.

edited Jun 30, 2020 at 15:59

answered Jun 30, 2020 at 15:52

Arkajyoti Banerjee

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Comments

Red · Accepted Answer · 2020-06-30 23:14:20Z

0

The sequence will end at 2, 2, 2, 2, 2... because no matter what n equals at first, it will go down to 2, and then:

n = 2

2 / 2 = 1     # sequence(1)
1 * 3 + 1 = 4 # sequence(4)
4 / 2 = 2     # sequence(2)
2 / 2 = 1     # sequence(1)
1 * 3 + 1 = 4 # sequence(4)
4 / 2 = 2     # sequence(2)
...           # ....

answered Jun 30, 2020 at 23:14

Red

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