[Solved] invalid syntax on int(number) on a try statement python

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So to fix your code as described:

def collatz(number):
    number = input()         # number is passed in - doesn't make sense to get it again
    try:
        number = int(number) # Just interpret once
    except ValueError:       # try must have an except clause
        return               # exit on error

    if number%2 == 0:
        print('hi')
    elif number%2 == 1:      # Don't really need an elif - %2 can only have 2 values
        print(number*3+1)

However, you may want to look into using a generator (yield):

def collatz(number):
    while number > 1:
        if number % 2 == 0:
            number //= 2
        else:
            number = 3*number + 1
        yield number

def collatz_series():
    try:
        number = int(input())
    except ValueError:
        return

    for n in collatz(number):
        print(n)

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solved invalid syntax on int(number) on a try statement python