Given an integer n, generate the nth count and say sequence
In each step of the "count-and-say sequence" (which is more usually called the "look-and-say sequence") you have to find the groups of consecutive runs of identical digits. So if you have the value 111221, these groups are 111, 22, and 1. Python has a built-in function itertools.groupby for finding groups in an iterator, and using this function, the look-and-say step becomes:

new_value = ''
for digit, group in groupby(value):
    count = sum(1 for _ in group)
    new_value += str(count) + digit
value = new_value


which can be rearranged into a single expression:

value = ''.join(str(sum(1 for _ in group)) + digit
                for digit, group in groupby(value))


and so the whole sequence can be generated like this:

from itertools import groupby

def look_and_say(value='1'):
    "Generate the look-and-say sequence starting at value."
    while True:
        yield value
        value = ''.join(str(sum(1 for _ in group)) + digit
                        for digit, group in groupby(value))


To select only the \$n\$th value of the sequence, use itertools.islice:

next(islice(look_and_say(), n - 1, None))