```
def countAndSay(self, n):
def count(i):
# base case for recursion, case = 2 b/c recursive step requires 2 character input
if i == 1: return "1"
if i == 2: return "11"
counter = 1
res = ""
# get previous output
prev = count(i-1)
for i in xrange(1, len(prev)):
if prev[i] == prev[i-1]:
counter += 1
else:
res += str(counter) + prev[i-1]
counter = 1
return res + str(counter) + prev[len(prev) - 1]
return count(n)
```