In Response to the Ackermann function answer, this is a pretty straightforward convert-the-call-stack-into-a-real-stack problem. This also shows one benefit of the iterative version.
on my platform (Python 3.1rc2/Vista32) the iterative version calculates ack(3,7) = 1021 fine, while the recursive version stackoverflows. NB: it didn't stackoverflow on python 2.6.2/Vista64 on a different machine, so it seems to be rather platform-dependant,
(Community wiki because this is really a comment to another answer [if only comments supported code formatting .... ])
def ack(m,n):
s = [m]
while len(s):
m = s.pop()
if m == 0:
n += 1
elif n == 0:
s.append(m-1)
n = 1
else:
s.append(m-1)
s.append(m)
n -= 1
return n
