Well, I guess it basically goes to show how hard it is to determine the actual kolmogorov complexity (even once specifying a particular encoding of a Turing Machine) of a string, but the following is a new upper bound on the complexity (in Python) of the second string.
In general, any string of 1s and 0s will be compressible using this method in python once the binary number is greater than 212 (you break even at 211). However, again, I only claim this to be an upperbound ;). (also assuming that invoking built-ins isn't cheating).
In the context of Turing machines, the output is really a _binary_ string, not a string of arbitrary characters. So in a sense, those two programs are the same thing. But of course, the Kolmogorov complexity of any number n is bounded by log(n) + c.
print '00011101001000101101001000101111010100000100111101'
print bin(128141569638717)[2:]
In general, any string of 1s and 0s will be compressible using this method in python once the binary number is greater than 212 (you break even at 211). However, again, I only claim this to be an upperbound ;). (also assuming that invoking built-ins isn't cheating).