The number of N-bit print statements whose output starts with the n-bit Hamlet is exactly 2^(N-n). If K(Hamlet)=n/6, then there are at least 2^(N-n/6) programs whose output starts with Hamlet. Probably more.
EDIT: Ooops, I didn’t prove anything about the structuredness of the remaining N-n bits, so this is not too useful in itself.
The number of N-bit print statements whose output starts with the n-bit Hamlet is exactly 2^(N-n). If K(Hamlet)=n/6, then there are at least 2^(N-n/6) programs whose output starts with Hamlet. Probably more.
EDIT: Ooops, I didn’t prove anything about the structuredness of the remaining N-n bits, so this is not too useful in itself.