And yet another way is this: every computable enumeration of all programs is equivalent to a UTM, and therefore generates the same measure of Kolmogorov complexity, up to the usual additive constant.

Proof: Computability of the enumeration means that there is a program which takes any integer N and returns the Nth in that enumeration. Represent this program as a Turing machine and cascade it with a second Turing machine that executes the program returned by the first TM.

Is this a novel observation? Does it go any distance towards answering Shalizi's question at the end of this note?