Kolmogorov complexity is uncomputable because it admits Turing complete languages, and reduces to the halting problem. If the language you admit isn't Turing co

Kolmogorov complexity is uncomputable because it admits Turing complete languages, and reduces to the halting problem. If the language you admit isn't Turing complete, then the Kolmogorov complexity of the thing is computable. It looks like OP's language is not Turing complete. It always terminates. You can just do a breadth first search on the program space. The first program you get that outputs the number you want is the shortest program. If it were Turing complete, you can't do this, because eventually you'll find a program that just keeps running for like a really long time. Is it running because the program never halts? Or is will it halt eventually and output the number you want? You can't know for sure.