Discussion on answer by JeffE: How can it be decidable whether $\pi$ has some sequence of digits?

@ShelbyMooreIII "f(n) requires a Turing-complete programming language because it requires unbounded recursion to expand the digits of π" -- that's a non-sequitur, and JeffE has proven that $f$ does not have to expand any digits. "If you mess this up, you need to go back to Compsci 101" -- agreed. See you there, I'll be your TA.