I suspect the answer is 3: SKI combinator calculus is Turing complete and you need 3 de Bruijn indices to define S. Good call! I got rid of all numbers above 2

I suspect the answer is 3: SKI combinator calculus is Turing complete and you need 3 de Bruijn indices to define S. Good call! I got rid of all numbers above 2, I can't count that high anyway ;-)