[reuben364]

Since the de Bruijn indices are limited (and presumably still Turing complete), I wonder how limited you can make them and still be Turing complete.

[jorkingit]

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 ;-)