In this Blog I am going to describe how to get past the 52nd hyperoperator trap or Z to do so we will need to employ some recursion.

Lets suppose we start with (3,f,3)=3^^^^3 this is a relatively small number as it is only requires 3 symbols to express.

(3(3,f,3)3) now you've got the (3,f,3)th hyper operator being applied to the two outermost 3's,this means that you have done 1 recursion. We write it like this (3,f,3#1). We can this make (3,f,3#1) the hyperoperator we will be using by recursing it again like so: (3(3(3,f,3)3)3)=(3,f,3#2). (3(3(3(3,f,3)3)3)3)=(3,f,3#3). (3(3(3(3(3,f,3)3)3)3)3)=(3,f,3#4). As you can see these numbers are going to be pretty big but they could still be beaten by something like (3,f,3#1000).

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