Comment by leni536

In your notation B(n)=A[n](n), where [] denotes the number of iterations. So C(n)=B[n](n)=(A[n])[n](n)=A[nn](n), so really iterA(n)=A[n^n](n). I you repeat this procedure with iterA, then you get iterA2(n)=A[n^(n^n)](n) . While n^n, n^n^n and so on are definitely fast growing you are better of putting A into []. So you can write an extremely fast growing function as f(n)=A[A(n)](n). So one can improve your strategy by using the [] notation, resolving the recursion and putting large functions inside.

And of course my method can improved with iterations that also can be resolved by creating a new notation. It's really never ending.

A[A[A(n)](n)](n)

And it's all computable, so the Busy Beaver grows faster than any* of these.