User blog comment:Eners49/New Notation Idea?/@comment-5529393-20180624004658

So, [n] is approximately \(f_{\omega+1}(n)\), i.e. it is at level w+1.

[n,n] iterates [n] [n] times, so it is at level w+2. (Technically, it is about \(f_{\omega+2}(f_{\omega+1}(n))\), which is a little better than \(f_{\omega+2}(n)\), but the \(f_{\omega+1}\) is weaker than the \f_{\omega+1}\), and won't make much difference in the long run.)

[n,n,n] iterates [n,n] twice, so it is about \(f_{\omega+2}^2(n)\). Not at w+3 yet.

[n,n,n,n] iterates [n,n,n] twice, so it basically iterates [n,n] four times for \(f_{\omega+2}^4(n)\). Still not at w+3.

2[n] = [1,1,1,...,1] with n 1's basically iterates [n,n] 2^(n-2)-1 times starting from [1,1]. So it is about \(f_{\omega+2}^{2^{n-2}-1}([1,1])\), which is roughly \(f_{\omega+3}(n)\). So we have reached w+3.

2[n,n] = 2[ [n,n,n,...n] ] is roughly iterating 2[n] twice, so it is about \(f_{\omega+3}^2(n)\).

2[n,n,n,...,n] with n+1 n's is 2[ [ [n,n,...,n], [n,n,...,n],..., [n,n,...,n] ] ] which is iterating 2[n] three times, for roughly \(f_{\omega+3}^3(n)\).

3[n] is thus at roughly \(f_{\omega+3}^3(n)\).

4[n] adds another two iterations, so we get roughly \(f_{\omega+3}^5(n)\).

n[n] therefore is roughly \(f_{\omega+3}^{2n-3}(n)\), or a little more than \(f_{\omega+4}(n)\).

{n} is [1][1][1]...[1], so we are iterating n[1] n-1 times starting from [1], so it is a little less than \(f_{\omega+5}(n)\).

We have advanced four steps going from [n] to {n}, so if we repeat the same definitions again with { } we will get

{1}{1}{1}...{1} with n 1's is approximately \(f_{\omega+9}(n)\).

Still a long way from w^w, I'm afraid. Keep trying!