User blog comment:Mh314159/A hopefully powerful new system/@comment-39585023-20190630030857/@comment-35470197-20190630050730

Let \(M\) denote the maximum of numbers in the left hand side. The symbols \(n\), \(a\), \(b\), and \(c\), denote positive integers. Then the large numbers described in your notation are bounded in the following way: \begin{eqnarray*} [0] \backslash [n] & = & M \\ [0] \backslash [n][1] & = & f_{0}(M) \\ [0] \backslash [1][a] & = & f_{0}(M) \\ [0] \backslash [2][a] & \sim & f_{2}(M) \\ [0] \backslash [n][a] & \sim & f_{2 \omega}(M) \\ [0] \backslash [n][a,b] & \leq & f_{2 \omega + \omega}(M) \\ [0] \backslash [n][a,b,c] & \leq & f_{2 \omega + \omega + \omega}(M) \\ [0] \backslash [n][n, \ldots,n] & \leq & f_{\omega \times \omega}(M) \\ [0] \backslash [n][n, \ldots,n][n, \ldots,n] & \leq & f_{\omega^{\omega}}(M) \\ [0] \backslash [n][n, \ldots,n] \cdots [n, \ldots,n] & \leq & f_{\omega^{\omega^2}}(M) \\ [1] \backslash [n] & \leq & f_{\omega^{\omega^2}}(M) \\ [1] \backslash [n][1] & \leq & [1] \backslash [n+1] & \leq & f_{\omega^{\omega^2}}(M) \\ [1] \backslash [1][a] & \leq & [1] \backslash [n+1] & \leq & f_{\omega^{\omega^2}}(M) \\ [1] \backslash [2][a] & \leq & f_{\omega^{\omega^2} +1}(M) \\ [1] \backslash [n][a] & \leq & f_{\omega^{\omega^2} + \omega}(M) \\ [1] \backslash [n][a,b] & \leq & f_{\omega^{\omega^2} + \omega + \omega}(M) \\ [1] \backslash [n][a,b,c] & \leq & f_{\omega^{\omega^2} + \omega + \omega + \omega}(M) \\ [1] \backslash [n][n, \ldots,n] & \leq & f_{\omega^{\omega^2} + \omega \times \omega}(M) \\ [1] \backslash [n][n, \ldots,n][n, \ldots,n] & \leq & f_{\omega^{\omega^2} + \omega^{\omega}}(M) \\ [1] \backslash [n][n, \ldots,n] \cdots [n, \ldots,n] & \leq & f_{\omega^{\omega^2} + \omega^{\omega^2}}(M) \\ [p+1] \backslash [n] & \leq & f_{\omega^{\omega^2} \times (p+1)}(M) \\ [p+1] \backslash [n][1] & \leq & [p+1] \backslash [f_{\omega^{\omega^2} \times p}(M)] & \sim & f_{\omega^{\omega^2} \times (p+1)}(M) \\ [p+1] \backslash [1][a] & \leq & [p+1] \backslash [f_{\omega^{\omega^2} \times p}(M)] & \sim & f_{\omega^{\omega^2} \times (p+1)}(M) \\ [p+1] \backslash [2][a] & \leq & f_{\omega^{\omega^2} \times (p+1) +1}(M) \\ [p+1] \backslash [n][a] & \leq & f_{\omega^{\omega^2} \times (p+1) + \omega}(M) \\ [p+1] \backslash [n][a,b] & \leq & f_{\omega^{\omega^2} \times (p+1) + \omega + \omega}(M) \\ [p+1] \backslash [n][a,b,c] & \leq & f_{\omega^{\omega^2} \times (p+1) + \omega + \omega + \omega}(M) \\ [p+1] \backslash [n][n, \ldots,n] & \leq & f_{\omega^{\omega^2} \times (p+1) + \omega \times \omega}(M) \\ [p+1] \backslash [n][n, \ldots,n][n, \ldots,n] & \leq & f_{\omega^{\omega^2} \times (p+1) + \omega^{\omega}}(M) \\ [p+1] \backslash [n][n, \ldots,n] \cdots [n, \ldots,n] & \leq & f_{\omega^{\omega^2} \times (p+1)}(M) \\ [n] \backslash [n] & \leq & f_{\omega^{\omega^2} \times \omega}(M) & \leq & f_{\omega ^{\omega ^2 + 1}}(n) \end{eqnarray*} Therefore the upper bound of your notation is \(\omega^{\omega^2+1}\) in FGH, if I am correct.

In my googological ruler, \([n][n, \ldots,n]\) yields a large number of level 7, \([n, \ldots,n] \cdots [n, \ldots, n]\) yields a large number of level 8, and \([n] \backslash [n]\) yields a large number of level 9. The next level 10 corresponds to ordinals greater than \(\omega^{\omega^{\omega}}\), which is a wall beyond which it is very difficult to go using simple nesting braces without rich structures.