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It is possible to accelerate growth of hierarchy if FGH will be nested in each step of this hierarchy. Let's define

$s_0(n)=n+1$,

$s_{\alpha+1}(n)=FGH_{\beta}(s_{\alpha}(n))$,

$s_{\alpha}(n)=s_{\alpha[n]}(n)$ iff $\alpha$ is a limit ordinal.

Let $\beta = \omega$ that means $FGH_{\omega}(s_{\alpha}(n))$ is equal to $f_{\omega}(n)$ for usual FGH with $f_0 (n)=s_{\alpha}(n)$

(in other words $f_0 (n)=s_{\alpha}(n)$, $f_{\alpha+1} (n)=f_{\alpha}^n (n)$, $f_{\alpha} (n)=f_{\alpha[n]} (n)$ iff $\alpha$ is a limit ordinal).

In this case $s_{\alpha+1}(n)=FGH_{\omega}(s_{\alpha}(n))\approx s_{\alpha}(n)\uparrow^n n$

where $s_{\alpha}(n)\uparrow b=s_{\alpha}^b(n)=\underbrace{s_{\alpha}(s_{\alpha}(\cdots(s_{\alpha}(s_{\alpha}(n)))\cdots))}_{b \quad s_{\alpha}'s}$

and $s_{\alpha}(n)\uparrow^{k+1} b=\underbrace{s_{\alpha}(n)\uparrow^{k}(s_{\alpha}(n)\uparrow^{k}(\cdots(s_{\alpha}(n)\uparrow^{k} n)\cdots))}_{b \quad s_{\alpha}'s}$.

Example n=3: then $s_0(3)=3+1=4$,

$s_1(n)=f_{\omega}(n) \approx2\uparrow^{n-1}n$ and $s_1(3)\approx 2 \uparrow \uparrow 3$.

For the calculation $s_2(3)$ we must start new FGH (let it be $f'_{\alpha} (n)$ to show difference) with $f'_0 (n)=s_{1}(n)\approx2\uparrow^{n-1}n$

then $f'_{1} (3) \approx \left. \begin{matrix} &&\underbrace{2 \uparrow \uparrow \cdots \uparrow \uparrow}3\\ & &\underbrace{2 \uparrow \uparrow \cdots \uparrow \uparrow} 3\\ & & 2 \uparrow \uparrow 3\\ \end{matrix} \right \} \text {3 layers}\approx\lbrace3,3,1,2\rbrace \approx f_{\omega+1} (3)$

$f'_{2} (3) \approx \lbrace3,3,2,2\rbrace \approx f_{\omega+2} (3)$

and $f'_{\omega} (3)=f'_{3} (3) \approx \lbrace3,3,3,2\rbrace \approx f_{\omega+3} (3)$

This way $s_2(n)=f'_{\omega} (n)=f'_{n} (n) \approx \lbrace n,n,n,2\rbrace \approx f_{\omega.2} (n)$

and $s_m (n) \approx \lbrace n,n,n,m\rbrace \approx f_{\omega.m} (n)$

After this we must overcome the limit ordinal $\omega$ (Indeed $FGH_{\omega}(f(x))$ can be expressed as huge amount of iterations of some function $f(x)$ )

$s_{\omega+1}(3)=FGH_{\omega}(s_{\omega}(3)) \approx s_{\omega}(3) \uparrow^3 3 \approx f_{\omega^2+\omega} (n)$

and so on.

So if $\beta=\omega$ then $s_{\omega^{\alpha}}(n) \approx f_{\omega^{\alpha+1}}(n)=f_{\omega^{\alpha}.\omega}(n)$ (comparing with usual FGH with $f_0(n)=n+1$) and that is why in this case $s_{\omega^{\omega}+1}(n) \approx f_{\omega^{\omega}+1}(n)$. As I remember, it is named "catching ordinal" ($\omega^\omega$ for this case)

To go further we can imagine for example $FGH_{\beta}=FGH_F$

$s_{\alpha+1}(n)=FGH_F(s_{\alpha}(n))$ where $FGH_F(s_{\alpha}(n))=f_{F}(n)$ for usual FGH with $f_0(n)=s_{\alpha}(n)$, where $F=\theta_{\varepsilon_{\Omega_\omega + 1}}(0)$ is Takeuti-Feferman-Buchholz ordinal - The supremum of the range of the Feferman theta function,

or even $FGH_{\beta}=FGH_R$

$s_{\alpha+1}(n)=FGH_R(s_{\alpha}(n))$ where $FGH_R(s_{\alpha}(n))=f_{R}(n)$ for usual FGH with $f_0(n)=s_{\alpha}(n)$, where $R= \psi_{\omega_1}(\chi_{\varepsilon_{\text {M+1}}}(0))$ is Rathjen's ordinal - The supremum of the range of the Rathjen's psi function (And, as I know, largest ordinal which is well-defined in professional math).

This way we can reach $s_{R}(n)$. Let $s_{R}(10^{100})$ is Roogol.

Extensions

Let's define

$s'_0(n)=n+1$,

$s'_{\alpha+1}(n)=SFGH_{R}(s'_{\alpha}(n))$,

$s'_{\alpha}(n)=s'_{\alpha[n]}(n)$ iff $\alpha$ is a limit ordinal.

where $SFGH_{R}(s'_{\alpha}(n))$ is equal to $s_{R}(n)$ for SFGH with $s_0 (n)=s'_{\alpha}(n)$

Let's name this hierachy as SFGH^1 and notate $s'_{\alpha}(n)=s_{\alpha}^{(1)}(n)$. After this same way we can define $s_{\alpha}^{(2)}(n)$ and $s_{\alpha}^{(3)}(n)$ and so on.

Let $m(n)=s_{R}^{(n)}(n)$ and Roogolplex is equal to $m(10^{100})$

One more definition of SFGH

Let's define SFGH as follows:

$$F_\alpha\uparrow^{\beta} 0(n)=n$$ if $$\beta=0$$ or $$\beta=\gamma+1$$

$$F_\alpha\uparrow^0 (m+1)(n)=F_\alpha(F_\alpha\uparrow^0 m(n))$$

$$F_\alpha\uparrow^{\beta+1} (m+1)(n)=F_\alpha\uparrow^{\beta}(F_\alpha\uparrow^{\beta+1} m(n))(n)$$

$$F_\alpha\uparrow^{\beta} m (n)=F_\alpha\uparrow^{\beta[m]}m(n)$$ if $$\beta$$ is a limit ordinal

$$F_0(n)=10^n$$

$$F_{\alpha+1}(n)=F_\alpha\uparrow^{\alpha+1} n(n)$$

$$F_\alpha(n)=F_{\alpha[n]}\uparrow^{\alpha}n(n)$$ if $$\alpha$$ is a limit ordinal.

Under this definition:

$$F_\alpha\uparrow^0m(n)=F_\alpha^m(n)=\underbrace{F_\alpha(F_\alpha(...(F_\alpha(n))...))}_{m\quad F's}$$

$$F_\alpha\uparrow^1m(n)=\underbrace{F_\alpha^{F_\alpha^{...^{F_\alpha^n(n)}...}(n)}}_{m\quad F's}(n)$$

$$F_\alpha\uparrow^2m(n)=\left. \begin{matrix} \underbrace{F_\alpha^{F_\alpha^{...^{F_\alpha^n(n)}...}(n)}(n)} \\ \underbrace{F_\alpha^{F_\alpha^{...^{F_\alpha^n(n)}...}(n)}(n)}\\ \vdots\\ \underbrace{F_\alpha^{F_\alpha^{...^{F_\alpha^n(n)}...}(n)}(n)}_{n}\\ \end{matrix} \right \} \text{m lower braces}$$

and so on.

$$F_\alpha\uparrow^{\omega+1}m+1(n)=F_\alpha\uparrow^\omega(F_\alpha\uparrow^{\omega+1}m(n))(n)=F_\alpha\uparrow^{F_\alpha\uparrow^{\omega+1}m(n)}(F_\alpha\uparrow^{\omega+1}m(n))(n)$$

Hmm, seems cool. But did it really gives significant acceleration for FGH? No. I guess:

$$F_\alpha(n)\approx f_{\alpha^2}(n)$$

where $$F$$ is the mentioned SFGH and $$f$$ is usual FGH.