User blog:Denis Maksudov/Super-FGH (SFGH)

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.