User blog comment:Simplicityaboveall/The Construction of Extremely Large Numbers/@comment-5529393-20160724085851

First off, props for putting some effort into this. However, it appears that you have a significant misunderstanding regarding the fast-growing hierarchy; namely, that you can evaluate $$f_\alpha(n)$$ by replacing $$\omega$$ with n everywhere in $$\alpha$$

On page 7, you appear to say that $$f_{20}(10) = f_{\omega 2}(10)$$ In fact $$f_{\omega 2}$$ is much, much greater. Even $$f_{\omega+1}(10)$$ is much greater, since $$f_{\omega+1}(10) = f_{\omega}^{10}(10)$$, and already $$f_{\omega}^2(10) = f_{\omega}(f_\omega(10)) = f_\omega(f_{10}(10)) = f_{f_{10}(10)}(f_{10}(10)}$$, and you can see that the subscript of $$f_{10}(10)$$ is much greater than 20. You make similar mistakes elsewhere, for example saying that $$f_{\varepsilon_0}(10) = f_{10\uparrow\uparrow 10}(10)$$; again, since $$ 10\uparrow \uparrow 10 < f_{10}(10)$$, $$f_{10\uparrow\uparrow 10}(10) < f_{\omega}^2(10) < f_{\omega+1}(10)$$.  Even with your expression for $$f_{\Gamma_0}(10)$$, we have $$f_{\lbrace 10, 10, 1, 2 \rbrace}(10) < f_{f_{\omega+1}(10)}(10) < f_\omega(f_{\omega+1}(10)) < f_{\omega+2}(10)$$.  So you can see that your method of evaluation severely underestimates how fast the hierarchy grows after $$\omega$$