User blog comment:Simplicityaboveall/Insanely Fast-Growing Functions/@comment-1605058-20170915175750/@comment-5529393-20170926020516

Simplicity,

I know that you never explicity said that $$F_{\omega\rightarrow\rightarrow\omega}(10)$$ is equal to $$F_{10\rightarrow\rightarrow 10}(10)$$, I gave it as an example of how things go very wrong if you just start replacing $$\omega$$ by 10 willy-nilly. As an example, you then go and say that

$$ F_{\omega\rightarrow\rightarrow\omega}(10) = H(10\rightarrow\rightarrow 10) = h(10\rightarrow\rightarrow 10 - 10, 10)$$

but we all agree that

$$h(10\rightarrow\rightarrow 10 - 10, 10) \sim F_{\omega + 10\rightarrow\rightarrow 10 - 10}( 10)$$

Saying that $$F_\varepsilon_0(10) = F_{\omega + 10\rightarrow\rightarrow 10 - 10}( 10)$$ is basically as bad as saying $$F_\varepsilon_0(10) =  F_{10\rightarrow\rightarrow 10}( 10)$$! $$F_\varepsilon_0(10)$$ is ENORMOUSLY bigger.

Please look through the examples given by me and Denis that demonstrate why you get very different numbers if you replace $$\omega$$ by 10. Examining $$F_{\omega 2 + 1}$$ is a prime example.

You then say, "H(100) is obtained from the expression at the beginning of page 3 of 61 as such :" and then give a description of changing 100 to $$\omega^2$$. But that is not how H(n) is defined on pages 2 and 3! You very clearly say

H(n) = h(n-10,10)

and

$$h(k,n) = h_k(n) = h_{k-1}^n(n)$$

where $$h(0,n) = 10 \uparrow^n 10$$.

And that's the reason why we've been saying that $$H(100,n) \sim F_{\omega + 90}(10)$$, because it follows from your definitions. And if you say that $$F_{\omega + 90}(10)$$ is the same as $$F_{\omega^2}(10)$$ because you can just make $$\omega$$ equal 10, we've explained many times why it does not.