User blog comment:Simplicityaboveall/Insanely Fast-Growing Functions/@comment-1605058-20170915175750/@comment-5529393-20170920234726

Simplicity, you're misunderstanding how the fast-growing hierarchy works. Just because you are using n=10 doesn't mean you can replace ω with 10 whenever you want and you'll get the same number. If that were the case, then

$$f_{\varepsilon_0}(10) = f_{\omega \uparrow\uparrow \omega}(10) = f_{10 \uparrow\uparrow 10}(10) < f_{\omega}(f_3(10))$$

What a horribly weak hierarchy that would be!

Even at ω2+1 your method goes off. You would get

$$f_{\omega 2+1}(10) = f_{\omega +10+1}(10) = f_{\omega +11}(10)$$

but in fact

$$f_{\omega 2+1}(10) = f_{\omega 2}(f_{\omega 2}(f_{\omega 2}(f_{\omega 2}(f_{\omega 2}(f_{\omega 2}(f_{\omega 2}(f_{\omega 2}(f_{\omega 2}(f_{\omega 2}(10))))))))))$$

where already

$$f_{\omega 2}(f_{\omega 2}(10)) = f_{\omega 2}(f_{\omega + 10}(10)) = f_{\omega + f_{\omega + 10}(10)}(f_{\omega + 10}(10))$$

and that is already much greater than what you would get with your method.

Put simply, you think that the fast=growing hierarchy is much, much slower than it actually is, so no wonder you are able to ascend it so fast.

You tell me that "nowhere have I defined h(n,n)". In fact right at the beginning of page 3 you define h(k,n), and of course we can set k = n.

You say "In my work, n is limited to 10". If this is true, then you can't get anywhere. Right from your definition of h(k,n)

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

that definition doesn't work at all unless n is allowed to be of all different sizes. So you can't limit n to 10. If you really wanted to limit n to 10, then you shouldn't use n at all and just use 10 everywhere.

For that matter, why do you think limiting n to 10 answers our objections? We've explained why your function doesn't go nearly as high up the fast-growing hierarchy as you think it does. Why would limiting n to 10 make your function grow so much faster, rather than having a more general function that allows n to take on all values?