User blog comment:Simplicityaboveall/Insanely Fast-Growing Functions/@comment-28606698-20171025182658

Glad to see you, Joe!

Let consider following comparison:

That is the fast-growing hierarchy


 * $$f_0(n) = n + 1$$
 * $$f_{\alpha+1}(n) = f^n_\alpha(n)$$,
 * $$f_\alpha(n) = f_{\alpha[n]}(n)$$ iff $$\alpha$$ is a limit

And this is your notation


 * $$f(0,k)=f_0(k)=10\cdot k$$
 * $$f(k,n)=f_{k}(n)=f_{k-1}^n(1)=10\uparrow k$$
 * $$h(0,n)=h_0(n)=f(n,10)=10\uparrow^n10 $$
 * $$h(k,n)=h_{k}(n)=h_{k-1}^n(1)$$
 * $$H(n)=\left\{\begin{array}{lcr} h(0,n) \text{ if }n\le 10\\ h(n-10,10) \text{ if }n> 10\\ \end{array}\right.$$

What is the difference? You said nothing about cases of limit ordinals. I suspect you meant something in this kind for case of a limit ordinal less than $$\varepsilon_0$$:

$$f(\Sigma_{i=1}^n \omega^{a_i}\cdot b_i,n)=f(\Sigma_{i=1}^n n^{a_i}\cdot b_i,n)$$ (*)

In this case -yes- $$f(100,10)=f(\omega^2,10)$$ no doubt

and $$f(11,10)=f(\omega+1,10)$$

But in the fast-growing hierarchy

$$f_{\omega+1}(10)$$ is much greater than $$f_{11}(10)$$

Actually in the fast-growing hierarchy

$$f_{\omega+1}(10)=f_\omega(f_\omega^9(10))=f_{f_\omega^9(10)}(f_\omega^9(10))\approx$$

$$\approx f_{10\rightarrow 10\rightarrow 9\rightarrow 2}(f_\omega^9(10))>$$

$$>f_{10\rightarrow 10\rightarrow 9\rightarrow 2}(10)$$

clearly  it is much greater than $$f_{11}(10)$$.

Thus, if we use expression (*) then you right $$h(100,10)=f(\omega^2,10)$$ but if we use standard fast-growing hierarchy then $$f {\omega^2}(10)$$   unimaginably larger  than $$h(100,10)$$