User blog comment:B1mb0w/Alpha Function and f epsilon nought/@comment-5529393-20150813003805

You are correct that $$D(m,m,0,\ldots,0) \approx f_{\omega^2}(m)$$ where there are m-1 zeroes. But the following conjectures are off.

Since $$D(m,m,0,\ldots,0) \approx f_{\omega^2}(m)$$, we will have $$f_{\omega^2}(f_{\omega^2}(m)) \approx D(m,m,0,\ldots, 0)$$ where there are $$f_{\omega^2(m) - 1$$ zeroes, or approximately $$D(m,m,0,\ldots,0)$$ zeroes. Call that number $$D_2$$. Then $$f_{\omega^2}(f_{\omega^2}(f_{\omega^2}(m)))$$ will be about $$D(m,m,0,\ldots,0)$$ where there are $$D_2$$ zeroes, and the pattern continues this way. In general, if f(m) = D(m,m,0,...,0) with m-1 zeroes, $$f_{\omega^2}^n(m) \approx f^n(m)$$; since $$f_{\omega^2+1}(m) f_{\omega^2}^m(m)$$, we have $$f_{\omega^2+1}(m) \approx f^m(m)$$.

In terms of the alpha function, $$f(m) \approx \alpha(2^m)$$, so f_{\omega^2+1(m) \approx f^m(m) \approx \alpha(2^{\alpha(2^{\alpha(\cdots)}} with m $$\alpha$$'s, which is a little bigger than $$\alpha^m(2^m)$$.

Going to the next step in the hierarchy, we have $$f_{\omega^2+1}^2(m) \approx f^{f^m(m)}(f^m(m)) > \alpha^{\alpha^m(2^m)}(\alpha^m(2^m)}$$, and $$f_{\omega^2+1}^3(m) \approx f^{f^{f^m(m)}(f^m(m))}(f^{f^m(m)}(f^m(m))} > \alpha^{\alpha^{\alpha^m(2^m)}(\alpha^m(2^m)}}(\alpha^{\alpha^m(2^m)}(\alpha^m(2^m)})$$. Thus $$f_{\omega^2 + 2}(m) = f_{\omega^2+1}^m(m) > \alpha^{\alpha^{\cdots^{\alpha^m(2^m)}\cdots}(2^m}(2^m)$$.

Note that $$\alpha^{\alpha^m(2^m)}(\alpha^m(2^m)}$$ grows faster than $$\alpha^{m\uparrow\uparrow m}(2^m)$$, since $$\alpha^m(2^m)$$ grows faster than $$m \uparrow \uparrow m$$, but the former is still less than $$f_{\omega^2+2}(m)$$.