User blog comment:B1mb0w/Strong D Function/@comment-5529393-20150805054856

I'm afraid that you have gotten off in your calculations. Hyp Cos's analysis is correct:

$$D(m,n) \approx f_m(n)$$ (ignoring some additive constants)

$$D(m,n,n) \approx f_{\omega+m)(n)$$ (ditto)

$$D(m,n,n,n) \approx f_{\omega \cdot 2 + m} (n)$$

$$D(m,n,\ldots,n) \approx f_{\omega \cdot (p-2) + m}(n)$$ where p is the number of entries in D.

so the strength of the D function tops off at $$\omega^2$$. The reason that your calculations go off course is your insistence on always applying the fast-growing hierarchy to the number 3. You should use a general n instead.

Let's take for example your calculation that $$D(3,0,1) > f_{\omega+3}(3)$$. That looks all right, and gives you the right idea about the growth rate of D(3,0,1), since D(3,0,n) grows at roughly $$f_{\omega+3}(n)$$. The problem is that you change the right hand side to $$f_{\omega*2}(3)$$. This is technically true, since $$f_{\omega*2}(3) = f_{\omega+3}(3)$$, but it is misleading because D(3,0,n) grows nowhere near as fast as $$f_{\omega*2}(3)$$. By replacing the ordinal, you have fooled yourself into thinking that you have somehow graduated to a new growth rate, but this is not the case; it is just because the variable is only 3 that D(3,0,n) can reach $$f_{\omega*2}(3)$$ for a reasonable n. $$f_{\omega*2}(4)$$ is equal to $$f_{\omega+4}(4)$$ and will exceed D(3,0,n) for any reasonably small n.

Indeed, in your next calculation you have $$D(4,0,1) > f_{\omega*2+1}(3)$$, but this is impossible, since D(4,0,n) is approximately $$f_{\omega+4}(n)$$, whereas $$f_{\omega*2+1}(3)$$ will exceed $$f_{\omega+n}(n)$$ for any reasonably small n.

To make a long story short; you go wrong by always using 3 as your variable, and then "graduating" your ordinals upwards just because they match when evaluated at 3.