Comparing conway chains See
https://en.wikipedia.org/wiki/Conway_chained_arrow_notation
for the details how conway chained arrow notation works.


*

*I want to calculate the approximate value $n$ such that


$$n\rightarrow n\rightarrow n\rightarrow n<a\rightarrow a\rightarrow a\rightarrow a\rightarrow a<(n+1)\rightarrow (n+1)\rightarrow (n+1) \rightarrow(n+1)$$
for $a\ge 3$
The number $n\rightarrow n\rightarrow ... \rightarrow n$ with $n$ $n's$ is 
approximately $\large f_{\omega^2}(n)$ (See https://en.wikipedia.org/wiki/Fast-growing_hierarchy for details of the fast growing hierarchy), but this does
not seem to help very much.
Is it true that $n$ must exceed $a\rightarrow a\rightarrow a\rightarrow a$ ?
I think it is, but I do not know how I can prove it.
 A: Unfortunately, such things are extermely hard to prove formally, because the smaller thing has a greater base. However I hope this approximation argument can help. We have $n\rightarrow n \rightarrow n \rightarrow n \approx f_{\omega2}(n)$, so if $n=a\rightarrow a \rightarrow a \rightarrow a$, then 
$$n\rightarrow n \rightarrow n \rightarrow n \approx f_{\omega2}(f_{\omega2}(a))$$
However we have
$$a\rightarrow a \rightarrow a \rightarrow a \rightarrow a \approx f_{\omega3}(a)=f_{\omega2+a}(a)>f_{\omega2+1}(a)=f_{\omega2}^a(a)\\>f_{\omega2}^2(a)\approx n\rightarrow n \rightarrow n \rightarrow n$$
Since we make quite large overestimations, it is quite clear.
The actual value of $n$ is actually larger than $k=a\rightarrow a \rightarrow a \rightarrow (a-1) \rightarrow a$ for sufficiently large $a$. Again, using the fast growing hierarchy, we have 
$$a\rightarrow a \rightarrow a \rightarrow a \rightarrow a \approx f_{\omega3}(a)=f_{\omega2+a}(a)=f_{\omega2+a-1}^a(a)>f_{\omega2+a-1}^a(a-1)=f_{\omega2+a-1}(f_{\omega2+a-1}^{a-1}(a-1))=f_{\omega2+a-1}(f_{\omega2+a}(a-1))\approx k\rightarrow k \rightarrow k \rightarrow k$$
This uses the approximations I gave you in another answer. 
