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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.

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  • $\begingroup$ 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 $n=a \rightarrow a \rightarrow a \rightarrow a \approx f_{\omega2}(a)$, then $$n \rightarrow n \rightarrow n \rightarrow n \approx f_{\omega2}(f_{\omega2}(a))$$ However $$a \rightarrow a \rightarrow a \rightarrow a \rightarrow a \approx f_{\omega3}(a)=f_{\omega2+a}(a)>f_{\omega2+1}(a)=f_{\omega2}^a(a)>f_{\omega2}^2(a)=f_{\omega2}(f_{\omega2}(a))$$ $\endgroup$ – wythagoras Jul 27 '15 at 9:53
  • $\begingroup$ And that was exactly 600 characters. However I don't think it is an answer. If you think it is an answer, I would be happy to post it as an answer. $\endgroup$ – wythagoras Jul 27 '15 at 9:55
  • $\begingroup$ Thanks again for your insights! Well, if you wish, you can formulate it as an answer. $\endgroup$ – Peter Jul 27 '15 at 16:16
  • $\begingroup$ Ok, so it is the actual intention that $b$ counts the number of arrows. I thought it would be an error... I will delete my last posts ... $\endgroup$ – Peter Jul 27 '15 at 16:39
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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.

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  • $\begingroup$ You can also look at this question : math.stackexchange.com/questions/1276161/… . I already got an answer with great insights, but maybe you can estimate the magnitude of this gigantic number with the fast-growing-hierarchy. Is it like $\large f_{\omega^{\omega^2}}(2)$ ? $\endgroup$ – Peter Jul 27 '15 at 16:47

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