# What is known about functions of type $f(n+1) = f(n)^ {f(n)}$?

Update : having looked at Knut's Double Arrow Notation ( Thank you DJC), it seems that this question is nothing more than a frivourless wondering that should be undertaken by whom ever wonders it, it is an example of type of questions that are easy to ask but seem to have no defined way of being answered or having a systematic use in mathematics, I have voted for it to close (if that is the correct thing todo in this situation).

$f(n+1) = \left(f(n)\right)^ {f(n)} , f(1)=2$ e.g. $f(2)=2^2 , f(3)=\left({2^2}\right)^{2^2},f(4)={\left({2^2}\right)^{2^2}}^{\left({2^2}\right)^{2^2}}$, ok it just another fast growing function, but it looks too simple that makes one wonder if there is an analytic version for it exists? e.g. what can be said about the $f(2.5)$ or $f(1+i)$ or it's inverse function or it's derivative, taylor series? of course even any result as relating to something as simple as $f(n+\frac{1}{2})$ seems intresting.

Edit : just looking for any known results for such a simple looking construction, tx

• Is this a question, or a fishing expedition? – Gerry Myerson Jun 10 '11 at 12:14
• @Gerry, tried to tidy it up as much as I could. – Arjang Jun 10 '11 at 13:34
• What's a fishing expedition? – futurebird Jun 10 '11 at 13:46
• en.wikipedia.org/wiki/Knuth's_up-arrow_notation – JavaMan Jun 10 '11 at 13:55
• J.E. Littlewood introduced analogue of the "double arrow" even earlier than Knuth, at 1948 and it is reprinted in a chapter "large numbers" in his "A Mathematician's Miscellany". He even used some idea about numbers like $N_{2.47}$. It is here archive.org/details/mathematiciansmi033496mbp page 100. – Alex 'qubeat' Jun 13 '11 at 12:46

Let $f(n)=a^{g(n)}$ , where $a\in\mathbb{R}^+$ and $a\neq1$ ,

Then $a^{g(n+1)}=(a^{g(n)})^{a^{g(n)}}$

$a^{g(n+1)}=a^{g(n)a^{g(n)}}$

$g(n+1)=g(n)a^{g(n)}$

Let $g(n)=a^{h(n)}$ , where $a\in\mathbb{R}^+$ and $a\neq1$ ,

Then $a^{h(n+1)}=a^{h(n)}a^{a^{h(n)}}$

$a^{h(n+1)}=a^{h(n)+a^{h(n)}}$

$h(n+1)=h(n)+a^{h(n)}$

which the solution is more complicated than the iterated exponentials.