Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to find the 2nd term of continued fraction for the power tower ${^5}e=e^{e^{e^{e^{e}}}}$ ( i.e. $\lfloor\{e^{e^{e^{e^{e}}}}\}^{-1}\rfloor$), or even higher towers. The number is too big to process in reasonable time with numerical libraries or algorithms known to me — the 1st term of the continued fraction has more than $10^{10^6}$ decimal digits. Is there a trick that allows to do such calculations faster?

UPDATE: I created a sequence A225053 in OEIS for 2nd terms of continued fractions for power towers $e,\,e^e,\,e^{e^e},\,\dots$ Please feel free to extend it if you find a way to calculate further terms.

share|improve this question
    
Have you looked at Euler's continued fraction formula for the exponential function? en.wikipedia.org/wiki/… It probably won't help you get an exact answer, but it may show you that it's not 0. –  Brian Rushton Apr 29 '13 at 22:04

1 Answer 1

up vote 16 down vote accepted
+200

It might be not a direct answer to your question, but it is possible that there is no 2nd term of the continued fraction in question.

I believe it is a long-standing open problem if $\,{^5 e}\in\mathbb{N}$, and, in general, for every integer $n \ge 5$, if $\,{^n e}\in\mathbb{N}$ (and also, for every integer $n \ge 4$, if $\,{^n \pi}\in\mathbb{N}$).

It is mentioned multiple times in Wikipedia, e.g. in Russian Wikipedia article Открытые математические проблемы (Open mathematical problems) and on some mailing lists. Similar questions were discussed here on Math.SE: $e^{e^{e^{79}}}$ and ultrafinitism, How to show $e^{e^{e^{79}}}$ is not an integer. Currently these questions look very far from being resolved and it is completely unclear how to approach them.

share|improve this answer
6  
I wonder, does anybody really believe it can be an integer? –  Liu Jin Tsai May 5 '13 at 19:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.