In this question, I needed to assume in my answer that $e^{e^{e^{79}}}$ is not an integer. Is there some standard result in number theory that applies to situations like this?

After several years, it appears this is an open problem. As a non-number theorist, I had assumed there would be known results that would answer the question. I was aware of the difficulty in proving various constants to be transcendental -- such as $e + \pi$, which is not known to be transcendental at present.

However, I was looking at a question that seems simpler, naively: whether a number is an integer, rather than whether it is transcendental. It seems that what appeared to be possibly simpler is actually not, with current techniques.

The main motivation for asking about this particular number is that it is very large. It is certainly possible to find a pair of very large numbers, at least one of which is transcendental. But the current lack of knowledge about this particular number is even an integer shows just how much progress remains to be made, in my opinion. Any answers that describe techniques that would suffice to solve the problem (perhaps with other, unproven assumptions) would be very welcome.

Carl Mummert
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    This is likely to be open, I think. – Qiaochu Yuan Dec 05 '10 at 08:21
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    If we don't even know if $\exp(e)$ is algebraic or not... – J. M. ain't a mathematician Dec 05 '10 at 09:41
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    @Qiaochu Yuan and J.M.: I wasn't aware of that. I knew (vaguely) that there are a lot of open problems in transcendence theory, but I was hoping that just the problem of being an integer was easier. – Carl Mummert Dec 05 '10 at 12:56
  • This may be useful: http://xxx.lanl.gov/pdf/1012.0822 – graveolensa Dec 06 '10 at 02:09
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    @Carl: hmm. Okay, so if we can compute the number to a precision of better than, say, 1/10, then we can probbly tell whether it's an integer or not. However you go about doing this it would seem to require an extremely high-precision calculation of some number, but I can't tell whether the amount of precision necessary is feasible or not. – Qiaochu Yuan Dec 06 '10 at 07:32
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    It's not particularly feasible; the number of decimal digits is approximately $\log_{10}$ of the number, which is much larger than $10^{80}$, which is supposed to be an estimate for the number of atoms in the observable universe. – Carl Mummert Dec 06 '10 at 23:03
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    And yet, the last ten digits of Graham's number (http://en.wikipedia.org/wiki/Graham%27s_number) are known. – graveolensa Dec 23 '10 at 10:34
  • @Carl Mummert: Do you think it would help if there's some BBP-type formula of this number? – Max Muller Jan 15 '11 at 19:16
  • Wait let me answer that question myself: I think it could. let's call your number $C$. We know that $C < 10^{10^{10^{40}}} = Q$ Then the only thing we have to do is to find a digit-extraction algorithm for $C$ to find the $Q+1$'th digit of this number. If it is not a zero, $C$ is not an integer. If it is, then we check the next digit and repeat this procedure until we find a digit that is not a zero. I think it could be possible to find such a digit-extraction algorithm, based on the earlier efforts of Bailey, Borwein and Plouffe, who discovered the BBP formula for $\pi$ (subsequently, ... – Max Muller Jan 15 '11 at 20:44
  • ... many more of such formulas were discovered for other constants) and used it to set up an algorithm to be able to find the value of any digit in $\pi$. (see the following paper for an overview of BBP formulas of several well-known constants: http://crd.lbl.gov/~dhbailey/dhbpapers/bbp-formulas.pdf .) – Max Muller Jan 15 '11 at 20:47
  • Never mind, it won't work. Even if there exists such a digit-extraction algorithm, it would still take much too long to compute the $10^{10^{10^{40}}}$th digit. See http://numbers.computation.free.fr/Constants/Algorithms/nthdigit.html – Max Muller Jan 26 '11 at 18:56
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    @Carl This is a really nice question. I don't think it needs to be "closed". There are some really innocuous questions running around the site just because the person who asked them never accepted an answer. But I believe that this is a question that deserves to show up from time to time. Who knows, maybe at some point someone will have something important to say about it, if not to answer it for good =) – Adrián Barquero Mar 20 '11 at 16:29
  • No conceptual answer. But I see some references to mathematica et al in the comments/answers below; for those who try to consider the number of digits or so they might be interested in Robert Munafo's "hypercalc" which implements a concept for the handling of such big numbers (actually I don't think it is helpful in any way for the original intention of the question, but - for what it's worth) – Gottfried Helms Mar 20 '11 at 23:29
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    I'll add to the general spirit if I-dunno: A while ago, in the 70's(?), at the problem session of the world famous West Coast Number Theory Conference, someone (I've forgotten who, but it's in the archives) proposed the question: He defined a _humdrum_ number to be an integer $N$ such that $e^{e^N}$ is an integer and asked for the existence of one. – B. Goddard Sep 19 '16 at 22:38
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    @B.G, it was at the 1984 problem session, and it was proposed by Jeff Lagarias and Carl Pomerance. – Gerry Myerson Feb 13 '18 at 21:44

2 Answers2


The paper Chuangxun Cheng, Brian Dietel, Mathilde Herblot, Jingjing Huang, Holly Krieger, Diego Marques, Jonathan Mason, Martin Mereb, and S. Robert Wilson, Some consequences of Schanuel’s conjecture, Journal of Number Theory 129 (2009) 1464–1467, shows that $e,e^e,e^{e^e},\dots$ is an algebraically independent set, on the assumption of Schanuel's Conjecture. Maybe a close reading of that paper will suggest a way of applying the result to the 79-question.

Gerry Myerson
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    Maybe writing to the authors and telling them about this question directly could prove useful. – Adrián Barquero Jun 16 '11 at 04:38
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    Unfortunately Schanuel's Conjecture (that if $n$ complex numbers are linearly independent over the rationals, there are at most $n$ algebraic dependencies between those $n$ numbers and their exponentials) is completely intractable. – HTFB Jun 03 '13 at 09:20

if $e^{e^{e^{79}}}$ is an integer then $e^{e^{e^{e^{79}}}}$ is not an integer (otherwise $e$ would be algebraic). Perhaps your arguments make sense with this number too.

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    This is a good point. Unfortunately, for my example I need to know for sure whether it's an integer or not; knowing that at least one of two numbers is not an integer is not as useful. – Carl Mummert Jun 16 '11 at 00:14