# Is there any result, that says that $\lfloor e^{n} \rfloor$ is never a prime for $n>2$?

Is there any result, that says that $\lfloor e^{n} \rfloor$ is never a prime for $n > 2$ ?

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$\lfloor e^{18} \rfloor = 65659969$ is a prime. –  achille hui Sep 19 '13 at 10:45
@achillehui Can I ask how you checked that? Did you just quickly write some code, check a website, or is there a remarkable insight hidden in your answer? –  snarski Sep 19 '13 at 12:56
@smarski I just use a computer algebra system (maxima in my case) to compute the first few terms. Whenever one see a claim like this, the first thing one should do is check the obvious cases before one investing time to prove it. –  achille hui Sep 19 '13 at 13:14
@achillehui And the obvious cases are the first several dozen? –  sasha Sep 19 '13 at 20:33
@sasha, Whatever that doesn't take too long to type and too long to run. In this case, the first several dozen. –  achille hui Sep 20 '13 at 2:32

## 3 Answers

The counterexamples up to 1000 are:

18, 50, 127, 141, 267, 310

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See also oeis.org/A050808. –  lhf Sep 19 '13 at 12:32

There is no such result, because $\lfloor e^{18} \rfloor = 65659969$ is prime. This is the smallest counterexample.

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Nope for $n=18$ we have

$$\lfloor \exp(18)\rfloor =65659969$$ which is prime

The counterexamples for $n\leq 10000$ you have \begin{array}{c} 18\\ 50\\ 127\\ 141\\ 267\\ 310\\ 2290\\ 4487\\ 5391\\ \end{array}

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See also oeis.org/A050808. –  lhf Sep 19 '13 at 13:01
Typo of 8 instead of 18 in your list I believe. –  Chris Sep 19 '13 at 16:11
@Chris in fact I wrote 18 but forgot to center the array so he took the 1 as the positioning letter (r,l,c) –  Dominic Michaelis Sep 19 '13 at 20:26