# 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 $n > 2$ ?

• $\lfloor e^{18} \rfloor = 65659969$ is a prime. 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?
– snar
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. Sep 19 '13 at 13:14
• @achillehui And the obvious cases are the first several dozen? 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. Sep 20 '13 at 2:32

The counterexamples up to 1000 are:

18, 50, 127, 141, 267, 310


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

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}

– lhf
Sep 19 '13 at 13:01
• Typo of 8 instead of 18 in your list I believe. 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) Sep 19 '13 at 20:26