Is it possible that this can be prime for infinitely many values of n?

Question

kainui · Answer

$$\Large \lfloor e^n \rfloor$$
For instance, when n=2 we have $$\lfloor e^2 \rfloor= \lfloor 7.389... \rfloor = 7$$
which is prime.

ganeshie8 · Answer

Looks hard to me
for n=18 it is prime

ganeshie8 · Answer

the next prime occurs for n=50

kainui · Answer

Some things I'm currently trying out:

Expand the binomial e^n=(2+.7)^n and try to wiggle something out of the fact that powers of numbers less than 1 go to zero.

All primes greater than 3 are 1 away from a multiple of 6.

Dirichlet's theorem says there are infinitely many values of n that make an+b into a prime number when gcd(a,b)=1.

When can I throw away extra terms in the power series expansion of $e^n$. It's sorta tricky cause while I want to remove stuff like 1/3, if you have 1/3 three times then it bumps the digit up.

zepdrix · Answer

You too smart Canoe-E.
Makes my head hurt >.<

kainui · Answer

Hahaha nooooo I'm not pls no, this is eBullying pls stop...!