Does there exist an algebraic number $\alpha$ such that
$$\left|\frac{\alpha^n+\alpha^n\_1}{n!}\right|\sim\_{n\to+\infty}\frac1{(n!)^2}$$ where $\alpha\_1$ is a conjugate of $\alpha$?
Obviously $\alpha$ can not be a rational number.


Thanks in advance for any answer.



algebraic numbers with small norms
For any fixed nonzero complex numbers $z\_1,\dotsc,z\_m$, there are infinitely many $n$'s such that the arguments of $z\_1^n,\dotsc,z\_m^n$ all lie in $[-\pi/4,\pi/4]$. This follows from Dirichlet's theorem on simultaneous diophantine approximation. For such $n$'s,
$$|z\_1^n+\dotsb+z\_m^n|\geq\Re(z\_1^n+\dotsb+z\_m^n)\geq\frac{|z\_1|^n+\dotsb+|z\_m|^n}{\sqrt{2}}.$$
In particular, the left hand side cannot be asymptotically $1/n!$, because the right hand side is exponentially small at worst.


In short, there is no $\alpha$ satisfying the requirements.


