Is integer part of $e^n$ infinitely often even and odd?

Question

Let $a_n:=\lfloor e^n\rfloor$. Are $\{n\mid 2|a_n\},\,\{n\mid 2|a_n-1\}$ both infinite sets?

More generally, for any irrational number $\alpha>1$, is each set of the form $\{n\mid p|\lfloor\alpha^n\rfloor-q\}$ infinite?

Answer 1

First, I don't know the answer to OPs first question and in fact believe it is an open question. However, in regard to the second question and indeed encompassing a majority of comments I can shed some light.

@Gerry_Myerson and @J.G. pointed out some interesting behavior regarding seed numbers $2+\sqrt{3}$ and $2+\sqrt{5}$. These numbers are examples of a special class of numbers known as Pisot numbers. A Pisot number $\alpha$ is an algebraic number such that $|\alpha|>1$ and all of its conjugate elements have absolute value less than $1$. It is well known that for any Pisot $\alpha$ we have

$$\lim_{n\to\infty}\min\{\alpha^n-\lfloor \alpha^n\rfloor,\lceil \alpha^n\rceil-\alpha^n\}=0$$

A nice detail I worked out in undergrad is that something stronger is also true. If we define $a_n=\alpha^n$ where $\alpha$ is a Pisot number, then there exists sequence $b_n\in\mathbb{N}$ such that

• The limit $$\lim_{n\to\infty}(a_n-b_n)=0$$

• The sequence $$c_n=\text{mod}(b_n,s)$$ is periodic for all $s\in\mathbb{N}$

Here's a sketch of the proof: Every Pisot number corresponds to some minimal polynomial

$$\alpha\Leftrightarrow P(x)=\sum_{n=0}^m r_n x^n$$

with roots $\{\alpha,\alpha_2,\alpha_3,...,\alpha_m\}$. You can then associate this polynomial with a Fibonacci like sequence

$$f_n=\sum_{i=0}^m r_{m-i}f_{n-i}$$

where the initial conditions for $f_n$ don't matter for our claim. You then show that this sequence $f_n$ has the properties of the sequence $b_n$ above and show that $f_n$ can also be written as

$$f_n=k\alpha^n +\sum_{i=2}^m k_i \alpha_i^n$$

where $k$ and $k_i$ are just constants. Since each $|\alpha_i|<1$ the summation part of the expression above goes to zero and you are basically left with

$$f_n\sim k \alpha^n$$

The rest is just bookkeeping to clean up details. QED

Although this doesn't answer your second question (indeed, the comments already pointed out a counterexample) it would let you find other counterexamples if you so chose. Something I also conjectured was that only Pisot numbers had the property above. That is, if you are given a number $\alpha$, and if there exists a sequence $b_n\in\mathbb{N}$ such that

• The limit $$\lim_{n\to\infty}(\alpha^n-b_n)=0$$

• The sequence $$c_n=\text{mod}(b_n,s)$$ is periodic for all $s\in\mathbb{N}$

then $\alpha$ is a Pisot number. If this were true, it would answer your first question in the affirmative. Simply take $s=2$ and then if there were a finite number with odd or even parity it would be obvious that $c_n$ would eventually be constant and therefore periodic. I couldn't prove this conjecture (it seems quite difficult) but there is at least one piece of evidence that supports it: there are a countable number of periodic sequences and there are obviously a countable number of Pisot numbers. Pretty slim evidence, but interesting nonetheless.