TITLE: What is the smallest $x$ such that $\lfloor x^n\rfloor$ has the same parity as n?
QUESTION [18 upvotes]: A previous MO question asked for information about a number $x$ such that $\lfloor x^n\rfloor$ has the same parity as $n$ for all positive $n$.  Answers to this post included two values of x which meet the requirements.  From Noam Elkies: the largest root of $x^3-3x^2-x+1$, giving $x=3.2143...$.  From Max Alekseyev: $(3+\sqrt{17})/2$, giving $x= 3.56155...$.
What is the smallest such positive number?  Failing this, what are the first few digits of the smallest such positive number?
This question was suggested to me by Moshe Newman

REPLY [4 votes]: In a sense, there is no smallest: the smaller root of $x^2+k x+2$ has the desired property, and this goes to $-\infty$ with $k$. I assume the problem is meant to ask for a positive $x$.
Here's some code (at the bottom) for Mathematica (infinite precision!) that computes the half-open set $[a_0,b_0) \cup [a_1,b_1) \cup \cdots$ of real numbers $x$ that are less than the root of $\alpha^3-3\alpha^2-\alpha+1$ and for which $\lfloor x^k \rfloor\equiv k \bmod 2$ for all integers $k$ between 1 and $n$, inclusive.
For $n=16$, the smallest $x$ is $\sqrt[16]{100801956}$, a number that appears in the answer by user35593. This lower bound on $x$ is improved, but only by a minuscule amount by taking $n=45$: we learn that $x$ is at least $\sqrt[45]{32341223721862945369971}$. This is still not large enough to provably dismiss user35593's claim.
To enrich the examples we have, I offer the following. The larger root $\alpha$ of $x^2 -(2k+1)x-2\ell$, where $1\leq \ell \leq k$, is slightly above $2k+1$, while the smaller root $\bar \alpha$ is between $-1$ and 0. Because of the recurrence satisfied by $\alpha^n+\bar\alpha^n$, the real number $\alpha$ has the desired property $\alpha^n \equiv n \bmod 2$ for all $n\geq 1$.  
alpha = Root[1 - #1 - 3 #1^2 + #1^3 &, 3];
NewUpInt[1] = {{1, 2}, {3, alpha}};
NewUpInt[n_] := NewUpInt[n] =
  Module[{int},
    If[EvenQ[n],
      int[{a_, b_}] := 
         Table[{Max[a, (2 k + 2)^(1/n)], Min[b, (2 k + 3)^(1/n)]},
           {k, Floor[(a^n - 3)/2] + 1, Ceiling[(b^n - 2)/2] - 1}],
      int[{a_, b_}] := 
         Table[{Max[a, (2 k + 1)^(1/n)], Min[b, (2 k + 2)^(1/n)]},
           {k, Floor[(a^n - 2)/2] + 1, Ceiling[(b^n - 1)/2] - 1}]];
    Flatten[Map[int, NewUpInt[n - 1]], 1]];