What is the smallest $x$ such that $\lfloor x^n\rfloor$ has the same parity as n? 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
 A: 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]];

A: As explained in the comments the smallest solution (if it exists) must be between $\sqrt[16]{100801956}$ and $\sqrt[16]{100801957}$. The first few digits are $3.16385673$.
