TITLE: Floors of powers of reals, how much do the first few determine the next?
QUESTION [17 upvotes]: Call an integer sequence $\mathbf{x}=\left( x_1,x_2,\cdots \right)$ feasible if it is $f(r)=\left(\lfloor r \rfloor, \lfloor r^2 \rfloor, \lfloor r^3 \rfloor, 
\ldots, \lfloor r^n \rfloor, \ldots \right)$. for some real $r \gt 1$.

Q: if the members of $f(r)$ all have the same parity, must $r$ be an integer?

I believe(d) so (I was wrong see below.) The motivation of the question is not so much about parity as about how much an initial segment of a feasible sequence limits the next term. An affirmative answer to the question  would imply that for every non-integer $r \gt 1$, there must be some $k$ so that $x_1,x_2,\cdots,x_k$ uniquely determines $x_{k+1}$.  
Given only the value $x_k$ we know the approximate value of $r$ and I claim that (provided $k$ is fairly large compared to $r$) this restricts the possible values of $x_{k+1}$ to an interval of length roughly $r.$.
In more detail: Given only $x_k$ we know that  $\sqrt[k]{x_k} \le r \lt \sqrt[k]{x_k+1}$ so $$(x_k)^{1+1/k}\le x_{k+1} \lt (x_k+1)^{1+1/k} \tag{ * }.$$ Suppose that $k$ is a good bit larger than $r$ and let $s=\sqrt[k]{x_k}.$ Then $\sqrt[k]{x_k+1} \approx s+\frac{1}{ks^{k-1}}$ and the bounds $( * )$are approximately $$s^{k+1} \le x_{k+1} \lt s^{k+1}+s(1+1/k)$$. 
Of course $x_{k+1}$ must be an integer. Also, we can say $$\max_{j \le k} \sqrt[j]{x_j}\le r \lt \min_{j \le k}\sqrt[j]{x_j+1}.$$

later Thanks for the good answers. I now realize that the answer to my question turns out to be

No, $r$ need not even be algebraic. No matter how many initial terms are given, Even if one term is forced by the previous ones, choices can then be made, one term at a time, which always allow several choices for the next term.

This might require  $x_1 \ge 3$ or a similar condition. I think that  one can maintain freedom of choice as long as one refrains from making the  greatest possible choice (unless forced to do so, in which case the next choice will not be forced.)
I will describe things in general terms here and add a graphic illustration as an answer:
In addition to $(*)$ above, showing that $x_k$ alone allows about $s(1+1/k)$ possible values for $x_{k+1}$ one has $$\lfloor s^{k-1} \rfloor \le x_{k-1} \le s^{k-1}+\frac{1-1/k}s.$$
Which means that there might be two possible values for $x_{k-1}$, but not more.

Here is an example with a forced term followed by a sketch of how to find such.  Typically a feasible a sequence $[235,x_2,x_3,x_4,x_5]$ will allow about $235+\frac{235}{5}=282$ possible values for $x_6$ and over $75000$ possible values for both $x_6,x_7.$ However the sequence  $[235, 55677, 13137758, 3100000161, 731479476931,x_6,x_7]$ has only $5$ completions: one must have $x_6=172600708777398$ and $| x_7-40727054702138152| \le 2.$   In this case $x_4 \approx 3100000000$ was the third thing I tried and I was fortunate to find out that  $3100000162^{\frac54}=731479476931.00004144$. Thus the restriction $$\sqrt[5]{731479476931} \le r \lt \sqrt[4]{3100000162}$$ forced by the sequence  is very strong.

REPLY [3 votes]: Here is an illustration of what things look like given that $x_1=\lfloor r \rfloor=3.$ I think that a larger value than $3$ would work even better. Then we know $3^k \le x_k \le 4^k-1$. So there are $781$ possible values for $x_5$ and, by my calculations, $991$ possibilities for the sequence $(x_1,x_2,x_3,x_4,x_5)$. This means that most of the time $x_5$ alone determines all the previous terms. In fact there are $196$ choices which can be arrived at in exactly two ways and only $7$ which can be arrived at in three ways. One is $x_5=569$ which can be due to any one of $$ [3,12,44,159,569],\ [3,12,44,160,569], \ [3,12,45,160,569]$$
One can imagine the successive terms $x_k$ as describing an upward path in a tree with a node labelled $x_k$ located at the point $(\sqrt[k]{x_k},k)$. There may be other nodes labelled $x_k$ at that level.  For example there are three nodes labelled $569$ at level $5$. They are at $(x,5)$ for $x=\sqrt[5]{569},\sqrt[4]{160},\sqrt[3]{45}$
Here are selected portions of that graph. The first three levels and above that portions of the next two levels above $46$ and $47$ in the third level. 
If the tree was extended to more levels eventually each node would have $3$ or $4$ continuations to the next level. If one takes an upward path which moves always to the right  (never directly vertically) as it goes up, then it will be consistent to keep doing this in any way desired. SO there is freedom to choose the parity.
If one avoids both the leftmost and rightmost upward edge leaving each node then each term will determine all the earlier ones.