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 $$\max\_{j \le k} \sqrt[j]{x\_j}\le r \lt \min\_{j \le k}\sqrt[j]{x\_j+1}.$$




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**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.




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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. 



Floors of powers of reals, how much do the first few determine the next?
This is related to [OEIS 014217 Floor(phi^n), where phi = (1+sqrt(5))/2 is the golden ratio](https://oeis.org/A014217).


From the comments:


$$a(n) =\lfloor \phi^n \rfloor = L(n)-(1+(-1)^n)/2$$


Where $L(n)$ are Lucas numbers.


Since $L(6n)$ are well known to be even, take $r=\phi^6$ where $\phi$ is the golden ratio.


$a(6n)=\lfloor r^n \rfloor = L(6n) - 1 \equiv 1 \pmod 2 $


$r$ is not an integer.


Experimentally $\lfloor r^n \rfloor$ is odd up to $n=1000$
(just to check).




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Since there are [multiplication formulas](https://oeis.org/A000032)
for Lucas numbers $L( k n) = f(L(n)) $, from $x\_n$ we can determine $x\_{kn}$.


$L(6n)=b(n)$ is [A087215](https://oeis.org/A087215) and
satisfies $b(n) =18 b(n-1) - b(n-2)$, so $x\_n,x\_{n+1}$ determine the
rest of the sequence $\lfloor r^n \rfloor$.


