Set of primes in the spectrum of a rational number. Let $x \in \mathbb{Q}^{+}$ (numerator and denominator are in the reduced form). Can we comment anything about the set $\left\{  \left\lfloor nx \right\rfloor \; : \; n \in \mathbb{N}  \right\}$? Is there any structure in this set of integers especially the possible set of prime integers?
Also, can anyone point me toward the topics in number theory which deals with such kind of problems? Any hint will be much appreciated.
 A: Let $m = \lfloor x \rfloor $.  Let $r = x-m$ so $0\le r < 1$.  $q$ is rational so let $j,k \in \mathbb N$ so that $\gcd(j,k) = 1$ and $r = \frac jk$.  In other words..... Let $x = m + \frac jk$ were $m$ is integers and $\frac jk$ is a non-negative rational in reduced terms less than $1$.
$ nx  =  nm + nj\cdot \frac 1k  = nm + q_n + \frac {a_n}k$ where $q_n = \lfloor {nj}\rfloor$ and $a_n \equiv nj \pmod k; 0 \le a < k$
So $\lfloor nx\rfloor= nm + q_n$ where $q_n = \lfloor {nj}\rfloor$.
....
Anyhow,... As $j, k$ are relatively prime $nj; n=1,...,k$ form a complete residue system $\mod k$. And $b'_n = \lfloor \frac {nj}k\rfloor;n=1,...,k$ (note $b'_1= 0$ and $b'_k = j$ always) will be a  sequence of $k$ integers.  We can define $b_{n+k} = b_n$ to extend this sequence to be repeating sequence of period $k$.
So  $nx =  nm + j\lfloor \frac nk\rfloor + b_n + \frac {a_n}k$ and $\lfloor nx\rfloor = nm + j\lfloor \frac nk\rfloor + b_n$
I gave, in the comments, and example of $x = 7\frac 35 = \frac {38}5$.
In this case $m=\lfloor \frac {38}5\rfloor = 7; r= \frac 35; j= 3; k=5; a_n \equiv 3,6,9,12,15 \pmod 5$ and $a_n = 3,1,4,2,0$ and $b_n = 0,1,1,2,3$
And $\lfloor n\cdot 7\frac 35\rfloor= 7n + 3\lfloor \frac n5\rfloor + b_n$.
Example:  $\lfloor 139\cdot 7\frac 35\rfloor =$
$\lfloor 139\cdot 7 + (27\cdot 5 + 4)\frac 35 \rfloor =$
$\lfloor 139\cdot 7 + 27\cdot 3 + \frac {4\times 3}5\rfloor =$
$\lfloor 139 \cdot 7 + 27\cdot 3 + \frac {2\cdot 5 + 2}5\rfloor = $
$\lfloor 139 \cdot 7 + 27\cdot 3 + 2 + \frac 25\rfloor =$
$139\cdot 7 + 27\cdot 3 + 2$
Where $7 = \lfloor x \rfloor = m$. and $3 =$ the numerator of $\frac 35 = $ numerator of $x-m = j$ and $2=b_{4}$ (as $139\equiv 4 \pmod 5$).
