Records in $Z$-numbers and a relaxation A Z-number
is a (non-zero) real number $x$ such that the fractional parts
$$\left\lbrace x \left(\frac 3 2\right)^ n \right\rbrace $$
are less than $\frac12$ for all natural numbers $n$.
It is not known whether $Z$-numbers exist.
First, I am interested in finite records, i.e. large $k$
such that for explicit $x > 1$ the fractional part is $ < \frac12$
for $n = 1  \ldots k$.
So far got $k=12$ for $x \approx 2.81365$ and don't think $ x < 2$ is possible
in this case.
Second, consider the following relaxation. Let $f$ be a strictly increasing
function $\mathbb{N} \to \mathbb{N} $
Is it possible:
$$\left\lbrace x \left(\frac 3 2\right)^ {f(n)} \right\rbrace < \frac12 $$
for all natural $n$?
 A: Suppose the fractional parts of $x, \frac{3}{2} x, (\frac{3}{2})^2 x, ... (\frac{3}{2})^kx$ are under $\frac{1}{2}$, but the fractional part of $(\frac{3}{2})^{k+1} x$ is between $\frac{1}{2}$ and $1$. Then consider $y = x + 2^k$. For $n \le k, \lbrace (\frac{3}{2})^n y\rbrace =  \lbrace (\frac{3}{2})^n x \rbrace$ while $\lbrace (\frac{3}{2})^{k+1} y \rbrace =  \lbrace  (\frac{3}{2})^{k+1} x \rbrace - \frac{1}{2} \lt \frac{1}{2}.$ So, if we start with an arbitrary $x$, then one of the $2^{k}$ numbers $x, x+1, ... x+2^{k}-1$ will have fractional parts under $\frac{1}{2}$ when you multiply by $\frac{3}{2}, ..., (\frac{3}{2})^k$.
Here is some Mathematica code which implements this:
Clear[j];
j[1] = 1;
j[n_] := j[n] = 
  If[FractionalPart[j[n - 1] (3/2)^n] < 1/2, 
   j[n - 1], 
   j[n - 1] + 2^(n - 1)]

For the choice $j[1]=1$, this computes the integer from $1$ to $2^n$ which has the right fractional parts when multiplied by $\frac{3}{2}, ..., (\frac{3}{2})^n$. $j[4]=1, j[5]=17, j[20] = 386737, j[100] = 719590229933913224019274229425$.

Yes, we can take $f(n)=3n$.
Lemma: For any integer $a$, $[(\frac{3}{2})^3 a, (\frac{3}{2})^3(a+\frac{1}{2})]$ contains at least one interval $[b,b+\frac{1}{2}]$ for some integer $b$.
This follows because $\frac{1}{2}(\frac{3}{2})^3 \gt 1+\frac{1}{2}$. 
Start with an interval $[a_1,a_1+\frac{1}{2}]$. Inductively pick $a_{n+1}$ so that $[a_{n+1},a_{n+1}+\frac{1}{2}] \subset (\frac{3}{2})^3 [a_n,a_n+\frac{1}{2}].$ Then the intervals $(\frac{3}{2})^{-3n}[a_n,a_n+\frac{1}{2}]$ are nested hence their intersection contains an element $x$ so that $\lbrace x (\frac{3}{2})^{3n} \rbrace \lt \frac{1}{2}$ for all $n$. You can choose $x\ne 0$.
