Problem:
Can every natural number be written as $\displaystyle \left\lfloor \frac{2^{2^s+1}}{3^k} \right\rfloor$ for some $s,k\in \mathbb{N}$?
Context and thoughts:
This is a generalization of these problems: Density of $2^a 3^b$ and Creating a number by multiplying by 2 and dividing by 3. In the second link there is a comment I did where I tried to confirm that when the exponent of the $2$ in the numerator can be any natural number, then the answer is yes. This procedure does not work, at least directly, in this case, because it would need that $2^n\alpha - \lfloor 2^n\alpha \rfloor$ to be dense in $[0,1]$, and in the question I linked it's proven to be false in general. I assume this has some connections with the normality in base $2$ of the irrational $\alpha$, which in our case would be $\frac{\log 2}{\log 3}$ if I'm not mistaken, but I don't know if it's even known whether it's normal or not. I also don't think both problems are equivalent, density and normality seem to be way stronger than the question only for naturals.
This new version comes from this question: Integer less than 7000 achievable by .... One can notice that if the answer to the question I posted is yes, then one can produce a method to obtain all integers in the linked question because the numerators are always achievable just with the first function and $4$. I understand the obvious overlap between the two, but I'm actually interested in methods for questions of density like this one, because I've seen them in different contexts, while the linked question can have other approaches. Moreover, I think the answer to this one could end up being negative.
My attempt at programming it for low numbers that you can try online here seems to get all decompositions from $1$ to $60$, due to the exponent the numbers get large quite fast so it starts to get slower the more you continue because the precision needs to be larger.
