What about a formula similar than Mill's formula, but producing positive integers without repeated prime factors? The Wikipedia's article for Prime number shows a known and curious formula for primes from its section Formula for primes, I say the Mills' theorem (please see also the Wikipedia Mills' constant).

Question. I wondered if one can to determine or calculate a constant and choose an arithmetic function, and using a floor or ceiling function, write a formula producing square-free integers in the same way that Mill's formula is a prime-generating formula. Many thanks.

Thus that I evoke is try to write a formula and try to determine unconditionally the constant. This formula should to generate a square-free integer for each $n\geq n_0$, for certain positive integer $n_0$. I would like to know if it is possible/feasible for a similar nice arithmetic function, see the exponential of Mill's formula (I don't know what is the statement of Wright's theorem, thus I am asking about a formula similar than Mill's formula, now for integers without repeated prime factors).
I don't know if my Question is in the literature. If in the literature there is such formula explicitly comment it refering the literature, and I try to search and reat it from the literature.
 A: Along the lines of the Wikipedia page, it is true that 
$|Q(x)-\frac{x}{\zeta(2)}|\leq2+\sqrt{x}$
where $Q(x)$ is the number of square-free numbers between $1$ and $x$.
So,
$Q(n^3)\leq\frac{n^3}{\zeta(2)}+2+n^\frac{3}{2}$
$Q((n+1)^3)\geq\frac{(n+1)^3}{\zeta(2)}-2-(n+1)^\frac{3}{2}$
The bounds above shows that there's a square-free number between $n^3$ and $(n+1)^3$ for every $n\geq 1$ (The bounds only works for $n\geq3$, but one could give examples for $n=1$ and $n=2$). 
This fact can be used to prove the existence of a Mill constant, i.e. $A$ such that $\lfloor{A^{3^n}}\rfloor$ is a square-free number. Namely, there's a sequence of positive integers, $a_n$, such that for all $n$,
$(((a_1^3+a_2)^3+a_3)^3+...a_{n-1})^3+a_n$ is square-free and $(((a_1^3+a_2)^3+a_3)^3+...a_{n-1})^3+a_n<(((a_1^3+a_2)^3+a_3)^3+...a_{n-1}+1)^3$.
In fact, we can take $A=\lim_{n→+\infty}{((((a_1^3+a_2)^3+a_3)^3+...a_{n-1})^3+a_n)^{3^{-n}}}$.
For a value of $A$, there is $1.30537247208307950327278835253<a<1.30537247208307950327278835254$ by exploiting the sequence $(((2^3+3)^3+2)^3+1)^3+2$.
