The Wikipedia's article for *Prime number* shows a known and curious formula for primes from its section [*Formula for primes*](https://en.wikipedia.org/wiki/Prime_number#Formulas_for_primes), I say the Mills' theorem (please see also the Wikipedia *Mills' constant*).



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> **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.**
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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.



What about a formula similar than Mill's formula, but producing positive integers without repeated prime factors?
Along the lines of the [Wikipedia page](https://en.wikipedia.org/wiki/Square-free_integer#Distribution), 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 $(((2^3+3)^3+2)^3+1)^3+2$.


