# 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.

• What do you mean by "determine unconditionally the constant"? Does that mean the constant should be in closed form, or it's OK for an algorithm that outputs each digit in finite time? – LeechLattice Jul 25 '19 at 13:39
• My thought @Bullet51 was that it is in closed-form, if it is possible determine/calculate the constant. Any case notice that I am not saying that other discussions or approaches to the problem are not valuable. Truly I am asking about the calculation of a constant $A>0$ and a nice arithmetic function $f(n)$ such that $\lfloor A^{f(n)}\rfloor$ is squarefree for each $n>N$, here $N$ a fixed positive integer (if possible $f(n)$ similar than Mill's sequence, I take the floor function but it is possible also work with the ceil function). – user142929 Jul 25 '19 at 14:31
• Since primes don't have repeated prime factors, Mill's formula already does what you want. But you have to know the primes to compute the value of $A$, rather than knowing $A$ some other way and using it to get the primes. I suspect the same woould be true for any formula "producing" squarefree numbers. – Gerry Myerson Jul 25 '19 at 23:04

## 1 Answer

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 $$1.30537247208307950327278835253 by exploiting the sequence $$(((2^3+3)^3+2)^3+1)^3+2$$.

• I think your answer is the best answer that can currently be provided, therefore I will accept the answer. Thank you for your great calculations and reasoning. – user142929 Jul 30 '19 at 20:29