# Bounds on squarefree numbers

Let $q_1,q_2,\ldots$ denote the squarefree integers 1, 2, 3, 5, .... What effective bounds are known for $q_n$? Clearly $$q_n\sim\zeta(2)n$$ but I need hard inequalities. Of course from the above there exist $\varepsilon,N$ with $$(\zeta(2)-\varepsilon)n < q_n < (\zeta(2)+\varepsilon)n$$ for all $n>N,$ but I do not have proven values for $\varepsilon,N$. Of course $$|q_n-\zeta(2)n| < f(n)$$ for sublinear $f$ would be preferable (and should be possible; squarefree numbers are fairly well-behaved).

I'm sure this is in some standard reference but I haven't found it. Ideas?

For those interested, my actual goal is to find a reasonable bound for the powerful (2-full) numbers for computational purposes. Their asymptotic growth is tightly constrained but for numerical computations I would prefer worst-case bounds that I can trust rather than a heuristic that says the error probably won't be much more than, say, 10 times the O-term.

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How big an n will you need? If n has 20 or fewer digits you can probably preprocess most of what you need and store it in a not very large table.. If you need n up to 100 digits, how sublinear/approximate do you want? Gerhard "Ask Me About System Design" Paseman, 2011.06.01 –  Gerhard Paseman Jun 2 '11 at 3:15
@Gerhard Paseman: 40 to 50 digits should suffice. –  Charles Jun 2 '11 at 15:03

## 3 Answers

Expanding upon the first part of Mark Lewko's answer: if $Q(x) = x/\zeta(2) + E(x)$, then we can quickly show that $|E(x)| < 2(\sqrt x + 1)$ for $x\ge4$. Start from the fact that $\sum_{d^2\mid n} \mu(d)$ equals 1 if $n$ is squarefree and 0 otherwise. Then $$Q(x) = \sum_{n\le x} \sum_{d^2\mid n} \mu(d) = \sum_{d\le\sqrt x} \mu(d) \sum_{n\le x,~d^2\mid n} 1 = \sum_{d\le\sqrt x} \mu(d) \Big\lfloor \frac x{d^2} \Big\rfloor.$$ Since $|y-\lfloor y\rfloor| \le 1$, $$\bigg|Q(x) - x \sum_{d\le\sqrt x} \frac{\mu(d)}{d^2} \bigg| \le \sum_{d\le\sqrt x} |\mu(d)| \le \sqrt x.$$ Also, $\sum_{d=1}^\infty \mu(d)/d^2 = 1/\zeta(2)$, and so $$\bigg| \frac x{\zeta(2)} - x \sum_{d\le\sqrt x} \frac{\mu(d)}{d^2} \bigg| \le x \sum_{d>\sqrt x} \frac{|\mu(d)|}{d^2} < x \int_{\sqrt x-1}^\infty \frac{dt}{t^2} = \frac x{\sqrt x-1} \le \sqrt x + 2$$ since $x\ge4$. Combining these two inequalities gives $|Q(x) - x/\zeta(2)| \le 2\sqrt x + 2$ as claimed.

Applying this with $x=q_n$ (for some $n>3$) gives $|n-q_n/\zeta(2)| = |Q(q_n)-q_n/\zeta(2)| \le 2\sqrt{q_n}+2$, or $$q_n - 2\zeta(2)\sqrt{q_n} - 2\zeta(2) \le \zeta(2)n \le q_n + 2\zeta(2)\sqrt{q_n} + 2\zeta(2).$$ From this various inequalities can be derived: for example, $$\zeta(2)n - 5\sqrt n < q_n < \zeta(2)n + 5\sqrt n$$ for $n\ge144$ (that is, for all squarefree numbers exceeding 230).

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Let $Q(x)$ denote the number of square-free numbers less than $x$ than $Q(x) = x/\zeta(2) + E(x)$. An elementary argument gives that $E(x) \ll x^{1/2}$. This can be found in Hardy and Wright's Introduction to the Theory of Numbers (it is Theorem 333 in my copy). One should be able to easily work out explicit constants in that argument if that is what you need. The truth is that the error term is probably much smaller. There have been a number of improvements assuming Riemann hypothesis. I believe the current such record is $E(x) \ll_{\epsilon} x^{17/54+\epsilon}$ (see: Jia, Chao Hua. "The distribution of square-free numbers", Science in China Series A: Mathematics 36:2 (1993), pp. 154–169).

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I know of the $\ll\sqrt x$ asymptotic, but not of Jia's paper. Thanks! –  Charles Jun 2 '11 at 14:49

The "Handbook of Number Theory" by Sandor, Mitrinovic and Crstici, page 201 gives $Q(x)\ge 53x/88$, where $Q(x)$ is the number of square free positive integers less than or equal to $x$. This implies $q_n\le 88n/53$. They cite "The Schnirelmann Density of the Squarefree Integers", K. Rogers, Proc. Am. Math. Soc. 15, 1964

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Thanks. I increasingly feel like I need to get a copy of that book... –  Charles Jun 2 '11 at 14:50