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Define $\phi(n,x)= \sum_{m\leq x,\gcd(m,n)=1} 1$, the number of elements in the interval $[1,x]$ that is relatively prime to $n$. $\omega(n)$ is the number of distinct prime factors of $n$.

It's not difficult to show that $\phi(n,x) \geq x \phi(n)/n - 2^{\omega(n)}$ (see this question). This shows for large $x$, say $x\geq n^\frac{1}{\log \log n}$, $\phi(n,x)$ is essentially $x\phi(n)/n$.

Are there better bounds known for smaller values of $x$, say $x=o(2^{\omega(n)})$? The bound is negative for $x$ in those values.

Even the naive bound $\phi(n,x)\geq 1+\pi(x)-\omega(n)$ performs better when $x$ is small.

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Chao Xu
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1 Answers1

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The fundamental lemma of sieve theory will give you good results here. Put $n_0 = \prod_{p|n, p< (\log n)^2} p$, so that $n_0$ is the product of the small prime factors of $n$. The fundamental lemma gives $$ \Big| \sum_{\substack{m \le x\\ (m,n_0)=1}} 1 - \frac{\phi(n_0)}{n_0} x \Big| \ll u^{-u(1+o(1))} x \frac{\phi(n_0)}{n_0}, $$ with $x= (\log n)^{2u}$ (and $u \ge 2$ say).

Next note that there cannot be too many primes dividing $n$ larger than $(\log n)^2$, and these make a negligible contribution: $$ \Big| \sum_{\substack{m\le x \\ (m,n)=1} }1 - \sum_{\substack{m\le x \\ (m,n_0)=1} }1 \Big| \le \sum_{\substack { p|n \\ p>(\log n)^2}} \frac xp = o\Big(\frac{x}{\log n}\Big). $$

Thus it follows that you can get a good asymptotic formula whenever $x$ is larger than $(\log n)^{2u}$ for any slowly growing quantity $u$. Consult any standard text on analytic number theory (for example Montgomery and Vaughan) for the fundamental lemma.

Lucia
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