Correlations of $\phi(n)/n$

Question

We know that the average of $\phi(n)/n$ is approximated by a constant. Here $\phi $ is the Euler quotient function. One can furthermore show asymptotics with a secondary main term, at least for the smooth sum $$ \sum_{n \in \mathbb{N} } \frac{\phi(n) } {n} w(n/x)=c_0(w) x+c_1(w) (\log x ) +o(\log x ) ,$$ where $w$ is a smooth weight and $c_0,c_1$ are constants depending on $w$.

Can we prove asymptotics for the secondary term regarding the shifted sum $$ \sum_{n \in \mathbb{N} } \frac{\phi(n) } {n}\frac{\phi(n+1) } {n+1} w(n/x) $$ for some $w$? It is not clear to me whether the secondary term here should be oscillating or like $\sim c \log^2 x$ or something else. The standard approach to prove the previous asymptotic relies on the fact that $\frac{\phi(n) } {n} $ is multiplicative, whereas $\frac{\phi(n) } {n}\frac{\phi(n+1) } {n+1}$ is clearly not.

Answer 1

In this answer, we will treat the case when the sum is over $n\leq x$ without the smooth weight function. The asymptotic formula should be suitably adapted to the weight function.

Ingham proved in p.208 (18) of Some Asymptotic Formulae in the Theory of Numbers (Journal of the London Mathematical Society, V. 1-2, Issue 3, 1927, pp. 202-208). that the following holds:

$$ \sum_{n\leq x} \phi(n)\phi(n+1) \sim \frac13 A x^3, \ \ (1) $$

where $A=\prod_p \left(1-\frac2{p^2}\right)$. By partial summation, we have $$ \sum_{n\leq x} \frac{\phi(n)}n \frac{\phi(n+1)}{n+1} \sim Ax. \ \ (2) $$

However, the derivation of (1) is not written in the paper. The method in the paper is adapted here to derive (2) directly with the secondary term estimate. By $\frac{\phi(n)}n=\sum_{d|n} \frac{\mu(d)}d$, we need to estimate the sum $$ \sum_{\substack{{ms\leq x}\\{lr-ms=1}}}\frac{\mu(l)}l \frac{\mu(m)}m. $$ Since $lmrs\leq x(x+1)=X^2$, the sum can be written as $$ \sum_{\substack{{lm\leq X^2}\\{lr-ms=1}\\{s\leq \frac xm}}}\frac{\mu(l)}l \frac{\mu(m)}m. $$ The condition $lr-ms=1$ restricts integers $l$ and $m$ into $(l,m)=1$, $s$ in single congruence modulo $l$, and $r$ is uniquely determined depending on $s$. Thus, it is enough to consider the sum $$ \sum_{\substack{{lm\leq X^2}\\{(l,m)=1}}}\frac{\mu(l)}l \frac{\mu(m)}m\left(\frac x{ml} + O(1)\right). $$ The error term $O(1)$ contributes at most $O(\log^2 X)$. The condition $(l,m)=1$ is treated with inserting $\sum_{d|l, d|m}\mu(d)$ into the sum. Then the main term without $x$ becomes $$ \begin{align} \sum_{d^2uv\leq X^2} &\frac{\mu(d)}{d^4}\frac{\mu(du)}{u^2}\frac{\mu(dv)}{v^2}\\ &=\sum_{d\leq X} \frac{\mu(d)}{d^4} \sum_{uv\leq \frac{X^2}{d^2}} \frac{\mu(du)}{u^2}\frac{\mu(dv)}{v^2}\\ &=\sum_{d\leq X} \frac{\mu(d)}{d^4} \sum_{u\leq \frac{X^2}{d^2}} \frac{\mu(du)}{u^2} \sum_{v\leq \frac{X^2}{ud^2}} \frac{\mu(dv)}{v^2}\\ &=\sum_{d\leq X} \frac{\mu(d)}{d^4} \sum_{u\leq \frac{X^2}{d^2}}\frac{\mu(du)}{u^2} \left( \frac{\mu(d)\prod_{p|d}\left(1-\frac1{p^2}\right)^{-1}}{\zeta(2)}+O\left(\frac{ud^2}{X^2}\right)\right)\\ &=\sum_{d\leq X} \frac{\mu(d)}{d^4} \sum_{u\leq \frac{X^2}{d^2}} \frac{\mu(du)}{u^2} \frac{\mu(d)\prod_{p|d}\left(1-\frac1{p^2}\right)^{-1}}{\zeta(2)}+O\left(\frac{\log X}{X^2}\right)\\ &=\sum_{d\leq X} \frac{\mu(d)}{d^4}\left(\frac{\mu(d)\prod_{p|d}\left(1-\frac1{p^2}\right)^{-1}}{\zeta(2)} \right)^2 + O\left( \frac1{X^2}\right)+O\left(\frac{\log X}{X^2}\right)\\ &=\prod_p \left(1-\frac2{p^2}\right) + O\left(\frac1{X^3}\right)+O\left( \frac1{X^2}\right)+O\left(\frac{\log X}{X^2}\right)\\ &=\prod_p \left(1-\frac2{p^2}\right)+O\left(\frac{\log X}{X^2}\right). \end{align} $$ Hence, we obtain the secondary term estimated by $O(\log^2 x)$, yielding $$ \sum_{n\leq x}\frac{\phi(n)}n \frac{\phi(n+1)}{n+1} = Ax + O(\log^2 x). $$

Answer 2

If $w$ is smooth and compactly supported in $[1/2,2]$, say, then $$\sum_{n}\frac{\varphi(n)}{n}\frac{\varphi(n+1)}{n+1}w(n/x) = Cx + O_A((\log x)^{-A}).$$ Write $$\frac{\varphi(n)}{n} = \sum_{d\mid n} \frac{\mu(d)}{d}$$ and $$\frac{\varphi(n+1)}{n+1} = \sum_{e \mid n+1} \frac{\mu(e)}{e},$$ then interchange the orders of summation. Perform Poisson summation on the $n$ variable to turn the sum into something like $$\sum_d \frac{\mu(d)}{d} \sum_{(e,d)=1}\frac{\mu(e)}{e} \frac{x}{d^2e^2}\sum_{|k|\leq de/x} e\left(-k\frac{\overline{d}}{e}\right)\widehat{w}(kx/de),$$ where $\overline{d}$ denotes the inverse of $d$ modulo $e$. The $k=0$ term gives the main term, which is seen to be $Cx + O_A((\log x)^{-A})$ by using the cancellation in the mobius function.

The nonzero frequencies contribute an error of size $O_A((\log x)^{-A})$. For these terms, break $d$ and $e$ into dyadic ranges $d \asymp D, e \asymp E$. Clearly we may assume $DE \gg x$. Also, we may assume without loss of generality that $E \ll D$, otherwise use the reciprocity relation $$\frac{\overline{d}}{e} + \frac{\overline{e}}{d}\equiv \frac{1}{de}\pmod{1}$$ to switch $d$ and $e$ in the exponential. After some work, Siegel-Walfisz and the large sieve cover the case when $E\leq D (\log x)^{-B}$, so we may assume $D \approx E$. The double sum over $d$ and $e$ may then be suitably bounded using results of Duke-Friedlander-Iwaniec on bilinear forms in Kloosterman fractions (see also work of Bettin and Chandee).

[Edit: March 15]

I'm adding a bit more detail, as requested. I'm assuming the smoothing $w$ is a smooth bump function, so the Fourier transform satisfies $|\widehat{w}(y)| \ll \exp(-|y|^{1/2})$, say. The arguments below might have to be adapted or substituted entirely for less smooth weights.

With $w$ as above, one finds that $$\sum_n \varphi(n)/n w(n/x) = Cx + O_A((\log x)^{-A})$$ by Mellin inversion, contour shifting, and the zero-free region for zeta.

Now, as to the correlation sum. Begin as above, and perform Poisson summation. We separate the $k=0$ term which gives the main term, and the contribution of the nonzero frequencies is $$\sum_{d\ll x}\frac{\mu(d)}{d}\sum_{\substack{e \ll x \\ (e,d)=1}}\frac{\mu(e)}{e}\frac{x}{de}\sum_{k \in \mathbb{Z}\backslash \{0\}} e\left(-k\frac{\overline{d}}{e}\right) \widehat{w}\left(k\frac{x}{de} \right).$$ Dyadically decompose $d\asymp D$ and $e \asymp E$. As mentioned above, up to changing the sign of $k$ in the exponential one may assume that $E \ll D$ by using reciprocity.

Now use the rapid decay of $\widehat{w}$ to truncate the sum on $k$. Up to a minor change in the coefficients and an acceptable error term, we have $$\frac{x}{D^2E^2}\sum_{e\asymp E} \mu(e) \sum_{\substack{d \asymp D \\ (d,e)=1}}\mu(d) \sum_{0<|k|\leq (\log x)^{10}DE/x} e\left(k\frac{\overline{d}}{e}\right)e\left(\frac{k}{de} \right) \widehat{w}\left(k\frac{x}{de} \right).$$ Clearly, we may assume $DE > x(\log x)^{-11}$, say.

We are going to use the large sieve inequality, but in order to do so we need to separate the $d$ and $e$ variables from each other inside $\widehat{w}$. Change variables in the definition of $\widehat{w}$ and interchange to obtain $$\int_{u \asymp 1} \frac{x}{D^2E^2}\sum_{0<|k|\leq (\log x)^{10}DE/x} \sum_{e \asymp E}\mu(e)w(\tfrac{eu}{E})\sum_{\substack{d \asymp D \\ (d,e)=1}}\mu(d) e\left(-k\frac{x}{dE}u \right) e\left(-k\frac{\overline{d}}{e}\right) du,$$ where this holds up to harmless changes in the $d$ and $e$ coefficients. We work uniformly in $u\asymp 1$, so we can drop the integral.

We give two arguments. The first works when $E$ is a bit smaller than $D$, and the other works when they are about the same size.

The first argument uses the large sieve. Split the sum over $d$ into arithmetic progressions modulo $e$ to separate the additive character, then apply multiplicative characters to detect the congruence condition modulo $e$. This turns the sum over $d$ into $$\frac{1}{\varphi(e)}\sum_{\chi (e)}\sum_{(a,e)=1}e\left(-k\frac{\overline{a}}{e}\right) \overline{\chi}(a)\sum_{\substack{d \asymp D}}\mu(d)\chi(d) e\left(-k\frac{x}{dE}u \right).$$ We take out the contribution of the principal character $\chi_0 \pmod{e}$. We get cancellation in the sum over $d$ using partial summation. The sum over $a\pmod{e}$ becomes a Ramanujan sum, which has size $\leq (|k|,e)$, the GCD of $|k|$ and $e$. It is easy to see that the total contribution from $\chi_0$ after summing over all the variables is $\ll_A (\log x)^{-A}$.

We have to bound the contribution of the non-principal characters: $$\frac{x}{D^2E^2}\sum_{0<|k|\leq (\log x)^{10}DE/x} \sum_{e \asymp E}\mu(e)w(\tfrac{eu}{E})\frac{1}{\varphi(e)}\sum_{\substack{\chi (e) \\ \chi \neq \chi_0}}\sum_{(a,e)=1}e\left(-k\frac{\overline{a}}{e}\right) \overline{\chi}(a)\sum_{\substack{d \asymp D}}\mu(d)\chi(d) e\left(-k\frac{x}{dE}u \right).$$ We fixed $k$, since we will not need to sum over this variable, and then apply Cauchy-Schwarz to the sum over $e$. This gives $$\sum_e \ll (\Sigma_1 \Sigma_2)^{1/2},$$ where $$\Sigma_1 = \sum_{e \asymp E} \frac{1}{\varphi(e)}\sum_{\substack{\chi (e) \\ \chi \neq \chi_0}} \left|\sum_{(a,e)=1}e\left(-k\frac{\overline{a}}{e}\right) \overline{\chi}(a) \right|^2$$ and $$\Sigma_2 = \sum_{e \asymp E} \frac{1}{\varphi(e)}\sum_{\substack{\chi (e) \\ \chi \neq \chi_0}} \left|\sum_{\substack{d \in I_D}}\mu(d)\chi(d) e\left(-k\frac{x}{dE}u \right) \right|^2.$$ We easily bound $\Sigma_1$ by using positivity to include $\chi_0$, then opening the square and using orthogonality of characters. It follows that $$\Sigma_1 \ll E^2.$$

The argument for $\Sigma_2$ is slightly more involved. First, we reduce to summing over primitive Dirichlet characters $\psi$ (cf. any standard proof of the Bombieri-Vinogradov theorem). We apply a dyadic decomposition to the conductor of the primitive characters, so we derive something like $$\Sigma_2 \ll \sum_{1 \ll R = 2^j\ll E}\sum_{f \ll E} \frac{1}{\varphi(f)}\sum_{r \asymp R} \frac{1}{\varphi(r)} \sum_{\substack{\psi( r) \\ \psi \text{ prim}}}\Big|\sum_{\substack{d \in I_D \\ (d,f)=1}}\mu(d)\psi(d) e\left(-k\frac{x}{dE}u \right) \Big|^2.$$ If $R \leq (\log x)^C$ then we use partial summation and the Siegel-Walfisz theorem to bound the sum over $d$. If $R > (\log x)^C$ we use the multiplicative large sieve inequality $$\sum_{r \asymp R} \frac{1}{\varphi(r)} \sum_{\substack{\psi( r) \\ \psi \text{ prim}}}\Big|\sum_{\substack{d \in I_D \\ (d,f)=1}}\mu(d)\psi(d) c(d)e\left(-k\frac{x}{dE}u \right) \Big|^2 \ll \frac{1}{R}\left(R^2 + D \right)D.$$ We deduce that $$\Sigma_2 \ll D^2 (\log x)^{-2A} + DE.$$

Recalling the application of Cauchy-Schwarz and the bounds for $\Sigma_1$ and $\Sigma_2$, we sum over $k$ to see that the total contribution is $$\ll (\log x)^{-A + O(1)} + (\log x)^{O(1)} \left(\frac{E}{D} \right)^{1/2}.$$ This is acceptably small if $E \leq D(\log x)^{-B}$.

Therefore, we may assume that $D x^{-o(1)} \ll E \ll D$. Since $DE \gg x^{1-o(1)}$ this implies $D \gg x^{1/2-o(1)}$. For this second argument we do not need any properties of the coefficients attached to the $d$ and $e$ variables other than that they are 1-bounded, so we wish to get a bound on $$\frac{x}{D^2E^2}\sum_{1\leq k\leq (\log x)^{10}DE/x}\Big| \mathop{\sum\sum}_{\substack{d \asymp D \\ e \asymp E \\ (d,e)=1}}\alpha_d \beta_e e \left(k \frac{\overline{d}}{e} \right)\Big|$$ for 1-bounded coefficients $\alpha_d,\beta_e$. Theorem 2 of the above-mentioned Duke-Friedlander-Iwaniec paper gives $$\Big| \mathop{\sum\sum}_{\substack{d \asymp D \\ e \asymp E \\ (d,e)=1}}\alpha_d \beta_e e \left(k \frac{\overline{d}}{e} \right)\Big| \ll (DE)^{1/2} (k+DE)^{3/8} (D+E)^{11/48}\ll D^{2-1/48+o(1)},$$ which saves $D^{1/48}$ over the trivial bound. Therefore $$\frac{x}{D^2E^2}\sum_{1\leq k\leq (\log x)^{10}DE/x}\Big| \mathop{\sum\sum}_{\substack{d \asymp D \\ e \asymp E \\ (d,e)=1}}\alpha_d \beta_e e \left(k \frac{\overline{d}}{e} \right)\Big| \ll \frac{D^{2 - \frac{1}{48} + o(1)}}{DE} \ll D^{-1/48+o(1)} \ll x^{-1/96+o(1)}.$$