---
abstract: 'Consider exponential Carmichael function $\lae$ such that $\lae$ is multiplicative and $\lae(p^a) \hm= \lambda(a)$, where $\lambda$ is usual Carmichael function. We discuss the value of $\sum \lae(n)$, where $n$ runs over certain subsets of $[1,x]$, and provide bounds on the error term, using analytic methods and especially estimates of $\int_1^T \bigl| \zeta(\sigma+it) \bigr|^m dt$.'
address: 'I. I. Mechnikov Odessa National University'
author:
- 'Andrew V. Lelechenko'
bibliography:
- 'taue.bib'
title: Exponential Carmichael function
---

Introduction
============

Consider an operator $E$ over arithmetic functions such that for every $f$ the function $Ef$ is multiplicative and $$(Ef)(p^a) = f(a), \qquad p \text{~is prime}.$$

For various functions $f$ (such as the divisor function, the sum-of-divisor function, Möbius function, the totient function and so on) the behaviour of $Ef$ was studied by many authors, starting from Subbarao [@subbarao1972]. The bibliography can be found in [@bibliography].

The notation for $Ef$, established by previous authors, is $f^{(e)}$.

Carmichael function $\lambda$ is an arithmetic function such that $$\lambda(p^a) = \begin{cases}
\phi(p^a), & p>2 \text{~or~} a=1,2, \\
\phi(p^a)/2, & p=2 \text{~and~} a>2,
\end{cases}$$ and if $n=p_1^{a_1} \cdots p_m^{a_m}$ is a canonical representation, then $$\lambda(n) = \lcm\bigl( \lambda(p_1^{a_1}), \ldots, \lambda(p_m^{a_m}) \bigr).$$

This function was introduced at the beginning of the XX century in [@carmichael1909], but intense studies started only in 1990-th, e. g. [@erdos1991]. Carmichael function finds applications in cryptography, e. g. [@friedlander1999].

Consider also the family of multiplicative functions $$\delta_r(p^a) = \begin{cases}
0, & a<r, \\
1, & a\ge r,
\end{cases}
\qquad r \text{~is integer.}$$

Function $\delta_2$ is a characteristic function of the set of square-full numbers, $\delta_3$ — of cube-full numbers and so on. Of course, $\delta_1 \equiv 1$.

Denote $\lae_r$ for the product of $\delta_r$ and $\lae$: $$\lae_r(n) = \delta_r(n) \lae(n).$$

The aim of our paper is to study asymptotic properties of $\lae \equiv \lae_1$, $\lae_2$, $\lae_3$ and $\lae_4$.

Note that all proofs below remains valid for $\phie_r(n) = \delta_r(n) \phie(n)$ instead of $\lae_r(n)$ for $r=1,2,3,4$.

Notations
=========

Letter $p$ with or without indexes denotes a prime number.

We write $f\star g$ for Dirichlet convolution $$(f \star g)(n) = \sum_{d|n} f(d) g(n/d).$$

Denote $$\tau(a_1,\ldots,a_k; n) := \sum_{d_1^{a_1}\cdots d_k^{a_k} = n} 1.$$

In asymptotic relations we use $\sim$, $\asymp$, Landau symbols $O$ and $o$, Vinogradov symbols $\ll$ and $\gg$ in their usual meanings. All asymptotic relations are given as an argument (usually $x$) tends to the infinity.

Everywhere $\eps>0$ is an arbitrarily small number (not always the same even in one equation).

As usual $\zeta(s)$ is Riemann zeta-function. Real and imaginary components of the complex $s$ are denoted as $\sigma:=\Re s$ and $t:=\Im s$, so $s=\sigma+it$.

For a fixed $\sigma\in[1/2,1]$ define $$m(\sigma) := \sup\biggl\{
m \biggm|
\int_1^T \bigl| \zeta(\sigma+it) \bigr|^m dt \ll T^{1+\eps}
\biggr\}.$$ and $$\mu(\sigma) := \limsup_{t\to\infty} {\log \bigl|\zeta(\sigma+it)\bigr| \over \log t}.$$

Below $H_{2005}=(32/205+\eps, 269/410+\eps)$ stands for Huxley’s exponent pair from [@huxley2005].

Preliminary lemmas
==================

\[l:rational-maximal-order\] Let $F\colon \Z\to\CC$ be a multiplicative function such that $F(p^a) \hm= f(a)$, where $f(n) \ll n^\beta$ for some $\beta>0$. Then $$%\label{eq:rational-maximal-order}
\limsup_{n\to\infty} {\log F(n)\llog n \over\log n} = \sup_{n\ge1} {\log f(n)\over n}.$$

See [@suryanarayana1975].

\[l:log-sum\] Let $f(t)\ge 0$. If $$\int_1^T f(t) \, dt \ll g(T),$$ where $g(T) = T^\alpha \log^\beta T$, $\alpha\ge 1$, then $$%\label{eq:log-summing}
I(T):= \int_1^T {f(t)\over t} dt \ll
\left\{ \begin{matrix}
\log^{\beta+1} T & \text{if } \alpha=1, \\
T^{\alpha-1} \log^{\beta} T & \text{if } \alpha>1.
\end{matrix} \right.$$

Let us divide the interval of integration into parts: $$I(T)
\le
\sum_{k=0}^{\log_2 T}
\int_{T/2^{k+1}}^{T/2^k} {f(t)\over t} dt
<
\sum_{k=0}^{\log_2 T} {1\over T/2^{k+1}}
\int_1^{T/2^k} f(t) dt
\ll
\sum_{k=0}^{\log_2 T} {g(T/2^{k})\over T/2^{k+1}}.$$ Now the lemma’s statement follows from elementary estimates.

\[l:order-in-the-critical-strip\] For $\sigma\ge1/2$ and for any exponent pair $(k,l)$ such that $l-k \ge \sigma$ we have $$\mu(\sigma) \le {k+l-\sigma\over2} + \eps.$$

See [@ivic2003 (7.57)].

A well-known application of Lemma \[l:order-in-the-critical-strip\] is $$\label{eq:mu-1/2}
\mu(1/2) \le 32/205,$$ following from the choice $(k,l) = H_{2005}$. Another (maybe new) application is $$\label{eq:mu-3/5}
\mu(3/5) \le 1409 / 12170,$$ following from $$(k,l) = \left( {269\over 2434}, {1755\over2434} \right) = ABAH_{2005},$$ where $A$ and $B$ stands for usual $A$- and $B$-processes [@kratzel1988 Ch. 2].

\[l:phragmen\] Let $\eta>0$ be arbitrarily small. Then for growing $|t|\ge3$ $$\label{eq:convexity}
\zeta(s) \ll \begin{cases}
|t|^{1/2 - (1-2\mu(1/2))\sigma}, & \sigma\in[0, 1/2],
\\
|t|^{2\mu(1/2)(1-\sigma)} , & \sigma\in[1/2, 1-\eta], \\
|t|^{2\mu(1/2)(1-\sigma)} \log^{2/3} |t| , & \sigma\in[1-\eta, 1], \\
\log^{2/3} |t|, & \sigma\ge1.
\end{cases}$$ More exact estimates for $\sigma\in[1/2, 1-\eta]$ are also available, e. g. $$\label{eq:convexity-3/5}
\mu(\sigma) \ll \begin{cases}
10\bigr( \mu(3/5)-\mu(1/2) \bigl) \sigma + \bigl( 6\mu(1/2) - 5\mu(3/5) \bigr), & \sigma\in[1/2, 3/5], \\
5\mu(3/5)(1-\sigma)/2 , & \sigma\in[3/5,1-\eta], \\
\end{cases}$$

Estimates follow from Phragmén—Lindelöf principle, exact and approximate functional equations for $\zeta(s)$ and convexity properties. See [@titchmarsh1986 Ch. 5] and [@ivic2003 Ch. 7.5] for details.

\[l:lae-maximal-order\] For any integer $r$ $$\max_{n\le x} \lae_r(n) \ll x^\eps.$$

Surely $\lae_r(n) \le \lae(n)$. By Lemma \[l:rational-maximal-order\] we have $$\limsup_{n\to\infty} {\log \lae(n) \log\log n \over \log n}
= \sup_m {\log \lambda(m) \over m} = {\log 4\over 5} =: c,$$ because $\lambda(m) \le m-1$. It implies $$\max_{n\le x} \lae(n) \ll x^{c/\log\log n} \ll x^\eps.$$

\[l:lae-series\] Let $L_r(s)$ be the Dirichlet series for $\lae_r$: $$L_r(s) := \sum_{n=1}^\infty \lae_r(n) n^{-s}.$$ Then for $r=1,2,3,4$ we have $L_r(s) = Z_r(s) G_r(s)$, where $$\begin{aligned}
\label{eq:lae-series-1}
Z_1(s) &= \zeta(s) \zeta(3s) \zeta^2(5s), \\
\label{eq:lae-series-2}
Z_2(s) &= \zeta(2s) \zeta^2(3s) \zeta(4s) \zeta^2(5s), \\
\label{eq:lae-series-3}
Z_3(s) &= \zeta^2(3s) \zeta^2(4s) \zeta^4(5s), \\
\label{eq:lae-series-4}
Z_4(s) &= \zeta^2(4s) \zeta^4(5s) \zeta^2(6s) \zeta^6(7s),\end{aligned}$$ Dirichlet series $G_1(s)$, $G_2(s)$, $G_3(s)$ converge absolutely for $\sigma>1/6$ and $G_4(s)$ converges absolutely for $\sigma>1/8$.

Follows from the identities $$\begin{aligned}
1+\sum_{a\ge1} \lae(p^a) x^a &=
    1+x+x^2+2x^3+2x^4+4x^5+2x^6+6x^7+O(x^8) \\
    &= {1+O(x^8) \over (1-x)(1-x^3)(1-x^5)^2}, \\
1+\sum_{a\ge2} \lae(p^a) x^a &=
    {1+O(x^6) \over (1-x^2)(1-x^3)^2(1-x^4)(1-x^5)^2}, \\
1+\sum_{a\ge3} \lae(p^a) x^a &=
    {1+O(x^6) \over (1-x^3)^2 (1-x^4)^2 (1-x^5)^4}, \\
1+\sum_{a\ge4} \lae(p^a) x^a &=
    {1+O(x^8) \over (1-x^4)^2 (1-x^5)^4 (1-x^6)^2 (1-x^7)^6}.\end{aligned}$$

\[l:kratzel\] Let $\Delta(x)$ be the error term in the well-known asymptotic formula for $\sum_{n\le x} \tau(a_1,a_2,a_3,a_4; n)$, let $A_4 = a_1+a_2+a_3+a_4$ and let $(k,l)$ be any exponent pair. Suppose that the following conditions are satisfied:

1.  $(k+l+2) a_4  < (k+l) a_1 + A_4$.

2.  $2(k+l+1) a_1 \le (2k+1) (a_2+a_3)$.

3.  $l a_1 \le k a_2$ and $(k+l+1)a_1 \ge k(a_2+a_3)$

4.  5.  $l a_1 \ge k a_2$ and $(l-k)(2k+1)a_3 \le (2l-2l-1)(k+l+1)a_1 + \bigl( 2k(k-l+1) + 1 \bigr) a_2$.

This is [@kratzel1992 Th. 3] with $p=4$.

\[l:ivic-moments-on-sigma\] $$m(\sigma) \ge \begin{cases}
\p\p\p\p4/(3-4\sigma), & \p1/2\p \le \sigma \le  \p5/8,  \\
\p\p\p10/(5-6\sigma), & \p5/8\p \le \sigma \le 35/54, \\
\p\p\p19/(6-6\sigma), & 35/54 \le \sigma \le 41/60, \\
\p2112/(859-948\sigma), & 41/60 \le \sigma \le \p3/4, \\
12408/(4537-4890\sigma), & \p3/4\p \le \sigma \le \p5/6, \\
\p4324/(1031-1044\sigma), & \p5/6\p \le \sigma \le \p7/8, \\
\p\p\p98/(31-32\sigma), & \p7/8\p \le \sigma \le 0.91591\ldots, \\
(24\sigma-9)/(4\sigma-1)(1-\sigma), & 0.91591\ldots \le \sigma \le 1-\eps.
\end{cases}$$

See [@ivic2003 Th. 8.4].

Main results
============

\[th:lae-1\] $$\sum_{n\le x} \lae(n)
= c_{11} x + c_{13} x^{1/3} + (c_{15}' \log x + c_{15}) x^{1/5}
+ O(x^{1153/6073+\eps}),$$ where $c_{11}$, $c_{13}$, $c_{15}$ and $c_{15}'$ are computable constants.

Lemma \[l:lae-series\] and equation implies that $
\lae = \tau(1,3,5,5;\cdot) \star g_1
$, where $\sum_{n\le x} g_1(n) \ll x^{1/6+\eps}$. Due to [@kratzel1988] $$\begin{gathered}
\sum_{n\le x} \tau(1,3,5,5; n)
= x \zeta(3) \zeta^2(5) \res_{s=1} \zeta(s)
+ 3 x^{1/3} \zeta(1/3) \zeta^2(5/3) \res_{s=1/3} \zeta(3s)
+
\\
+ 5 x^{1/5} \zeta(1/5) \zeta(3/5) \res_{s=1/5} \zeta^2(5s)
+ R(x).\end{gathered}$$ To estimate $R(x)$ we use Lemma \[l:kratzel\] with $a_1=1$, $a_2=3$, $a_3=a_4=5$. Exponent pair $
(k,l) = H_{2005}$ satisfies conditions 1, 2 and 3.2 and thus $$R(x) \ll x^{(k+l+2)/(k+l+14)} = x^{1153/6073+\eps},
\qquad
1/6 < 1153/6073 < 1/5.$$ Now the convolution argument completes the proof.

Exponential totient function $\phie$ has similar to $\lae$ Dirichlet series: $$\sum_{n=1}^\infty \phie(n) = \zeta(s) \zeta(3s) \zeta^2(5s) H(s),$$ where $H(s)$ converges absolutely for $\sigma>1/6$. Theorem \[th:lae-1\] can be extended to this case without any changes, so $$\sum_{n\le x} \phie(n)
= c_{11} x + c_{13} x^{1/3} + (c_{15}' \log x + c_{15}) x^{1/5}
+ O(x^{1153/6073+\eps}).$$ This improves the result of Pétermann [@petermann2010], who obtained $
\sum_{n\le x} \phie(n)
\hm= c_{11} x \hm+ c_{13} x^{1/3} + O(x^{1/5} \log x)
$. Recently Cao and Zhai [@caozhai2013 (1.13)] obtained $R(x) \hm\ll x^{18/95+\eps}$, which is better than both Pétermann’s $x^{1/5} \log x$ and our $x^{1153/6073+\eps}$.

\[th:lae-2\] $$\sum_{n\le x} \lae_2(n)
= c_{22} x^{1/2} + (c_{23}'\log x + c_{23}) x^{1/3} + c_{24} x^{1/4}
+ O(x^{1153/5586+\eps}),$$ where $c_{22}$, $c_{23}$, $c_{23}'$ and $c_{24}$ are computable constants.

Similar to Theorem \[th:lae-1\] with following changes: now by $$\lae_2 = \tau(2,3,3,4; \cdot) \star g_2,$$ where $\sum_{n\le x} g_2(n) \ll x^{1/6+\eps}$. But $$\begin{gathered}
\sum_{n\le x} \tau(2,3,3,4; n)
= 2 x^{1/2} \zeta^2(3/2) \zeta(2) \res_{s=1/2} \zeta(2s)
+
\\
+ 3 x^{1/3} \zeta(2/3) \zeta(4/3) \res_{s=1/3} \zeta^2(3s)
+ 4 x^{1/4} \zeta(1/2) \zeta^2(3/4) \res_{s=1/4} \zeta(4s) + R(s).\end{gathered}$$ Again by Lemma \[l:kratzel\] with $a_1=2$, $a_2=a_3=3$, $a_4=4$, $(k,l)=H_{2005}$ we get $$R(x) \ll x^{(k+l+2)/(k+l+12)} = x^{1153/5586+\eps},
\qquad
1/5 < 1153/5586 < 1/4.$$

\[th:lae-3\] $$\begin{gathered}
\label{eq:lae-asymp-3}
\sum_{n\le x} \lae_3(n) =
(c_{33}' \log x + c_{33}) x^{1/3}
+ (c_{34}' \log x + c_{34}) x^{1/4}
+\\+ P_{35}(\log x) x^{1/5} + O(x^{1/6+\eps}),\end{gathered}$$ where $c_{33}$, $c_{33}'$, $c_{34}$ and $c_{34}'$ are computable constants, $P_{35}$ is a polynomial of degree 3 with computable coefficients.

Lemma \[l:lae-series\] and equation implies that $
\lae_3 = z_3 \star g_3
$, where $z_3$ is defined implicitly by $$\sum_{n=1}^\infty z_3(n) n^{-s} = Z_3(s) = \zeta^2(3s) \zeta^2(4s) \zeta^4(5s),$$ and $g_3$ is a multiplicative function such that $\sum_{n\le x} g_3(n) \ll x^{1/6+\eps}$.

The main term at the right side of equals to $$M_3(x) := \left( \res_{s=1/3} + \res_{s=1/4} + \res_{s=1/5}  \right) \bigl( \zeta^2(3s) \zeta^2(4s) \zeta^4(5s) x^s s^{-1} \bigr).$$ To obtain the desirable error term it is enough to prove that $$\sum_{n\le x} z_3(n) = M_3(x) + O(x^{1/6+\eps}).$$ By Perron formula for $c:=1/3+1/\log x$ we have $$\sum_{n\le x} z_3(n)
= {1\over 2\pi i} \int_{c-iT}^{c+iT} Z_3(s) x^s s^{-1} ds
+ O(x^{1+\eps} T^{-1}).$$ Substituting $T=x$ and moving the contour of the integration till $[1/6 \hm -ix, 1/6 \hm+ ix]$ we get $$\sum_{n\le x} f_3(n) = M_3(x) + O(I_0 + I_- + I_+ + x^\eps),$$ where $$I_0 := \int_{1/6-ix}^{1/6+ix} Z_3(s) x^s s^{-1} ds,
\qquad
I_\pm := \int_{1/6\pm ix}^{c\pm ix} Z_3(s) x^s s^{-1} ds.$$

Firstly, $$I_+ \ll x^{-1} \int_{1/6}^c Z_3(\sigma+ix) x^\sigma d\sigma.$$ Let $\alpha(\sigma)$ be a function such that $Z_3(\sigma+ix) \ll x^{\alpha(\sigma)+\eps}$. By we have $$\alpha(\sigma) \le \begin{cases}
(16-68\sigma)\mu(1/2) < 4/5, & \sigma \in [1/6, 1/5), \\
\p(8-28\sigma)\mu(1/2) < 3/4, & \sigma \in [1/5, 1/4), \\
\p(4-12\sigma)\mu(1/2) < 2/3, & \sigma \in [1/4, 1/3), \\
\phantom{(0}0, & \sigma \in [1/3, c].
\end{cases}$$ This means that $I_+ \ll x^\eps$. Plainly, the same estimate holds for $I_-$.

Secondly, it remains to prove that $I_0 \ll x^{1/6+\eps}$. Here $$I_0 \ll x^{1/6} \int_1^x Z_3(1/6+it) t^{-1} dt$$ and taking into account Lemma \[l:log-sum\] it is enough to show $\int_1^x Z_3(1/6+it)  dt \hm\ll x^{1+\eps}$. Applying Cauchy inequality twice we obtain $$\begin{gathered}
\int_1^x Z_3(1/6+it)  dt \ll
\left( \int_1^x \bigl|\zeta^4(1/2+it)\bigr| dt \right)^{1/2}
\times \\ \times
\left( \int_1^x \bigl|\zeta^8(2/3+it)\bigr| dt \right)^{1/4}
\left( \int_1^x \bigl|\zeta^{16}(5/6+it)\bigr| dt \right)^{1/4}
\ll \\
\ll x^{(1+\eps)\cdot 1/2} x^{(1+\eps)\cdot 1/4} x^{(1+\eps)\cdot 1/4}
\ll x^{1+\eps}\end{gathered}$$ since by Lemma \[l:ivic-moments-on-sigma\] $m(1/2)\ge4$, $m(2/3)\ge 8$ and $m(5/6)\hm\ge 16$.

\[th:lae-4\] $$\begin{gathered}
\sum_{n\le x} \lae_4(n)
= (c_{44}' \log x + c_{44}) x^{1/4}
+ P_{45}(\log x) x^{1/5}
+ (c_{46}' \log x + c_{46}) x^{1/6}
+ \\
+ P_{47}(\log x) x^{1/7}
+ O(x^{C_4+\eps}),\end{gathered}$$ where $c_{44}$, $c_{44}'$, $c_{46}$ and $c_{46}'$ are computable constants, $P_{45}$ and $P_{47}$ are computable polynomials, $\deg P_{45} = 3$, $\deg P_{47} = 5$, $$\label{eq:C_4}
C_4 = {7863059 - \sqrt{13780693090921} \over 85962240} = 0.134656\ldots, \qquad 1/8 < C_4 < 1/7.$$

We shall follow the outline of Theorem \[th:lae-3\]. Let us prove that for $c:=1/4 \hm+ 1/\log x$ we can estimate $$I_+ := \int_{C_4+ix}^{c+ix} Z_4(s) x^s s^{-1} ds \ll x^{C_4+\eps}$$ and $$I_0 := \int_{C_4-ix}^{C_4+ix} Z_4(s) x^s s^{-1} ds \ll x^{C_4+\eps}.$$

We start with $I_+ \ll x^{-1} \int_{C_4}^c Z_4(\sigma+ix) x^\sigma d\sigma$. Now let $\alpha(\sigma)$ be a function such that $Z_4(\sigma+ix) \ll x^{\alpha(\sigma)+\eps}$. By and we have $$\alpha(\sigma) \le \begin{cases}
%(28-164\sigma)\mu(1/2), & \sigma \in [1/8, 1/7), \\
(16-80\sigma)\mu(1/2) < 5/6, & \sigma \in [1/7, 1/6), \\
(12-56\sigma)\mu(1/2) < 4/5, & \sigma \in [1/6, 1/5), \\
\p(4-16\sigma)\mu(1/2) < 3/4, & \sigma \in [1/5, 1/4), \\
\phantom{(0}0, & \sigma \in [1/4, c].
\end{cases}$$ So $\int_{1/7}^c Z_4(\sigma+ix) x^{\sigma-1} d\sigma \ll x^\eps$ and the only case that requires further investigations is $\sigma\in[C_4, 1/7)$. Instead of we apply together with and to obtain $$\alpha(\sigma) \le {1045018\over249485} - {2459357\over99794} \sigma, \qquad \sigma\in[1/8,1/7],$$ which implies $\int_{C_4}^{1/7} x^{\alpha(\sigma) + \sigma - 1} d\sigma \ll x^{C_4+\eps}$ as soon as $$C_4 \ge 1591066/12296785 = 0.129388\ldots$$ Our choice of $C_4$ in is certainly the case.

Let us move on $I_0$ and prove that $\int_1^x Z_4(C_4+it)\,dt \ll x^{1+\eps}$. For $q_1$, $q_2$, $q_3$, $q_4$ such that $$\label{eq:holder-conditions}
1/q_1+1/q_2+1/q_3+1/q_4=1 \qquad\text{and}\qquad q_1,q_2,q_3,q_4 \ge 1$$ by Hölder inequality we have $$\begin{gathered}
\int_1^x Z_4(C_4+it)\,dt
\ll
\left( \int_1^x \bigl| \zeta^{2q_1}(4s+it) \bigr| dt \right)^{1/q_1}
\left( \int_1^x \bigl| \zeta^{4q_2}(5s+it) \bigr| dt \right)^{1/q_2}
\sp
\times \\ \times
\left( \int_1^x \bigl| \zeta^{2q_3}(6s+it) \bigr| dt \right)^{1/q_3}
\left( \int_1^x \bigl| \zeta^{6q_4}(7s+it) \bigr| dt \right)^{1/q_4}
\sp
.\end{gathered}$$ Choose $$\label{eq:choice-of-qk}
q_1 = m(4C_4)/2, ~~
q_2 = m(5C_4)/4, ~~
q_3 = m(6C_4)/2, ~~
q_4 = m(7C_4)/6$$ One can make sure by substituting the value of $C_4$ from into Lemma \[l:ivic-moments-on-sigma\] that such choice of $q_k$ satisfies . Thus we obtain $$\int_1^x Z_4(C_4+it)\,dt \ll x^{(1+\eps)/q1} x^{(1+\eps)/q2} x^{(1+\eps)/q3} x^{(1+\eps)/q4} \ll x^{1+\eps},$$ which finishes the proof.

Decrease of C4
==============

In this section we obtain lower value of $C_4$ by improving lower bounds of $m(\sigma)$ from Lemma \[l:ivic-moments-on-sigma\].

Estimates below depend on values of $$\label{eq:linear-programming}
\inf_{(k,l)} {ak+bl+c \over dk+el+f},$$ where $(k,l)$ runs over the set of exponent pairs and satisfies certain linear inequalities. A method to estimate without linear constrains was given by Graham [@graham1986]. In the recent paper [@lelechenko2013-acta] we have presented an effective algorithm to deal with under a nonempty set of linear constrains.

Let $c$ be an arbitrary function such that $c(\sigma) \ge \mu(\sigma)$. Define $\theta$ by an implicit equation $$2 c\bigl(\theta(\sigma)\bigr) + 1 + \theta(\sigma) - 2 \bigl( 1+c\bigl(\theta(\sigma)\bigr) \bigr) \sigma = 0.$$ Finally, define $$f(\sigma) = 2{1+c\bigl(\theta(\sigma)\bigr) \over c\bigl(\theta(\sigma)\bigr)}.$$ Due to Lemma \[l:order-in-the-critical-strip\] one can take $c(\sigma) = \inf_{l-k\ge\sigma} (k+l-\sigma)/2$, where $(k,l)$ runs over the set of exponent pairs. However even rougher choice of $c$ leads to satisfiable values of $f$ such as in [@ivic2003 (8.71)].

\[l:pointwise-moments-on-sigma\] Let $\sigma\ge5/8$. Compute $$\alpha_1 = {4-4\sigma \over 1+2\sigma},
\qquad
\beta_1 = - {12\over 1+2\sigma},
\qquad
m_1 = {1-\alpha_1 \over \mu(\sigma)} - \beta_1,$$ $$\alpha_2(k,l)
= {4(1-\sigma)(k+l) \over (2+4l)\sigma-1+2k-2l},
\qquad
\beta_2(k,l)
= -{4(1+2k+2l) \over (2+4l)\sigma-1+2k-2l},$$ $$m_2(k,l) = {1-\alpha_2(k,l) \over \mu(\sigma)} - \beta_2(k,l),
\qquad
m_2 = \inf_{\alpha_2(k,l) \le 1  } m_2(k,l),$$ where $(k,l)$ runs over the set of exponent pairs. Then $$m(\sigma) \ge \min\bigl(m_1, m_2, 2f(\sigma)\bigr).$$ Note that for $\sigma\ge2/3$ the condition $\alpha_2(k,l) \le 1$ is always satisfied.

Follows from [@ivic2003 (8.97)] and from $
T^\alpha V^\beta \ll T V^{\beta + (\alpha-1) / \mu(\sigma)}
$ for $\alpha<1$ and $V \ll T^{\mu(\sigma)}$.

Substituting pointwise estimates of $m(\sigma)$ from Lemma \[l:pointwise-moments-on-sigma\] instead of segmentwise from Lemma \[l:ivic-moments-on-sigma\] into we obtain following result.

The statement of Theorem \[th:lae-4\] remains valid for $$C_4 = 0.133437785\ldots$$

Conclusion
==========

We have obtained nontrivial error terms in asymptotic estimates of $\sum_{n\le x} \lae_r(n)$ for $r=1,2,3,4$. Cases of $r=1$ and $r=2$ depend on the method of exponent pairs. Cases of $r=3$ and $r=4$ depend on lower bounds of $m(\sigma)$. Note that case of $r=4$ may be improved under Riemann hypothesis up to $C_4=1/8$, because Riemann hypothesis implies $\mu(\sigma)=0$ and $m(\sigma)=\infty$ for $\sigma\in[1/2,1]$.
