1 Introduction

Let \(n\ge 1\) be an integer. Denote by \(\phi (n)\) and \(\lambda (n)\) the Euler Phi function and the Carmichael Lambda function, which output the order and the exponent of the group \((\mathbb {Z}/n\mathbb {Z})^{*}\), respectively. We use \(p(\mathrm {or} \ p_i)\) and \(q(\mathrm {or } \ q_i)\) to denote the prime divisors of n and \(\phi (n)\), respectively. Then it is clear that \(\lambda (n)|\phi (n)\) and the set of prime divisors q of \(\phi (n)\) and that of \(\lambda (n)\) are identical. Let \(n=p_1^{e_1}\cdots p_r^{e_r}\) be a prime factorization of n. Then we can compute \(\phi (n)\) and \(\lambda (n)\) as follows:

$$\begin{aligned} \phi (n)=\prod _{i=1}^r \phi (p_i^{e_i}), \ \mathrm {and} \ \lambda (n)=\mathrm {lcm}\left( \lambda (p_1^{e_1}), \ldots , \lambda (p_r^{e_r})\right) , \end{aligned}$$

where \(\phi (p_i^{e_i})=p_i^{e_i-1}(p_i-1)\) and \(\lambda (p_i^{e_i})=\phi (p_i^{e_i})\) if \(p_i>2\) or \(p_i=2\) and \(e_i=1, 2\), and \(\lambda (2^e)=2^{e-2}\) if \(e\ge 3\).

From the work of Hardy and Ramanujan  [4], it is well known that the normal order of \(\tau (n)\) is \((\log n)^{\log 2 + o(1)}\). On the other hand, the average order \(\frac{1}{x}\sum \nolimits _{n\le x} \tau (n)\) is known to be \(\log x + O(1)\) which is somewhat larger than the normal order. For \(\tau (\lambda (n))\) and \(\tau (\phi (n))\), the normal orders of these follow from  [2] that they are \(2^{(\frac{1}{2} +o(1))(\log \log n)^2}\). On the contrary, the work of Luca and Pomerance  [5] showed that their average order is significantly larger than the normal order. Define \(F(x) = \exp \left( \sqrt{\frac{\log x}{\log \log x}}\right) \). In  [5, Theorem 1,2], they proved that

$$\begin{aligned} F(x)^{b_1+o(1)} \le \frac{1}{x}\sum _{n\le x}\tau (\lambda (n)) \le \frac{1}{x}\sum _{n\le x}\tau (\phi (n)) \le F(x)^{b_2+o(1)} \end{aligned}$$

as \(x\rightarrow \infty \), where \(b_1 = \frac{1}{7} e^{-\gamma /2}\) and \(b_2 = 2\sqrt{2} e^{-\gamma /2}\).

In this paper, we are able to raise the constant \(b_1\) so that it is almost \(b_2\), differing only by a factor \(\sqrt{2}\). Here, we take advantage of the inequalities of Bombieri–Vinogradov type regarding primes in arithmetic progression (see  [1, Theorem 9], also  [3, Theorem 2.1]). In this paper, we apply the following version which can be obtained from  [3, Theorem 2.1]: For \((a,n)=1\), we write \(E(x;n,a):=\pi (x;n,a)-\frac{\pi (x)}{\phi (n)}\). Let \(0<\lambda <1/10\). Let \(R\le x^{\lambda }\). For some \(B=B(A)>0\), \(M=\log ^B x\), and \(Q=x/M\),

$$\begin{aligned} \sum _{\begin{array}{c} {r\le R}\\ {(r,a)=1} \end{array}}\left| \sum _{\begin{array}{c} {q\le \frac{Q}{ r}}\\ {(q,a)=1} \end{array}}E(x;qr,a) \right| \ll _{A,\lambda } x \log ^{-A} x. \end{aligned}$$

In fact,  [3, Theorem 2.1] builds on  [1, Theorem 9] and obtains a more accurate estimate, but we only need the above form for our purpose. Note that one of the important differences between  [1, Theorem 9] and  [3, Theorem 2.1] is the presence of \(\frac{Q}{r}\) in the inner sum. This will be essential in the proof of our lemmas (see Lemmas 2.2 and 2.3).

It is interesting to note that one of these improvements is related to a Poisson distribution that we can obtain from prime numbers. Another point of improvement comes from the idea in the proof of Gauss’ Circle Problem.

Theorem 1.1

As \(x\rightarrow \infty \), we have

$$\begin{aligned} \sum _{n\le x}\tau (\phi (n))\ge \sum _{n\le x} \tau (\lambda (n))\ge x\exp \left( 2 e^{-\frac{\gamma }{2}}\sqrt{\frac{\log x}{\log \log x}}(1+o(1))\right) . \end{aligned}$$

It is clear from \(\lambda (n)|\phi (n)\) that \(\sum _{n\le x}\tau (\lambda (n))\le \sum _{n\le x}\tau (\phi (n))\). A natural question to ask is how large is the latter compared to the former. Luca and Pomerance proved in  [5, Theorem 2] that

$$\begin{aligned} \frac{1}{x} \sum _{n\le x}\tau (\lambda (n))=o\left( \max _{y\le x} \frac{1}{y} \sum _{n\le y}\tau (\phi (n)) \right) . \end{aligned}$$

Moreover, they mentioned that a stronger statement

$$\begin{aligned} \frac{1}{x}\sum _{n\le x}\tau (\lambda (n)) = o \left( \frac{1}{x}\sum _{n\le x}\tau (\phi (n))\right) \end{aligned}$$

is probably true, but they did not have the proof. Here, we prove that this statement is indeed true. As in the proof of  [5, Theorem 2], we take advantage of the fact that prime 2 appears rarely in the factorization of \(\lambda (n)\) than in the factorization of \(\phi (n)\).

Theorem 1.2

As \(x\rightarrow \infty \), we have

$$\begin{aligned} \sum _{n\le x}\tau (\lambda (n)) = o \left( \sum _{n\le x}\tau (\phi (n))\right) . \end{aligned}$$

Finally, we provide a heuristic argument suggesting that the constant in the upper bound is indeed optimal. Here, we try to extend the method in the proof of Theorem 1.1 by devising a binomial distribution model. However, we were unable to prove it. The main difficulty is due to the short range of u (\(u<\log ^{A_1} x\)) in the lemmas (see Lemmas 2.1, 2.3 and Corollaries 2.1, 2.2).

Conjecture 1.1

As \(x\rightarrow \infty \), we have

$$\begin{aligned} \sum _{n\le x} \tau (\lambda (n))=x\exp \left( 2\sqrt{2} e^{-\frac{\gamma }{2}} \sqrt{\frac{\log x}{\log \log x}}(1+o(1))\right) . \end{aligned}$$

Throughout this paper, x is a positive real number, n, k are positive integers, and p, q are prime numbers. We use Landau symbols O and o. Also, we write \(f(x)\asymp g(x)\) for positive functions f and g, if \(f(x)=O(g(x))\) and \(g(x)=O(f(x))\). We will also use Vinogradov symbols \(\ll \) and \(\gg \). We write the iterated logarithms as \(\log _2 x = \log \log x\) and \(\log _3 x = \log \log \log x\). The notations (a, b) and [a, b] mean the greatest common divisor and the least common multiple of a and b, respectively. We write \(P_z=\prod _{p\le z} p\). We also use the following restricted divisor functions:

$$\begin{aligned} \tau _z(n):=\prod _{\begin{array}{c} {p^e||n}\\ {p>z} \end{array}}\tau (p^e), \ \ \tau _{z,w}(n):=\prod _{\begin{array}{c} {p^e||n}\\ {z<p\le w} \end{array}}\tau (p^e), \ \ \text {and } \ \tau _z'(n):=\prod _{\begin{array}{c} {p^e||n}\\ {p\le z} \end{array}}\tau (p^e). \end{aligned}$$

Moreover, for \(n>1\), denote by p(n) the smallest prime factor of n.

2 Lemmas

The following lemma is  [5, Lemma 3] with a slightly relaxed z, and it is essential toward proving the theorem. This is stated and proved with the Chebyshev functions \(\psi (x):=\sum \nolimits _{n\le x} \Lambda (n)\) and \(\psi (x;q,a):=\sum \nolimits _{n\le x, \ n\equiv a \ \mathrm {mod} \ q} \Lambda (n)\) in  [6]. Here, we use the prime counting functions \(\pi (x):=\sum \nolimits _{p\le x} 1\) and \(\pi (x;q,a):=\sum \nolimits _{p\le x, \ p\equiv a \ \mathrm {mod} \ q} 1\) instead. We are allowed to do these replacements by applying the partial summation.

Lemma 2.1

Let \(0<\lambda <\frac{1}{10}\). Assume that \(z\le \lambda \log x\). Then, for any \(A>0\), there is \(B=B(A)>0\) such that, for \(M=\log ^B x\) and \(Q=\frac{x}{M}\),

$$\begin{aligned} E_z(x):=\sum _{r|P_z}\mu (r)\sum _{\begin{array}{c} {n\le Q}\\ {r|n} \end{array}}\left( \pi (x;n,1)-\frac{\pi (x)}{\phi (n)}\right) \ll _{A, \lambda }\frac{x}{\log ^A x}. \end{aligned}$$
(1)

Let \(0<\lambda <\frac{1}{10}\). Assume that u is a positive integer with \(p(u)>z\), \(u<(\log x)^{A_1}\), and \(\tau (u)<A_1\). Then, for any \(A>0\), there is \(B=B(A,A_1)>0\) such, that for \(M=\log ^B x\) and \(Q=\frac{x}{M}\),

$$\begin{aligned} E_{u,z}(x):=\sum _{r|P_z}\mu (r)\sum _{\begin{array}{c} {n\le Q}\\ {r|n} \end{array}}\left( \pi (x;[u,n],1)-\frac{\pi (x)}{\phi ([u,n])}\right) \ll _{A, A_1,\lambda } \frac{x}{\log ^A x}. \end{aligned}$$
(2)

Proof

of (1) For \((a,n)=1\), we write \(E(x;n,a):=\pi (x;n,a)-\frac{\pi (x)}{\phi (n)}\). If \(r|P_z\), we have by the Prime Number Theorem \(r\le R:=P_z=\exp (z+o(z))\le x^{\lambda '}\) with \(0<\lambda '<1/10\). By partial summation and diadically applying  [3, Theorem 2.1], we have, for \(B=B(A)>0\), \(M=\log ^B x\), and \(Q=x/M\),

$$\begin{aligned} \sum _{\begin{array}{c} {r\le R}\\ {(r,a)=1} \end{array}}\left| \sum _{\begin{array}{c} {q\le \frac{Q}{ r}}\\ {(q,a)=1} \end{array}}E(x;qr,a) \right| \ll _{A,\lambda } \frac{x}{\log ^A x}. \end{aligned}$$
(3)

Taking \(a=1\) and \(|\mu (r)|\le 1\), (1) follows. \(\square \)

Proof

of (2) Let \(d\le x^{\epsilon }\) so that \(dR\le x^{\lambda '}\) with \(0<\lambda '<1/10\). By (3), there exists \(B=B(A)>0\) such that we have, for \(M=\log ^B x\) and \(Q=x/M\),

$$\begin{aligned} \sum _{ r\le R }\left| \sum _{ q\le \frac{Q}{ r} }E(x;dqr,1)\right|&=\sum _{\begin{array}{c} {r\le dR}\\ {r\equiv 0 \ \mathrm { mod } \ d} \end{array}}\left| \sum _{ q\le \frac{Q}{ r} }E(x;qr,1)\right| \nonumber \\&\le \sum _{ r\le dR} \left| \sum _{ q\le \frac{Q}{ r}} E(x;qr,1)\right| \ll _{A,\lambda } \frac{x}{\log ^A x}. \end{aligned}$$
(4)

By \((u,r)=1\), we have \([u,n]=[u,qr]=r[u,q]=ruq/(u,q)\). We partition the set of \(q\le \frac{Q}{r}\) as \(\bigcup _{d|u} A_d\), where \(q\in A_d\) if and only if \((u,q)=d\). Let \(B_{Q,d}=\{q\le \frac{Q}{r}: q \equiv 0 \ \mathrm { mod } \ d\}\). By inclusion–exclusion, we have, for any d|u,

$$\begin{aligned} \sum _{q\in A_d}E\left( x;\frac{ruq}{d},1\right) = \sum _{s|\frac{u}{d}} \mu (s) \sum _{q\in B_{Q,\mathrm{d}s}}E\left( x;\frac{ruq}{d},1\right) . \end{aligned}$$

It is clear that

$$\begin{aligned} \sum _{q\in B_{Q,\mathrm{d}s}}E\left( x;\frac{ruq}{d},1\right) =\sum _{q\in B_{\frac{uQ}{d},us}}E(x;qr,1). \end{aligned}$$

Since \(r\le R:=P_z<x^{\lambda '}\) with \(\lambda '<\frac{1}{10}\), \(\frac{uQ}{d}\le Q\log ^{A_1}x\), and \(us<\log ^{2A_1} x < x^{\epsilon }\), we have, by (4),

$$\begin{aligned} \sum _{r\le R} \left| \sum _{q\in B_{\frac{uQ}{d},us}}E(x;qr,1)\right| \ll _{A, A_1, \lambda } \frac{x}{\log ^{A} x} \end{aligned}$$

with a suitable choice of \(B=B(A, A_1)\). Then

$$\begin{aligned} \sum _{r\le R}\left| \sum _{q\in A_d}E\left( x;\frac{ruq}{d},1\right) \right|&=\sum _{r\le R}\left| \sum _{s|\frac{u}{d}}\mu (s)\sum _{q\in B_{Q,\mathrm{d}s}}E\left( x;\frac{ruq}{d},1\right) \right| \\&\le \sum _{s|\frac{u}{d}}\sum _{r\le R}\left| \sum _{q\in B_{Q,\mathrm{d}s}}E\left( x;\frac{ruq}{d},1\right) \right| \\&\ll _{A,A_1,\lambda } \tau \left( \frac{u}{d}\right) \frac{x}{\log ^{A} x}. \end{aligned}$$

Thus, summing over d|u, we have

$$\begin{aligned} \left| \sum _{r|P_z}\mu (r) \sum _{q\le \frac{Q}{r}} E(x;[u,qr],1) \right|&\le \sum _{d|u}\sum _{r\le R}\left| \sum _{q\in A_d}E\left( x;\frac{ruq}{d},1\right) \right| \\&\ll _{A,A_1,\lambda } (\tau (u))^2 \frac{x}{\log ^{A} x}\ll _{A, A_1,\lambda } \frac{x}{\log ^{A} x}. \end{aligned}$$

Thus, we have the result (2). \(\square \)

The following is  [5, Lemma 5] with a slightly relaxed z.

Lemma 2.2

Let \(0<\lambda <\frac{1}{10}\) and \(1<z\le \lambda \log x\). Let \(c_1=e^{-\gamma }\). Then we have

$$\begin{aligned} R_z(x):=\sum _{p\le x} \tau _z(p-1)=c_1 \frac{x}{\log z} + O\left( \frac{x}{\log ^2 z}\right) , \end{aligned}$$
(5)

and, for \(1<z\le \frac{\log x}{\log _2^2 x}\),

$$\begin{aligned} S_z(x):=\sum _{p\le x}\frac{\tau _z(p-1)}{p}=c_1\frac{\log x}{\log z} + O\left( \frac{\log x}{\log ^2 z}\right) . \end{aligned}$$
(6)

Proof

of (5) Take \(A=2\) and the corresponding B(A) and M in Lemma 2.1(1). Then by inclusion–exclusion,

$$\begin{aligned} R_z(x)&= \sum _{d\in D_z(x)} \pi (x;d,1)\\&= \sum _{d\in D_z\left( \frac{x}{M}\right) } \pi (x;d,1)+\sum _{r|P_z}\mu (r)\sum _{\frac{x}{rM}<q\le \frac{x}{r}}\pi (x;qr,1)=R_1+R_2, \text { say.} \end{aligned}$$

By  [5, Lemma 4] and Lemma 2.1(1),

$$\begin{aligned} R_1&=\sum _{d\in D_z\left( \frac{x}{M}\right) } \frac{\pi (x)}{\phi (d)} + \sum _{r|P_z}\mu (r)\sum _{q\le \frac{x}{rM}} E(x;qr,1)\\&=c_1\frac{x}{\log z}+O\left( \frac{x}{\log ^2 z}\right) + O\left( \frac{x}{\log ^2 x}\right) . \end{aligned}$$

By divisor-switching technique and Brun–Titchmarsh inequality as in  [6], we have

$$\begin{aligned} R_2\ll \sum _{r|P_z}\sum _{k\le M}\pi (x;rk,1)\ll \sum _{r|P_z}\sum _{k\le M} \frac{x}{\phi (rk) \log x}\ll \frac{x\log z\log M}{\log x}\ll \frac{x}{\log ^2 z}. \end{aligned}$$

Therefore, (5) follows. \(\square \)

Proof

of (6) By partial summation,

$$\begin{aligned} S_z(x)=\frac{R_z(t)}{t}|_2^x + \int _2^x \frac{R_z(t)}{t^2}\mathrm{d}t. \end{aligned}$$

We split the integral at \(z = \lambda \log t\). Then, by (4),

$$\begin{aligned} \int _{z\le \lambda \log t} \frac{R_z(t)}{t^2}dt = \int _{e^{z/\lambda }}^x \left( c_1 \frac{t}{\log z} + O\left( \frac{t}{\log ^2 z}\right) \right) \frac{\mathrm{d}t}{t^2}=c_1\frac{\log x}{\log z} + O\left( \frac{\log x}{\log ^2 z}\right) . \end{aligned}$$

On the other hand, by the trivial bound \(R_z(t)\ll t\),

$$\begin{aligned} \int _{z>\lambda \log t} \frac{R_z(t)}{t^2}\mathrm{d}t \ll \int _2^{e^{z/\lambda }} t \frac{\mathrm{d}t}{t^2} \ll z. \end{aligned}$$

Since \(z\log ^2 z\ll \log x\), (6) follows. \(\square \)

The following is  [5, Lemma 6] with a wider range of z. This relaxes the rather severe restriction \(z\le \frac{\sqrt{\log x}}{\log _2^6 x}\).

Lemma 2.3

Let \(1\le u \le x\) be any positive integer. Then

$$\begin{aligned} R_{u,z}(x):&=\sum _{\begin{array}{c} {p\le x}\\ {p \equiv 1 \ \mathrm { mod } \ u} \end{array}}\tau _z(p-1) \ll \frac{\tau (u)}{\phi (u)}x, \ \ S_{u,z}(x):\nonumber \\&=\sum _{\begin{array}{c} {p\le x}\\ {p \equiv 1 \ \mathrm { mod } \ u} \end{array}}\frac{\tau _z(p-1)}{p}\ll \frac{\tau (u)}{\phi (u)}\log x, \end{aligned}$$
(7)

and \(\phi (u)\) can be replaced by u if \(p(u)>z\) and \(\tau (u)<A_1\).

Assume that u is a positive integer with \(p(u)>z\), \(u< (\log x)^{A_1}\), and \(\tau (u)<A_1\). Then, for \(z\le \lambda \log x\),

$$\begin{aligned} R_{u,z}(x) =\frac{\tau (u)}{u}R_z(x) \left( 1+O\left( \frac{1}{\log z}\right) \right) , \end{aligned}$$
(8)

and, for \(z\le \frac{\log x}{\log _2^2 x}\),

$$\begin{aligned} S_{u,z}(x) =\frac{\tau (u)}{u} S_z(x)\left( 1+O\left( \frac{1}{\log z}\right) \right) . \end{aligned}$$
(9)

Proof

of (7) This is a uniform version of [8, Lemma 3.7]. We apply Dirichlet’s hyperbola method as it was done in  [8, Lemma 3.7]. First, we see that

$$\begin{aligned} R_{u,z}(x)\le & {} \sum _{\begin{array}{c} {p\le x}\\ {p \equiv 1 \ \mathrm { mod } \ u} \end{array}}\tau (p-1)\\\le & {} \sum _{\begin{array}{c} {p\le x}\\ {p \equiv 1 \ \mathrm { mod } \ u} \end{array}}\tau \left( \frac{p-1}{u}\right) \tau (u) \le 2\tau (u) \sum _{k\le \sqrt{\frac{x}{u}}} \pi (x;ku,1). \end{aligned}$$

Since the sum is zero for \(x\le u\), we may assume that \(x>u\). By Brun–Titchmarsh inequality,

$$\begin{aligned} \pi (x;ku,1)\le \frac{2x}{\phi (ku)\log \left( \frac{x}{ku}\right) }\le \frac{4x}{\phi (u)\phi (k)\log \frac{x}{u}}. \end{aligned}$$

Thus, summing over k gives

$$\begin{aligned} \sum _{k\le \sqrt{\frac{x}{u}}} \pi (x;ku,1)\le \frac{8x}{\phi (u)}\sum _{d=1}^{\infty }\frac{\mu ^2(d)}{d\phi (d)}. \end{aligned}$$

Therefore, we have the result. The estimate for \(S_{u,z}\) follows from partial summation.

We remark that, for u with \(p(u)>z\),

$$\begin{aligned} \frac{u\phi (d)}{\phi (ud)}&=\prod _{p|u, p\not \mid d} \left( 1-\frac{1}{p}\right) ^{-1}=1+O\left( \frac{\tau (u)}{z}\right) , \ \ \ \frac{1}{\phi (u)}\\&=\frac{1}{u} \prod _{p|u}\left( 1-\frac{1}{p}\right) ^{-1}=\frac{1}{u} \left( 1+O\left( \frac{\tau (u)}{z}\right) \right) . \end{aligned}$$

Therefore, \(\phi (u)\) can be replaced by u if \(p(u)>z\) and \(\tau (u)<A_1\). \(\square \)

Proof

of (8) We begin with

$$\begin{aligned} R_{u,z}(x)=\sum _{d\in D_z(x)}\pi (x;[u,d],1). \end{aligned}$$

Let \(A>0\) be a positive number such that \(\frac{x}{\log ^A x} \ll \frac{\tau (u)}{u} \frac{x}{\log ^2 x}\), and B(A) and M be the corresponding parameters depending on A in Lemma 2.1(2). By inclusion–exclusion,

$$\begin{aligned} \sum _{d\in D_z(x)}\pi (x;[u,d],1)&=\sum _{d\in D_z\left( \frac{x}{M}\right) }\pi (x;[u,d],1)\\&\quad +\sum _{r|P_z}\mu (r)\sum _{\frac{x}{rM}<q\le \frac{x}{r}}\pi (x;[u,qr],1)=R_1+R_2, \ \text { say}. \end{aligned}$$

By Lemma 2.1(2), we have

$$\begin{aligned} R_1&=\sum _{d\in D_z\left( \frac{x}{M}\right) } \frac{\pi (x)}{\phi ([u,d])}+\sum _{r|P_z}\mu (r)\sum _{q\le \frac{x}{rM}} E(x;[u,qr],1)\\&=\sum _{d\in D_z\left( \frac{x}{M}\right) } \frac{\pi (x)}{\phi ([u,d])}+O\left( \frac{\tau (u)}{u} \frac{x}{\log ^2 x}\right) . \end{aligned}$$

The first sum is treated as follows:

$$\begin{aligned} \sum _{d\in D_z\left( \frac{x}{M}\right) } \frac{\pi (x)}{\phi ([u,d])}&=\sum _{d_1\in D_z\left( \frac{x}{uM}\right) } \frac{\pi (x)N_{d_1}}{\phi (ud_1)}+O\left( \pi (x)\sum _{\begin{array}{c} {\frac{x}{uM}<d_1\le \frac{x}{M} }\\ {p(d_1)>z} \end{array}}\frac{\tau (u)}{\phi (ud_1)} \right) \\&=\sum _{d_1\in D_z\left( \frac{x}{uM}\right) } \frac{\pi (x)N_{d_1}}{\phi (ud_1)}+O\left( \pi (x)\frac{\tau (u) \log u }{\phi (u)\log z}\right) \\&= \sum _{d_1\in D_z\left( \frac{x}{uM}\right) } \frac{\pi (x)N_{d_1}}{\phi (ud_1)}+O\left( \frac{\tau (u)}{u} \frac{x}{\log ^2 z}\right) , \end{aligned}$$

where \(N_{d_1}=\left| \{d\in D_z\left( \frac{x}{M}\right) : [u, d]=ud_1\} \right| \). Since \(N_{d_1}\le \tau (u)\) and \(\phi (ud_1)\ge \phi (u)\phi (d_1)\), by  [5, Lemma 4],

$$\begin{aligned} \sum _{d_1\in D_z\left( \frac{x}{uM}\right) } \frac{\pi (x)N_{d_1}}{\phi (ud_1)}\le \frac{\tau (u)}{\phi (u)}\left( c_1\frac{x}{\log z}+O\left( \frac{x}{\log ^2 z}\right) \right) . \end{aligned}$$

Thus, we have the upper bound

$$\begin{aligned} \sum _{d_1\in D_z\left( \frac{x}{uM}\right) } \frac{\pi (x)N_{d_1}}{\phi (ud_1)}\le \frac{\tau (u)}{u}\left( c_1\frac{x}{\log z}+O\left( \frac{x}{\log ^2 z}\right) \right) . \end{aligned}$$

On the other hand, \(N_{d_1}=\tau (u)\) if \((u,d_1)=1\). Then, we may apply  [5, Lemma 4] since \(P(u)\le \log ^{A_1} x\), and we obtain

$$\begin{aligned} \sum _{d_1\in D_z\left( \frac{x}{uM}\right) } \frac{\pi (x)N_{d_1}}{\phi (ud_1)}&\ge \frac{\tau (u)}{u} \left( \sum _{\begin{array}{c} {d_1\in D_z\left( \frac{x}{uM}\right) }\\ {(u,d_1)=1} \end{array}} \frac{\pi (x)}{\phi (d_1)}+O\left( \frac{x}{\log ^2 z}\right) \right) \\ {}&\ge \frac{\tau (u)}{u} \frac{\phi (u)}{u} \left( c_1\frac{x}{\log z}+O\left( \frac{x}{\log ^2 z}\right) \right) . \end{aligned}$$

Thus, we have the lower bound

$$\begin{aligned} \sum _{d_1\in D_z\left( \frac{x}{uM}\right) } \frac{\pi (x)N_{d_1}}{\phi (ud_1)}\ge \frac{\tau (u)}{u}\left( c_1\frac{x}{\log z}+O\left( \frac{x}{\log ^2 z}\right) \right) . \end{aligned}$$

This shows that

$$\begin{aligned} R_1=\frac{\tau (u)}{u}\left( c_1\frac{x}{\log z}+O\left( \frac{x}{\log ^2 z}\right) \right) . \end{aligned}$$

By divisor-switching technique and Brun–Titchmarsh inequality as in  [6], we have

$$\begin{aligned} R_2&\ll \sum _{r|P_z} \sum _{d|u}\sum _{s|\frac{u}{d}}\sum _{\begin{array}{c} {\frac{x}{rM}<q\le \frac{x}{r}}\\ {ds|q} \end{array}}\pi \left( x;\frac{uqr}{d},1\right) \\&\ll \sum _{r|P_z} \sum _{d|u}\sum _{s|\frac{u}{d}}\sum _{\frac{x}{dsrM}<q\le \frac{x}{dsr}}\pi \left( x;rusq,1\right) \\&\ll \sum _{r|P_z}\sum _{d|u}\sum _{s|\frac{u}{d}} \sum _{k\le \frac{dM}{u}} \pi (x;rusk ,1)\\&\ll \sum _{r|P_z}\sum _{d|u}\sum _{s|\frac{u}{d}} \sum _{k\le \frac{dM}{u}}\frac{x}{\phi (rusk) \log x}\ll \tau (u)\frac{x\log z\log u \log M}{\phi (u)\log x}\ll \frac{\tau (u)}{u} \frac{x}{\log ^2 z}. \end{aligned}$$

This completes the proof of (8). \(\square \)

Proof

of (9) We use (7) and (8) and apply partial summation as in (6). \(\square \)

The following is used with inequality in  [5, Lemma 7]. Here, we obtain an equality that will be used frequently in this paper.

Lemma 2.4

Let \(0<\lambda <\frac{1}{10}\). Fix \(a>1\) and an integer \(0\le B<\infty \). We use \(z=\lambda \log x\) for the formula for \(R_B\) and \(z=\frac{\log x}{\log _2^2 x}\) for the formula for \(S_B\). Let \(I_a(x)=[z,z^a]\). Define

$$\begin{aligned} \mathcal {U}_B&= \{ u{:}\, u \text { is a positive square-free integer consisted} \\&\quad {\text {of exactly { B} prime divisors in }}I_a(x)\}. \end{aligned}$$

Then we have

$$\begin{aligned} R_B:=\sum _{u\in \mathcal {U}_B} R_{u,z}(x)=\frac{(2\log a)^B}{B!}R_z(x)\left( 1+O\left( \frac{1}{\log z}\right) \right) \end{aligned}$$

and

$$\begin{aligned} S_B:=\sum _{u\in \mathcal {U}_B} S_{u,z}(x)=\frac{(2\log a)^B}{B!}S_z(x)\left( 1+O\left( \frac{1}{\log z}\right) \right) . \end{aligned}$$

Proof

We apply Lemma 2.3 with \(u\in \mathcal {U}_B\). Note that \(u\in \mathcal {U}_B\) satisfies the conditions for u in Lemma 2.3(8), (9). Then

$$\begin{aligned} \sum _{u\in \mathcal {U}_B} R_{u,z}(x)&=\sum _{u\in \mathcal {U}_B} \frac{\tau (u)}{u} R_z(x)\left( 1+O\left( \frac{1}{\log z}\right) \right) \\&= \left( \frac{1}{B!}\left( \sum _{p\in I_a(x)} \frac{2}{p}\right) ^B\right. \\&\quad \left. +\,O\left( \frac{1}{(B-2)!}\left( \sum _{p\in I_a(x)}\frac{4}{p^2}\right) \left( \sum _{p\in I_a(x)} \frac{2}{p}\right) ^{B-2}\right) \right) R_z(x)\left( 1+O\left( \frac{1}{\log z}\right) \right) \\&=\left( \frac{1}{B!}\left( \sum _{p\in I_a(x)} \frac{2}{p}\right) ^B+O\left( \frac{1}{z}\right) \right) R_z(x)\left( 1+O\left( \frac{1}{\log z}\right) \right) \\&=\frac{2^B}{B!}\left( \log \log z^a - \log \log z + O\left( \frac{1}{\log z}\right) \right) ^BR_z(x)\left( 1+O\left( \frac{1}{\log z}\right) \right) \\&=\frac{(2\log a)^B}{B!}R_z(x)\left( 1+O\left( \frac{1}{\log z}\right) \right) . \end{aligned}$$

The result for \(S_B\) can be obtained similarly. \(\square \)

Although we relaxed \(z\le \frac{\sqrt{\log x}}{\log _2^6 x}\) to \(z\le \frac{\log x}{\log _2^2 x}\), the range is still not enough for further use. We will see how this range can be relaxed to \(\log ^{\frac{1}{A}} x<z\le \log ^A x\) in Lemma 2.5. A probability mass function of a Poisson distribution comes up as certain densities.

Lemma 2.5

Let \(0<\lambda <\frac{1}{10}\). Fix \(a>1\) and an integer \(0\le B<\infty \). We use \(z=\lambda \log x\) for the formula for \(R'_B\) and \(z=\frac{\log x}{\log _2^2 x}\) for the formula for \(S'_B\). Let \(I_a(x)=(z,z^a]\). Define

$$\begin{aligned} \tau _{z,z^a} (n) = \prod _{\begin{array}{c} {p^e||n}\\ {p\in I_a(x)} \end{array}} \tau (p^e), \ \ w_{z,z^a}(n)=|\{p|n \ : \ p\in I_a(x)\}|, \end{aligned}$$

and

$$\begin{aligned} R'_B: =\sum _{\begin{array}{c} {p\le x}\\ {w_{z,z^a}(p-1)=B} \end{array}} \tau _z(p-1), \ \ S'_B: =\sum _{\begin{array}{c} {p\le x}\\ {w_{z,z^a}(p-1)=B} \end{array}} \frac{\tau _z(p-1)}{p}. \end{aligned}$$

Then, as \(x\rightarrow \infty \), we have

$$\begin{aligned} R'_B=\frac{(2\log a)^B}{B!a^2} R_z(x)(1+o(1)), \ \ S'_B=\frac{(2\log a)^B}{B!a^2} S_z(x)(1+o(1)), \end{aligned}$$
(10)

and we have

$$\begin{aligned} R_{z^a}(x)=\frac{1}{a} R_z(x) (1+o(1)), \ \ S_{z^a}(x)=\frac{1}{a} S_z(x) (1+o(1)). \end{aligned}$$
(11)

Proof

of (10) We remark that by (7), (8), and (9), the contribution of primes p such that \(p-1\) is divisible by a square of a prime \(q>z\) is negligible. In fact, those contributions to \(R_z(x)\) and \(S_z(x)\) are \(O(R_z(x)/z)\) and \(O(S_z(x)/z)\), respectively. Thus, we assume that \(p-1\) is not divisible by square of any prime \(q>z\). By Lemma 2.4 and inclusion–exclusion principle,

$$\begin{aligned} R'_B= R_B - \left( {\begin{array}{c}B+1\\ 1\end{array}}\right) R_{B+1} + \left( {\begin{array}{c}B+2\\ 2\end{array}}\right) R_{B+2} - \left( {\begin{array}{c}B+3\\ 3\end{array}}\right) R_{B+3} +-\cdots . \end{aligned}$$

Moreover, for any \(k\ge 1\),

$$\begin{aligned} \sum _{j=0}^{2k-1} (-1)^j \left( {\begin{array}{c}B+j\\ j\end{array}}\right) R_{B+j} \le R'_B \le \sum _{j=0}^{2k} (-1)^j\left( {\begin{array}{c}B+j\\ j\end{array}}\right) R_{B+j}. \end{aligned}$$

Then dividing by \(R_z(x)\) gives

$$\begin{aligned} \sum _{j=0}^{2k-1} (-1)^j \left( {\begin{array}{c}B+j\\ j\end{array}}\right) \frac{R_{B+j}}{R_z(x)} \le \frac{R'_B}{R_z(x)} \le \sum _{j=0}^{2k} (-1)^j\left( {\begin{array}{c}B+j\\ j\end{array}}\right) \frac{R_{B+j}}{R_z(x)}. \end{aligned}$$

By Lemma 2.4, we have

$$\begin{aligned}&\frac{(2\log a)^B}{B!}\sum _{j=0}^{2k-1} (-1)^j \frac{(2\log a)^j}{j!} \left( 1+O\left( \frac{1}{\log z}\right) \right) \\&\quad \le \frac{R'_B}{R_z(x)}\le \frac{(2\log a)^B}{B!}\sum _{j=0}^{2k} (-1)^j \frac{(2\log a)^j}{j!} \left( 1+O\left( \frac{1}{\log z}\right) \right) . \end{aligned}$$

Taking \(x\rightarrow \infty \), we have

$$\begin{aligned}&\frac{(2\log a)^B}{B!}\sum _{j=0}^{2k-1} (-1)^j \frac{(2\log a)^j}{j!} \le \liminf _{x\rightarrow \infty } \frac{R'_B}{R_z(x)}\\&\quad \le \limsup _{x\rightarrow \infty } \frac{R'_B}{R_z(x)}\le \frac{(2\log a)^B}{B!}\sum _{j=0}^{2k } (-1)^j \frac{(2\log a)^j}{j!}. \end{aligned}$$

Letting \(k\rightarrow \infty \), we obtain

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{R'_B}{R_z(x)}=\frac{(2\log a)^B}{B!a^2}. \end{aligned}$$

The result for \(S'_B\) can be obtained similarly. \(\square \)

Proof

of (11) As in the proof of (10), we assume that \(p-1\) is not divisible by square of any prime \(q>z\). Note that \(\tau _z(p-1)=\tau _{z^a}(p-1)\tau _{z,z^a}(p-1)\). Let \(0\le B<\infty \) be a fixed integer. If \(w_{z,z^a}(p-1)=B\) then \(\tau _{z,z^a}(p-1)=2^B\). Then we have, by (10),

$$\begin{aligned} \sum _{\begin{array}{c} {p\le x}\\ {w_{z,z^a}(p-1)=B} \end{array}} \tau _{z^a}(p-1)= & {} \sum _{\begin{array}{c} {p\le x}\\ {w_{z,z^a}(p-1)=B} \end{array}} \frac{\tau _z(p-1)}{2^B}\\= & {} \frac{R'_B}{2^B}=\frac{(\log a)^B}{B!a^2} R_z(x)(1+o(1)) . \end{aligned}$$

Then, by Lemma 2.4,

$$\begin{aligned} \frac{R_{z^a}(x)}{R_z(x)}&= \sum _{j< B} \frac{(\log a)^j}{j!a^2} (1+o(1))+ \frac{1}{R_z(x)}\sum _{ j\ge B} \frac{1}{2^j}\sum _{\begin{array}{c} {p\le x}\\ {w_{z,z^a}(p-1)=j} \end{array}} \tau _{z }(p-1)\\&=\sum _{j<B} \frac{(\log a)^j}{j!a^2} (1+o(1))+O\left( \frac{1}{2^BR_z(x)} \sum _{\begin{array}{c} {p\le x}\\ {w_{z,z^a}(p-1)\ge B} \end{array}} \tau _{z }(p-1) \right) \\&=\sum _{j< B} \frac{(\log a)^j}{j!a^2} (1+o(1))+O\left( \frac{R_B}{2^BR_z(x)} \right) \\&=\sum _{j< B} \frac{(\log a)^j}{j!a^2} (1+o(1))+O\left( \frac{(2\log a)^B}{2^BB!}\left( 1+O\left( \frac{1}{\log z}\right) \right) \right) . \end{aligned}$$

Thus, both \(\liminf \limits _{x\rightarrow \infty } \frac{R_{z^a}(x)}{R_z(x)}\) and \(\limsup \limits _{x\rightarrow \infty } \frac{R_{z^a}(x)}{R_z(x)}\) are

$$\begin{aligned} \sum _{j\le B} \frac{(\log a)^j}{j!a^2} + O\left( \frac{(\log a)^B}{B!}\right) \end{aligned}$$

and the constant implied in O does not depend on B. Therefore, letting \(B\rightarrow \infty \), we obtain

$$\begin{aligned} \lim _{x\rightarrow \infty } \frac{R_{z^a}(x)}{R_z(x)} = \frac{1}{a}. \end{aligned}$$

The result for \(S_{z^a}(x)\) can be obtained similarly. \(\square \)

Lemma 2.5 allows us to have an extended range of z, and the same method applied to \(R_{u,z}(x)\); we can also extend the range of z for \(R_{u,z}(x)\) and \(S_{u,z}(x)\).

Corollary 2.1

Fix any \(A>1\). Let \(\log ^{\frac{1}{A}} x<z\le \log ^A x\). Then, as \(x\rightarrow \infty \), we have

$$\begin{aligned} R_z(x) = c_1 \frac{x}{\log z} (1+o(1)), \ \ S_z(x) = c_1 \frac{\log x}{\log z} (1+o(1)). \end{aligned}$$
(12)

Assume that u is a positive integer with \(p(u)>z\), \(u< (\log x)^{A_1}\), and \(\tau (u)<A_1\). Then, as \(x\rightarrow \infty \), we have

$$\begin{aligned} R_{u,z}(x) = \frac{\tau (u)}{u} R_z(x)(1+o(1)), \ \ S_{u,z}(x) = \frac{\tau (u)}{u} S_z(x)(1+o(1)). \end{aligned}$$
(13)

We apply Corollary 2.1 to obtain the following uniform distribution result:

Corollary 2.2

Let \(2\le v\le x\) and \(r:={(v^{\frac{3}{2}}\log v)}^{-1}\). Suppose also that \(r\ge \log ^{-\frac{4}{5}} x\), \(0\le \alpha \le \beta \le 1\), and \(\beta -\alpha \ge r\). Then, for \(z\le \frac{\log x^r}{\log _2^2 x^r}\),

$$\begin{aligned} \sum _{\alpha \le \frac{\log p}{\log x} < \beta } \frac{\tau _z(p-1)}{p}=(\beta -\alpha )S_z(x)\left( 1+O\left( \frac{1}{\log z}\right) \right) . \end{aligned}$$
(14)

For \(\log ^{\frac{1}{A}} x<z\le \log ^A x\), we have, as \(x\rightarrow \infty \),

$$\begin{aligned} \sum _{\alpha \le \frac{\log p}{\log x} < \beta } \frac{\tau _z(p-1)}{p}=(\beta -\alpha )S_z(x)\left( 1+o(1)\right) . \end{aligned}$$
(15)

Assume that u is a positive integer with \(p(u)>z\), \(u< (\log x)^{A_1}\), and \(\tau (u)<A_1\). Then we have, for \(z\le \frac{\log x^r}{\log _2^2 x^r}\),

$$\begin{aligned} \sum _{\begin{array}{c} {\alpha \le \frac{\log p}{\log x} < \beta } \\ {p\equiv 1\ \mathrm { mod } \ u} \end{array}} \frac{\tau _z(p-1)}{p}=(\beta -\alpha )\frac{\tau (u)}{u}S_z(x)\left( 1+O\left( \frac{1}{\log z}\right) \right) , \end{aligned}$$
(16)

and, for \(\log ^{\frac{1}{A}} x<z\le \log ^A x\), we have, as \(x\rightarrow \infty \),

$$\begin{aligned} \sum _{\begin{array}{c} {\alpha \le \frac{\log p}{\log x} < \beta } \\ {p\equiv 1\ \mathrm { mod } \ u} \end{array}} \frac{\tau _z(p-1)}{p}=(\beta -\alpha )\frac{\tau (u)}{u}S_z(x)\left( 1+o(1)\right) . \end{aligned}$$
(17)

Proof

By Lemma 2.2(5) and partial summation, we have, for \(\beta -\alpha \ge r\),

$$\begin{aligned} \sum _{\alpha \le \frac{\log p}{\log x} < \beta } \frac{\tau _z(p-1)}{p}&=\frac{R_z(t)}{t}|_{x^{\alpha }}^{x^{\beta }}+\int _{x^{\alpha }}^{x^{\beta }} \frac{R_z(t)}{t^2}dt\\&=c_1(\beta -\alpha )\frac{\log x}{\log z}\left( 1+O\left( \frac{1}{\log z}\right) \right) +O\left( \frac{1}{\log ^2 z}\right) . \end{aligned}$$

Clearly, \(r\log x \gg 1\). Thus, the second O-term can be included in the first O-term. Then (14) follows.

Since \(r\log x \ge \log ^{\frac{1}{5}} x\), the range \(\log ^{\frac{1}{A}} x<z\le \log ^A x\) can be obtained from taking powers of \(\frac{\log x^r}{\log _2^2 x^r}\). We have by (12), as \(x\rightarrow \infty \),

$$\begin{aligned} \sum _{\alpha \le \frac{\log p}{\log x} < \beta } \frac{\tau _z(p-1)}{p}&=\frac{R_z(t)}{t}|_{x^{\alpha }}^{x^{\beta }}+\int _{x^{\alpha }}^{x^{\beta }} \frac{R_z(t)}{t^2}dt\\&=c_1(\beta -\alpha )\frac{\log x}{\log z}\left( 1+o(1)\right) +o\left( \frac{1}{\log z}\right) . \end{aligned}$$

Also, by \(r\log x \gg 1\), the second o-term can be included in the first o-term. Therefore, (15) follows. Similarly, (16) follows from Lemma 2.3 (8) and (17) follows from (13). \(\square \)

We use \(p_1\), \(p_2\), \(\ldots \) , \(p_v\) to denote prime numbers. We define the following multiple sums for \(2\le v\le x\):

$$\begin{aligned} \mathfrak {T}_{v,z}(x):=\sum _{p_1p_2\cdots p_v\le x} \frac{\tau _z(p_1-1)\tau _z(p_2-1)\cdots \tau _z(p_v-1)}{p_1p_2\cdots p_v}, \end{aligned}$$

and, for \(\mathbf {u}=(u_1,\ldots , u_v)\) with \(1\le u_i\le x\),

$$\begin{aligned} \mathfrak {T}_{\mathbf {u},v,z}(x):=\sum _{\begin{array}{c} {p_1p_2\cdots p_v\le x}\\ {\forall _i, \ p_i \equiv 1 \ \mathrm { mod } \ u_i} \end{array}} \frac{\tau _z(p_1-1)\tau _z(p_2-1)\cdots \tau _z(p_v-1)}{p_1p_2\cdots p_v}. \end{aligned}$$

Define \(\mathbb {T}_v:=\{(t_1,\ldots , t_v) : \forall _i, \ t_i\in [0,1], \ t_1+\cdots + t_v \le 1 \}\). We adopt the idea from Gauss’ Circle Problem. Recall that \(r=(v^{\frac{3}{2}}\log v)^{-1}\). Consider a covering of \(\mathbb {T}_v\) by v-cubes of side length r of the form:

Let \(s_1, \ldots , s_v\) be nonnegative integers and let

$$\begin{aligned} B_{s_1, \ldots , s_v}:=\{(t_1,\ldots , t_v) : \forall _i, \ r s_i\le t_i< r (s_i+1) \}. \end{aligned}$$

Let \(M_v\) be the set of those v-cubes lying completely inside \(\mathbb {T}_v\). Then the sum \(\mathfrak {T}_{v,z}(x)\) is over the primes satisfying

$$\begin{aligned} \left( \frac{\log p_1}{\log x}, \ldots , \frac{\log p_v}{\log x}\right) \in \mathbb {T}_v. \end{aligned}$$

Instead of the whole \(\mathbb {T}_v\), we consider the contribution of the sum over primes satisfying

$$\begin{aligned} \left( \frac{\log p_1}{\log x}, \ldots , \frac{\log p_v}{\log x}\right) \in \cup M_v, \end{aligned}$$

which come from the v-cubes lying completely inside \(\mathbb {T}_v\). We define

$$\begin{aligned} {\mathfrak {S}_{v,z}}(x):=\sum _{\left( \frac{\log p_1}{\log x}, \ldots , \frac{\log p_v}{\log x}\right) \in \cup M_v} \frac{\tau _z(p_1-1)\tau _z(p_2-1)\cdots \tau _z(p_v-1)}{p_1p_2\cdots p_v}, \end{aligned}$$

and similarly, for \(\mathbf {u}=(u_1,\cdots , u_v)\) with \(1\le u_i\le x\),

$$\begin{aligned} {\mathfrak {S}_{\mathbf {u},v,z}}(x):=\sum _{\begin{array}{c} {\left( \frac{\log p_1}{\log x}, \ldots , \frac{\log p_v}{\log x}\right) \in \cup M_v}\\ {\forall _i, \ p_i \equiv 1 \ \mathrm { mod } \ u_i} \end{array}} \frac{\tau _z(p_1-1)\tau _z(p_2-1)\cdots \tau _z(p_v-1)}{p_1p_2\cdots p_v}, \end{aligned}$$

Let \(v=\left\lfloor c \sqrt{\frac{\log x}{\log _2 x}}\right\rfloor \) for some positive constant c to be determined. Then v satisfies the conditions in Corollary 2.2. Then we have:

Lemma 2.6

Let \(\log ^{\frac{1}{A}} x< z \le \log ^A x\). Then, as \(x\rightarrow \infty \),

$$\begin{aligned} {\mathfrak {S}_{v,z}}(x)=\frac{1}{v!}S_z(x)^v(1+o(1))^v. \end{aligned}$$
(18)

For \(\mathbf {u}=(u_1, u_2, 1, \ldots , 1)\) with \(1\le u_i\le x\),

$$\begin{aligned} {\mathfrak {S}_{\mathbf {u},v,z}}(x) \ll \frac{\tau (u_1)\tau (u_2)}{\phi (u_1)\phi (u_2)} {\mathfrak {S}_{v,z}}(x)\log ^k z, \end{aligned}$$
(19)

where \(0\le k\le 2\) is the number of \(u_i\)’s that are not 1.

Assume that each \(u_i\), \(i=1, 2\), is a positive integer with \(p(u_i)>z\), \(u_i< (\log x)^{A_1}\) and \(\tau (u_i)<A_1\). Then, as \(x\rightarrow \infty \), we have

$$\begin{aligned} {\mathfrak {S}_{\mathbf {u},v,z}}(x)=\frac{\tau (u_1)\tau (u_2)}{u_1u_2} {\mathfrak {S}_{v,z}}(x)\left( 1+o(1)\right) .\end{aligned}$$
(20)

Proof

of (18) It is clear that

$$\begin{aligned} \mathrm {vol}\left( (1-r\sqrt{v})\mathbb {T}_v\right) \le |M_v| \mathrm {vol} (B_{0, \ldots , 0}) \le \mathrm {vol} (\mathbb {T}_v). \end{aligned}$$

We have \(\mathrm {vol} (\mathbb {T}_v)=\frac{1}{v!}\), \(\mathrm {vol} (B_{0, \ldots , 0})=r^v\), and \(\mathrm {vol} \left( (1- r\sqrt{v})\mathbb {T}_v\right) =\frac{1}{v!}\left( 1- r\sqrt{v}\right) ^v\). Also, recall that \(r:={(v^{\frac{3}{2}}\log v)}^{-1}\). Then

$$\begin{aligned} \frac{\frac{1}{v!}\left( 1-\frac{1}{v\log v}\right) ^v}{(v^{\frac{3}{2}}\log v)^{-v}}\le |M_v|\le \frac{\frac{1}{v!}}{(v^{\frac{3}{2}}\log v)^{-v}}. \end{aligned}$$

On the other hand, by Corollary 2.2 (15), the contribution of each v-cube \([\alpha _1,\beta _1]\times \cdots \times [\alpha _v, \beta _v]\subseteq [0,1]^v\) of side length r to the sum is

$$\begin{aligned}&\sum _{\forall _i, \ \alpha _i\le \frac{\log p_i}{\log x}<\beta _i} \frac{\tau _z(p_1-1)\tau _z(p_2-1)\cdots \tau _z(p_v-1)}{p_1p_2\cdots p_v}\\&\qquad \qquad \qquad =\,\left( \prod _{i=1}^v (\beta _i-\alpha _i) \right) S_z(x)^v ( 1+o(1))^v\\&\qquad \qquad \qquad =\,r^v S_z(x)^v(1+o(1))^v. \end{aligned}$$

Combining this with the bounds for \(|M_v|\), we obtain the result. \(\square \)

Proof

of (19), (20) Let v and r be as defined in Corollary 2.2. We write (15) and (17) in the form of

$$\begin{aligned} \sum _{\alpha \le \frac{\log p}{\log x} < \beta } \frac{\tau _z(p-1)}{p}=(\beta -\alpha )S_z(x)\left( 1+f_{\alpha ,\beta }(x)\right) \end{aligned}$$
(21)

and

$$\begin{aligned} \sum _{\begin{array}{c} {\alpha \le \frac{\log p}{\log x} < \beta } \\ {p\equiv 1\ \mathrm { mod } \ u} \end{array}} \frac{\tau _z(p-1)}{p}=(\beta -\alpha )\frac{\tau (u)}{u}S_z(x)\left( 1+g_{\alpha , \beta }(x)\right) . \end{aligned}$$
(22)

We note that there is a function \(f(x)=o(1)\) such that uniformly, for \(0\le \alpha \le \beta \le 1\) and \(\beta -\alpha \ge r\),

$$\begin{aligned} \text {max}(|f_{\alpha ,\beta }(x)|, |g_{\alpha ,\beta }(x)|)\le f(x). \end{aligned}$$

Then we can write

$$\begin{aligned} \sum _{\begin{array}{c} {\alpha \le \frac{\log p}{\log x}< \beta } \\ {p\equiv 1\ \mathrm { mod } \ u} \end{array}} \frac{\tau _z(p-1)}{p}&=(\beta -\alpha )\frac{\tau (u)}{u}S_z(x)\left( 1+g_{\alpha , \beta }(x)\right) \\&=\frac{\tau (u)}{u}\sum _{\alpha \le \frac{\log p}{\log x}< \beta } \frac{\tau _z(p-1)}{p} \left( \frac{1+g_{\alpha ,\beta }(x)}{1+f_{\alpha ,\beta }(x)}\right) \\&=\frac{\tau (u)}{u}\sum _{\alpha \le \frac{\log p}{\log x} < \beta } \frac{\tau _z(p-1)}{p} \left( 1+O(f(x))\right) . \end{aligned}$$

Consider any v-cube \([\alpha _1,\beta _1]\times \cdots \times [\alpha _v, \beta _v]\subseteq [0,1]^v\) of side length r. Then, by the above observation,

$$\begin{aligned}&\sum _{\begin{array}{c} {\forall _i, \ \alpha _i\le \frac{\log p_i}{\log x}<\beta _i}\\ {p_i \equiv 1 \ \mathrm { mod } \ u_i \ \mathrm { for }\ i=1, \ 2} \end{array}}\frac{\tau _z(p_1-1)\tau _z(p_2-1)\cdots \tau _z(p_v-1)}{p_1p_2\cdots p_v}\\&\quad =\frac{\tau (u_1)\tau (u_2)}{u_1u_2}\sum _{\forall _i, \ \alpha _i\le \frac{\log p_i}{\log x}<\beta _i} \frac{\tau _z(p_1-1)\tau _z(p_2-1)\cdots \tau _z(p_v-1)}{p_1p_2\cdots p_v}(1+O(f(x)))^2. \end{aligned}$$

This proves (20). For the proof of (19), we use instead

$$\begin{aligned} \sum _{\begin{array}{c} {\alpha \le \frac{\log p}{\log x}< \beta } \\ {p\equiv 1\ \mathrm { mod } \ u} \end{array}} \frac{\tau _z(p-1)}{p}&=\frac{R_{u,z}(t)}{t}|_{x^{\alpha }}^{x^{\beta }}+\int _{x^{\alpha }}^{x^{\beta }} \frac{R_{u,z}(t)}{t^2}dt\\&\ll \frac{\tau (u)}{\phi (u)}\left( (\beta -\alpha )\log x + O(1)\right) \ll \frac{\tau (u)}{\phi (u)}(\beta -\alpha )\log x \\&\ll \frac{\tau (u)}{\phi (u)}(\beta -\alpha ) S_z(x) \log z\ll \frac{\tau (u)}{\phi (u)} \sum _{\alpha \le \frac{\log p}{\log x}<\beta } \frac{\tau _z(p-1)}{p} \log z, \end{aligned}$$

which follows from Lemma 2.3(7). \(\square \)

We impose some restrictions on the primes \(p_1\), \(\ldots \) , \(p_v\):

R1. \(p_1, \ldots , p_v\) are distinct.

R2. For each i, \(q^2\not \mid p_i-1\) for any prime \(q>z\).

R3. \(q^2\not \mid \phi (p_1\cdots p_v)\) for any prime \(q>z^2\).

Recall that we chose

$$\begin{aligned} v=\left\lfloor c \sqrt{\frac{\log x}{\log _2 x}} \right\rfloor \end{aligned}$$

for some positive constant c to be determined. Let \( {\mathfrak {S}_{v,z}}^{(1)}(x)\) be the contribution of primes to \( {\mathfrak {S}_{v,z}}(x)\) not satisfying R1. Note that if R1 is not satisfied, then some primes among \(p_1\), \(\ldots \) , \(p_v\) are repeated. Then, by Lemma 2.6(18),

$$\begin{aligned} {\mathfrak {S}_{v,z}}^{(1)}(x)&\ll \left( {\begin{array}{c}v\\ 2\end{array}}\right) \left( \sum _{z<p\le x} \frac{\tau _z(p-1)^2}{p^2} \right) {\mathfrak {S}_{v-2, z}}(x)\\&\ll v^2 \frac{\log ^3 z}{z} \frac{v(v-1)}{S_z(x)^2} {\mathfrak {S}_{v,z}}(x)\\&\ll \frac{v^4\log ^5 z}{z\log ^2 x} {\mathfrak {S}_{v,z}}(x)\ll \frac{\log ^3 z}{z} {\mathfrak {S}_{v,z}}(x). \end{aligned}$$

Let \( {\mathfrak {S}_{v,z}}^{(2)}(x)\) be the contribution of primes to \( {\mathfrak {S}_{v,z}}(x)\) not satisfying R2. Note that if R2 is not satisfied, then \(q^2|p_i-1\) for some primes \(p_i\) and \(q>z\). Let \(\mathbf {u}_{q^2}:=(q^2, 1, \ldots , 1)\). Suppose that \(q^2|p_i-1\) for some \(p_i\) and \(q>z^2\). Then the contribution of those primes to \( {\mathfrak {S}_{v,z}}^{(2)}(x)\) is, by (19),

$$\begin{aligned}&\ll \sum _{q>z^2} \left( {\begin{array}{c}v\\ 1\end{array}}\right) {\mathfrak {S}_{\mathbf {u}_{q^2}, v,z}}(x)\\&\ll \sum _{q>z^2}\frac{v}{\phi (q^2)} {\mathfrak {S}_{v,z}}(x) \log z\\&\ll \sum _{q>z^2} \frac{v}{q^2} {\mathfrak {S}_{v,z}}(x)\log z \ll \frac{v}{z^2} {\mathfrak {S}_{v,z}}(x). \end{aligned}$$

Suppose that \(q^2|p_i-1\) for some \(p_i\) and \(z<q\le z^2\). Then we have, by (20),

$$\begin{aligned} \ll \sum _{z<q\le z^2} \left( {\begin{array}{c}v\\ 1\end{array}}\right) {\mathfrak {S}_{\mathbf {u}_{q^2}, v,z}}(x)&\ll \sum _{z<q\le z^2}\frac{v}{q^2} {\mathfrak {S}_{v,z}}(x)\ll \frac{v}{z\log z} {\mathfrak {S}_{v,z}}(x). \end{aligned}$$

Thus, we have

$$\begin{aligned} {\mathfrak {S}_{v,z}}^{(2)}(x)\ll \frac{v}{z\log z} {\mathfrak {S}_{v,z}}(x). \end{aligned}$$

Let \( {\mathfrak {S}_{v,z}}^{(3)}(x)\) be the contribution of primes to \( {\mathfrak {S}_{v,z}}(x)\) satisfying R1 and R2, but not satisfying R3. Note that if R1, R2 are satisfied and R3 is not satisfied, then there are at least two distinct primes \(p_i\), \(p_j\) such that \(q|p_i-1\) and \(q|p_j-1\). Let \(\mathbf {u}_{q,q}:=(q,q,1, \ldots , 1)\). Suppose first that this happens with \(q>z^4\). Then, by (19), the contribution is

$$\begin{aligned}&\ll \sum _{q>z^4} \left( {\begin{array}{c}v\\ 2\end{array}}\right) {\mathfrak {S}_{\mathbf {u}_{q,q}, v,z}}(x)\ll \sum _{q>z^4} \frac{v^2}{\phi (q)^2} {\mathfrak {S}_{v,z}}(x) \log ^2 z \ll \frac{v^2 \log z}{z^4} {\mathfrak {S}_{v,z}}(x). \end{aligned}$$

Suppose that this happens with \(z^2<q\le z^4\). Then, by (20), the contribution is

$$\begin{aligned}&\ll \sum _{z^2<q\le z^4} \left( {\begin{array}{c}v\\ 2\end{array}}\right) {\mathfrak {S}_{\mathbf {u}_{q,q}, v,z}}(x)\ll \sum _{z^2<q\le z^4} \frac{v^2}{q^2} {\mathfrak {S}_{v,z}}(x) \ll \frac{v^2}{z^2\log z} {\mathfrak {S}_{v,z}}(x). \end{aligned}$$

Thus, we have

$$\begin{aligned} {\mathfrak {S}_{v,z}}^{(3)}(x)\ll \frac{v^2}{z^2\log z} {\mathfrak {S}_{v,z}}(x). \end{aligned}$$

We write \( {\mathfrak {S}_{v,z}}^{(0)}(x)\) to denote the contribution of those primes to \( {\mathfrak {S}_{v,z}}(x)\) satisfying all three restrictions R1, R2, and R3. By the above estimates, we have

$$\begin{aligned} {\mathfrak {S}_{v,z}}^{(0)}(x)&\ge {\mathfrak {S}_{v,z}}(x)- {\mathfrak {S}_{v,z}}^{(1)}(x)- {\mathfrak {S}_{v,z}}^{(2)}(x)- {\mathfrak {S}_{v,z}}^{(3)}(x) \\&= {\mathfrak {S}_{v,z}}(x)\left( 1+O\left( \frac{\log ^3 z}{z}\right) +O\left( \frac{v}{z\log z}\right) +O\left( \frac{v^2}{z^2\log z}\right) \right) . \end{aligned}$$

Therefore,

$$\begin{aligned} {\mathfrak {S}_{v,z}}^{(0)}(x)= {\mathfrak {S}_{v,z}}(x)\left( 1+O\left( \frac{\log ^3 z}{z}\right) +O\left( \frac{v}{z\log z}\right) +O\left( \frac{v^2}{z^2\log z}\right) \right) . \end{aligned}$$
(23)

3 Proof of Theorem 1.1

We set

$$\begin{aligned} v=v(x):=\left\lfloor c \sqrt{\frac{\log x}{\log _2 x}} \right\rfloor , \ \ z=z(x):=\sqrt{\log x}, \end{aligned}$$
$$\begin{aligned} y:=\exp \left( \sqrt{\log x}\right) \end{aligned}$$

with a positive constant c to be determined.

Consider a subset \(Q_z(x)\) of primes defined by

$$\begin{aligned} Q=Q_z(x):=\{p : p \le x, \ q^2 \not \mid p-1 \ \text { for any prime } q >z \}. \end{aligned}$$

We define \(\mathcal {N}\), \(\mathcal {M}\) by

$$\begin{aligned} \mathcal {N}=\mathcal {N}_v(x):=\{n\le x : n \ \text { is square-free, } p|n \ \Rightarrow p\in Q, \ w(n)=v \}, \end{aligned}$$
$$\begin{aligned} \mathcal {M}=\mathcal {M}_v(x):=\{n\le x : n\in \mathcal {N}, \ q^2 \not \mid \phi (n) \ \text { for any prime } q>z^2 \}. \end{aligned}$$

We write

$$\begin{aligned} V_{\mathcal {M}}(x):=\sum _{n\in \mathcal {M}} \frac{\tau _z(\lambda (n))}{n}, \ \ \tau ''_z(n):=\prod _{p|n} \tau _z(p-1). \end{aligned}$$

We also write

$$\begin{aligned} W_{\mathcal {M}} :=\sum _{n\in \mathcal {M}} \frac{\tau ''_z(n)}{n}, \ \ W_{\mathcal {M}}':=\sum _{n\in \mathcal {M}} \frac{\tau ''_{z^2}(n)}{n}. \end{aligned}$$

By (23), the contribution of those primes satisfying R1, R2, and R3 to \( {\mathfrak {S}_{v,z}}(x)\), which we wrote as \( {\mathfrak {S}_{v,z}}^{(0)}(x)\) satisfies

$$\begin{aligned} {\mathfrak {S}_{v,z}}^{(0)}(x)&= {\mathfrak {S}_{v,z}}(x)\left( 1+O\left( \frac{\log ^3 z}{z}\right) +O\left( \frac{v}{z\log z}\right) +O\left( \frac{v^2}{z^2\log z}\right) \right) .\\&= {\mathfrak {S}_{v,z}}(x)\left( 1+O\left( \frac{1}{\log _2 x}\right) \right) . \end{aligned}$$

Then, by Lemma 2.6(18) and Stirling’s formula,

$$\begin{aligned} W_{\mathcal {M}}\ge \frac{1}{v!} {\mathfrak {S}_{v,z}}^{(0)}(x)\asymp \frac{1}{v} \left( \frac{e}{v}\right) ^{2v} \left( c_1\frac{\log x}{\log z}\right) ^v(1+o(1))^{v} \end{aligned}$$

Thus,

$$\begin{aligned} W_{\mathcal {M}}\gg \exp \left( \sqrt{\frac{\log x}{\log _2 x}} \left( 2c + c \log c_1 - 2 c\log c + c \log 2+o(1)\right) \right) . \end{aligned}$$

Maximizing \(2c + c \log c_1 - 2 c\log c + c \log 2\) by the first derivative, we have \(c=\sqrt{2} e^{-\gamma /2}\), and hence

$$\begin{aligned} W_{\mathcal {M}}\gg \exp \left( 2\sqrt{2} e^{-\frac{\gamma }{2}} \sqrt{\frac{\log x}{\log _2 x}} (1+o(1)) \right) . \end{aligned}$$

For \(W_{\mathcal {M}}'\), we have, by (23), the contribution of those primes satisfying R1, R2, and R3 to \( {\mathfrak {S}_{v,z^2}}(x)\), say \( {\mathfrak {S}_{v,z^2}}^{(0')}(x)\) satisfies

$$\begin{aligned} {\mathfrak {S}_{v,z^2}}^{(0')}(x)&= {\mathfrak {S}_{v,z^2}}(x)\left( 1+O\left( \frac{\log ^3 z}{z^2}\right) +O\left( \frac{v}{z\log z}\right) +O\left( \frac{v^2}{z^2\log z}\right) \right) \\&= {\mathfrak {S}_{v,z^2}}(x)\left( 1+O\left( \frac{1}{\log _2 x}\right) \right) . \end{aligned}$$

Then by Lemma 2.6(18) and Stirling’s formula, as \(x\rightarrow \infty \),

$$\begin{aligned} W_{\mathcal {M}}'\ge \frac{1}{v!} {\mathfrak {S}_{v,z^2}}^{(0')}(x)\asymp \frac{1}{v} \left( \frac{e}{v}\right) ^{2v} \left( c_1\frac{\log x}{\log z^2}\right) ^v(1+o(1))^{v} \end{aligned}$$

Thus,

$$\begin{aligned} W_{\mathcal {M}}'\gg \exp \left( \sqrt{\frac{\log x}{\log _2 x}} \left( 2c + c \log c_1 - 2 c\log c +o(1)\right) \right) . \end{aligned}$$

Maximizing \(2c + c \log c_1 - 2 c\log c\) by the first derivative, we have \(c= e^{-\gamma /2}\), and hence, as \(x\rightarrow \infty \),

$$\begin{aligned} W_{\mathcal {M}}'\gg \exp \left( 2 e^{-\frac{\gamma }{2}} \sqrt{\frac{\log x}{\log _2 x}} (1+o(1)) \right) . \end{aligned}$$

Therefore, we have just proved the lower bounds of the following:

Theorem 3.1

For \(z=\sqrt{\log x}\), as \(x\rightarrow \infty \),

$$\begin{aligned} \sum _{n\le x}\mu ^2 (n) \frac{\tau ''_z(n)}{n} =\exp \left( 2\sqrt{2} e^{-\frac{\gamma }{2}}\sqrt{\frac{\log x}{\log _2 x}}(1+o(1)) \right) \end{aligned}$$
(24)

and

$$\begin{aligned} \sum _{n\le x} \mu ^2(n)\frac{\tau ''_{z^2}(n)}{n} =\exp \left( 2e^{-\frac{\gamma }{2}}\sqrt{\frac{\log x}{\log _2 x}}(1+o(1)) \right) . \end{aligned}$$
(25)

Note that the upper bounds follow from Rankin’s method as in  [5, Theorem 1].

We proceed the similar argument as in  [5]. Let \(\mathcal {M}=\mathcal {M}_v(x)\) be as above with the choice \(c=e^{-\gamma /2}\). Now, for \(n\in \mathcal {M}\), we have

$$\begin{aligned} \tau _z(\phi (n))=\tau _{z,z^2}(\phi (n)) \tau _{z^2} (\phi (n)) \ge \tau _{z^2}(\phi (n))= \tau ''_{z^2}(n), \end{aligned}$$
$$\begin{aligned} \tau _z(\lambda (n))=\tau _{z,z^2}(\lambda (n)) \tau _{z^2} (\lambda (n)) \ge \tau _{z^2}(\lambda (n)) =\tau ''_{z^2}(n). \end{aligned}$$

Then, as \(x\rightarrow \infty \),

$$\begin{aligned} V_{\mathcal {M}}(x)\ge W_{\mathcal {M}}'\gg \exp \left( 2e^{-\frac{\gamma }{2}} \sqrt{\frac{\log x}{\log _2 x}}(1+o(1))\right) . \end{aligned}$$

The argument proceeds as in  [5]. Let \(\mathcal {M'}\) be defined by

$$\begin{aligned} \mathcal {M'}:=\left\{ np : n\in \mathcal {M}_v(xy^{-1}), \ \ p \ \text {is a prime}, \ \ p \le \frac{x}{n} \right\} . \end{aligned}$$

For those \(n'=np\in \mathcal {M'}\), we have

$$\begin{aligned} \tau (\lambda (np))\ge \tau (\lambda (n))\ge \tau _z(\lambda (n)), \end{aligned}$$

and a given \(n'\in \mathcal {M'}\) has at most \(v+1\) decompositions of the form \(n'=np\) with \(n\in \mathcal {M}_v(xy^{-1})\), \( p\le \frac{x}{n}\).

Since \(n\le x y^{-1}\) for \(n\in \mathcal {M}_v(xy^{-1})\), the number of p in \( p\le \frac{x}{n}\) is

$$\begin{aligned} \pi \left( \frac{x}{n}\right) \gg \frac{x}{n\log x}. \end{aligned}$$

Note that \(\log y = \sqrt{\log x}=o(\log x)\). This gives

$$\begin{aligned} V_{\mathcal {M}}(xy^{-1})\gg \exp \left( 2e^{-\frac{\gamma }{2}} \sqrt{\frac{\log x}{\log _2 x}}(1+o(1))\right) . \end{aligned}$$

Then

$$\begin{aligned} \sum _{n\le x}\tau (\lambda (n))\ge & {} \sum _{n\in \mathcal {M'}}\tau (\lambda (n)) \gg V_{\mathcal {M}}(xy^{-1}) \frac{x}{v\log x}\\\gg & {} x\exp \left( 2e^{-\frac{\gamma }{2}} \sqrt{\frac{\log x}{\log _2 x}}(1+o(1))\right) . \end{aligned}$$

This completes the proof of Theorem 1.1.

Remarks

  1. 1.

    In the proof of Theorem 1.1, we dropped \(\tau _{z,z^2}(\phi (n))\). This is where a prime \(z<q\le z^2\) can divide multiple \(p_i-1\) for \(i=1, 2, \cdots , v\), and that is the main difficulty in obtaining more precise formulas for \(\sum _{n\le x}\tau (\phi (n))\) and \(\sum _{n\le x}\tau (\lambda (n))\).

  2. 2.

    We will see a heuristic argument suggesting that, as \(x\rightarrow \infty \),

    $$\begin{aligned} \sum _{n\le x} \tau (\lambda (n))=x\exp \left( 2\sqrt{2} e^{-\frac{\gamma }{2}}\sqrt{\frac{\log x}{\log _2 x}}(1+o(1)) \right) , \end{aligned}$$

    and hence

    $$\begin{aligned} \sum _{n\le x} \tau (\phi (n))=x\exp \left( 2\sqrt{2} e^{-\frac{\gamma }{2}}\sqrt{\frac{\log x}{\log _2 x}}(1+o(1)) \right) . \end{aligned}$$

    However, we have

    $$\begin{aligned} \sum _{n\le x}\tau (\lambda (n)) = o\left( \sum _{n\le x}\tau (\phi (n))\right) . \end{aligned}$$

    We will prove this in the following section. The prime 2 plays a crucial role in the proof of Theorem 1.2.

4 Proof of Theorem 1.2

We put k and w as in  [5]:

$$\begin{aligned} k=\lfloor A\log _2 x \rfloor , \ \ \omega =\left\lfloor \frac{\sqrt{\log x}}{\log _2^2 x}\right\rfloor . \end{aligned}$$

Here, A is a positive constant to be determined. Also, define \(\mathcal {E}_1(x)\), \(\mathcal {E}_2(x)\), and \(\mathcal {E}_3(x)\) in the same way:

$$\begin{aligned} \mathcal {E}_1(x):= \{n\le x : 2^k | n \ \text {or} \ \text {there is a prime { p}|{ n} with} \ p\equiv 1 \ \mathrm {mod} \ 2^k \}, \end{aligned}$$
$$\begin{aligned} \mathcal {E}_2(x):=\{ n\le x : \omega (n)\le \omega \}, \end{aligned}$$

and

$$\begin{aligned} \mathcal {E}_3(x):=\{ n\le x \} - \left( \mathcal {E}_1(x)\cup \mathcal {E}_2(x)\right) . \end{aligned}$$

We need the following lemma.

Lemma 4.1

For any \(2\le y\le x\), we have

$$\begin{aligned} \sum _{n\le \frac{x}{y}} \frac{\tau (\phi (n))}{n} \ll \frac{\log ^5 x}{x} \sum _{n\le x} \tau (\phi (n)). \end{aligned}$$

Proof

As in the proof of  [5, Theorem 1], we use the square-free kernel \(k=k(n)\) (if a prime p divides n, then p|k, and k is a square-free positive integer which divides n) and the factorization \(n=mk\) to rewrite the sum as

$$\begin{aligned} \sum _{n\le \frac{x}{y}} \frac{\tau (\phi (n))}{n}&\le \sum _{k\le \frac{x}{y}}\mu ^2(k) \sum _{m\le \frac{x}{ky}} \frac{\tau (m)\tau (\phi (k))}{mk}\\&\ll \sum _{k\le \frac{x}{y}}\mu ^2(k) \frac{\tau (\phi (k))}{k} \log ^2 x. \end{aligned}$$

Note that we have uniformly \(w(n)\ll \log x\). Find v such that

$$\begin{aligned} \sum _{\begin{array}{c} {k\le \frac{x}{y}}\\ {\omega (k)=v} \end{array}}\mu ^2(k) \frac{\tau (\phi (k))}{k} \end{aligned}$$

is maximal. Then we have

$$\begin{aligned} \sum _{k\le \frac{x}{y}}\mu ^2(k) \frac{\tau (\phi (k))}{k} \ll \log x \sum _{\begin{array}{c} {k\le \frac{x}{y}}\\ {\omega (k)=v} \end{array}}\mu ^2(k) \frac{\tau (\phi (k))}{k}. \end{aligned}$$

We adopt an idea from the proof of Theorem 1.1. Let \(\mathcal {M}=\mathcal {M}_v(xy^{-1})\) be the set of square-free numbers \(k\le xy^{-1}\) with \(\omega (k)=v\). Define

$$\begin{aligned} \mathcal {M}':= \left\{ kp : k\in \mathcal {M}_v(xy^{-1}), \ \ p \ \text {is a prime}, \ \ p \le \frac{x}{k} \right\} . \end{aligned}$$

For those \(n'=kp\in \mathcal {M'}\) with \(k\in \mathcal {M}\), we have

$$\begin{aligned} \tau (\phi (kp))\ge \tau (\phi (k)), \end{aligned}$$

and any given \(n'\in \mathcal {M'}\) has at most \(v+1\) decompositions of the form \(n'=kp\) with \(k\in \mathcal {M}\), \(p\le \frac{x}{k}\).

Since the number of p satisfying \(p\le \frac{x}{k}\) is

$$\begin{aligned} \pi \left( \frac{x}{k}\right) \gg \frac{x}{k\log x}, \end{aligned}$$

it follows that

$$\begin{aligned} \sum _{n\le x}\tau (\phi (n))\ge \sum _{n\in \mathcal {M'}}\tau (\phi (n)) \gg \sum _{\begin{array}{c} {k\le \frac{x}{y}}\\ {\omega (k)=v} \end{array}}\mu ^2(k) \frac{\tau (\phi (k))}{k}\frac{x}{v\log x}. \end{aligned}$$

Since \(v\ll \log x\), we have

$$\begin{aligned} \sum _{\begin{array}{c} {k\le \frac{x}{y}}\\ {w(k)=v} \end{array}}\mu ^2(k) \frac{\tau (\phi (k))}{k} \ll \frac{\log ^2 x}{x} \sum _{n\le x}\tau (\phi (n)). \end{aligned}$$

This gives

$$\begin{aligned} \sum _{k\le \frac{x}{y}}\mu ^2(k) \frac{\tau (\phi (k))}{k} \ll \frac{\log ^3 x}{x} \sum _{n\le x}\tau (\phi (n)). \end{aligned}$$

Then the result follows. \(\square \)

For \(n\in \mathcal {E}_1(x)\), we have, by Lemmas 2.3 and 4.1,

$$\begin{aligned} \sum _{n\in \mathcal {E}_1(x)} \tau (\lambda (n))&\le x\sum _{n\in \mathcal {E}_1(x)}\frac{\tau (\phi (n))}{n}\\&\le x \frac{\tau (2^k)}{2^k}\sum _{m\le \frac{x}{2^k}}\frac{\tau (\phi (m))}{m}+ x \sum _{\begin{array}{c} {p\le x}\\ {p \equiv 1 \ \mathrm {mod} \ 2^k} \end{array}}\frac{\tau (p-1)}{p}\sum _{m\le \frac{x}{p}} \frac{\tau (\phi (m))}{m} \\&\ll \log ^5 x \left( \frac{\tau (2^k)}{\phi (2^k)} \log x \sum _{n\le x}\tau (\phi (n))\right) \\&\ll \log ^6 x \frac{A\log _2 x}{\log ^{A\log 2} x} \sum _{n\le x}\tau (\phi (n)). \end{aligned}$$

If we take \(A\log 2 > 7\), then we obtain

$$\begin{aligned} \sum _{n\in \mathcal {E}_1(x)} \tau (\lambda (n)) = o\left( \sum _{n\le x} \tau (\phi (n))\right) . \end{aligned}$$

For \(n\in \mathcal {E}_2(x)\), we use the square-free kernel \(k=k(n)\) and the factorization \(n=mk\) as before,

$$\begin{aligned} \sum _{n\in \mathcal {E}_2(x)}\tau (\lambda (n))&\le \sum _{n\in \mathcal {E}_2(x)}\tau (\phi (n))\\&\ll \sum _{\begin{array}{c} {k\le x}\\ {\omega (k)\le \omega } \end{array}}\mu ^2(k)\sum _{m\le \frac{x}{k}} \tau (m)\tau (\phi (k))\\&\ll \sum _{\begin{array}{c} {k\le x}\\ {\omega (k)\le \omega } \end{array}}\mu ^2(k) \frac{x}{k} (\log x) \tau (\phi (k))\\&\ll x\omega \log x\left( \sum _{p\le x}\frac{\tau (p-1)}{p}\right) ^{\omega }\\&\ll x (\log x)^{\frac{3}{2}} (C\log x)^{\omega } \ll x \exp \left( 2\frac{\sqrt{\log x}}{\log _2 x}\right) . \end{aligned}$$

Thus, by Theorem 1.1,

$$\begin{aligned} \sum _{n\in \mathcal {E}_2(x)}\tau (\lambda (n))=o\left( \sum _{n\le x}\tau (\phi (n))\right) . \end{aligned}$$

For \(n\in \mathcal {E}_3(x)\), we follow the method of  [5]. We have

$$\begin{aligned} \frac{\tau (\phi (n))}{\tau (\lambda (n))}\ge \frac{\omega }{k} \gg \frac{\sqrt{\log x}}{\log _2^3 x}. \end{aligned}$$

Then

$$\begin{aligned} \sum _{n\in \mathcal {E}_3(x)} \tau (\lambda (n))\ll \frac{\log _2^3 x}{\sqrt{\log x}}\sum _{n\in \mathcal {E}_3(x)} \tau (\phi (n))\le \frac{\log _2^3 x}{\sqrt{\log x}} \sum _{n\le x}\tau (\phi (n)). \end{aligned}$$

Therefore, putting these together, we have

$$\begin{aligned} \sum _{n\le x}\tau (\lambda (n))\ll \frac{\log _2^3 x}{\sqrt{\log x}}\sum _{n\le x}\tau (\phi (n)), \end{aligned}$$

and Theorem 1.2 follows.

5 Heuristics

Recall that \(\tau _z(\lambda (n))=\tau _{z,z^2}(\lambda (n)) \tau _{z^2} (\lambda (n))\). Let \(\mathcal {M}\) be the set defined in Sect. 3 with the choice of \(v=\big \lfloor \sqrt{2} e^{-\gamma /2} \sqrt{\frac{\log x}{\log _2 x}}\big \rfloor \). As in Sect. 3, we have \(\tau _{z^2}(\lambda (n))=\tau ''_{z^2}(n)\) for \(n\in \mathcal {M}\). It is important to note that \(q^2\not \mid p_i-1\) for any primes \(p_i|n\) and \(q>z\). Also, we have \(q^2\not \mid \phi (n)\) for \(q>z^2\). Thus, it is enough to focus on the sum \(V_{\mathcal {M}}(x)\). If we could prove that \(V_{\mathcal {M}}(x)=\sum _{n\in \mathcal {M}}\frac{\tau _z(\lambda (n))}{n}\gg \exp \big (2\sqrt{2} e^{-\frac{\gamma }{2}} \sqrt{\frac{\log x}{\log _2 x}}(1+o(1))\big )\), then the same argument as in Theorem 1.1 would allow \(\sum _{n\le x}\tau (\lambda (n))\gg x \exp \big (2\sqrt{2} e^{-\frac{\gamma }{2}} \sqrt{\frac{\log x}{\log _2 x}}(1+o(1))\big )\). We need the contribution of \(\tau _{z,z^2}(\lambda (n))\) over \(n\in \mathcal {M}\). Let \(\mathfrak {S}_{v,z}(x)\) be the sum defined in Sect. 2, and define

$$\begin{aligned} \mathfrak {U}_{v,z}(x):= & {} \sum _{\left( \frac{\log p_1}{\log x}, \ldots , \frac{\log p_v}{\log x}\right) \in \cup M_v} \frac{\tau _{z,z^2}(\mathrm {lcm}(p_1-1,p_2-1, \ldots ,p_v-1))}{\tau _{z,z^2}(p_1-1)\tau _{z,z^2}(p_2-1)\cdots \tau _{z,z^2}(p_v-1)}\\&\times \frac{\tau _z(p_1-1)\tau _z(p_2-1)\cdots \tau _z(p_v-1)}{p_1p_2\cdots p_v}. \end{aligned}$$

We have also defined in Sect. 2 that, for \(\mathbf {u}=(u_1,\ldots , u_v)\) with \(1\le u_i\le x\),

$$\begin{aligned} {\mathfrak {S}_{\mathbf {u},v,z}}(x):=\sum _{\begin{array}{c} {\left( \frac{\log p_1}{\log x}, \ldots , \frac{\log p_v}{\log x}\right) \in \cup M_v}\\ {\forall _i, \ p_i \equiv 1 \ \mathrm { mod } \ u_i} \end{array}} \frac{\tau _z(p_1-1)\tau _z(p_2-1)\cdots \tau _z(p_v-1)}{p_1p_2\cdots p_v}. \end{aligned}$$

We need to extend Lemma 2.6 to cover all components of \(\mathbf {u}\).

Lemma 5.1

Let \(\log ^{\frac{1}{A}} x< z \le \log ^A x\). Then, for \(\mathbf {u}=(u_1, u_2, \ldots , u_v)\) with \(1\le u_i\le x\),

$$\begin{aligned} {\mathfrak {S}_{\mathbf {u},v,z}}(x) \ll \frac{\tau (u_1)\tau (u_2)\cdots \tau (u_v)}{\phi (u_1)\phi (u_2)\cdots \phi (u_v)} {\mathfrak {S}_{v,z}}(x)(1+o(1))^k\log ^k z, \end{aligned}$$
(26)

where \(0\le k\le v\) is the number of \(u_i\)’s that are not 1.

Assume that each \(u_i\), \(1\le i\le v\), is either 1 or a positive integer with \(p(u_i)>z\), \(u_i< (\log x)^{A_1}\), and \(\tau (u_i)<A_1\). Then

$$\begin{aligned} {\mathfrak {S}_{\mathbf {u},v,z}}(x)=\frac{\tau (u_1)\tau (u_2)\cdots \tau (u_v)}{u_1u_2\cdots u_v} {\mathfrak {S}_{v,z}}(x)\left( 1+o(1)\right) ^k, \end{aligned}$$
(27)

where \(0\le k\le v\) is the number of \(u_i\)’s that are not 1.

The same proof as in Lemma 2.6 applies with the need of considering all components of \(\mathbf {u}\).

Fix a prime \(z<q\le z^2\). Consider the number \(X_q\) of primes \(p_1, \ldots , p_v\) such that q divides \(p_i-1\). By Lemma 5.1, it is natural to model \(X_q\) by a binomial distribution with parameters v and \(\frac{2}{q}\). In fact, Lemma 5.1 implies that

Lemma 5.2

For any \(0\le k\le v\), as \(x\rightarrow \infty \),

$$\begin{aligned} P(X_q=k)&:=\frac{1}{\mathfrak {S}_{v,z}(x)}\\&\quad \times \,\sum _{\begin{array}{c} {\left( \frac{\log p_1}{\log x}, \ldots , \frac{\log p_v}{\log x}\right) \in \cup M_v}\\ {\text {Exactly } k \text { primes }p_i\text { satisfy }q|p_i-1} \end{array}} \frac{\tau _z(p_1-1)\tau _z(p_2-1)\cdots \tau _z(p_v-1)}{p_1p_2\cdots p_v}\\&=\left( {\begin{array}{c}v\\ k\end{array}}\right) \left( \frac{2}{q}\right) ^k \left( 1-\frac{2}{q}\right) ^{v-k}(1+o(1))^v. \end{aligned}$$

Here, the functions implied in \(1+o(1)\) only depend on x and do not depend on k.

Denote by \(A_q\) the contribution of a power of q in

$$\begin{aligned} \frac{\tau _{z,z^2}(\mathrm {lcm}(p_1-1,p_2-1, \ldots ,p_v-1))}{\tau _{z,z^2}(p_1-1)\tau _{z,z^2}(p_2-1)\cdots \tau _{z,z^2}(p_v-1)}. \end{aligned}$$

Similarly, denote by \(A_{q_1, \cdots , q_j}\) the contribution of powers of \(q_1, \cdots , q_j\) in the above. Let

$$\begin{aligned} B_{z,v}:=\frac{\tau _z(p_1-1)\tau _z(p_2-1)\cdots \tau _z(p_v-1)}{p_1p_2\cdots p_v}. \end{aligned}$$

We can combine the contributions of finite number of primes \(q_1, \ldots , q_j\) in \((z,z^2]\). For these multiple primes, Lemma 5.2 becomes:

Lemma 5.3

For any \(0\le k_1, \ldots , k_j \le v\), as \(x\rightarrow \infty \),

$$\begin{aligned}&P(X_{q_1}=k_1, \ldots , X_{q_j}=k_j)\\&\quad :=\frac{1}{\mathfrak {S}_{v,z}(x)}\sum _{\begin{array}{c} {\left( \frac{\log p_1}{\log x}, \ldots , \frac{\log p_v}{\log x}\right) \in \cup M_v}\\ {{For each s=1,\ldots , j,}}\\ {{ exactly }k_s { primes }p_i{ satisfy }q_s|p_i-1} \end{array}} \frac{\tau _z(p_1-1)\tau _z(p_2-1)\cdots \tau _z(p_v-1)}{p_1p_2\cdots p_v}\\&\quad =\prod _{s\le j} \left( {\begin{array}{c}v\\ k_s\end{array}}\right) \left( \frac{2}{q_s}\right) ^{k_s} \left( 1-\frac{2}{q_s}\right) ^{v-k_s}(1+o(1))^v. \end{aligned}$$

Here, the functions implied in \(1+o(1)\) only depend on j, x and they do not depend on \(k_s\).

This shows that the random variables \(X_{q_i}\) behave similar as independent binomial distributions. For \(z<q\le z^2\), we have \(A_q=\frac{2}{2^k}\) for \(k\ge 1\), and \(A_q=1\) for \(k=0\). Thus, the contribution of this prime q is

$$\begin{aligned} \mathbf {E}[A_q]=\left( 2\left( 1-\frac{1}{q}\right) ^v - \left( 1-\frac{2}{q}\right) ^v\right) (1+o(1))^v. \end{aligned}$$

For distinct primes \(q_1, \ldots , q_j\) in \((z,z^2]\), the contribution of these primes is

$$\begin{aligned} \mathbf {E}[A_{q_1,\ldots ,q_j}]=\prod _{s\le j} \left( 2\left( 1-\frac{1}{q_s}\right) ^v - \left( 1-\frac{2}{q_s}\right) ^v\right) (1+o(1))^v, \end{aligned}$$

where the function implied in \(1+o(1)\) only depends on j, x.

Then, we conjecture that the contribution of all primes in \(z<q\le z^2\) will be

Conjecture 5.1

As \(x\rightarrow \infty \), we have

$$\begin{aligned} \mathfrak {U}_{v,z}(x)=\prod _{z<q\le z^2} \left( 2\left( 1-\frac{1}{q}\right) ^v - \left( 1-\frac{2}{q}\right) ^v\right) \mathfrak {S}_{v,z}(x)(1+o(1))^v. \end{aligned}$$

It is clear that

$$\begin{aligned} 2\left( 1-\frac{1}{q}\right) ^v - \left( 1-\frac{2}{q}\right) ^v= 1+o\left( \frac{v}{q}\right) . \end{aligned}$$

Thus, we have, as \(x\rightarrow \infty \),

$$\begin{aligned} \prod _{z<q\le z^2}\left( 2\left( 1-\frac{1}{q}\right) ^v - \left( 1-\frac{2}{q}\right) ^v\right) =(1+o(1))^v. \end{aligned}$$

Therefore, we obtain the following heuristic result according to Conjecture 5.1.

Conjecture 5.2

As \(x\rightarrow \infty \), we have

$$\begin{aligned} \mathfrak {U}_{v,z}(x)=\mathfrak {S}_{v,z}(x)(1+o(1))^v. \end{aligned}$$

Then Conjecture 1.1 follows from Lemma 2.6.

Remarks

We were unable to prove Conjecture 1.1. The main difficulty is due to the short range of u in Corollary 2.1. Because of the range of u, we could not extend Lemma 5.3 to all primes in \(z<q\le z^2\).