# Multiplicative arithmetic functions and the Ewens measure

Dor Elboim, Ofir Gorodetsky

###### Abstract

Random integers, sampled uniformly from $[1,x]$, share similarities with
random permutations, sampled uniformly from $S_{n}$. These similarities
include the Erdős-Kac theorem on the distribution of the number of prime
factors of a random integer, and Billingsley theorem on the largest prime
factors of a random integer.

Given a multiplicative function $\alpha\colon\mathbb{N}\to\mathbb{R}_{\geq
0}$, one may associate with it a measure on the integers in $[1,x]$, where $n$
is sampled with probability proportional to the value $\alpha(n)$. When the
mean value of $\alpha$ over primes tends to some positive $\theta$, we draw an
analogy between these measures and the Ewens measure with parameter $\theta$.
In particular, we extend the results of Erdős-Kac and Billingsley to these
measures, obtaining results which are in perfect agreement with the ones known
for the Ewens measure.

We also study a generalized Ewens measure on the integers arising from
multiplicative $\alpha$ with $\alpha(p)=(\log p)^{\gamma}$, and obtain results
in agreement with those in the permutation setting.

##  1 Introduction

Given a positive integer $m\in\mathbb{N}$, denote by
$\mathcal{p}_{1}(m)\geq\mathcal{p}_{2}(m)\geq\ldots$ the prime factors of $m$
(repeated according to their multiplicity), arranged in non-increasing order.
We have

$\log\mathcal{p}_{1}(m)+\log\mathcal{p}_{2}(m)+\ldots=\log m.$ (1.1)

We denote by $\Omega(m)$ the number of prime factors of $m$, counted with
multiplicity. If $k$ is greater than $\Omega(m)$, we let
$\mathcal{p}_{k}(m)=1$. Given a prime $p$, we denote by $\nu_{p}(n)$ the
multiplicity of $p$ in the prime factorization of $n$. If $p^{k}\mid n$ and
$k=\nu_{p}(n)$, we write $p^{k}\mid\mid n$.

A function $\alpha\colon\mathbb{N}\to\mathbb{C}$ is called multiplicative if
$\alpha(1)=1$ and $\alpha(nm)=\alpha(n)\alpha(m)$ for every coprime
$n,m\in\mathbb{N}$. Given a non-negative multiplicative function $\alpha$, we
define a measure on the positive integers up to $x$ by

$\mathbb{P}_{x,\alpha}(m)=\frac{1}{S(x)}\alpha(m)=\frac{1}{S(x)}\prod_{p^{k}\mid\mid
n}\alpha(p^{k})$ (1.2)

where the product is over (maximal) prime powers dividing $n$, and $S(x)$ is
the normalization constant

$S(x)=\sum_{m\leq x}\alpha(m).$ (1.3)

We let $N_{x}:=N_{x,\alpha}$ be the random integer whose probability
distribution is (1.2). We consider multiplicative functions
$\alpha\colon\mathbb{N}\to\mathbb{R}_{\geq 0}$ satisfying the following two
conditions for some real $d>-1$:

$\displaystyle\text{(I) }\sum_{p\leq x}\frac{\alpha(p)\log p}{p^{d}}=\theta
x+O_{A}\left(\frac{x}{\log^{A}x}\right)\mbox{for all $A>0$,}$ (1.4)
$\displaystyle\text{(II) }\frac{\alpha(p^{k})}{p^{dk}}=O({r}^{k})\mbox{ for
some $1\leq{r}<\sqrt{2}$ and all $k\geq 1$.}$ (1.5)

Here $p$ denotes a prime number. Recall that the prime number theorem says
that $\sum_{p\leq x}\log p\sim x$. Thus, (1.4) should be interpreted as
$\alpha(p)/p^{d}$ being, on average, of size $\theta$, and it is a common
condition in multiplicative number theory. Condition (1.5) is of a more
technical nature. Examples of common functions satisfying the conditions are
given in §1.2 below. Our main result is the following.

######  Theorem 1.1.

Let $\alpha\colon\mathbb{N}\to\mathbb{R}_{\geq 0}$ be a multiplicative
function satisfying (1.4)–(1.5) with parameters $\theta>0$ and $d>-1$.

  1. 1.

As $x\to\infty$ we have

$\frac{\Omega(N_{x})-\theta\log\log x}{\sqrt{\theta\log\log
x}}\overset{d}{\longrightarrow}N(0,1).$ (1.6)

  2. 2.

As $x\to\infty$ we have

$\left(\frac{\log\mathcal{p}_{1}(N_{x})}{\log
x},\frac{\log\mathcal{p}_{2}(N_{x})}{\log
x},\ldots\right)\overset{d}{\longrightarrow}\mathrm{PD}(\theta),$ (1.7)

where $\mathrm{PD}(\theta)$ is the Poisson-Dirichlet distribution with
parameter $\theta$ (defined in §3 below).

Here the arrows indicate convergence in distribution. The proof of the first
part of Theorem 1.1 applies to $\omega(N_{x})$ as well, where $\omega(n)$
counts the number of prime factors of $n$ _without_ multiplicities, and the
result is the same.

The case $\alpha\equiv 1$ of (1.6) is the Erdős-Kac theorem [EK40]. Our proof
of it is a generalization of a proof given by Billingsley [Bil69] to the
original Erdős-Kac.

The case $\alpha\equiv 1$ of (1.7) is Billingsley theorem [Bil72]. Our proof
of it is a generalization of Donnelly and Grimmett’s proof [DG93], who
elucidated Billingsley’s result.

In the proofs of both parts of Theorem 1.1, we use a recent result of
Granville and Koukoulopoulos on the asymptotics of $\sum_{n\leq x}\alpha(n)$
[GS07]; see Theorem 4.2 below and the discussion preceding it.

###  1.1 Comparison with permutations

Given a permutation $\pi\in S_{n}$, we denote by
$\ell_{1}(\pi)\geq\ell_{2}(\pi)\geq\ldots$ the lengths of the disjoint cycles
of $\pi$, arranged in non-increasing order. They satisfy

$\ell_{1}(\pi)+\ell_{2}(\pi)+\ldots=n.$ (1.8)

We denote by $C(\pi)$ the number of cycles in $\pi$. If $k$ is greater than
$C(\pi)$, we let $\ell_{k}(\pi)=0$. We let $C_{i}(\pi)$ be the number of
cycles of $\pi$ of length $i$.

The Ewens measure with parameter $\theta$ ($>0$) is a measure on $S_{n}$,
defined as follows:

$\mathbb{P}_{n,\theta}(\pi)=\frac{1}{h_{n}n!}\theta^{C(\pi)},$ (1.9)

where $h_{n}$ is the normalization constant

$h_{n}=\frac{1}{n!}\sum_{\pi\in S_{n}}\theta^{C(\pi)}=\binom{n+\theta-1}{n}.$
(1.10)

This measure has first appeared in the study of population genetics [Ewe72].
The Ewens measure has found many practical applications, through its
connection with Kingman’s coalescent process [Kin82] and its occurrence in
non-parametric Bayesian statistics [Ant74].

Theorem 1.1 should be compared with two results on the Ewens measure, one on
$C(\pi_{n,\theta})$ by Hansen, and another on
$(\ell_{1}(\pi_{n,\theta})/n,\ell_{2}(\pi_{n,\theta})/n,\ldots)$ by Watterson.

######  Theorem 1.2.

Let $\theta>0$.

  1. 1.

[Han90] As $n\to\infty$ we have

$\frac{C(\pi_{n,\theta})-\theta\log n}{\sqrt{\theta\log
n}}\overset{d}{\longrightarrow}N(0,1).$ (1.11)

  2. 2.

[Wat76] As $n\to\infty$ we have

$\left(\frac{\log\ell_{1}(\pi_{n,\theta})}{n},\frac{\log\ell_{2}(\pi_{n,\theta})}{n},\ldots\right)\overset{d}{\longrightarrow}\mathrm{PD}(\theta).$
(1.12)

The similarity of Theorem 1.1 and Theorem 1.2 is most apparent for functions
$\alpha$ where $\alpha(p)\approx\theta$. It suggests an analogy between
permutations chosen according to the Ewens measure and integers chosen
according to multiplicative weights.

The analogy between permutations and integers is a well-established one, see
the surveys [ABT03, Ch. 1] and [Gra08]. The analogy leads to advancements both
in permutations and in integers, see e.g. [Gra06, GS07, EFG16, EFG17]. This
analogy always involves comparing a uniformly drawn integer in $[1,x]$ and a
uniformly drawn permutation from $S_{n}$, where $n\approx\log x$. Our results
suggest that the analogy persists even when the chosen measures are not
uniform.

###  1.2 Examples

The prototypical example of a multiplicative arithmetic function satisfying
(1.4)–(1.5) is

$\alpha(n)=\theta^{\omega(n)}$ (1.13)

for some $\theta>0$. Condition (1.4) with $d=0$ holds due to the prime number
theorem in the form [dlVP56]

$\sum_{p\leq x}\log p=x+O\left(xe^{-a\sqrt{\log x}}\right)$ (1.14)

for some $a>0$. Condition (1.5) holds with $r=1$. A related example is the
$k$-fold divisor function

$d_{k}(n):=\\{(n_{1},n_{2},\ldots,n_{k})\in\mathbb{N}:n_{1}n_{2}\cdots
n_{k}=n\\},$ (1.15)

where $k$ is a positive integer. In fact, we may define $d_{k}$ for arbitrary
real $k>0$ by noting that for positive integers $k$ one has

$d_{k}(p^{i})=\binom{i+k-1}{i},$ (1.16)

which makes sense for any real $k$ and defines a unique multiplicative
function. Condition (1.4) holds again from (1.14), while (1.5) is a
consequence of (1.16) and e.g. Stirling’s approximation.

We may use (1.14) to verify many other examples that satisfy (1.4)–(1.5) with
$\theta=1$ and $d=0$, for instance

  * •

The indicator function of $k$-powerfree integers (integers not divisible by
$p^{k}$ for any prime $p$). The case $k=2$ corresponds to squarefree integers.

  * •

$\phi(n)/n$, where $\phi$ is Euler’s totient function
$\phi(n)=\left|(\mathbb{Z}/n\mathbb{Z})^{\times}\right|=n\prod_{p\mid
n}(1-1/p)$.

  * •

$\sigma_{z}(n)=\sum_{d\mid n}d^{z}$ for real $z<0$, the sum of $z$-th powers
of the divisors of $n$.

A more delicate example with $\theta=1$ and $d=0$ is $a(n)$, the number of
non-isomorphic abelian groups of order $n$. It is known that $a(n)$ is
multiplicative and that $a(p^{k})$ is the number of partitions of $k$ [Ivi85,
p. 37]. As $a(p)=1$, (1.4) is easily verified. As the number of partitions of
$k$ grows subexponentially, (1.5) holds as well. Examples with $\theta=1$ and
$d\neq 0$ include

  * •

Euler’s totient function $\phi(n)$, more generally Jordan’s function
$\phi_{k}(n)=n^{k}\prod_{p\mid n}(1-1/p^{k})$ ($k\geq 1$ a positive integer),
and Dedekind’s $\psi$ function $\psi(n)=n\prod_{p\mid n}(1+1/p)$.

  * •

$\sigma_{z}(n)=\sum_{d\mid n}d^{z}$ for real $z>0$.

  * •

$n\mapsto n^{z}$ for real $z>-1$.

We now discuss some examples related to number fields. Let $K/\mathbb{Q}$ be a
number field with ring of integers $\mathcal{O}_{K}$. Let

$r_{K}(n)=\\#\\{I\subseteq\mathcal{O}_{K}:\mathrm{Nm}_{K/\mathbb{Q}}I=(n)\\},$
(1.17)

where $I$ is an ideal of $\mathcal{O}_{K}$. It is known that $r_{K}$ is
multiplicative and that $r_{K}(p^{k})$ is bounded by a polynomial in $k$,
whose degree depends on $[K:\mathbb{Q}]$ [CN63, Lem. 9], establishing (1.5)
with $d=0$. Condition (1.4) with $d=0$, $\theta=1$ follows from Landau’s prime
ideal theorem [Lan03].

A related example is $b_{K}(n)=\mathds{1}_{r_{K}(n)>0}$, the indicator
function of integers whose ideal is a norm. Condition (1.5) holds trivially
with $d=0$. As the condition $r_{K}(p)>0$ is determined by the Frobenius
conjugacy class of $p$ in the Galois group of the normal closure of
$K/\mathbb{Q}$, a standard application of Chebotarev Density Theorem [LO77]
implies (1.4) for some rational $\theta>0$.

Given an irreducible $P\in\mathbb{Z}[x]$, we may define interesting variants
of $r_{K}$ and $b_{K}$ which often appear in sieve theory [HR74]:
$\rho_{P}(n):=\\#\\{a\in\mathbb{Z}/n\mathbb{Z}:P(a)\equiv 0\bmod n\\}$ and
$\mathds{1}_{P}(n):=\mathds{1}_{\rho_{P}(n)>0}$. These are multiplicative
functions which again satisfy (1.4)–(1.5), since $\rho_{P}(p)=\rho_{K}(p)$ for
all but finitely many primes, where $K=\mathbb{Q}[x]/P(x)$. Realizing the
Galois group $G=\mathrm{Gal}(P/\mathbb{Q})$ of the splitting field of $P$ as a
permutation group on the roots of $P$, the parameter $\theta$ associated to
$\mathds{1}_{P}$ is the proportion of permutations in $G$ with a fixed point,
see e.g. [Ser03].

###  1.3 Previous works

####  1.3.1 Erdős-Kac

In a series of works, Alladi [All82, All84, All85a, All85b] proved a
generalization of Erdős-Kac involving weights $\alpha$ as well. As opposed to
Theorem 1.1, Alladi’s conditions for having a normal distribution are in the
spirit of Bombieri–Vinogradov’s theorem, and come from his use of the
combinatorial sieve. They do not involve the values of the function $\alpha$
on primes, and require instead an upper bound on a sum of _congruence sums_.
In practice, verifying this bound in the case of multiplicative weights such
as $\alpha(n)=\theta^{\omega(n)}$ is highly non-trivial, and we do not know
how to do it in general (in some specific families of $\alpha$, such as those
discussed in [FI10, App. A.8], the verification is easier). A related (but
simpler) sieve-theoretic approach to Erdős-Kac and its generalizations was
introduced by Granville and Soundararajan [GS07]. If one adapts it to the
weighted case, with multiplicative weights $\alpha(n)$, again one ends up with
the highly non-trivial problem of bounding congruence sums.

Liu [Liu04] proved an Erdős-Kac result for certain semigroups other than the
positive integers, e.g. ideals in number fields, which is related to (1.17).

####  1.3.2 Billingsley

Arratia, Kochman and Miller proved an analogue of Billingsley theorem for
_normed arithmetic semigroups_ satisfying certain growth conditions [AKM14,
Thm. 2]. A commutative semigroup $S$ is called normed arithmetic semigroup if
it contains an identity element and admits unique factorization into ‘prime’
elements. Furthermore, it should come equipped with a multiplicative norm
function $s\mapsto|s|\in\mathbb{R}_{>0}$, such that $N(x)=\\#\\{s\in S:|s|\leq
x\\}$ is a finite number for each $x>0$.

There is some overlap between the second part of Theorem 1.1 and [AKM14, Thm.
2], for the following reason. There are examples where a multiplicative
function $\alpha$ satisfying the conditions (1.4)–(1.5) coincides with
$\alpha_{S}(n):=\\#\\{s\in S:|s|=n\\}$ for some semigroup $S$ with integer-
valued norm (e.g. $a(n)$ from §1.2 coincides with $\alpha_{S}$ where $S$ is
the semigroup of finite abelian groups, with the group operation being direct
product of groups, and the norm function being the order of the group). In
such cases, (1.7) is not new. However, since for any $S$, $\alpha_{S}$ is
necessarily integer-valued, then for most of the arithmetic functions $\alpha$
satisfying (1.4)–(1.5), one cannot find a semigroup $S$ with
$\alpha_{S}=\alpha$, e.g. $\alpha_{S}(n)$ cannot coincide with
$\theta^{\omega(n)}$ for $\theta\notin\mathbb{N}$. In fact, with some work it
can be shown that unless $\theta=1$, $\theta^{\omega(n)}$ is not of the form
$\alpha_{S}$.

####  1.3.3 Smoothness

Recall that an integer $x$ is $y$-smooth if all the prime factors of $x$ are
of size at most $y$. By considering only the first coordinate in (1.7) we
obtain the probability that $N_{x}$ is $x^{1/u}$-smooth, which turns out to
tend to $\rho_{\theta}(u)$, where
$\rho_{\theta}\colon[0,\infty)\to\mathbb{R}_{\geq 0}$ may be defined by the
integral delay equation

$\rho_{\theta}(x)=\frac{1}{x^{\theta}}\int_{x-1}^{x}\theta
y^{\theta-1}\rho_{\theta}(y)\,dy$ (1.18)

and the initial conditions $\rho_{\theta}(x)=1$ for $x\in[0,1]$. For further
formulas for $\rho_{\theta}$ see [Hol01, Prop. 3.1] in Holst’s survey on the
Poisson-Dirichlet distribution. Although smoothness probabilities for $N_{x}$
were studied extensively in the literature, by de Bruijn and van Lint
[dBvL64], Song [Son02] and Tenenbaum and Wu [TW03], the connection with the
Poisson-Dirichlet distribution seems to have been missed.

### Conventions

In the arguments below, we think of the function $\alpha$ as fixed, and write
$N_{x}$ for $N_{x,\alpha}$. We denote the set of prime numbers by
$\mathcal{P}$, and reserve the letter $p$ for primes. The letters $C$ and $c$
always denote positive constants, which may vary from line to line. However,
$C$ and $c$ depend only on the arithmetic function $\alpha$ considered unless
otherwise stated. The arguments in the proofs always hold for sufficiently
large $x$.

### Acknowledgments

We thank Andrew Granville, Dimitris Koukoulopoulos, Ron Peled and Ze’ev
Rudnick for useful discussions. The second author was supported by the
European Research Council under the European Union’s Seventh Framework
Programme (FP7/2007-2013) / ERC grant agreement n${}^{\text{o}}$ 786758.

##  2 Further results

###  2.1 Polynomially-growing weights

In recent years, there has been significant activity in the study of
permutations sampled according to _cycle weights_ [Yak07, Tim08, Lug09, BUV11,
MNZ12, NSZ13, EU14, CZ15, DM15, SZ15, RZ18]; this model is related to the
study of the quantum Bose gas in statistical mechanics, see e.g. [BU09, BU11,
EP19]. To state the model, let $\theta_{1},\ldots,\theta_{n}$ be non-negative
reals (not all zero). The probability of a permutation $\pi$ with respect to
the weights $\theta_{i}$ is defined to be

$\mathbb{P}_{n,\theta_{i}}(\pi)=\frac{1}{h_{n}n!}\prod_{i=1}^{n}\theta_{i}^{C_{i}(\pi)}=\frac{1}{h_{n}n!}\prod_{\begin{subarray}{c}c\in\pi\\\
\text{cycle of }\pi\end{subarray}}\theta_{|c|},$ (2.1)

where $C_{i}(\pi)$ is the number of cycles of $\pi$ of length $i$, and $h_{n}$
is the normalization constant

$h_{n}=\frac{1}{n!}\sum_{\pi\in S_{n}}\prod_{i=1}^{n}\theta_{i}^{C_{i}(\pi)},$
(2.2)

the so-called partition function. We let $\pi_{n,\theta_{i}}$ be the random
permutation whose probability distribution is (2.1). The measure
$\mathbb{P}_{n,\theta_{i}}$ is called a generalized Ewens measure. The measure
$\mathbb{P}_{n,\theta_{i}}$ was studied extensively in the case of
polynomially-growing cycle weights, that is

$\theta_{n}\approx An^{\gamma},$ (2.3)

see [EG08, MNZ12, EU14, DM15, CZ15]. Ercolani and Ueltschi proved the
following.

######  Theorem 2.1.

[EU14, Thm. 5.1] Let $\gamma>0$, and take

$\theta_{n}=\frac{\Gamma(\gamma+n+1)}{n!}=(1+o(1))n^{\gamma}.$ (2.4)

As $n\to\infty$ we have

$\mathbb{E}C(\pi_{n,\theta_{i}})\sim
n^{\frac{\gamma}{\gamma+1}}\left(\frac{\Gamma(\gamma)}{\gamma^{\gamma}}\right)^{\frac{1}{\gamma+1}}.$
(2.5)

The second equality in (2.4) follows from Stirling’s approximation. The
specific choice (2.4) simplifies the computations greatly. Maples, Nikeghbali
and Zeindler [MNZ12, Cor. 1.2] were able to prove that $C(\pi_{n,\theta_{i}})$
converges, after an explicit normalization, to a normal distribution (for
$\theta_{n}$-s as in (2.4) and also scalar multiples of it).

Next we describe a result of Ercolani and Ueltschi about a permutation
statistic which we have yet to discuss, $L_{1}(\pi)$. This is the length of
the cycle of a permutation $\pi$ which contains the element $1$. In the Ewens
case $\theta_{i}=\theta$, it is known that $L_{1}(\pi_{n,\theta_{i}})/n$
converges in distribution to a beta distribution [EU14, §6]. For polynomially-
growing weights, Ercolani and Ueltschi proved that $L_{1}(\pi_{n,\theta_{i}})$
exhibits a very different behavior. First, the order of magnitude of
$L_{1}(\pi_{n,\theta_{i}})$ in this case is $n^{1/(\gamma+1)}=o(n)$ and not
$n$. Second, the limiting distribution is a gamma distribution, whose
definition is recalled in §3.

######  Theorem 2.2.

[EU14, Thm. 5.1] Let $\gamma>0$, and take $\theta_{n}$-s as in (2.4). Then, as
$n\to\infty$,

$\frac{L_{1}(\pi_{n,\theta_{i}})}{n^{\frac{1}{\gamma+1}}}\overset{d}{\longrightarrow}\operatorname{gamma}(\gamma+1,\Gamma(\gamma+1)^{1/(\gamma+1)}).$
(2.6)

See Dereich and Mörters [DM15] for finer results about
$L_{1}(\pi_{n,\theta_{i}})$ for similar weights.

We derive number-theoretic analogues of Theorems 2.1 and 2.2. Since there is
no such thing as ‘a prime divisor of $n$ containing a fixed element’, we must
turn to a different interpretation of $L_{1}(\pi)$.

######  Definition 2.3.

Let $\mathbf{a}=\\{a_{j}\\}_{j\geq 1}$ be a sequence of non-negative reals
summing to $0<S<\infty$. A size-biased sampling of an element from
$\mathbf{a}$ is a random variable $X$ whose distribution is given by

$\mathbb{P}(X=a_{j})=\frac{a_{j}\cdot\left|\left\\{i\geq 1\ \big{|}\
a_{i}=a_{j}\right\\}\right|}{S}.$ (2.7)

Suppose that $\mathbb{P}$ is some conjugation-invariant measure on $S_{n}$
(e.g. $\mathbb{P}_{n,\theta_{i}}$). If $\pi\in S_{n}$ is sampled according to
$\mathbb{P}$, then the distribution of $L_{1}(\pi)$ coincides with the
distribution of a typical cycle of $\pi$, that is: of a size-biased sampling
of an element from $\\{\ell_{i}(\pi)\\}_{i\geq 1}$. See Lemma 3.3 below for
the proof. It is now clear how to define an integer analogue of
$L_{1}(\pi_{n,\theta_{i}})$: given $N_{x}$, we define $P_{1}(N_{x})$ by
letting $\log P_{1}(N_{x})$ be a size-biased sampling of an element from
$\\{\log\mathcal{p}_{i}(N_{x})\\}_{i\geq 1}$. We think of $P_{1}(N_{x})$ as a
typical prime divisor of $N_{x}$.

In the integer setting, the polynomial weights (2.3) correspond to

$\alpha(p)\approx K\log^{\gamma}p.$ (2.8)

For our results, we require specifically

$\displaystyle\text{(I) }\alpha(p)=K\log^{\gamma}p+O\left(\log^{-2}p\right),$
(2.9) $\displaystyle\text{(II) }\sum_{k\geq
2}\frac{k\alpha(p^{k})}{p^{k}}=O\left(\frac{1}{p\log^{2}p}\right)$ (2.10)

for all primes $p$.

######  Theorem 2.4.

Let $\alpha\colon\mathbb{N}\to\mathbb{R}_{\geq 0}$ be a multiplicative
function satisfying (2.9)–(2.10) for some $K>0$, $\gamma>0$.

  1. 1.

As $x\to\infty$ we have

$\frac{\log P_{1}(N_{x})}{(\log
x)^{\frac{1}{\gamma+1}}}\overset{d}{\longrightarrow}\operatorname{gamma}(\gamma+1,(K\Gamma(\gamma+1))^{1/(\gamma+1)}).$
(2.11)

  2. 2.

As $x\to\infty$ we have

$\mathbb{E}\Omega(N_{x})\sim(\log
x)^{\frac{\gamma}{\gamma+1}}\left(\frac{K\Gamma(\gamma)}{\gamma^{\gamma}}\right)^{\frac{1}{\gamma+1}}.$
(2.12)

The proof of the second part of Theorem 2.4 applies to $\omega(N_{x})$ as
well. It is quite interesting that although the measure
$\mathbb{P}_{x,\alpha}$ gives larger weights to larger primes, the typical
prime factors of $N_{x}$ are much smaller than in the case of uniformly drawn
integers between $1$ and $x$. Indeed, it follows from the first part of
Theorem 2.4 that $\log P_{1}(N_{x})=o(\log x)$, while from Theorem 1.1 it
follows that $\log P_{1}(N_{x})=\Theta(\log x)$ for uniform integers.

The main ingredient in the proof of Theorem 2.4 is the asymptotics of
$S(x)=\sum_{n\leq x}\alpha(n)$ for multiplicative $\alpha$ obeying
(2.9)–(2.10). This unusual sum was studied by Schwarz [Sch65] and Marenich
[Mar83]. As they both appeal to the same Tauberian theorem [Ing41, Thm. 1], no
error term is obtained. We prove the following estimate.

######  Theorem 2.5.

Let $\alpha\colon\mathbb{N}\to\mathbb{R}_{\geq 0}$ be a multiplicative
function satisfying (2.9)–(2.10) for some $K>0$, $\gamma>0$. There exists
$\varepsilon>0$ for which

$\frac{1}{x}\sum_{n\leq
x}\alpha(n)=\left(1+O\left(\frac{1}{\log^{\varepsilon}x}\right)\right)A_{\alpha}\frac{\exp(B(\log
x)^{\frac{\gamma}{\gamma+1}})}{(\log x)^{\frac{\gamma+2}{2(\gamma+1)}}}$
(2.13)

as $x\to\infty$, where $A_{\alpha}$ is a positive constant and

$B=\left(1+\frac{1}{\gamma}\right)(K\Gamma(\gamma+1))^{\frac{1}{\gamma+1}}.$
(2.14)

######  Remark 2.6.

The proof of Theorem 2.5 shows that

$A_{\alpha}=A\prod_{p}\frac{\sum_{k\geq
0}\frac{\alpha(p^{k})}{p^{k}}}{\exp(\frac{K\log^{\gamma}p}{p})}$ (2.15)

for a constant $A$ depending only on $K$ and $\gamma$. Additionally, one may
take $\varepsilon=1/(\gamma+1)$ if $\gamma>2$, and $\varepsilon$ arbitrarily
close to $\gamma/(2(\gamma+1))$ otherwise.

The proof of Theorem 2.5 uses Perron’s formula to relate the mean value of
$\alpha$ to estimation of a complex integral. We use the saddle point method
to study the resulting integral, and obtain the main term quite easily. Most
of the work goes into bounding the contribution of the integral away from the
saddle point. This is achieved using Theorem 5.5, a classical yet non-trivial
result about the distribution of primes in short intervals (a result that goes
beyond the prime number theorem).

###  2.2 Small cycles

The limiting distribution of
$C_{1}(\pi_{n,\theta}),C_{2}(\pi_{n,\theta}),\ldots$ (small cycles) was
studied by Arratia, Barbour and Tavaré.

######  Theorem 2.7.

[ABT92] Let $\theta>0$. As $n\to\infty$ the random variables
$C_{1}(\pi_{n,\theta})$, $C_{2}(\pi_{n,\theta})$,
$C_{3}(\pi_{n,\theta}),\ldots$ converge in distribution to independent Poisson
with respective means $\theta$, $\theta/2$, $\theta/3,\ldots$

Recently, Ercolani and Ueltschi extended this result as follows.

######  Theorem 2.8.

[EU14, Thms. 5.1, 6.1] Suppose that $\lim_{i\to\infty}\theta_{i}=\theta$, or
that $\theta_{i}$ are defined by (2.4). Then As $n\to\infty$ the random
variables $C_{1}(\pi_{n,\theta_{i}})$, $C_{2}(\pi_{n,\theta_{i}})$,
$C_{3}(\pi_{n,\theta_{i}}),\ldots$ converge in distribution to independent
Poisson with respective means $\theta_{1}$, $\theta_{2}/2$,
$\theta_{3}/3,\ldots$

An analogue of $C_{i}(\pi)$ in the integers is $\nu_{p}(n)$, the multiplicity
of a prime $p$ in $n$. We prove the following.

######  Theorem 2.9.

Let $\alpha\colon\mathbb{N}\to\mathbb{R}_{\geq 0}$ be a non-negative
multiplicative function, satisfying (1.4)–(1.5) or (2.9)–(2.10). For any prime
$p$ let $X_{p}$ be an integer-valued random variable with

$\mathbb{P}(X_{p}=k)=\left(\sum_{i=0}^{\infty}\frac{\alpha(p^{i})}{p^{i(d+1)}}\right)^{-1}\frac{\alpha(p^{k})}{p^{k(d+1)}}$
(2.16)

for any $k\geq 0$ (here $d=0$ if $\alpha$ satisfies (2.9)–(2.10)). Then
$\nu_{p}(N_{x})\overset{d}{\longrightarrow}X_{p}$ as $x\to\infty$. More
generally, let $p_{1},p_{2},\ldots,p_{m}$ be a sequence of distinct primes,
and suppose that $(X_{p_{i}})_{1\leq i\leq m}$ are independent. Then the joint
distribution of $(\nu_{p_{i}}(N_{x}))_{1\leq i\leq k}$ satisfies

$\left(\nu_{p_{i}}(N_{x})\right)_{1\leq i\leq
k}\overset{d}{\longrightarrow}\left(X_{p_{i}}\right)_{1\leq i\leq k},\quad
x\to\infty.$ (2.17)

For $\alpha\equiv 1$, the random variables $X_{p}$ in Theorem 2.9 are
geometrically distributed with parameter $1/p$, and in this case the result
dates back at least to Billingsley [Bil74].

##  3 Preliminaries from probability theory

We denote by $\operatorname{beta}(\alpha,\beta)$ the beta distribution with
shape parameters $\alpha$ and $\beta$ whose density with respect to Lebesgue
measure on $[0,1]$ is given by

$\frac{\Gamma(\alpha+\beta)x^{\alpha-1}\left(1-x\right)^{\beta-1}}{\Gamma(\alpha)\Gamma(\beta)},\quad
x\in[0,1].$

We denote by $\operatorname{gamma}(\alpha,\beta)$ the gamma distribution with
shape parameters $\alpha$ and $\beta$ whose density with respect to Lebesgue
measure on $[0,\infty)$ is given by

$\frac{\beta^{\alpha}x^{\alpha-1}e^{-\beta x}}{\Gamma(\alpha)},\quad
x\in[0,\infty).$

We define the Poisson-Dirichlet distribution with parameter $\theta$, denoted
by $\operatorname{PD}(\theta)$. Let $Y_{1},Y_{2},\dots$ be an i.i.d. sequence
of random variables with $\operatorname{beta}(1,\theta)$ distribution. Define
the sequence

$Z_{j}=(1-Y_{1})\cdots(1-Y_{j-1})Y_{j},\quad j\geq 1.$ (3.1)

Intuitively $Z_{1}$ takes a $\operatorname{beta}(1,\theta)$ distributed
fraction of the unit interval. Conditioned on $Z_{1}$, $Z_{2}$ takes a
$\operatorname{beta}(1,\theta)$-distributed fraction of the remaining part of
the interval, etc. Finally, the $\operatorname{PD}(\theta)$ distribution is
defined to be the distribution of the sequence $(Z_{j})_{j\geq 1}$ arranged in
non-increasing order.

In the proof of Theorem 1.1, we establish the convergence in distribution to
$\operatorname{PD}(\theta)$ by convergence of a certain sequence to a sequence
of independent $\operatorname{beta}(1,\theta)$ random variables. To this end
we define the size-biased permutation of a sequence of random variables. Let
$X_{1},X_{2},\dots$ be a non-increasing sequence of random variables such that

$\sum_{j=1}^{\infty}X_{j}=1,\quad\text{almost surely.}$ (3.2)

A sized-biased permutation $(\tilde{X_{i}})_{i}$ of the sequence $(X_{i})_{i}$
is a random reordering of the elements of the sequence such that for any
$j\geq 1$

$\mathbb{P}\left(\tilde{X}_{1}=X_{j}\ |\
X_{1},X_{2},\dots\right)=X_{j}\cdot\left|\left\\{j^{\prime}\geq 1\ \big{|}\
X_{j^{\prime}}=X_{j}\right\\}\right|$ (3.3)

and inductively for $k\geq 1$

$\mathbb{P}\left(\tilde{X}_{k}=X_{j}\ |\
\tilde{X}_{1},\dots,\tilde{X}_{k-1},X_{1},X_{2},\dots\right)=\frac{X_{j}\cdot\left(\left|\left\\{j^{\prime}\geq
1\ \big{|}\ X_{j^{\prime}}=X_{j}\right\\}\right|-\left|\left\\{j^{\prime}<k\
\big{|}\
\tilde{X}_{j^{\prime}}=X_{j}\right\\}\right|\right)}{1-\tilde{X}_{1}-\ldots-\tilde{X}_{k-1}}.$
(3.4)

The sized-biased permutation can be used to reconstruct the
$\operatorname{beta}(1,\theta)$ random variables from the
$\operatorname{PD}(\theta)$ distribution in the following sense.

######  Proposition 3.1.

Let $X_{1},X_{2},\dots$ be a non-increasing sequence of random variables with

$\sum_{j=1}^{\infty}X_{j}=1,\quad\text{almost surely.}$ (3.5)

Then, the sequence $(X_{1},X_{2},\dots)$ has the $\operatorname{PD}(\theta)$
distribution if and only if the sequence

$\left(\frac{\tilde{X}_{j}}{1-\tilde{X}_{1}-\ldots-\tilde{X}_{j-1}}\right)_{j=1}^{\infty}$
(3.6)

is an i.i.d. sequence of $\operatorname{beta}(1,\theta)$ random variables.

For a discussion and references for Proposition 3.1 see the introduction of
[PY97]. We use the following result in order to prove the convergence to
$\operatorname{PD}(\theta)$ in Theorem 1.1.

######  Proposition 3.2.

For any $n\geq 1$, let $X_{1}^{(n)},X_{2}^{(n)},\dots$ be a non-increasing
sequence of random variables with

$\sum_{j=1}^{\infty}X_{j}^{(n)}=1,\quad\text{almost surely.}$ (3.7)

Suppose that

$\left(\frac{\tilde{X}^{(n)}_{j}}{1-\tilde{X}^{(n)}_{1}-\ldots-\tilde{X}^{(n)}_{j-1}}\right)_{j=1}^{\infty}\overset{d}{\longrightarrow}(Y_{1},Y_{2},\dots),\quad
n\to\infty,$ (3.8)

where $Y_{1},Y_{2},\dots$ is a sequence of i.i.d.
$\operatorname{beta}(1,\theta)$ random variables. Then, we have

$(X_{1},X_{2},\dots)\overset{d}{\longrightarrow}\operatorname{PD}(\theta),\quad
n\to\infty.$ (3.9)

###### Proof.

Consider the function $g\colon[0,1]^{\mathbb{N}}\to[0,1]^{\mathbb{N}}$ defined
by

$g(y_{1},y_{2},\dots)=\big{(}(1-y_{1})\cdots(1-y_{k-1})y_{k+1}\big{)}_{k=1}^{\infty},$
(3.10)

and the function

$r\colon\Big{\\{}(z_{1},z_{2},\dots)\in[0,1]^{\mathbb{N}}:\sum_{j=1}^{\infty}z_{j}\leq
1\Big{\\}}\to[0,1]^{\mathbb{N}}$ (3.11)

that takes a sequence and returns the same sequence in non-increasing order.
Since $r\circ g$ is continuous in the product topology on $[0,1]^{\mathbb{N}}$
we have by (3.8) that

$(r\circ
g)\left(\bigg{(}\frac{\tilde{X}^{(n)}_{j}}{1-\tilde{X}^{(n)}_{1}-\ldots-\tilde{X}^{(n)}_{j-1}}\bigg{)}_{j=1}^{\infty}\right)\overset{d}{\longrightarrow}(r\circ
g)(Y_{1},Y_{2},\dots),\quad n\to\infty.$ (3.12)

By the definition of the Poisson-Dirichlet distribution $(r\circ
g)(Y_{1},Y_{2},\dots)\sim\operatorname{PD}(\theta)$ and moreover, since

$g^{-1}(z_{1},z_{2},\dots)=\left(\frac{z_{j}}{1-z_{1}-\ldots-
z_{j-1}}\right)_{j=1}^{\infty},$ (3.13)

we have that

$(r\circ
g)\left(\bigg{(}\frac{\tilde{X}^{(n)}_{j}}{1-\tilde{X}^{(n)}_{1}-\ldots-\tilde{X}^{(n)}_{j-1}}\bigg{)}_{j=1}^{\infty}\right)=r(\tilde{X}_{1},\tilde{X}_{2},\ldots)=(X_{1},X_{2},\dots).$
(3.14)

Thus, (3.12) simplifies to (3.9), as needed. ∎

######  Lemma 3.3.

Let $\mathbb{P}$ be a conjugation-invariant measure on $S_{n}$. Given $\pi\in
S_{n}$, let $L_{1}(\pi)$ be the size of the cycle containing $1$, and let
$\mathrm{Typ}(\pi)$ be a random variable which is a size-biased sampling of a
cycle of $\pi$, according to Definition 2.3. Then $L_{1}(\pi)$ and
$\mathrm{Typ}(\pi)$ have the same distribution, where $\pi$ is a permutation
drawn according to $\mathbb{P}$.

###### Proof.

By using the conjugation-invariance, for any $k\in\mathbb{N}$ we have

$\mathbb{P}\left(L_{1}(\pi)=k\right)=\frac{1}{n}\sum_{i=1}^{n}\mathbb{P}\left(L_{i}(\pi)=k\right)$
(3.15)

where $L_{i}(\pi)$ is the size of the cycle of $\pi$ containing $i$. Letting
$U_{n}$ be a uniformly drawn integer from $\\{1,2,\ldots,n\\}$, we have just
shown that

$\mathbb{P}\left(L_{1}(\pi)=k\right)=\mathbb{P}\left(L_{U_{n}}(\pi)=k\right).$
(3.16)

The size of the cycle of $\pi$ which contains $U_{n}$ is a size-biased
sampling of a cycle of $\pi$, which concludes the proof. ∎

##  4 Asymptotically Ewens measure

###  4.1 Multiplicative number theory

######  Lemma 4.1.

Fix $d>-1$ and $s\in\mathbb{R}$. Then

$\int_{2}^{x}t^{d}\log^{s}t\,dt=\left(1+O\left(\frac{1}{\log
x}\right)\right)\frac{x^{d+1}\log^{s}x}{d+1}$ (4.1)

as $x\to\infty$. In particular,

$\int_{2}^{x}t^{d}\log^{s}t\,dt=O\left(x^{d+1}\log^{s}x\right).$ (4.2)

###### Proof.

For $s=0$ the lemma is straightforward. For $s\neq 0$, integration by parts
gives

$\begin{split}\int_{2}^{x}t^{d}\log^{s}t\,dt&=\frac{t^{d+1}}{d+1}\log^{s}t\Big{|}^{t=x}_{t=2}-\int_{2}^{x}\frac{t^{d+1}}{d+1}\left(\log^{s}t\right)^{\prime}\,dt\\\
&=\frac{x^{d+1}}{d+1}\log^{s}x+O(1)-\frac{s}{d+1}\int_{2}^{x}t^{d}\log^{s-1}t\,dt,\end{split}$
(4.3)

so it suffices to establish (4.2) (and apply it with $s-1$ in place of $s$).
The bound (4.2) follows by

$\begin{split}\int_{2}^{x}t^{d}\log^{s}t\,dt&=\int_{2}^{\sqrt{x}}t^{d}\log^{s}t\,dt+\int_{\sqrt{x}}^{x}t^{d}\log^{s}t\,dt\\\
&\leq\left(\max_{2\leq
t\leq\sqrt{x}}\log^{s}t\right)\int_{2}^{\sqrt{x}}t^{d}\,dt+\left(\max_{\sqrt{x}\leq
t\leq
x}\log^{s}t\right)\int_{\sqrt{x}}^{x}t^{d}\,dt=O(x^{d+1}\log^{s}x),\end{split}$
(4.4)

as needed. ∎

An important ingredient in our proofs is estimating $S(x)=\sum_{n\leq
x}\alpha(n)$ for nice multiplicative functions $\alpha$. There are three
common approaches to this problem. One approach involves Wirsing’s theorems
[Wir61, Wir67], which yield (somewhat unwieldy) asymptotic information under
very mild conditions on $\alpha$, but give no error term. A second approach is
the Landau-Selberg-Delange method [Ten15, Ch. II.5], which gives an error term
but requires an analytic continuation of $D_{\alpha}(s):=\sum_{n\geq
1}\alpha(n)/n^{s}$ in a certain region. The third approach, which we use, is a
recent result of Granville and Koukoulopoulos [GS07]. They obtain an
asymptotic formula for $S(x)$ with an error term, under conditions which are
weaker than an analytic continuation. See Theorem 4.2 for a precise statement
of their result (cf. [Odo91, Odo02]).

######  Theorem 4.2.

[GK, Thm. 1; §7] Let $\alpha\colon\mathbb{N}\to\mathbb{C}$ be a multiplicative
function satisfying

$\sum_{n\leq x}\alpha(p)\log p=\theta x+O\left(\frac{x}{(\log
x)^{A}}\right)\qquad(x\geq 2)$ (4.5)

for some $\theta\in\mathbb{C}$ and $A>0$. Suppose that for some positive real
$k$ we have

$\sum_{p\leq x,\,\nu\geq 1}\frac{|\alpha(p^{\nu})|}{p^{\nu}}\leq k\log\log
x+O(1),$ (4.6)

and that

$\sum_{p\leq x}\frac{|\alpha(p)|\log p}{p}=O(\log x)\quad\text{ and
}\quad\sum_{p\leq x,\,\nu\geq 1}\frac{|\alpha(p^{\nu})|^{2}}{p^{\nu}}=o(\log
x).$ (4.7)

Then, letting $J$ be the largest integer less than $A$, we have

$\sum_{n\leq
x}\alpha(n)=x\sum_{j=0}^{J}\frac{\widetilde{c_{j}}}{\Gamma(\theta-j)}(\log
x)^{\theta-j-1}+O\left(x(\log x)^{k-1-A}(\log\log
x)^{\mathds{1}_{A=J+1}}\right)$ (4.8)

for some explicit constants $\widetilde{c_{j}}$, where in particular
$\widetilde{c_{0}}=\prod_{p}(\sum_{i\geq
0}\alpha(p^{i})/p^{i})(1-1/p)^{\theta}$.

(If $\theta-j$ is a pole of $\Gamma$, then $1/\Gamma(\theta-j)$ is to be
interpreted as $0$.) The following corollary of Theorem 4.2 suits our needs.

######  Corollary 4.3.

Suppose a multiplicative function $\alpha\colon\mathbb{N}\to\mathbb{R}_{\geq
0}$ satisfies (1.4)–(1.5) for some $\theta>0$ and $d>-1$. Then

$\sum_{n\leq x}\alpha(n)=\left(1+O\left(\frac{1}{\log
x}\right)\right)A_{\alpha}x^{d+1}\log^{\theta-1}x,\qquad x\to\infty,$ (4.9)

for some positive $A_{\alpha}$, namely
$A_{\alpha}=(d+1)^{-1}\Gamma(\theta)^{-1}\prod_{p}(\sum_{i\geq
0}\alpha(p^{i})/p^{i(d+1)})(1-1/p)^{\theta}$.

###### Proof.

Let $\alpha_{d}(n)=\alpha(n)/n^{d}$. This function is still multiplicative. If
$\alpha$ satisfies (1.4)–(1.5), then $\alpha_{d}$ satisfies these conditions
as well with the same $\theta$, and with parameter $d$ equal now to $0$.

We claim that the conditions of Theorem 4.2 hold for $\alpha_{d}$, with
arbitrarily large $A$ and some $k$ (which depends on the implied constant in
(1.5)). Indeed, (1.5) implies that $\alpha_{d}(p)$ is bounded, so that

$\sum_{p\leq x}\frac{|\alpha_{d}(p)|\log p}{p}\leq C\sum_{p\leq x}\frac{\log
p}{p}\leq C\log x$ (4.10)

by Mertens’ first theorem [Ten15, Thm. 1.8]. Moreover, as
$\alpha_{d}(p^{\nu})=O({r}^{\nu})$ for ${r}$ with ${r}<\sqrt{2}$, we have

$\sum_{p\leq x,\,\nu\geq
1}\frac{\left|\alpha_{d}(p^{\nu})\right|^{2}}{p^{\nu}}\leq C\sum_{p\leq
x}\sum_{\nu\geq 1}\left(\frac{{r}^{2}}{p}\right)^{\nu}\leq C\sum_{p\leq
x}\frac{1}{p}\leq C\log\log x$ (4.11)

by Mertens’ second theorem [Ten15, Thm. 1.10], and for the same reasons

$\sum_{p\leq x,\,\nu\geq 1}\frac{|\alpha_{d}(p^{\nu})|}{p^{\nu}}=\sum_{p\leq
x}\frac{\alpha_{d}(p)}{p}+C\leq\max_{p}\alpha_{d}(p)\cdot\log\log x+C.$ (4.12)

Thus, we conclude from Theorem 4.2 that

$\sum_{n\leq
x}\alpha_{d}(n)=A_{\alpha_{d}}x\log^{\theta-1}x+O\left(x\log^{\theta-2}x\right),$
(4.13)

where $A_{\alpha_{d}}=\Gamma(\theta)^{-1}\prod_{p}(\sum_{i\geq
0}\alpha(p^{i})/p^{i(d+1)})(1-1/p)^{\theta}$ is positive since $\alpha\geq 0$,
$\alpha(1)=1$.

Using integration by parts and (4.13) we have

$\begin{split}\sum_{n\leq x}\alpha(n)&=\sum_{n\leq
x}n^{d}\alpha_{d}(n)=x^{d}\bigg{(}\sum_{n\leq
x}\alpha_{d}(n)\bigg{)}-\int_{1}^{x}\bigg{(}\sum_{n\leq
t}\alpha_{d}(n)\bigg{)}(t^{d})^{\prime}\,dt\\\
&=A_{\alpha_{d}}x^{d+1}\log^{\theta-1}x-\int_{2}^{x}A_{\alpha_{d}}t\log^{\theta-1}t\cdot
dt^{d-1}\,dt+O\left(x^{d+1}\log^{\theta-2}x+\int_{2}^{x}t^{d}\log^{\theta-2}t\,dt+1\right),\end{split}$
(4.14)

which gives (4.9) with $A_{\alpha}=A_{\alpha_{d}}/(d+1)$ by Lemma 4.1. ∎

######  Lemma 4.4.

Suppose a function $\alpha\colon\mathbb{N}\to\mathbb{R}_{\geq 0}$ satisfies
(1.4)–(1.5) for some $\theta>0$ and $d>-1$. Let $p$ be a prime. We have

$\sum_{\begin{subarray}{c}n\leq x\\\ p\mid
n\end{subarray}}\alpha(n)\leq\frac{C}{p}x^{d+1}\log^{\theta}x.$ (4.15)

If furthermore $p\leq\sqrt{x}$ we have

$\sum_{\begin{subarray}{c}n\leq x\\\ p\mid
n\end{subarray}}\alpha(n)\leq\frac{C}{p}x^{d+1}\log^{\theta-1}x.$ (4.16)

###### Proof.

By multiplicativity of $\alpha$, we have

$\sum_{\begin{subarray}{c}n\leq x\\\ p\mid
n\end{subarray}}\alpha(n)=\sum_{k=1}^{\lfloor\log_{p}x\rfloor}\sum_{\begin{subarray}{c}n\leq
x\\\ p^{k}\mid\mid
n\end{subarray}}\alpha(n)\leq\sum_{k=1}^{\lfloor\log_{p}x\rfloor}\alpha(p^{k})\sum_{\begin{subarray}{c}l=1\\\
p\nmid
n\end{subarray}}^{\left\lfloor\frac{x}{p^{k}}\right\rfloor}\alpha(l)\leq\sum_{k=1}^{\lfloor\log_{p}x\rfloor}\alpha(p^{k})\sum_{l=1}^{\left\lfloor\frac{x}{p^{k}}\right\rfloor}\alpha(l).$
(4.17)

By Corollary 4.3 there exists a constant $C$ such that

$\sum_{n\leq y}\alpha(n)\leq Cy^{d+1}\log^{\theta-1}y\leq
Cy^{d+1}\log^{\theta}y$ (4.18)

for all $y\geq 2$. By (4.17) and (4.18) we have

$\sum_{\begin{subarray}{c}n\leq x\\\ p\mid n\end{subarray}}\alpha(n)\leq
C\sum_{k=1}^{\lfloor\log_{p}x\rfloor}\frac{\alpha(p^{k})}{p^{k(d+1)}}x^{d+1}\log^{\theta}x+\alpha(p^{\lfloor\log_{p}x\rfloor}),$
(4.19)

where the last term comes from the possible contribution of
$k=\lfloor\log_{p}x\rfloor$ in case $\lfloor x/p^{k}\rfloor<2$. From (1.5) and
(4.19) we obtain

$\sum_{\begin{subarray}{c}n\leq x\\\ p\mid n\end{subarray}}\alpha(n)\leq
Cx^{d+1}\log^{\theta}x\sum_{k\geq
1}\left(\frac{{r}}{p}\right)^{k}+Cx^{d}c^{\lfloor\log_{p}x\rfloor}\leq\frac{C}{p}x^{d+1}\log^{\theta}x$
(4.20)

for some ${r}<\sqrt{2}$, and so (4.15) holds. To prove (4.16), we split the
right-hand side of (4.17) into two sums, one for terms with
$p^{k}\leq\sqrt{x}$ and another for the rest. Set
$L:=\lfloor\log_{p}\sqrt{x}\rfloor+1\geq 2$. For $1\leq k<L$, we have
$\log^{\theta-1}(x/p^{k})\leq C\log^{\theta-1}x$, and so Corollary 4.3 gives

$\sum_{l=1}^{\left\lfloor\frac{x}{p^{k}}\right\rfloor}\alpha(l)\leq
C\left(\frac{x}{p^{k}}\right)^{d+1}\log^{\theta-1}x.$ (4.21)

From (1.5) we obtain

$\begin{split}\sum_{k=1}^{L-1}\alpha(p^{k})\sum_{l=1}^{\left\lfloor\frac{x}{p^{k}}\right\rfloor}\alpha(l)\leq
Cx^{d+1}\log^{\theta-1}x\sum_{k=1}^{L-1}\frac{\alpha(p^{k})}{p^{k(d+1)}}\leq
Cx^{d+1}\log^{\theta-1}x\sum_{k=1}^{\infty}\left(\frac{{r}}{p}\right)^{k}\leq\frac{C}{p}x^{d+1}\log^{\theta-1}x.\end{split}$
(4.22)

When $L\leq k\leq\lfloor\log_{p}x\rfloor$ we use the bound

$\sum_{l=1}^{\left\lfloor\frac{x}{p^{k}}\right\rfloor}\alpha(l)\leq
C\left(\frac{x}{p^{k}}\right)^{d+1}\log^{\theta}x$ (4.23)

to obtain, from (1.5) again, that

$\begin{split}\sum_{k=L}^{\lfloor\log_{p}x\rfloor}\alpha(p^{k})\sum_{l=1}^{\left\lfloor\frac{x}{p^{k}}\right\rfloor}\alpha(l)&\leq
Cx^{d+1}\log^{\theta}x\sum_{k=L}^{\infty}\frac{\alpha(p^{k})}{p^{k(d+1)}}\leq
Cx^{d+1}\log^{\theta}x\sum_{k=L}^{\infty}\left(\frac{{r}}{p}\right)^{k}\leq
Cx^{d+1}\log^{\theta}x\left(\frac{{r}}{p}\right)^{L}\\\
&\leq\frac{C}{p}x^{d+1}\log^{\theta}x\left(\frac{{r}}{p}\right)^{L/2}\leq\frac{C}{p}x^{d+1}\log^{\theta}x\left(\frac{\sqrt{2}}{p}\right)^{\log_{p}\sqrt{x}/2}\leq\frac{C}{p}x^{d+1}\log^{\theta-1}x.\end{split}$
(4.24)

From (4.22) and (4.24) we obtain (4.16). ∎

######  Remark 4.5.

Lemma 4.4 gives upper bounds for $\mathbb{P}(p\mid N_{x})$ which are uniform
in $p$. For fixed $p$ we can obtain asymptotics for this probability from
Theorem 2.9, but it will lack uniformity.

######  Lemma 4.6.

Suppose that $\alpha\colon\mathbb{N}\to\mathbb{R}_{\geq 0}$ satisfies (1.4)
with $\theta>0$, $d>-1$. Let $I\subseteq[2,\infty)$ be an interval. Let $M>0$
and $g\colon I\to\mathbb{R}$ be a differentiable function such that

$\left|\frac{g(t)}{\log
t}\right|\leq\frac{M}{t}\quad\mbox{and}\quad\left|\left(\frac{g(t)}{\log
t}\right)^{\prime}\right|\leq\frac{M}{t^{2}}$ (4.25)

for all $t\in I$. Then

$\left|\sum_{p\in I}\frac{\alpha(p)g(p)}{p^{d}}-\theta\int_{I}\frac{g(t)}{\log
t}\,dt\right|=O(M),$ (4.26)

where the implied constant depends only on $\alpha$.

###### Proof.

Write $I=[x,y]$. Using integration by parts and the assumption on $g$ we get
that

$\begin{split}\theta\intop_{x}^{y}\frac{g(t)}{\log t}\,dt=\frac{\theta
g(y)}{\log
y}\intop_{x}^{y}1\,dt-\intop_{x}^{y}\theta(t-x)\left(\frac{g(t)}{\log
t}\right)^{\prime}\,dt=O(M)-\intop_{x}^{y}\theta(t-x)\left(\frac{g(t)}{\log
t}\right)^{\prime}\,dt\end{split}$ (4.27)

and

$\begin{split}\sum_{x\leq p\leq y}\frac{\alpha(p)g(p)}{p^{d}}&=\sum_{x\leq
n\leq y}\mathds{1}_{\left\\{n\mbox{ is prime}\right\\}}\frac{\alpha(n)\log
n}{n^{d}}\cdot\frac{g(n)}{\log n}\\\ &=\frac{g(y)}{\log y}\left(\sum_{x\leq
p\leq y}\frac{\alpha(p)\log p}{p^{d}}\right)-\int_{x}^{y}\left(\sum_{x\leq
p\leq t}\frac{\alpha(p)\log p}{p^{d}}\right)\left(\frac{g(t)}{\log
t}\right)^{\prime}\,dt\\\ &=O(M)-\intop_{x}^{y}\left(\sum_{x\leq p\leq
t}\frac{\alpha(p)\log p}{p^{d}}\right)\left(\frac{g(t)}{\log
t}\right)^{\prime}\,dt,\end{split}$ (4.28)

where in the last equality we bound the first term using the assumptions about
$\alpha$ and $g$. From (4.27), (4.28), the assumptions on $g$ and (1.4) we
obtain

$\left|\sum_{x\leq p\leq
y}\frac{\alpha(p)g(p)}{p^{d}}-\theta\intop_{x}^{y}\frac{g(t)}{\log
t}\,dt\right|\leq
O(M)+C\intop_{x}^{y}\frac{t}{\log^{2}t}\left|\left(\frac{g(t)}{\log
t}\right)^{\prime}\right|\,dt\leq
O(M)+M\intop_{x}^{y}\frac{dt}{t\log^{2}t}=O(M),$ (4.29)

as needed. ∎

###  4.2 Proof of second part of Theorem 1.1

####  4.2.1 Auxiliary lemma

######  Lemma 4.7.

Fix $k\geq 1$. Let $\delta\in(0,1)$ and suppose that $x$ is sufficiently large
in terms of $\delta$. Let $p_{1},\dots,p_{k}$ be distinct primes such that
$p_{1}\cdots p_{k}\leq x^{1-\delta}$ and $p_{i}\geq x^{\delta}$ for all $1\leq
i\leq k$. Given a multiplicative function
$\alpha\colon\mathbb{N}\to\mathbb{R}_{\geq 0}$ satisfying (1.4)–(1.5) for some
$\theta>0$ and $d>-1$, we have

$\sum_{\begin{subarray}{c}n=\lceil\delta x\rceil\\\ p_{1}\cdots
p_{k}|n\end{subarray}}^{x}\alpha(n)\geq
A_{\alpha}(1-3\delta^{d+1})\left(\prod_{i=1}^{k}\frac{\alpha(p_{i})}{p_{i}^{d+1}}\right)x^{d+1}\log^{\theta-1}\left(\frac{x}{p_{1}\cdots
p_{k}}\right),$ (4.30)

where $A_{\alpha}$ is the constant given in Corollary 4.3.

###### Proof.

By the multiplicativity of $\alpha$,

$\sum_{\begin{subarray}{c}n=\lceil\delta x\rceil\\\ p_{1}\cdots
p_{k}|n\end{subarray}}^{x}\alpha(n)\geq\sum_{l=\lceil\frac{\delta
x}{p_{1}\cdots p_{k}}\rceil}^{\lfloor\frac{x}{p_{1}\cdots
p_{k}}\rfloor}\alpha\left(l\cdot p_{1}\cdots
p_{k}\right)\geq\sum_{\begin{subarray}{c}l=\lceil\frac{\delta x}{p_{1}\cdots
p_{k}}\rceil\\\ \forall i\ p_{i}\nmid
l\end{subarray}}^{\lfloor\frac{x}{p_{1}\cdots p_{k}}\rfloor}\alpha\left(l\cdot
p_{1}\cdots
p_{k}\right)=\alpha(p_{1})\cdots\alpha(p_{k})\sum_{\begin{subarray}{c}l=\lceil\frac{\delta
x}{p_{1}\cdots p_{k}}\rceil\\\ \forall i\ p_{i}\nmid
l\end{subarray}}^{\lfloor\frac{x}{p_{1}\cdots p_{k}}\rfloor}\alpha(l).$ (4.31)

For the sum in the right-hand side we have the naive lower bound

$\sum_{\begin{subarray}{c}l=\lceil\frac{\delta x}{p_{1}\cdots p_{k}}\rceil\\\
\forall i\ p_{i}\nmid l\end{subarray}}^{\lfloor\frac{x}{p_{1}\cdots
p_{k}}\rfloor}\alpha(l)\geq\sum_{l=\lceil\frac{\delta x}{p_{1}\cdots
p_{k}}\rceil}^{\lfloor\frac{x}{p_{1}\cdots
p_{k}}\rfloor}\alpha(l)-\sum_{i=1}^{k}\sum_{\begin{subarray}{c}l=1\\\ \
p_{i}|l\end{subarray}}^{\lfloor\frac{x}{p_{1}\cdots p_{k}}\rfloor}\alpha(l).$
(4.32)

The first sum in the right-hand side of (4.32) can be estimated by applying
Corollary 4.3 twice, and the second sum can be bounded from above by (4.15).
We thus obtain

$\begin{split}\sum_{\begin{subarray}{c}l=\lceil\frac{\delta x}{p_{1}\cdots
p_{k}}\rceil\\\ \forall i\ p_{i}\nmid
l\end{subarray}}^{\lfloor\frac{x}{p_{1}\cdots p_{k}}\rfloor}\alpha(l)&\geq
A_{\alpha}\left(1-2\delta^{d+1}\right)\left(\frac{x}{p_{1}\cdots
p_{k}}\right)^{d+1}\log^{\theta-1}\left(\frac{x}{p_{1}\cdots
p_{k}}\right)-Cx^{-\delta}\left(\frac{x}{p_{1}\cdots
p_{k}}\right)^{d+1}\log^{\theta}\left(\frac{x}{p_{1}\cdots p_{k}}\right)\\\
&\geq A_{\alpha}\left(1-3\delta^{d+1}\right)\left(\frac{x}{p_{1}\cdots
p_{k}}\right)^{d+1}\log^{\theta-1}\left(\frac{x}{p_{1}\cdots
p_{k}}\right)\end{split}$ (4.33)

for sufficiently large $x$, as needed. ∎

####  4.2.2 Conclusion of proof

From Corollary 4.3 it follows that

$\frac{\log N_{x}}{\log x}\overset{d}{\longrightarrow}1$ (4.34)

as $x\to\infty$. Thus, it suffices to prove that

$\left(X_{j}\right)_{j=1}^{\infty}:=\left(\frac{\log\mathcal{p}_{j}(N_{x})}{\log
N_{x}}\right)_{j=1}^{\infty}\overset{d}{\longrightarrow}\mathrm{PD}(\theta)$
(4.35)

as $x\to\infty$. By Proposition 3.2, it suffices to prove that

$\left(B_{j}\right)_{j=1}^{\infty}:=\left(\frac{\tilde{X}_{j}}{1-\tilde{X}_{1}-\ldots-\tilde{X}_{j-1}}\right)_{j=1}^{\infty}\overset{d}{\longrightarrow}(Y_{1},Y_{2},\dots)$
(4.36)

as $x\to\infty$, where $Y_{1},Y_{2},\dots$ is a sequence of i.i.d.
$\operatorname{beta}(1,\theta)$ random variables and where
$(\tilde{X}_{j})_{j=1}^{\infty}$ is a sized-biased permutation of the sequence
$(X_{j})_{j=1}^{\infty}$ as defined in §3. To this end, fix $k\geq 1$ and
$0<a_{j}<b_{j}<1$ for any $1\leq j\leq k$. We shall show that

$\liminf_{x\to\infty}\mathbb{P}\left(\forall j\ \ a_{j}\leq B_{j}\leq
b_{j}\right)\geq\mathbb{P}\left(\forall j\ \ a_{j}\leq Y_{j}\leq
b_{j}\right).$ (4.37)

We let $P_{j}$ be the sequence of typical primes of $N_{x}$, that is
$P_{j}:={N_{x}}^{\tilde{X}_{j}}$. We have that

$B_{j}=\frac{\tilde{X}_{j}}{1-\tilde{X}_{1}-\ldots-\tilde{X}_{j-1}}=\frac{\log(P_{j})}{\log\left(\frac{N_{x}}{P_{1}\cdots
P_{j-1}}\right)}$ (4.38)

and therefore

$\mathbb{P}\left(\forall j\ \ a_{j}\leq B_{j}\leq b_{j}\right)=\sum_{n\leq
x}\sum_{p_{1},\dots,p_{k}}\mathbb{P}\left(N_{x}=n,\,\forall j\ \
P_{j}=p_{j}\right),$ (4.39)

where the inner sum is over a sequence of $k$ primes $p_{1},\dots,p_{k}$ such
that for any $1\leq j\leq k$ we have $\left(n/(p_{1}\cdots
p_{j-1})\right)^{a_{j}}\leq p_{j}\leq\left(n/(p_{1}\cdots
p_{j-1})\right)^{b_{j}}$. Let $0<\delta<\min\\{a_{1},\dots
a_{k}\\}\prod_{i=1}^{k}(1-b_{i})$ and put $x_{j}:=x/(p_{1}\cdots p_{j-1})$ for
any $1\leq j\leq k+1$. We have the following lower bound:

$\mathbb{P}\left(\forall j\ \ a_{j}\leq B_{j}\leq
b_{j}\right)\geq\sum_{\lceil\delta x\rceil\leq n\leq
x}\sum_{\begin{subarray}{c}p_{1},\dots,p_{k}\mbox{ distinct}\\\
x_{j}^{a_{j}}\leq p_{j}\leq(\delta x_{j})^{b_{j}},\ p_{j}\mid\mid
n\end{subarray}}\mathbb{P}\left(N_{x}=n,\,\forall j\ \ P_{j}=p_{j}\right).$
(4.40)

By the definition of a sized-biased permutation, when $p_{1},\dots,p_{k}$ are
distinct and $p_{j}\mid\mid n$ we have

$\mathbb{P}\left(N_{x}=n,\,\forall j\ \
P_{j}=p_{j}\right)=\frac{\alpha(n)}{\sum_{m\leq
x}\alpha(m)}\prod_{j=1}^{k}\frac{\log p_{j}}{\log\left(\frac{n}{p_{1}\cdots
p_{j-1}}\right)}\geq\frac{\alpha(n)}{\sum_{m\leq
x}\alpha(m)}\prod_{j=1}^{k}\frac{\log p_{j}}{\log x_{j}}$ (4.41)

for any $n\leq x$. Thus, changing the order of summation in (4.40) we obtain

$\begin{split}\mathbb{P}\left(\forall j\ \ a_{j}\leq B_{j}\leq
b_{j}\right)&\geq\sum_{\begin{subarray}{c}x_{1}^{a_{1}}\leq p_{1}\leq(\delta
x_{1})^{b_{1}}\end{subarray}}\cdots\sum_{\begin{subarray}{c}x_{k}^{a_{k}}\leq
p_{k}\leq(\delta x_{k})^{b_{k}}\\\ p_{k}\notin\\{p_{1},\dots
p_{k-1}\\}\end{subarray}}\sum_{\begin{subarray}{c}\lceil\delta x\rceil\leq
n\leq x\\\ p_{i}\mid\mid n\end{subarray}}\frac{\alpha(n)}{\sum_{m\leq
x}\alpha(m)}\prod_{j=1}^{k}\frac{\log p_{j}}{\log x_{j}}\\\
&\geq\frac{1}{\sum_{m\leq
x}\alpha(m)}\sum_{\begin{subarray}{c}x_{1}^{a_{1}}\leq p_{1}\leq(\delta
x_{1})^{b_{1}}\end{subarray}}\frac{\log p_{1}}{\log
x_{1}}\cdots\\!\\!\sum_{\begin{subarray}{c}x_{k}^{a_{k}}\leq p_{k}\leq(\delta
x_{k})^{b_{k}}\\\ p_{k}\notin\\{p_{1},\dots
p_{k-1}\\}\end{subarray}}\frac{\log p_{k}}{\log
x_{k}}\sum_{\begin{subarray}{c}\lceil\delta x\rceil\leq n\leq x\\\
p_{i}\mid\mid n\end{subarray}}\alpha(n).\end{split}$ (4.42)

Next, we would like to use Lemma 4.7 in order to lower bound the inner sum in
(4.42). We have $x_{j}=p_{j}x_{j+1}\leq(\delta x_{j})^{b_{j}}x_{j+1}\leq
x_{j}^{b_{j}}x_{j+1}$ and therefore, for any $1\leq j\leq k$ we have
$x_{j+1}\geq x_{j}^{1-b_{j}}\geq\cdots\geq x^{\nu}$ where
$\nu:=\prod_{j=1}^{k}(1-b_{j})$. Thus $p_{j}\geq x^{a_{j}\nu}\geq x^{\delta}$
and moreover $p_{1}\cdots p_{k}=x/x_{k+1}\leq x^{1-\nu}\leq x^{1-\delta}$. We
get that the assumptions of Lemma 4.7 hold and therefore using also Corollary
4.3 we obtain

$\begin{split}&\mathbb{P}\left(\forall j\ \ a_{j}\leq B_{j}\leq
b_{j}\right)\geq\frac{1-4\delta^{d+1}}{\log^{\theta-1}x}\sum_{\begin{subarray}{c}x_{1}^{a_{1}}\leq
p_{1}\leq(\delta x_{1})^{b_{1}}\end{subarray}}\frac{\alpha(p_{1})\log
p_{1}}{p_{1}^{d+1}\log
x_{1}}\cdots\\!\\!\sum_{\begin{subarray}{c}x_{k}^{a_{k}}\leq p_{k}\leq(\delta
x_{k})^{b_{k}}\\\ p_{k}\notin\\{p_{1},\dots
p_{k-1}\\}\end{subarray}}\frac{\alpha(p_{k})\log p_{k}}{p_{k}^{d+1}\log
x_{k}}\log^{\theta-1}x_{k+1}\\\
&=\left(1-4\delta^{d+1}\right)\sum_{\begin{subarray}{c}x_{1}^{a_{1}}\leq
p_{1}\leq(\delta x_{1})^{b_{1}}\end{subarray}}\frac{\alpha(p_{1})\log
p_{1}}{p_{1}^{d+1}\log x_{1}}\left(1-\frac{\log p_{1}}{\log
x_{1}}\right)^{\theta-1}\cdots\\!\\!\sum_{\begin{subarray}{c}x_{k}^{a_{k}}\leq
p_{k}\leq(\delta x_{k})^{b_{k}}\\\ p_{k}\notin\\{p_{1},\dots
p_{k-1}\\}\end{subarray}}\frac{\alpha(p_{k})\log p_{k}}{p_{k}^{d+1}\log
x_{k}}\left(1-\frac{\log p_{k}}{\log x_{k}}\right)^{\theta-1}\end{split}$
(4.43)

for sufficiently large $x$. Consider the innermost sum. We define the function
$g=g_{n_{k}}\colon[x_{k}^{a_{k}},(\delta x_{k})^{b_{k}}]\to\mathbb{R}$ by

$g(t):=\frac{\log t}{t}\left(1-\frac{\log t}{\log x_{k}}\right)^{\theta-1}.$
(4.44)

We let the reader check that $g$ satisfies the assumptions of Lemma 4.6 with
$M$ which is independent of $x_{k}$, and so we obtain

$\begin{split}\sum_{\begin{subarray}{c}x_{k}^{a_{k}}\leq p_{k}\leq(\delta
x_{k})^{b_{k}}\\\ p_{k}\notin\\{p_{1},\dots
p_{k-1}\\}\end{subarray}}\frac{\alpha(p_{k})\log p_{k}}{p_{k}^{d+1}\log
x_{k}}\left(1-\frac{\log p_{k}}{\log
x_{k}}\right)^{\theta-1}=O_{\delta}\left(\frac{1}{\log x}\right)+\frac{1}{\log
x_{k}}\sum_{\begin{subarray}{c}x_{k}^{a_{k}}\leq p_{k}\leq(\delta
x_{k})^{b_{k}}\end{subarray}}\frac{\alpha(p_{k})}{p_{k}^{d}}g(p_{k})\\\
=O_{\delta}\left(\frac{1}{\log x}\right)+\frac{1}{\log
x_{k}}\intop_{x_{k}^{a_{k}}}^{(\delta
x_{k})^{b_{k}}}\frac{\theta}{t}\left(1-\frac{\log t}{\log
x_{k}}\right)^{\theta-1}\,dt=O_{\delta}\left(\frac{1}{\log
x}\right)+\intop_{a_{k}}^{b_{k}}\theta(1-y)^{\theta-1}\,dy\end{split}$ (4.45)

where in the last equality we used the change of variables $y=\log t/\log
x_{k}$. Substituting the last estimate into (4.43) we get the same expression
with $k$ replaced by $k-1$. Thus, for sufficiently large $x$ depending on
$\delta$,

$\mathbb{P}\left(\forall j\ \ a_{j}\leq B_{j}\leq
b_{j}\right)\geq(1-5\delta^{d+1})\prod_{j=1}^{k}\intop_{a_{j}}^{b_{j}}\theta(1-y)^{\theta-1}\,dy=(1-5\delta^{d+1})\mathbb{P}\left(\forall
j\ a_{j}\leq Y_{j}\leq b_{j}\right),$ (4.46)

and so

$\liminf_{x\to\infty}\mathbb{P}\left(\forall j\ \ a_{j}\leq B_{j}\leq
b_{j}\right)\geq(1-5\delta^{d+1})\mathbb{P}\left(\forall j\ \ a_{j}\leq
X_{j}\leq b_{j}\right).$ (4.47)

Since $\delta$ is arbitrary it follows that (4.37) holds, as needed. ∎

###  4.3 Proof of first part of Theorem 1.1

####  4.3.1 Preparatory results

We need the following results from probability, which are given as Remarks 1,
2 and 3 in [Bil69].

######  Lemma 4.8.

Let $D_{x}$, $E_{x}$ be random variables defined for any $x\gg 1$.

  1. 1.

If $D_{x}\overset{d}{\longrightarrow}1$ and
$E_{x}\overset{d}{\longrightarrow}0$, then
$U_{x}\overset{d}{\longrightarrow}N(0,1)$ if and only if
$D_{x}U_{x}+E_{x}\overset{d}{\longrightarrow}N(0,1)$.

  2. 2.

Let $X\sim N(0,1)$. If $\mathbb{E}D^{k}_{x}\to\mathbb{E}X^{k}$ for each $k\geq
1$ then $D_{x}\overset{d}{\longrightarrow}N(0,1)$.

  3. 3.

Let $X\sim N(0,1)$. If $D_{x}\overset{d}{\longrightarrow}N(0,1)$ and if
$\sup_{x}\mathbb{E}|D_{x}|^{k+\varepsilon}<\infty$ for some $\varepsilon>0$
then $\mathbb{E}D_{x}^{k}\to\mathbb{E}X^{k}$.

Recall that $\mathcal{P}$ is the set of primes. By definition,

$\omega(N_{x})=\sum_{p\in\mathcal{P}}\mathds{1}_{\\{p\mid N_{x}\\}}.$ (4.48)

######  Lemma 4.9.

We have $\mathbb{E}\left|\Omega(N_{x})-\omega(N_{x})\right|=O(1)$.

###### Proof.

Consider the multiplicative function
$\widetilde{\alpha}(n):=\alpha(n)t^{\Omega(n)-\omega(n)}$, where we choose
$1<t<\sqrt{2}/{r}$, so that $\widetilde{\alpha}$ will still satisfy
(1.4)–(1.5) with the same $\theta$ and $d$, and ${r}$ replaced with ${r}t$.
Applying Corollary 4.3 with $\alpha$ and $\widetilde{\alpha}$, we obtain that

$\mathbb{E}t^{\Omega(N_{x})-\omega(N_{x})}=\frac{\sum_{n\leq
x}\widetilde{\alpha}(n)}{\sum_{n\leq
x}\alpha(n)}=\frac{A_{\widetilde{\alpha}}}{A_{\alpha}}\left(1+O\left(\frac{1}{\log
x}\right)\right)=O_{t}(1)$ (4.49)

which, by Jensen’s inequality for instance, implies that
$\mathbb{E}\left|\Omega(N_{x})-\omega(N_{x})\right|$ is bounded as
$x\to\infty$. ∎

For $x\gg 1$, we define the set

$\mathcal{P}_{x}:=\left\\{p\in\mathcal{P}:\ \log^{2}x\leq
p\leq\exp\left(\exp\left(\log\log x-\log^{\frac{1}{3}}\log
x\right)\right)\right\\}.$ (4.50)

For each prime $p\gg 1$, define a Bernoulli random variable $X_{p}$ such that
$\mathbb{P}(X_{p}=1)=\alpha(p)/p^{d+1}$ (for $p\gg 1$, this is in $[0,1]$) and
such that the different $X_{p}$-s are independent. Define $\sigma_{x}$ by

$\sigma^{2}_{x}:=\sum_{p\in\mathcal{P}_{x}}\frac{\alpha(p)}{p^{d+1}}\left(1-\frac{\alpha(p)}{p^{d+1}}\right).$
(4.51)

We define the random variables

$A_{x}:=\frac{\omega(N_{x})-\theta\log\log x}{\sqrt{\theta\log\log x}},\quad
B_{x}:=\frac{\sum_{p\in\mathcal{P}_{x}}\left(\mathds{1}_{\\{p\mid
N_{x}\\}}-\frac{\alpha(p)}{p^{d+1}}\right)}{\sigma_{x}},\quad
C_{x}:=\frac{\sum_{p\in\mathcal{P}_{x}}\left(X_{p}-\frac{\alpha(p)}{p^{d+1}}\right)}{\sigma_{x}}.$
(4.52)

######  Lemma 4.10.

We have $\sum_{p\in\mathcal{P}_{x}}\alpha(p)/p^{d+1}=\theta\log\log
x+O(\log^{\frac{1}{3}}\log x)$, as well as $\sigma_{x}^{2}=\theta\log\log
x+O(\log^{\frac{1}{3}}\log x)$.

###### Proof.

Applying Lemma 4.6 with $g(t)=1/t$ and the interval
$[\log^{2}x,\exp(\exp(\log\log x-\log^{1/3}\log x))]$, we find that
$\sum_{p\in\mathcal{P}_{x}}\alpha(p)/p^{d+1}=\theta\log\log
x+O(\log^{\frac{1}{3}}\log x)$. Since
$\sum_{p\in\mathcal{P}}\alpha^{2}(p)/p^{2(d+1)}\leq\sum_{p\in\mathcal{P}}C/p^{2}\leq
C$, the estimate for $\sigma_{x}^{2}$ follows from the first estimate. ∎

######  Lemma 4.11.

For each integer $k\geq 1$, we have $\sup_{x}|\mathbb{E}C_{x}^{k}|<\infty$. In
particular, $\sup_{x}\mathbb{E}|C_{x}|^{2k}<\infty$.

###### Proof.

Let $Y_{p}:=X_{p}-\alpha(p)/p^{d+1}$. We have the following expansion:

$\mathbb{E}C_{x}^{k}=\frac{1}{\sigma_{x}^{k}}\sum_{p_{1},\dots,p_{k}\in\mathcal{P}_{x}}\mathbb{E}\left[Y_{p_{1}}\cdots
Y_{p_{k}}\right]=\frac{1}{\sigma_{x}^{k}}\sum_{m=1}^{k}\sum_{\begin{subarray}{c}\
l_{1},\dots,l_{m}\geq 1\\\ \sum
l_{i}=k\end{subarray}}\binom{k}{l_{1},\dots,l_{m}}S(l_{1},\dots,l_{m}),$
(4.53)

where

$S(l_{1},\dots,l_{m}):=\sum_{\begin{subarray}{c}q_{1},\dots,q_{m}\in\mathcal{P}_{x}\\\
q_{1}<\cdots<q_{m}\end{subarray}}\mathbb{E}\left[Y_{q_{1}}^{l_{1}}\right]\mathbb{E}\left[Y_{q_{2}}^{l_{2}}\right]\cdots\mathbb{E}\left[Y_{q_{m}}^{l_{m}}\right].$
(4.54)

As we have $\mathbb{E}Y_{p}=0$, it follows that $S(l_{1},\dots,l_{m})$
vanishes if $l_{i}=1$ for some $i$, and so we may restrict the summation in
(4.53) to $l_{i}\geq 2$. Since $|Y_{p}|\leq 1$ (for $p\gg 1$), it follows that
if $l_{i}\geq 2$ then
$|\mathbb{E}\left[Y_{q_{i}}^{l_{i}}\right]|\leq\mathbb{E}\left[Y_{q_{i}}^{2}\right]$,
so that

$S(l_{1},\dots,l_{m})\leq\left(\sum_{p\in\mathcal{P}_{x}}\mathbb{E}\left[Y_{p}^{2}\right]\right)^{m}=\sigma_{x}^{2m}.$
(4.55)

As $\sum_{i=1}^{m}l_{i}=k$ and $l_{i}\geq 2$, it follows that $2m\leq k$. For
$x\gg 1$ we have $\sigma_{x}\geq 1$ by Lemma 4.10, and so

$\left|\mathbb{E}C_{x}^{k}\right|\leq\sum_{m=1}^{k}\frac{\sigma_{x}^{2m}}{\sigma_{x}^{k}}\sum_{\begin{subarray}{c}\
l_{1},\dots,l_{m}\geq 2\\\ \sum
l_{i}=k\end{subarray}}\binom{k}{l_{1},\dots,l_{m}}\leq\sum_{m=1}^{k}\sum_{\begin{subarray}{c}\
l_{1},\dots,l_{m}\geq 2\\\ \sum
l_{i}=k\end{subarray}}\binom{k}{l_{1},\dots,l_{m}}$ (4.56)

as needed. ∎

######  Lemma 4.12.

Fix $m\geq 1$. Let $q_{1},\dots,q_{m}\in\mathcal{P}_{x}$ be distinct primes.
We have for $x\gg 1$

$\left|\mathbb{P}\left(\forall j,\ q_{j}\mid
N_{x}\right)-\prod_{j=1}^{m}\frac{\alpha(q_{j})}{q_{j}^{d+1}}\right|\leq
C_{m}\exp\left(-\log^{\frac{1}{3}}\log
x\right)\prod_{j=1}^{m}\frac{1}{q_{j}}.$ (4.57)

###### Proof.

By multiplicativity of $\alpha$ we have

$\sum_{\begin{subarray}{c}n\leq x\\\ q_{1}\cdots q_{m}\mid
n\end{subarray}}\alpha(n)=\sum_{l_{1}=1}^{\infty}\cdots\sum_{l_{m}=1}^{\infty}\sum_{\begin{subarray}{c}n\leq
x\\\ \forall j,\ q_{j}^{l_{j}}\mid\mid
n\end{subarray}}\alpha(n)=\sum_{l_{1}=1}^{\infty}\cdots\sum_{l_{m}=1}^{\infty}\prod_{j=1}^{m}\alpha(q_{j}^{l_{j}})\sum_{\begin{subarray}{c}l=1\\\
\forall j,\ q_{j}\nmid l\end{subarray}}^{\lfloor\frac{x}{q_{1}^{l_{1}}\cdots
q_{m}^{l_{m}}}\rfloor}\alpha(l).$ (4.58)

The term corresponding to $(l_{1},\ldots,l_{m})=(1,\ldots,1)$ in (4.58) may be
estimated as

$\begin{split}\left|\prod_{j=1}^{m}\alpha(q_{j})\sum_{\begin{subarray}{c}l=1\\\
\forall j,\ q_{j}\nmid l\end{subarray}}^{\lfloor\frac{x}{q_{1}\cdots
q_{m}}\rfloor}\alpha(l)-\prod_{j=1}^{m}\alpha(q_{j})\sum_{l=1}^{\lfloor\frac{x}{q_{1}\cdots
q_{m}}\rfloor}\alpha(l)\right|&\leq\prod_{j=1}^{m}\alpha(q_{j})\sum_{k=1}^{m}\sum_{\begin{subarray}{c}l=1\\\
\ q_{k}\mid l\end{subarray}}^{\lfloor\frac{x}{q_{1}\cdots
q_{m}}\rfloor}\alpha(l)\\\ \leq
Cx^{d+1}\log^{\theta-1}x&\prod_{j=1}^{m}\frac{\alpha(q_{j})}{q_{j}^{d+1}}\sum_{k=1}^{m}\frac{1}{q_{k}}\leq
C_{m}x^{d+1}\log^{\theta-3}x\prod_{j=1}^{m}\frac{1}{q_{j}},\end{split}$ (4.59)

where in the second inequality we used (4.16), which is applicable when $x\gg
1$, and (1.5). Using Corollary 4.3 we get

$\begin{split}\prod_{j=1}^{m}\alpha(q_{j})\sum_{l=1}^{\lfloor\frac{x}{q_{1}\cdots
q_{m}}\rfloor}\alpha(l)&=\left(1+O\left(\frac{1}{\log
x}\right)\right)A_{\alpha}x^{d+1}\log^{\theta-1}\left(\frac{x}{q_{1}\cdots
q_{m}}\right)\prod_{j=1}^{m}\frac{\alpha(q_{j})}{q_{j}^{d+1}}\\\
&=\left(1+O\left(\frac{\log(q_{1}\cdots q_{m})}{\log
x}\right)\right)A_{\alpha}x^{d+1}\log^{\theta-1}x\prod_{j=1}^{m}\frac{\alpha(q_{j})}{q_{j}^{d+1}}\\\
&=\left(1+O_{m}\left(\exp\left(-\log^{\frac{1}{3}}\log
x\right)\right)\right)A_{\alpha}x^{d+1}\log^{\theta-1}x\prod_{j=1}^{m}\frac{\alpha(q_{j})}{q_{j}^{d+1}}.\end{split}$
(4.60)

The contribution of terms with $(l_{1},\dots,l_{m})\neq(1,\dots,1)$ to (4.58)
may be bounded using Corollary 4.3 as follows:

$\begin{split}\sum_{(l_{1},\dots,l_{m})\neq(1,\dots,1)}&\prod_{j=1}^{m}\alpha(q_{j}^{l_{j}})\sum_{\begin{subarray}{c}l=1\\\
\forall j,\ q_{j}\nmid l\end{subarray}}^{\lfloor\frac{x}{q_{1}^{l_{1}}\cdots
q_{m}^{l_{m}}}\rfloor}\alpha(l)\leq
Cx^{d+1}\log^{\theta}x\sum_{(l_{1},\dots,l_{m})\neq(1,\dots,1)}\prod_{j=1}^{m}\frac{\alpha(q_{j}^{l_{j}})}{q_{j}^{(d+1)l_{j}}}\\\
&\leq
Cx^{d+1}\log^{\theta}x\sum_{j=1}^{m}\left(\sum_{l_{j}=2}^{\infty}\frac{\alpha(q_{j}^{l_{j}})}{q_{j}^{(d+1)l_{j}}}\right)\prod_{\begin{subarray}{c}i=1\\\
i\neq
j\end{subarray}}^{m}\left(\sum_{l_{i}=1}^{\infty}\frac{\alpha(q_{i}^{l_{i}})}{q_{i}^{(d+1)l_{i}}}\right)\\\
&\leq
C_{m}x^{d+1}\log^{\theta}x\sum_{j=1}^{m}\frac{1}{q_{j}^{2}}\prod_{\begin{subarray}{c}i=1\\\
i\neq j\end{subarray}}^{m}\frac{1}{q_{i}}\leq\frac{C_{m}}{\log
x}x^{d+1}\log^{\theta-1}x\prod_{j=1}^{m}\frac{1}{q_{j}}.\end{split}$ (4.61)

To conclude the proof, we observe that $\mathbb{P}\left(\forall j,\ q_{j}\mid
N_{x}\right)=\sum_{\begin{subarray}{c}n\leq x\\\ q_{1}\cdots q_{m}\mid
n\end{subarray}}\alpha(n)/S(x)$, which combined with (4.58), (4.59), (4.60),
(4.61) and Corollary 4.3 gives the desired bound. ∎

######  Lemma 4.13.

For each integer $k\geq 1$, we have
$\mathbb{E}B_{x}^{k}-\mathbb{E}C_{x}^{k}\to 0$.

###### Proof.

As in (4.53), we have

$\mathbb{E}\bigg{(}\sum_{p\in\mathcal{P}_{x}}X_{p}\bigg{)}^{k}=\sum_{m=1}^{k}\sum_{\begin{subarray}{c}\
l_{1},\dots,l_{m}\geq 1\\\ \sum
l_{i}=k\end{subarray}}\binom{k}{l_{1},\dots,l_{m}}\sum_{\begin{subarray}{c}q_{1},\dots,q_{m}\in\mathcal{P}_{x}\\\
q_{1}<\cdots<q_{m}\end{subarray}}\mathbb{E}\left[X_{q_{1}}^{l_{1}}X_{q_{2}}^{l_{2}}\cdots
X_{q_{m}}^{l_{m}}\right].$ (4.62)

As $X_{p}$ assumes the values $1$ and $0$ only, we have
$X_{q_{i}}^{l_{i}}=X_{q_{i}}$ and then $\mathbb{E}[X_{q_{1}}^{l_{1}}\cdots
x_{q_{m}}^{l_{m}}]=\prod_{j=1}^{m}\frac{\alpha(q_{j})}{q_{j}^{d+1}}$, so that

$\mathbb{E}\bigg{(}\sum_{p\in\mathcal{P}_{x}}X_{p}\bigg{)}^{k}=\sum_{m=1}^{k}\sum_{\begin{subarray}{c}\
l_{1},\dots,l_{m}\geq 1\\\ \sum
l_{i}=k\end{subarray}}\binom{k}{l_{1},\dots,l_{m}}\sum_{\begin{subarray}{c}q_{1},\dots,q_{m}\in\mathcal{P}_{x}\\\
q_{1}<\cdots<q_{m}\end{subarray}}\prod_{j=1}^{m}\frac{\alpha(q_{j})}{q_{j}^{d+1}}.$
(4.63)

Similarly,

$\mathbb{E}\bigg{(}\sum_{p\in\mathcal{P}_{x}}\mathds{1}_{\\{p\mid
N_{x}\\}}\bigg{)}^{k}=\sum_{m=1}^{k}\sum_{\begin{subarray}{c}\
l_{1},\dots,l_{m}\geq 1\\\ \sum
l_{i}=k\end{subarray}}\binom{k}{l_{1},\dots,l_{m}}\sum_{\begin{subarray}{c}q_{1},\dots,q_{m}\in\mathcal{P}_{x}\\\
q_{1}<\cdots<q_{m}\end{subarray}}\mathbb{P}\left(\forall j,\ q_{j}\mid
N_{x}\right).$ (4.64)

By Lemma 4.12, we obtain

$\begin{split}\left|\mathbb{E}\bigg{(}\sum_{p\in\mathcal{P}_{x}}\mathds{1}_{\\{p\mid
N_{x}\\}}\bigg{)}^{k}-\mathbb{E}\bigg{(}\sum_{p\in\mathcal{P}_{x}}X_{p}\bigg{)}^{k}\right|&\leq
C_{m}\exp\left(-\log^{\frac{1}{3}}\log
x\right)\sum_{m=1}^{k}\sum_{\begin{subarray}{c}\ l_{1},\dots,l_{m}\geq 1\\\
\sum
l_{i}=k\end{subarray}}\binom{k}{l_{1},\dots,l_{m}}\sum_{\begin{subarray}{c}q_{1},\dots,q_{m}\in\mathcal{P}_{x}\\\
q_{1}<\cdots<q_{m}\end{subarray}}\prod_{j=1}^{m}\frac{1}{q_{j}}\\\ &\leq
C_{m}\exp\left(-\log^{\frac{1}{3}}\log
x\right)\sum_{m=1}^{k}\sum_{\begin{subarray}{c}\ l_{1},\dots,l_{m}\geq 1\\\
\sum l_{i}=k\end{subarray}}\binom{k}{l_{1},\dots,l_{m}}\left(\sum_{p\leq
x}\frac{1}{p}\right)^{m}\\\ &\leq C_{k}\exp\left(-\log^{\frac{1}{3}}\log
x\right)(\log\log x)^{k}.\end{split}$ (4.65)

By the binomial theorem, (4.65) and Lemma 4.10, we have

$\begin{split}\left|\mathbb{E}B_{x}^{k}-\mathbb{E}C_{x}^{k}\right|&\leq\sigma_{x}^{-k}\sum_{i=0}^{k}\binom{k}{i}\left|\mathbb{E}\bigg{(}\sum_{p\in\mathcal{P}_{x}}\mathds{1}_{\\{p\mid
N_{x}\\}}\bigg{)}^{i}-\mathbb{E}\bigg{(}\sum_{p\in\mathcal{P}_{x}}X_{p}\bigg{)}^{i}\right|\left(\sum_{p\in\mathcal{P}_{x}}\frac{\alpha(p)}{p^{d+1}}\right)^{k-i}\\\
&\leq\sigma_{x}^{-k}C_{k}\exp\left(-\log^{\frac{1}{3}}\log
x\right)\sum_{i=0}^{k}\binom{k}{i}(\log\log x)^{i}(\log\log x)^{k-i}=(\log\log
x)^{k/2}\exp\left(-\log^{\frac{1}{3}}\log x\right)\to 0\end{split}$ (4.66)

as needed. ∎

####  4.3.2 Conclusion of proof

By Lemma 4.9 and the first part of Lemma 4.8 with $D_{x}=1$ and
$E_{x}=(\Omega(N_{x})-\omega(N_{x}))/\sqrt{\theta\log\log x}$, it follows that
(1.6) holds if and only if $A_{x}\overset{d}{\longrightarrow}N(0,1)$. Now let
$D_{x}=\sqrt{\theta\log\log x}/\sigma_{x}$ and

$E_{x}=\frac{\theta\log\log
x-\sum_{p\in\mathcal{P}_{x}}\frac{\alpha(p)}{p^{d+1}}}{\sigma_{x}}-\frac{\sum_{p\in\mathcal{P}\setminus\mathcal{P}_{x}}\mathds{1}_{p\mid
N_{x}}}{\sigma_{x}}.$ (4.67)

We have $D_{x}\overset{d}{\longrightarrow}1$. Moreover, in the sum in the
second fraction in (4.67), there can be at most one non-zero term with $p$
greater than $\sqrt{x}$, and so we may use Lemma 4.10 and (4.16) to obtain
that

$\mathbb{E}|E_{x}|\leq C(\log\log
x)^{-\frac{1}{6}}+\frac{1}{\sigma_{x}}\sum_{\begin{subarray}{c}p\in\mathcal{P}\setminus\mathcal{P}_{x}\\\
p\leq\sqrt{x}\end{subarray}}\mathbb{P}(p\mid N_{x})+\frac{1}{\sigma_{x}}\leq
o(1)+\frac{C}{\sigma_{x}}\sum_{\begin{subarray}{c}p\in\mathcal{P}\setminus\mathcal{P}_{x}\\\
p\leq\sqrt{x}\end{subarray}}\frac{1}{p}\to 0,$ (4.68)

where the last expression tends to $0$ by Lemma 4.6 with $g(t)=1/t$ and
$\alpha\equiv 1$. Hence $E_{x}\overset{d}{\longrightarrow}0$. Since
$B_{x}=D_{x}A_{x}+E_{x}$, it follows by Lemma 4.10 and the first part of Lemma
4.8 that $A_{x}\overset{d}{\longrightarrow}N(0,1)$ if and only if
$B_{x}\overset{d}{\longrightarrow}N(0,1)$. By the second part of Lemma 4.8, to
establish $B_{x}\overset{d}{\longrightarrow}N(0,1)$ it suffices to show that
$\mathbb{E}B_{x}^{k}\to\mathbb{E}X^{k}$ for each $k$, where $X\sim N(0,1)$. By
Lemma 4.13 this is equivalent to $\mathbb{E}C_{x}^{k}\to\mathbb{E}X^{k}$.
Since the random variables $X_{p}-\alpha(p)/p^{d+1}$ are uniformly bounded as
$p$ varies, and since the denominator of $C_{x}$ tends to infinity by Lemma
4.10, we have $C_{x}\overset{d}{\longrightarrow}N(0,1)$ by the Lindeberg-
Feller theorem (also known as Central Limit Theorem for triangular arrays)
[Pet95, Thm. 4.7]. Thus, the moments of $C_{x}$ converge to those of $X$ by
Lemma 4.11 and the last part of Lemma 4.8. ∎

##  5 Polynomially-growing weights

######  Lemma 5.1.

Fix a function $f\colon\mathcal{P}\to\mathbb{R}_{>0}$ on primes such that
$\log f(p)=o(\log p)$, and let

$G_{f}(s)=\sum_{p}\frac{f(p)}{p^{s}},\quad\Re s>1.$ (5.1)

Let $\alpha\colon\mathbb{N}\to\mathbb{R}_{\geq 0}$ be a multiplicative
function satisfying

$\begin{split}&\text{(I) }\alpha(p)=f(p)+O\left(\log^{-2}p\right),\\\
&\text{(II) }\sum_{k\geq
2}\frac{k\alpha(p^{k})}{p^{k}}=O\left(\frac{1}{p\log^{2}p}\right),\end{split}$
(5.2)

and define

$F(s)=\sum_{n\geq 1}\frac{\alpha(n)}{n^{s}},$ (5.3)

the Dirichlet series of $\alpha$. For $\Re s>1$ we have

$F(s)=\varphi(s)e^{G_{f}(s)}$ (5.4)

where $\varphi$ is differentiable, bounded and has bounded derivative on $\Re
s\geq 1$.

###### Proof.

By multiplicativity of $\alpha$, we may write

$F(s)=\prod_{p}\left(\sum_{k=0}^{\infty}\frac{\alpha(p^{k})}{p^{ks}}\right)=e^{G_{f}(s)}\prod_{p}\left(1+E_{p}(s)\right),$
(5.5)

where

$E_{p}(s)=-1+\exp\left[\frac{-f(p)}{p^{s}}\right]\cdot\sum_{k=0}^{\infty}\frac{\alpha(p^{k})}{p^{ks}}.$
(5.6)

Using (LABEL:eq:alpha_cond_gen) and the triangle inequality, we obtain

$\begin{split}\left|\exp\left[\frac{f(p)}{p^{s}}\right]-\sum_{k=0}^{\infty}\frac{\alpha(p^{k})}{p^{ks}}\right|&\leq\left|\exp\left[\frac{f(p)}{p^{s}}\right]-\left(1+\frac{f(p)}{p^{s}}\right)\right|+\left|\frac{\alpha(p)}{p^{s}}-\frac{f(p)}{p^{s}}\right|+\sum_{k=2}^{\infty}\left|\frac{\alpha(p^{k})}{p^{ks}}\right|\\\
&\leq\frac{Cf^{2}(p)}{p^{2}}+\frac{C}{p\log^{2}p}+\sum_{k=2}^{\infty}\frac{\alpha(p^{k})}{p^{k}}\leq\frac{C}{p\log^{2}p}\end{split}$
(5.7)

for $\Re s\geq 1$. Thus, $|E_{p}(s)|\leq C/(p\log^{2}p)$ for $\Re s\geq 1$. We
turn to bound the derivative of $E_{p}$. We have

$E_{p}^{\prime}(s)=\log
p\cdot\exp\left[-\frac{f(p)}{p^{s}}\right]\left(\frac{f(p)}{p^{s}}\sum_{k=0}^{\infty}\frac{\alpha(p^{k})}{p^{ks}}-\sum_{k=1}^{\infty}\frac{k\alpha(p^{k})}{p^{ks}}\right)$
(5.8)

and therefore

$\left|E_{p}^{\prime}(s)\right|\leq C\log
p\left(\left|\frac{f(p)}{p^{s}}-\frac{\alpha(p)}{p^{s}}\right|+\frac{f(p)}{p}\sum_{k=1}^{\infty}\frac{\alpha(p^{k})}{p^{k}}+\sum_{k=2}^{\infty}\frac{k\alpha(p^{k})}{p^{k}}\right)\leq\frac{C}{p\log
p}.$ (5.9)

Let $p_{0}>0$ so that for $p\geq p_{0}$ we have $|E_{p}(s)|\leq 1/2$ for any
$s$ with $\Re s\geq 1$. We have that

$F(s)=e^{G_{f}(s)}\cdot\psi_{1}(s)\cdot\psi_{2}(s),$ (5.10)

where

$\psi_{1}(s):=\prod_{p<p_{0}}\left(1+E_{p}(s)\right),\quad\psi_{2}(s):=\prod_{p\geq
p_{0}}\left(1+E_{p}(s)\right)=\exp\left[\sum_{p\geq
p_{0}}\log\left(1+E_{p}(s)\right)\right]$ (5.11)

and $\log z$ is the principal branch of the logarithm. The function $\psi_{1}$
is trivially differentiable, bounded and has bounded derivative on $\Re s\geq
1$. As for $\psi_{2}$, observe that

$\left|\log(\psi_{2}(s))\right|=|\sum_{p\geq
p_{0}}\log\left(1+E_{p}(s)\right)|\leq\sum_{p\geq
p_{0}}|E_{p}(s)|\leq\sum_{p\geq p_{0}}\frac{C}{p\log^{2}p}<\infty$ (5.12)

and

$\left|\frac{d}{ds}\log(\psi_{2}(s))\right|\leq\sum_{p\geq
p_{0}}\left|\frac{E_{p}^{\prime}(s)}{1+E_{p}(s)}\right|\leq 2\sum_{p\geq
p_{0}}\left|E_{p}^{\prime}(s)\right|\leq\sum_{p\geq p_{0}}\frac{C}{p\log
p}<\infty.$ (5.13)

It follows that $\psi_{2}$ is also differentiable, bounded and has bounded
derivative on $\Re s\geq 1$. Thus, the same is true for
$\varphi:=\psi_{1}\cdot\psi_{2}$, as needed. ∎

######  Lemma 5.2.

Let $\alpha\colon\mathbb{N}\to\mathbb{R}_{\geq 0}$ be a multiplicative
function satisfying (2.10) and

$S(y/h)h\leq C\cdot S(y)$ (5.14)

for all $y>0$, $h\geq 2$. Then
$|\mathbb{E}\Omega(N_{x})-\mathbb{E}\omega(N_{x})|\leq C$.

###### Proof.

Writing $\omega(N_{x})$ as $\sum_{p\leq x}\mathds{1}_{p\mid N_{x}}$ and
$\Omega(N_{x})$ as $\sum_{p\leq x}\sum_{k\geq 1}\mathds{1}_{p^{k}\mid N_{x}}$,
we have

$\begin{split}0\leq\mathbb{E}(\Omega(N_{x})-\omega(N_{x}))&=\sum_{\begin{subarray}{c}k\geq
2\\\ p\end{subarray}}\mathbb{P}(p^{k}\mid
N_{x})\leq\sum_{\begin{subarray}{c}k\geq 2\\\ p\end{subarray}}\sum_{i\geq
0}\alpha(p^{k+i})\frac{S\big{(}x/p^{k+i}\big{)}}{S(x)}\\\ &\leq
C\sum_{\begin{subarray}{c}k\geq 2\\\ p\end{subarray}}\sum_{i\geq
0}\frac{\alpha(p^{k+i})}{p^{k+i}}\leq C\sum_{p}\sum_{k\geq
2}\frac{k\alpha(p^{k})}{p^{k}}\leq C\sum_{p}\frac{1}{p\log^{2}p}\leq
C,\end{split}$ (5.15)

as needed. ∎

Fix $\gamma>0$ and define the Dirichlet series

$G(s)=\sum_{p}\frac{\log^{\gamma}p}{p^{s}},\quad\Re s>1.$ (5.16)

######  Lemma 5.3.

Fix a non-negative integer $k$. We have

$G^{(k)}(s)=(-1)^{k}\frac{\Gamma(\gamma+k)}{(s-1)^{\gamma+k}}+B_{k}+O(|s-1||s|)$
(5.17)

for $\Re s>1$, where $B_{k}$ is a real constant depending on $\gamma$ and $k$.
Here $(s-1)^{\gamma}=\exp(\gamma\log(s-1))$ is defined using the natural
branch of the logarithm.

###### Proof.

We start with the case $k=0$. Let $E(t)=(\sum_{p\leq t}\log p)-t$ be the error
term in the prime number theorem. Using integration by parts we obtain, for
$\Re s>1$, that

$\begin{split}\sum_{p}\frac{\log^{\gamma}p}{p^{s}}&=\sum_{n=2}^{\infty}\mathds{1}_{\\{n\mbox{
is prime}\\}}\log
n\cdot\frac{\log^{\gamma-1}n}{n^{s}}=-\intop_{2}^{\infty}\bigg{(}\sum_{p\leq
t}\log p\bigg{)}\left(\frac{\log^{\gamma-1}t}{t^{s}}\right)^{\prime}\,dt\\\
&=\intop_{2}^{\infty}\left(t+E(t)\right)\frac{s\log^{\gamma-1}t-(\gamma-1)\log^{\gamma-2}t}{t^{s+1}}\,dt=\psi_{1}(s)+\psi_{2}(s),\end{split}$
(5.18)

where

$\psi_{1}(s):=\intop_{2}^{\infty}E(t)\frac{s\log^{\gamma-1}t-(\gamma-1)\log^{\gamma-2}t}{t^{s+1}}\,dt,\quad\psi_{2}(s):=\intop_{2}^{\infty}\frac{s\log^{\gamma-1}t-(\gamma-1)\log^{\gamma-2}t}{t^{s}}\,dt.$
(5.19)

When $\Re s\geq 1$ we may bound $\psi_{1}$ using (1.14) as follows:

$\left|\psi_{1}(s)\right|\leq C|s|\intop_{2}^{\infty}\left|e^{-a\sqrt{\log
t}}\frac{\log^{\gamma-1}t}{t^{s}}\right|\,dt\leq
C|s|\intop_{2}^{\infty}e^{-a\sqrt{\log t}}\frac{\log^{\gamma-1}t}{t}\,dt\leq
C|s|\intop_{0}^{\infty}e^{-a\sqrt{y}}y^{\gamma-1}\,dy\leq C|s|,$ (5.20)

and similarly $\psi_{1}^{\prime}$ may be bounded by

$\left|\psi_{1}^{\prime}(s)\right|\leq
C|s|\intop_{2}^{\infty}\left|e^{-a\sqrt{\log
t}}\frac{\log^{\gamma}t}{t^{s}}\right|\,dt\leq
C|s|\intop_{0}^{\infty}e^{-a\sqrt{y}}y^{\gamma}\,dy\leq C|s|,$ (5.21)

and so

$\psi_{1}(s)=\psi_{1}(1)+O(|s-1||s|)$ (5.22)

in $\Re s\geq 1$. We turn to estimate $\psi_{2}$. Using integration by parts
we obtain

$\intop_{2}^{\infty}\frac{\log^{\gamma-1}t}{t^{s}}\,dt=\frac{\log^{\gamma-1}t}{(1-s)t^{s-1}}\
\bigg{|}_{t=2}^{t=\infty}-\int_{2}^{\infty}\frac{(\gamma-1)\log^{\gamma-2}t}{(1-s)t^{s}}\,dt=\frac{\log^{\gamma-1}2}{(s-1)2^{s-1}}+\frac{1}{s-1}(\int_{2}^{\infty}\frac{s\log^{\gamma-1}t}{t^{s}}\,dt-\psi_{2}(s))$
(5.23)

for $\Re s>1$, so that

$\psi_{2}(s)=\intop_{2}^{\infty}\frac{\log^{\gamma-1}t}{t^{s}}\,dt+\frac{\log^{\gamma-1}2}{2^{s-1}}.$
(5.24)

Setting

$\psi_{3}(s):=\intop_{1}^{2}\frac{\log^{\gamma-1}t}{t^{s}}\,dt$ (5.25)

we obtain

$\psi_{2}(s)=\intop_{1}^{\infty}\frac{\log^{\gamma-1}t}{t^{s}}\,dt-\psi_{3}(s)+\log^{\gamma-1}2+O(|s-1|)=\intop_{1}^{\infty}\frac{\log^{\gamma-1}t}{t^{s}}\,dt-\psi_{3}(1)+\log^{\gamma-1}2+O(|s-1|),$
(5.26)

where in the last equality we used that
$|\psi_{3}(s)|,|\psi_{3}^{\prime}(s)|\leq C$ for $\Re s\geq 1$. In order to
compute the integral in the right-hand side of (5.26) we perform the change of
variables $w=(s-1)\log t$, obtaining

$\intop_{1}^{\infty}\frac{\log^{\gamma-1}t}{t^{s}}\,dt=\frac{1}{(s-1)^{\gamma}}\intop_{0}^{(s-1)\infty}w^{\gamma-1}e^{-w}\,dw.$
(5.27)

Letting $C_{R}$ be the circular contour from $R$ to $R\frac{s-1}{|s-1|}$, we
compute that $\lim_{R\to\infty}\int_{C_{R}}w^{\gamma-1}e^{-w}\,dw=0$ when $\Re
s>1$. Thus, we may deform the contour $\\{(s-1)w:w\geq 0\\}$ in (5.27) to the
positive real line, obtaining

$\intop_{1}^{\infty}\frac{\log^{\gamma-1}t}{t^{s}}\,dt=\frac{1}{(s-1)^{\gamma}}\intop_{0}^{\infty}w^{\gamma-1}e^{-w}\,dw=\frac{\Gamma(\gamma)}{(s-1)^{\gamma}}.$
(5.28)

Substituting (5.28) into (5.26) and then substituting (5.22) and (5.26) into
(5.18), we obtain (5.17) with $k=0$ and

$B_{0}=\log^{\gamma-1}2+\psi_{1}(1)-\psi_{3}(1).$ (5.29)

Next we consider the case $k\geq 1$. We have

$G^{(k)}(s)=(-1)^{k}\sum_{p}\frac{\log^{\gamma+k}p}{p^{s}}.$ (5.30)

Hence, repeating the above arguments with $\gamma$ replaced by $\gamma+k$
gives the desired result for any $k\geq 1$. ∎

The following lemma bounds $\Re\left(G(\sigma+it)\right)$ when $t$ is not too
large.

######  Lemma 5.4.

There exists $c>0$ with the following property. For $\sigma>1$ sufficiently
close to $1$, and for any $t\in\mathbb{R}$ with $1\leq|t|\leq
e^{1/(\sigma-1)}$ we have

$\Re\left(G(\sigma+it)\right)\leq(1-c)\cdot G(\sigma).$ (5.31)

To prove Lemma 5.4 we use bounds on primes in short intervals. Information of
the form we need is already found in a work of Hoheisel [Hoh30]. For ease of
presentation, we use in fact a stronger result, which follows e.g. from the
work of Heath-Brown.

######  Theorem 5.5.

[HB88] For any sufficiently large $m$, and any $n\geq m$ with $n-m\geq
n^{3/4}$, we have

$\left|\\{p:p\in[m,n]\text{ prime}\\}\right|\geq\frac{n-m}{2\log n}.$ (5.32)

###### Proof of Lemma 5.4.

Let $t\in\mathbb{R}$ with $1\leq|t|\leq e^{1/(\sigma-1)}$. As
$\Re\left(G(\sigma+it)\right)$ is an even function of $t$, we may assume that
$t>0$. Consider the set of integers

$M:=\left\\{n\geq
0:\frac{10}{\sigma-1}\leq\frac{2\pi}{t}\left(n+\frac{3}{4}\right)\leq\frac{20}{\sigma-1}\right\\}.$
(5.33)

Clearly, for $\sigma$ close enough to $1$ we have $|M|\geq t/(\sigma-1)$. For
any $n\in M$ we define

$A_{n}:=\left\\{p:t\log p\in\left[2\pi n+\frac{\pi}{2},2\pi
n+\frac{3\pi}{2}\right]\right\\}=\left\\{p:p\in\left[e^{-\frac{\pi}{t}}x_{n},x_{n}\right]\right\\},$
(5.34)

where

$x_{n}:=\exp\left(\frac{2\pi}{t}\left(n+\frac{3}{4}\right)\right).$ (5.35)

For any $n\in M$ and $p\in A_{n}$ we have that $\cos(t\log p)\leq 0$ and
therefore

$\Re\left(G(\sigma+it)\right)=\sum_{p}\frac{\log^{\gamma}p}{p^{\sigma}}\cos(t\log
p)\leq G(\sigma)-\sum_{n\in M}\sum_{p\in
A_{n}}\frac{\log^{\gamma}p}{p^{\sigma}}.$ (5.36)

Since $\log^{\gamma}p/p^{\sigma}$ is decreasing for sufficiently large $p$
(independently of $\sigma\geq 1$) and as $\min_{n\in M}\min A_{n}\to\infty$ as
$\sigma\to 1^{+}$ (uniformly in $t\geq 1$) we get that, when $\sigma$ is close
enough to $1$,

$\sum_{p\in
A_{n}}\frac{\log^{\gamma}p}{p^{\sigma}}\geq|A_{n}|\frac{\log^{\gamma}x_{n}}{x_{n}^{\sigma}}\geq\frac{\left(1-e^{-\frac{\pi}{t}}\right)x_{n}}{2\log
x_{n}}\frac{\log^{\gamma}x_{n}}{x_{n}^{\sigma}}\geq
c\frac{\log^{\gamma-1}x_{n}}{t\cdot
x_{n}^{\sigma-1}}\geq\frac{c}{t}\left(\frac{1}{\sigma-1}\right)^{\gamma-1}$
(5.37)

where in the second inequality we used (5.32). Indeed the conditions of
Theorem 5.5 hold for $[e^{-\pi/t}x_{n},x_{n}]$ as for any $n\in M$ we have

$x_{n}\geq\exp\left(\frac{10}{\sigma-1}\right)\geq t^{10}$ (5.38)

and so

$x_{n}-e^{-\frac{\pi}{t}}x_{n}=x_{n}\left(1-e^{-\frac{\pi}{t}}\right)\geq\frac{c}{t}x_{n}\geq
cx_{n}^{0.9}.$ (5.39)

Summing (5.37) over $n\in M$ we get

$\sum_{n\in M}\sum_{p\in
A_{n}}\frac{\log^{\gamma}p}{p^{\sigma}}\geq\frac{c|M|}{t}\left(\frac{1}{\sigma-1}\right)^{\gamma-1}\geq
c\left(\frac{1}{\sigma-1}\right)^{\gamma}\geq c\cdot G(\sigma),$ (5.40)

where in the last inequality we used Lemma 5.3 with $k=0$. From (5.36) and
(5.40) we obtain the desired bound. ∎

It turns out that when $|t|\geq e^{1/(\sigma-1)}$, the result of Lemma 5.4
does not hold and $\Re\left(G(\sigma+it)\right)$ might be as large as
$G(\sigma)$. We shall show that the reals $t\in\mathbb{R}$ for which
$\Re\left(G(\sigma+it)\right)$ is as large as $G(\sigma)$ are quite rare. More
precisely, we will show in the following lemma that if $t_{1}$, $t_{2}$ are
such that $\Re\left(G(\sigma+it)\right)$ is large then the same holds for
$t_{1}-t_{2}$, and therefore by Lemma 5.4 $t_{1}$ and $t_{2}$ must be far away
from each other.

######  Lemma 5.6.

Let $\\{a_{n}\\}_{n\geq 1}$ be a sequence of non-negative reals with
$\sum_{n=1}^{\infty}a_{n}<\infty$. Consider the function

$f(t)=\sum_{n=1}^{\infty}a_{n}\cos(t\log n),\quad t\in\mathbb{R}.$ (5.41)

For any $0<\varepsilon<1$ and any $t_{1},t_{2}\in\mathbb{R}$ with

$f(t_{1})\geq(1-\varepsilon)f(0),\quad f(t_{2})\geq(1-\varepsilon)f(0)$ (5.42)

we have

$f(t_{1}-t_{2})\geq(1-8\sqrt{\varepsilon})f(0).$ (5.43)

###### Proof.

Let $0<\varepsilon<1$ and define

$A_{t}:=\\{n\geq 1:\cos(t\log n)>1-\sqrt{\varepsilon}\\}$ (5.44)

for $t\in\\{t_{1},t_{2}\\}$. By the assumption on $t_{1}$ we have

$(1-\varepsilon)f(0)\leq f(t_{1})=\sum_{n=1}^{\infty}a_{n}\cos(t_{1}\log
n)\leq\sum_{n\in A_{t_{1}}}a_{n}+(1-\sqrt{\varepsilon})\sum_{n\notin
A_{t_{1}}}a_{n}=f(0)-\sqrt{\varepsilon}\sum_{n\notin A_{t_{1}}}a_{n},$ (5.45)

so that

$\sum_{n\notin A_{t_{1}}}a_{n}\leq\sqrt{\varepsilon}f(0).$ (5.46)

The same argument shows that (5.46) holds for $t_{2}$ in place of $t_{1}$ as
well. Therefore, by the union bound,

$\sum_{n\notin A_{t_{2}}\cap A_{t_{2}}}a_{n}\leq
2\sqrt{\varepsilon}f(0)\quad\mbox{and}\quad\sum_{n\in A_{t_{2}}\cap
A_{t_{2}}}a_{n}\geq(1-2\sqrt{\varepsilon})f(0).$ (5.47)

Now, for any $n\in A_{t_{1}}\cap A_{t_{2}}$ and $i=1,2$ we have

$|\sin(t_{i}\log n)|=\sqrt{1-\cos^{2}(t_{i}\log
n)}\leq\sqrt{1-(1-\sqrt{\varepsilon})^{2}}\leq\sqrt{1-(1-2\sqrt{\varepsilon})}=\sqrt{2}\varepsilon^{\frac{1}{4}}.$
(5.48)

Thus, for any $n\in A_{t_{1}}\cap A_{t_{2}}$,

$\cos\left((t_{1}-t_{2})\log n\right)=\cos(t_{1}\log n)\cos(t_{2}\log
n)+\sin(t_{1}\log n)\sin(t_{2}\log
n)\geq(1-\sqrt{\varepsilon})^{2}-2\sqrt{\varepsilon}\geq
1-4\sqrt{\varepsilon}.$ (5.49)

From (5.47) and (5.49) we obtain that

$\begin{split}f(t_{1}-t_{2})=\sum_{n=1}^{\infty}a_{n}\cos\left((t_{1}-t_{2})\log
n\right)&\geq(1-4\sqrt{\varepsilon})\sum_{n\in A_{t_{1}}\cap
A_{t_{2}}}a_{n}-\sum_{n\notin A_{t_{1}}\cap A_{t_{2}}}a_{n}\\\
&\geq(1-4\sqrt{\varepsilon})(1-2\sqrt{\varepsilon})f(0)-2\sqrt{\varepsilon}f(0)\geq(1-8\sqrt{\varepsilon})f(0),\end{split}$
(5.50)

as needed. ∎

Fix $K>0$. The function $G^{\prime}(s)$ is monotone-increasing for real $s>1$,
with $\lim_{s\to\infty}G^{\prime}(s)=0$ and $\lim_{s\to
1^{+}}G^{\prime}(s)=-\infty$. For $x>1$, we let $\sigma=\sigma_{x}$ be the
unique real solution to

$K\cdot G^{\prime}(\sigma)=-\log x.$ (5.51)

The point $\sigma$ plays the role of the saddle point in the proof of Theorem
2.5. The following is a corollary of Lemma 5.3.

######  Corollary 5.7.

As $x\to\infty$ we have

$\sigma-1=\left(1+O\left(\frac{1}{\log
x}\right)\right)\left(K\Gamma(\gamma+1)\right)^{\frac{1}{\gamma+1}}\left(\log
x\right)^{-\frac{1}{\gamma+1}},$ (5.52)

and moreover

$\begin{split}G(\sigma)&=\frac{\Gamma(\gamma)}{(K\Gamma(\gamma+1))^{\frac{\gamma}{\gamma+1}}}\left(\log
x\right)^{\frac{\gamma}{\gamma+1}}+B_{0}+O\left((\log
x)^{-\frac{1}{\gamma+1}}\right),\\\
G^{\prime\prime}(\sigma)&=\frac{\Gamma(\gamma+2)}{(K\Gamma(\gamma+1))^{\frac{\gamma+2}{\gamma+1}}}\left(\log
x\right)^{\frac{\gamma+2}{\gamma+1}}+O\left((\log
x)^{\frac{1}{\gamma+1}}\right),\\\
G^{\prime\prime\prime}(\sigma)&\sim-\frac{\Gamma(\gamma+3)}{(K\Gamma(\gamma+1))^{\frac{\gamma+3}{\gamma+1}}}\left(\log
x\right)^{\frac{\gamma+3}{\gamma+1}}.\end{split}$ (5.53)

###### Proof.

Since $\lim_{s\to 1^{+}}G^{\prime}(s)=-\infty$, it follows that
$\sigma_{x}=O(1)$ for $x\geq 2$. Using Lemma 5.3, (5.51) becomes

$K\frac{\Gamma(\gamma+1)}{(\sigma-1)^{\gamma+1}}=\log x+O(1),$ (5.54)

from which we deduce (5.52). Applying the estimates for $G^{(k)}(s)$ in Lemma
5.3 for $s=\sigma$ and $k=0,2,3$, and using (5.52), we obtain the stated
estimates for $G(\sigma)$, $G^{\prime\prime}(\sigma)$,
$G^{\prime\prime\prime}(\sigma)$. ∎

###  5.1 Proof of Theorem 2.5

If we replace $x$ with $\lfloor x\rfloor+\frac{1}{2}$ in (2.13), then the
left-hand side does not change, while the function in the right-hand side is
changed by a factor of $1+O(1/x)$, which can be absorbed in the relative error
term. Thus, in proving Theorem 2.5 we may assume without loss of generality
that $x$ is of the form $m+\frac{1}{2}$ for some positive integer $m$ (i.e.
half-integer). We denote by $F(s)$ the Dirichlet series of $\alpha$, which by
Lemma 5.1 is of the form $F(s)=\varphi(s)e^{K\cdot G(s)}$ where $\varphi$ is
differentiable, bounded and has bounded derivative on $\Re s\geq 1$. By an
effective version of Perron’s formula [Ten15, Ch. II.2, Thm. 2.3] we have

$\sum_{n\leq x}\alpha(n)=\frac{1}{2\pi i}\intop_{\sigma-
iT}^{\sigma+iT}F(s)x^{s}\,\frac{ds}{s}+O\left(x^{\sigma}\sum_{n\geq
1}\frac{\alpha(n)}{n^{\sigma}(1+T|\log(x/n)|)}\right)$ (5.55)

for any $T\geq 1$, where $\sigma=\sigma_{x}$ is defined in (5.51). We choose
$T=x^{2}$, obtaining

$\sum_{n\leq
x}\alpha(n)=\frac{1}{2\pi}\intop_{-x^{2}}^{x^{2}}\varphi(\sigma+it)e^{K\cdot
G(\sigma+it)}\frac{x^{\sigma+it}}{\sigma+it}\,dt+O(E(x)),$ (5.56)

where

$E(x)=x^{\sigma}\sum_{n\geq
1}\frac{\alpha(n)}{n^{\sigma}(1+x^{2}|\log(x/n)|)}\leq
C\frac{x^{\sigma}}{x}\sum_{n\geq
1}\frac{\alpha(n)}{n^{\sigma}}=C\frac{x^{\sigma}F(\sigma)}{x}\leq
C\frac{x^{\sigma}e^{K\cdot G(\sigma)}}{x}.$ (5.57)

(We have used the fact that $x$ is an half-integer and so
$\left|\log(x/n)\right|\geq C/x$.) Set

$t_{x}:=\left(\log
x\right)^{\frac{\delta}{6(\gamma+1)}-\frac{\gamma+2}{2(\gamma+1)}}$ (5.58)

for some sufficiently small $\delta>0$. We decompose the integral in the
right-hand side of (5.56) into three parts, to be estimated in the following
ways:

$\begin{split}|t|\leq t_{x}:\quad&\mbox{estimated using
Corollary~{}\ref{cor:derivatives at he saddle point},}\\\ t_{x}\leq|t|\leq
1:\quad&\mbox{bounded using Lemma~{}\ref{lem:derivatives of G next to 1},}\\\
1\leq|t|\leq x^{2}:\quad&\mbox{bounded using Lemmas~{}\ref{lem:bound for large
t} and \ref{lem:t_1 t_2}.}\end{split}$ (5.59)

We denote by $I_{1},I_{2},I_{3}$ the integrals over these respective domains.
We begin by computing the asymptotics for $I_{1}$, which gives the main term.
When $|t|\leq t_{x}$ we have, by Corollary 5.7,

$\frac{\varphi(\sigma+it)}{\sigma+it}=\varphi(1)\left(1+O\left(t_{x}+\sigma-1\right)\right)=\varphi(1)\left(1+O\left(\log^{-\frac{1}{\gamma+1}}x\right)\right)$
(5.60)

since $\varphi$ has bounded derivative. A second-order Taylor approximation of
$G(\sigma+it)$ around $t=0$ gives

$\begin{split}K\cdot G(\sigma+it)&=K\cdot G(\sigma)+itK\cdot
G^{\prime}(\sigma)-\frac{t^{2}}{2}K\cdot
G^{\prime\prime}(\sigma)+O\left(|t_{x}^{3}G^{\prime\prime\prime}(\sigma)|\right)\\\
&=K\cdot G(\sigma)-it\log x-\frac{t^{2}}{2}K\cdot
G^{\prime\prime}(\sigma)+O\left(\log^{-\frac{\gamma-\delta}{2(\gamma+1)}}x\right)\end{split}$
(5.61)

for $|t|\leq t_{x}$, where in the first equality we used the fact that
$|G^{\prime\prime\prime}(\sigma+it)|\leq|G^{\prime\prime\prime}(\sigma)|$ and
in the second equality we used Corollary 5.7 and the definition of $\sigma$ in
(5.51). From (5.60) and (5.61), we get

$\varphi(\sigma+it)e^{K\cdot
G(\sigma+it)}\frac{x^{\sigma+it}}{\sigma+it}=\varphi(1)x^{\sigma}e^{K\cdot
G(\sigma)-\frac{t^{2}}{2}K\cdot
G^{\prime\prime}(\sigma)}\left(1+O\left(\log^{-z}x\right)\right)$ (5.62)

where $z:=\min\\{\gamma-\delta,2\\}/(2(\gamma+1))$. We thus have

$I_{1}=\left(1+O\left(\log^{-z}x\right)\right)\frac{\varphi(1)x^{\sigma}e^{K\cdot
G(\sigma)}}{2\pi}\intop_{-t_{x}}^{t_{x}}e^{-\frac{K\cdot
G^{\prime\prime}(\sigma)}{2}t^{2}}\,dt=\left(1+O\left(\log^{-z}x\right)\right)\frac{\varphi(1)x^{\sigma}e^{K\cdot
G(\sigma)}}{\sqrt{2\pi K\cdot G^{\prime\prime}(\sigma)}},$ (5.63)

which by Corollary 5.7 can be simplified to

$I_{1}=\left(1+O\left(\log^{-z}x\right)\right)A_{\alpha}x(\log
x)^{-\frac{\gamma+2}{2(\gamma+1)}}\exp\left[B(\log
x)^{\frac{\gamma}{\gamma+1}}\right],$ (5.64)

where $B$ is defined in (2.14), and $A_{\alpha}$ is defined in (2.6). Next we
bound $I_{2}$. Using Lemma 5.3 with $s=\sigma+it$ where $t_{x}\leq|t|\leq 1$,
we get

$\Re\left(G(\sigma+it)\right)\leq\left|G(\sigma+it)\right|\leq\frac{\Gamma(\gamma)}{\left|\sigma-1+it\right|^{\gamma}}+C\leq\frac{\Gamma(\gamma)}{\left|\sigma-1+it_{x}\right|^{\gamma}}+C\leq\left|G(\sigma+it_{x})\right|+C,$
(5.65)

and so a second-order Taylor approximation shows that

$\begin{split}\Re\left(G(\sigma+it)\right)\leq\left|G(\sigma)+it_{x}G^{\prime}(\sigma)-\frac{t_{x}^{2}}{2}G^{\prime\prime}(\sigma)\right|+C&=\sqrt{G(\sigma)^{2}-t_{x}^{2}\left(G(\sigma)G^{\prime\prime}(\sigma)-G^{\prime}(\sigma)^{2}\right)+t_{x}^{4}G^{\prime\prime}(\sigma)^{2}/4}+C\\\
&\leq\sqrt{G(\sigma)^{2}-c(\log x)^{\frac{3\gamma+\delta}{3(\gamma+1)}}}+C\leq
G(\sigma)-c(\log x)^{\frac{\delta}{3(\gamma+1)}},\end{split}$ (5.66)

where the second inequality holds for sufficiently small $\delta$ and follows
from Corollary 5.7. Thus

$|I_{2}|\leq Cx^{\sigma}e^{K\cdot G(\sigma)-c(\log
x)^{\frac{\delta}{3(\gamma+1)}}}\leq\frac{C}{\log x}|I_{1}|.$ (5.67)

We now show that the contribution from $I_{3}$ is negligible as well. Fix
$\varepsilon\in(0,1)$ such that $8\sqrt{\varepsilon}$ is strictly less than
the constant $c$ from Lemma 5.4. Consider the set

$S:=\\{t>0:\Re\left(G(\sigma+it)\right)>(1-\varepsilon)G(\sigma)\\}.$ (5.68)

We have, by definition of $S$,

$\begin{split}\intop_{\left[1,x^{2}\right]\setminus
S}\left|\varphi(\sigma+it)e^{K\cdot
G(\sigma+it)}\frac{x^{\sigma+it}}{\sigma+it}\right|\,dt\leq
Cx^{\sigma}e^{(1-\varepsilon)K\cdot
G(\sigma)}\intop_{\left[1,x^{2}\right]\setminus S}\frac{dt}{t+1}\leq
Cx^{\sigma}e^{(1-\varepsilon)K\cdot G(\sigma)}\log x\leq\frac{C}{\log
x}|I_{1}|,\end{split}$ (5.69)

where in the last inequality we used Corollary 5.7. We now study the integral
over $S$. Applying Lemma 5.6 with the sequence

$a_{n}:=\mathds{1}_{\\{n\text{ is prime}\\}}\frac{\log^{\gamma}n}{n^{\sigma}}$
(5.70)

we find that for any $t_{1},t_{2}\in S$ we have that
$\Re\left(G(\sigma+i(t_{1}-t_{2}))\right)\geq(1-8\sqrt{\varepsilon})G(\sigma)$
and therefore, by Lemma 5.4 and the choice of $\varepsilon$, either
$|t_{1}-t_{2}|\leq 1$ or $|t_{1}-t_{2}|\geq e^{1/(\sigma-1)}$. It follows that

$S\subseteq\bigcup_{j=0}^{\infty}[a_{j},b_{j}]$ (5.71)

for some $a_{j+1}>b_{j}>a_{j}\geq 0$ with

$b_{j}-a_{j}\leq 1,\quad a_{j}\geq j\cdot e^{\frac{1}{\sigma-1}}\quad\text{
and }\quad a_{0}=0.$ (5.72)

Thus, for sufficiently large $x$,

$\begin{split}\intop_{\left[1,x^{2}\right]\cap
S}\left|\varphi(\sigma+it)e^{K\cdot
G(\sigma+it)}\frac{x^{\sigma+it}}{\sigma+it}\,dt\right|&\leq
Cx^{\sigma}e^{K\cdot G(\sigma)}\sum_{1\leq j\leq
x^{2}}\intop_{a_{j}}^{b_{j}}\frac{dt}{t}\leq Cx^{\sigma}e^{K\cdot
G(\sigma)}\sum_{1\leq j\leq x^{2}}\frac{b_{j}-a_{j}}{a_{j}}\\\ &\leq
Cx^{\sigma}e^{K\cdot G(\sigma)}e^{-\frac{1}{\sigma-1}}\sum_{1\leq j\leq
x^{2}}\frac{1}{j}\leq Cx^{\sigma}e^{K\cdot
G(\sigma)}e^{-\frac{1}{\sigma-1}}\log x\leq\frac{C}{\log
x}|I_{1}|.\end{split}$ (5.73)

Combining the estimates for the integrals over $\left[1,x^{2}\right]\setminus
S$ and $\left[1,x^{2}\right]\cap S$, we obtain

$|I_{3}|\leq\frac{C}{\log x}|I_{1}|.$ (5.74)

We conclude the proof by plugging the estimates (5.57), (5.64), (5.67) and
(5.74) in (5.56). ∎

###  5.2 Proof of Theorem 2.4

####  5.2.1 Auxiliary results

An important step in the proof is understanding the asymptotic behavior of
$\mathbb{P}(p\mid N_{x})$. We shall see that $\mathbb{P}(p\mid
N_{x})\approx\alpha(p)S(x/p)/S(x)$, and so begin by studying the ratio
$S(x/h)/S(x)$. Observe that

$S(x/h)h\leq C\cdot S(x)$ (5.75)

for $x\geq 1$, $h\geq 1$ by Theorem 2.5.

######  Lemma 5.8.

Let $\alpha\colon\mathbb{N}\to\mathbb{R}_{\geq 0}$ be a multiplicative
function satisfying (2.9)–(2.10) and suppose that $x$ is sufficiently large.
Let $2\leq h\leq x$. When $\log h\leq(\log x)^{(\gamma+4)/(4\gamma+4)}$, we
have that

$\frac{S(x/h)}{S(x)}=\frac{1}{h}\exp\bigg{[}-\frac{B\gamma}{\gamma+1}\frac{\log
h}{(\log x)^{\frac{1}{\gamma+1}}}+O\bigg{(}\frac{1}{(\log
x)^{c}}\bigg{)}\bigg{]}.$ (5.76)

When $\log h\geq(\log x)^{(\gamma+4)/(4\gamma+4)}$, we have that

$\frac{S(x/h)}{S(x)}\leq\frac{1}{h}e^{-(\log x)^{c}}.$ (5.77)

###### Proof.

Let $h_{x}:=\exp((\log x)^{(\gamma+4)/(4\gamma+4)})$. Suppose that $h\leq
h_{x}$. By a first-order Taylor approximation, we get

$(\log(x/h))^{\frac{\gamma}{\gamma+1}}=(\log
x)^{\frac{\gamma}{\gamma+1}}\left(1-\frac{\log h}{\log
x}\right)^{\frac{\gamma}{\gamma+1}}=(\log
x)^{\frac{\gamma}{\gamma+1}}-\frac{\gamma}{\gamma+1}\frac{\log h}{(\log
x)^{\frac{1}{\gamma+1}}}+O\left(\frac{1}{(\log x)^{c}}\right).$ (5.78)

Thus, by Theorem 2.5 applied with $x$ and $x/h$, we obtain the first part of
the lemma. We turn to prove the second part of the lemma. Using the first part
of the lemma and (5.75) we get that when $h\geq h_{x}$

$\frac{S(x/h)}{S(x)}=\frac{S(x/h)}{S(x/h_{x})}\frac{S(x/h_{x})}{S(x)}\leq\frac{Ch_{x}}{h}\frac{1}{h_{x}}\exp\left[-c\frac{\log
h_{x}}{(\log x)^{\frac{1}{\gamma+1}}}\right]\leq\frac{1}{h}e^{-(\log x)^{c}},$
(5.79)

as needed. ∎

######  Lemma 5.9.

Let $\alpha\colon\mathbb{N}\to\mathbb{R}_{\geq 0}$ be a multiplicative
function satisfying $\log\alpha(p)=o(\log p)$, (2.10) and (5.75). For
sufficiently large $x$ and any prime $p\leq x$ we have that

$\mathbb{P}\left(p\mid
N_{x}\right)=\frac{\alpha(p)S(x/p)}{S(x)}+O\bigg{(}\frac{1}{p\log^{2}p}\bigg{)}.$
(5.80)

###### Proof.

By (5.75) we have for any $y>1$

$\sum_{\begin{subarray}{c}n\leq y\\\ p\mid
n\end{subarray}}\alpha(n)\leq\sum_{k\geq 1}\sum_{\begin{subarray}{c}n\leq y\\\
p^{k}\mid\mid n\end{subarray}}\alpha(n)=\sum_{k\geq
1}\alpha(p^{k})S\Big{(}\frac{y}{p^{k}}\Big{)}\leq C\cdot S(y)\sum_{k\geq
1}\frac{\alpha(p^{k})}{p^{k}}\leq\frac{C\cdot S(y)}{\sqrt{p}}.$ (5.81)

Hence

$\begin{split}\mathbb{P}\left(p\mid\mid
N_{x}\right)=\frac{1}{S(x)}\sum_{\begin{subarray}{c}n\leq x\\\ p\mid\mid
n\end{subarray}}\alpha(n)=\frac{\alpha(p)}{S(x)}\sum_{\begin{subarray}{c}m\leq
x/p\\\ p\nmid
m\end{subarray}}\alpha(m)&=\frac{\alpha(p)S(x/p)}{S(x)}+O\left(\frac{\alpha(p)S(x/p)}{\sqrt{p}S(x)}\right)\\\
&=\frac{\alpha(p)S(x/p)}{S(x)}+O\left(\frac{1}{p\log^{2}p}\right),\end{split}$
(5.82)

where in the last inequality we use (5.75) again. Using the same arguments as
in (5.81), we have that

$\mathbb{P}(p^{2}\mid N_{x})\leq
C\sum_{k=2}^{\infty}\frac{\alpha(p^{k})}{p^{k}}\leq\frac{C}{p\log^{2}p}.$
(5.83)

As $\mathbb{P}(p\mid N_{x})=\mathbb{P}(p\mid\mid N_{x})+\mathbb{P}(p^{2}\mid
N_{x})$, the proof is concluded. ∎

####  5.2.2 Conclusion of proof

We begin with the first part of the theorem. We abbreviate $P_{1}(N_{x})$ as
$P_{1}$. Fix $0<a<b<\infty$. It suffices to show that

$\liminf_{x\to\infty}\mathbb{P}\left(a\leq\frac{\log P_{1}}{(\log
x)^{\frac{1}{\gamma+1}}}\leq b\right)\geq\mathbb{P}\left(a\leq Y\leq b\right)$
(5.84)

where $Y$ has
$\operatorname{gamma}(\gamma+1,(K\Gamma(\gamma+1))^{1/(\gamma+1)})$
distribution.

For any prime $p$ such that $a(\log x)^{\frac{1}{\gamma+1}}\leq\log p\leq
b(\log x)^{\frac{1}{\gamma+1}}$, we have, by Lemmas 5.8 and 5.9,

$\begin{split}\mathbb{P}(P_{1}=p)&=\sum_{\begin{subarray}{c}n\leq x\\\ p\mid
n\end{subarray}}\frac{\alpha(n)}{S(x)}\frac{\nu_{p}(n)\log p}{\log
n}\geq\frac{\log p}{\log x}\mathbb{P}(p\mid N_{x})\geq\frac{\log p}{\log
x}\left(\frac{K\log^{\gamma}pS(x/p)}{S(x)}+O\bigg{(}\frac{1}{p\log^{2}p}\bigg{)}\right)\\\
&\geq\left(1+O\left(\frac{1}{(\log
x)^{c}}\right)\right)\frac{K\log^{\gamma+1}p}{p\log
x}\exp\left[-\frac{B\gamma}{\gamma+1}\frac{\log p}{(\log
x)^{\frac{1}{\gamma+1}}}\right],\end{split}$ (5.85)

where the error term $1/(p\log^{2}p)$ is absorbed in the last error term,
since the main term is of order $p^{-1}$. Thus,

$\mathbb{P}\left(a\leq\frac{\log P_{1}}{(\log x)^{1/(\gamma+1)}}\leq
b\right)\geq\left(1+O\left(\frac{1}{(\log
x)^{c}}\right)\right)\sum_{n_{x}^{a}\leq p\leq
n_{x}^{b}}\frac{K\log^{\gamma+1}p}{p\log
x}\exp\left[-\frac{B\gamma}{\gamma+1}\frac{\log p}{(\log
x)^{\frac{1}{\gamma+1}}}\right]$ (5.86)

where $n_{x}=\exp((\log x)^{1/(\gamma+1)})$. By Lemma 4.6 with $\alpha\equiv
1$, the interval $[n_{x}^{a},n_{x}^{b}]$ and

$g(t)=\frac{K\log^{\gamma+1}t}{t\log
x}\exp\bigg{[}-\frac{B\gamma}{\gamma+1}\frac{\log t}{(\log
x)^{\frac{1}{\gamma+1}}}\bigg{]}$ (5.87)

we have

$\mathbb{P}\left(a\leq\frac{\log P_{1}}{(\log x)^{1/(\gamma+1)}}\leq
b\right)\geq\left(1+O\left(\frac{1}{(\log
x)^{c}}\right)\right)\left(\intop_{n_{x}^{a}}^{n_{x}^{b}}\frac{K\log^{\gamma}t}{t\log
x}\exp\left[-\frac{B\gamma}{\gamma+1}\frac{\log t}{(\log
x)^{\frac{1}{\gamma+1}}}\right]\,dt+O(\log^{-\frac{1}{\gamma+1}}x)\right).$
(5.88)

The change of variables $t=n_{x}^{z}$ in the last integral shows that it
equals $\mathbb{P}(a\leq Y\leq b)$. Taking $x$ to infinity in (5.88) we obtain
(5.84), as needed.

We turn to the second part of the theorem. By Lemma 5.9 and (2.9) we have

$\mathbb{E}\omega(N_{x})=\sum_{p\leq x}\mathbb{P}(p\mid N_{x})=\sum_{p\leq
x}\frac{K\log^{\gamma}pS(x/p)}{S(x)}+O\bigg{(}\sum_{p}\frac{1}{p\log^{2}p}\bigg{)}.$
(5.89)

The error term is bounded by a constant. In order to estimate the sum, we we
split it into three sums $S_{1}$, $S_{2}$ and $S_{3}$, over the respective
ranges $p<\exp((\log x)^{\delta_{1}})$, $\exp((\log x)^{\delta_{1}})\leq
p\leq\exp((\log x)^{\delta_{2}})$ and $\exp((\log x)^{\delta_{2}})<p\leq x$,
where $\delta_{1}=1/(2(\gamma+1))$ and $\delta_{2}=(\gamma+4)/(4\gamma+4)$. We
bound $S_{1}$ using (5.75):

$S_{1}\leq C\sum_{p\leq\exp((\log
x)^{\delta_{1}})}\frac{\log^{\gamma}p}{p}\leq C\log^{\delta_{1}\gamma}x,$
(5.90)

where in the last inequality we used Lemma 4.6 with $\alpha\equiv 1$ and
$g(t)=\log^{\gamma}t/t$. We bound $S_{3}$ using the second part of Lemma 5.8,
which gives

$S_{3}\leq e^{-(\log x)^{c}}\sum_{\exp((\log x)^{\delta_{2}})<p\leq
x}\frac{K\log^{\gamma}p}{p}\leq e^{-(\log x)^{c}}.$ (5.91)

We now estimate $S_{2}$. By Lemma 5.8,

$S_{2}=\left(1+O\left(\frac{1}{(\log x)^{c}}\right)\right)\sum_{(\log
x)^{\delta_{1}}\leq\log p\leq(\log
x)^{\delta_{2}}}\frac{K\log^{\gamma}p}{p}\exp\left[-\frac{B\gamma}{\gamma+1}\frac{\log
p}{(\log x)^{\frac{1}{\gamma+1}}}\right].$ (5.92)

From (5.92) and Lemma 4.6 with $\alpha\equiv 1$ and

$g(t)=\frac{K\log^{\gamma}t}{t}\exp\left[-\frac{B\gamma}{\gamma+1}\frac{\log
t}{(\log x)^{\frac{1}{\gamma+1}}}\right]$ (5.93)

we obtain

$S_{2}=\left(1+O\left(\frac{1}{(\log
x)^{c}}\right)\right)\left(\int_{\exp((\log x)^{\delta_{1}})}^{\exp((\log
x)^{\delta_{2}})}\frac{g(t)}{\log
t}\,dt+O\left(\log^{\max\\{0,\frac{\gamma-1}{\gamma+1}\\}}x\right)\right).$
(5.94)

The change of variables $t=n_{x}^{z}$ in the last integral shows that it
equals

$(\log x)^{\frac{\gamma}{\gamma+1}}\intop_{(\log
x)^{\delta_{1}-\frac{1}{\gamma+1}}}^{(\log
x)^{\delta_{2}-\frac{1}{\gamma+1}}}Kz^{\gamma-1}\exp\left(-\frac{B\gamma
z}{\gamma+1}\right)\,dz=(\log
x)^{\frac{\gamma}{\gamma+1}}K\Gamma(\gamma)\left(\frac{\gamma+1}{B\gamma}\right)^{\gamma}\left(1+O\left((\log
x)^{\gamma(\delta_{1}-\frac{1}{\gamma+1})}\right)\right).$ (5.95)

Since all the accumulated error terms are of order smaller than $(\log
x)^{\gamma/(\gamma+1)}$, the expectation of $\omega$ is estimated. The
expectation of $\Omega$ behaves the same by Lemma 5.2. ∎

##  6 Small primes: proof of Theorem 2.9

Assume that $\alpha$ satisfies (1.4)–(1.5). Let $m\in\mathbb{N}$ and let
$p_{1},\ldots,p_{m}$ be distinct primes. We have

$\mathbb{P}\left(\forall 1\leq j\leq m,\
\nu_{p_{j}}(N_{x})=0\right)=\bigg{(}\sum_{n\leq
x}\alpha(n)\bigg{)}^{-1}\sum_{{\begin{subarray}{c}n\leq x\\\
p_{1},\dots,p_{m}\nmid n\end{subarray}}}\alpha(n)=\bigg{(}\sum_{n\leq
x}\alpha(n)\bigg{)}^{-1}\sum_{n\leq x}\alpha^{\prime}(n),$ (6.1)

where the function $\alpha^{\prime}\colon\mathbb{N}\to\mathbb{R}_{\geq 0}$ is
the multiplicative function defined on prime powers by

$\alpha^{\prime}(p^{k}):=\begin{cases}\alpha(p^{k})&\mbox{if
}p\notin\\{p_{1},\dots,p_{m}\\},\\\ \quad 0&\mbox{if
}p\in\\{p_{1},\dots,p_{m}\\}.\end{cases}$ (6.2)

Since $\alpha^{\prime}$ satisfies (1.4)–(1.5) with the same parameters as
$\alpha$, we may use Corollary 4.3 with $\alpha$ and $\alpha^{\prime}$ to
obtain that

$\begin{split}\mathbb{P}\left(\forall 1\leq j\leq m,\
\nu_{p_{j}}(N_{x})=0\right)&=\frac{\sum_{n\leq
x}\alpha^{\prime}(n)}{\sum_{n\leq
x}\alpha(n)}=\frac{A_{\alpha^{\prime}}x^{d+1}\log^{\theta-1}x\left(1+O\left(\frac{1}{\log
x}\right)\right)}{A_{\alpha}x^{d+1}\log^{\theta-1}x\left(1+O\left(\frac{1}{\log
x}\right)\right)}\\\
&=\frac{A_{\alpha^{\prime}}}{A_{\alpha}}\left(1+O\left(\frac{1}{\log
x}\right)\right)\to\frac{A_{\alpha^{\prime}}}{A_{\alpha}}\end{split}$ (6.3)

as $x\to\infty$. We may compute the ratio of the constants as follows:

$\frac{A_{\alpha^{\prime}}}{A_{\alpha}}=\frac{\frac{1}{(d+1)\Gamma(\theta)}\prod_{p}(\sum_{i\geq
0}\alpha^{\prime}(p^{i})/p^{i(d+1)})(1-1/p)^{\theta}}{\frac{1}{(d+1)\Gamma(\theta)}\prod_{p}(\sum_{i\geq
0}\alpha(p^{i})/p^{i(d+1)})(1-1/p)^{\theta}}=\prod_{j=1}^{m}\left(\sum_{i\geq
0}\frac{\alpha(p_{j}^{i})}{p_{j}^{i(d+1)}}\right)^{-1}.$ (6.4)

Let $l_{1},\dots,l_{m}\geq 0$ and set $r=\prod_{j=1}^{m}p_{j}^{l_{j}}$. We
have

$\begin{split}\mathbb{P}\left(\forall 1\leq j\leq m,\
\nu_{p_{j}}(N_{x})=l_{j}\right)&=\bigg{(}\sum_{n\leq
x}\alpha(n)\bigg{)}^{-1}\sum_{{\begin{subarray}{c}n\leq x\\\
p_{1}^{l_{1}}\mid\mid n,\dots,p_{m}^{l_{m}}\mid\mid
n\end{subarray}}}\alpha(n)=\bigg{(}\sum_{n\leq
x}\alpha(n)\bigg{)}^{-1}\prod_{j=1}^{m}\alpha(p_{j}^{l_{j}})\sum_{{\begin{subarray}{c}n\leq
x/r\\\ p_{1},\dots,p_{m}\nmid n\end{subarray}}}\alpha(n)\\\
&=\frac{\sum_{n\leq x/r}\alpha(n)}{\sum_{n\leq
x}\alpha(n)}\prod_{j=1}^{m}\alpha(p_{j}^{l_{j}})\mathbb{P}\left(\forall 1\leq
j\leq m,\
\nu_{p_{j}}(N_{x/k})=l_{j}\right)\to\frac{1}{r^{d+1}}\prod_{j=1}^{m}\alpha(p_{j}^{l_{j}})\frac{A_{\alpha^{\prime}}}{A_{\alpha}},\end{split}$
(6.5)

as $x\to\infty$, where the last limit follows from (6.3), as well as Corollary
4.3 with $\alpha$ and the values $x$ and $x/r$. Combined with the value of
$A_{\alpha^{\prime}}/A_{\alpha}$ this concludes the proof in the Ewens case.

If $\alpha$ satisfies (2.9)–(2.10), the proof is essentially the same, with
the left-hand side of (6.3) tending to the right-hand side by Theorem 2.5, and
the evaluation of the left-hand side of (6.4) is done by (2.15), giving the
right-hand side result. In addition, the limit (6.5) is computed using Theorem
2.5 with $\alpha$ and the values $x$ and $x/r$. ∎

## References

  * [ABT92] Richard Arratia, A. D. Barbour, and Simon Tavaré.  Poisson process approximations for the Ewens sampling formula.  Ann. Appl. Probab., 2(3):519–535, 1992.
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Department of Mathematics, Princeton University, Princeton, NJ 08544, USA

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Raymond and Beverly Sackler School of Mathematical Sciences, Tel Aviv
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