[a,b]Dian-Jun Zhao

# Low energy constant and mixed-action effect

Yi-Bo Yang

###### Abstract

We present the pion mass and decay constant using the overlap fermion valence
on domain wall (DW) fermion sea at several lattice spacings. The mixed action
effect in the lattice calculation is also studied, and the result suggests
that the mixed action effect with the overlap valence on DW sea would be
proportional to the fourth power of the lattice spacing. We obtain the pion
decay constant at the physical pion mass and $N_{f}=2$ chiral limit to be
$92.4(3)(2)~{}{\rm MeV}$ and $87.0(5)(7)~{}{\rm MeV}$ respectively; and the
physical u/d averaged quark mass at $\overline{\textrm{MS}}$ 2 GeV is
$3.74(4)(5)(5)(3)~{}{\rm MeV}$ with the linear ${\cal O}(a^{2})$ continuum
extrapolation. Using the FLAG value of the NLO low energy constant, we obtain
the $N_{f}=2$ chiral condensate to be $\Sigma^{\overline{\textrm{MS}}(2~{}{\rm
GeV})}=\big{(}266(2)(1)~{}{\rm MeV}\big{)}^{3}$.

##  1 Introduction

The low energy feature of the strong interaction binds the quarks into the
mesons and hadrons, and creates most of the mass of the visible matter. At the
leading order of the chiral perturbative theory, there are just two additional
parameters beside the quark mass: the chiral condensate and pion decay
constant. Since those values are quark mass dependent, it is essential to
determine them from the first principle lattice QCD calculation.

In this proceeding, we provide the preliminary lattice QCD result of the quark
mass and decay constant at the physical pion mass, and also the leading order
(LO) low energy constants.

##  2 Numerical setup and the mixed action effect

The basic information of the RBC/UKQCD 2+1 flaovr dynomcial ensembles we used
in this project is summarized in Table 1, which “DW” refers to domain wall
fermion action, while “I” indicates that the gauge field is Iwasaki. For more
information about these two ensembles, see [1]. We used the overlap fermion
action [9] for the valence quark, with 1-step HYP smearing on the gauge field
and $\rho=1.5$.

Table 1: Basic information of the ensembles we used. Action | Symbol | $L^{3}\times T$ | $a({\rm fm})$ | $m_{\pi,ss}({\rm MeV})$
---|---|---|---|---
Overlap | 24J | $24^{3}\times\ 48$ | $0.112$ | $290$
DW+I | 48I | $48^{3}\times\ 96$ | $0.114$ | $139$
DW+I | 64I | $64^{3}\times 128$ | $0.084$ | $139$

Using different valence and sea actions can introduce the mixed action effect.
For example, if we calculate the valence quark propagator using discretized
fermion action A on a gauge ensemble with dynamical light quark using another
action B, and tune the valence quark mass to make the corresponding pion mass
$m_{\pi,vv}=m_{\pi,ss}$, then the pion mass $m_{\pi,vs}$ of the pion
correlator formed by one valence quark with one action and one sea
(anti-)quark with another action, can suffer from a discretization error and
larger than $m_{\pi,vv}$.

Figure 1: Mixed action effect on the JLQCD ensemble with $m_{\pi}\sim$ 290 MeV
and $a\sim 0.11$ fm [11]. The red dots and purple boxes show the cases with
and without HYP smearing on the valence quark action.

Fig. 1 shows the mixed action effect defined by

$\Delta_{mix}=m_{\pi,vs}^{2}-\frac{m_{\pi,vv}^{2}+m_{\pi,ss}^{2}}{2},$ (1)

on the JLQCD ensemble at 0.11 fm with $\sim$ 300 MeV pion mass. The purple
data points in the figure show the case with the same valence and sea actions
(No HYP smearing, $\rho=1.3$), and $\Delta_{mix}$ is consistent with zero
within the statistical uncertainty which is relatively larger when the valence
pion mass is far away from the sea one. But if we apply the HYP smearing in
the definition of the valence quark action, $\Delta_{mix}$ becomes obvious
non-zero, as the red data points in Fig. 1. It suggests that $\Delta_{mix}$
can actually provide a good reference on the mixed action effect.

Figure 2: Mixed action effect on the RBC/UKQCD ensembles with $m_{\pi}\sim$
139 MeV and $a\sim 0.08$ fm (green dots) and 0.11 fm (orange triangles)
respectively.

On the RBC/UKQCD ensembles at physical pion mass, the mixed action effect is
much larger, while it is still much smaller than our previous estimate [7]
which is around 0.01 GeV2 at a=0.111 fm. At the same time, the values at
$a\sim 0.08$ fm are much smaller than those at a$\sim$0.11 fm, and the lattice
spacing dependence seems to be ${\cal O}(a^{4})$ instead of the widely used
${\cal O}(a^{2})$ estimate.

But even though the mixed action effect here has been much smaller than those
in the literature with the other setups (likes the Overlap on Clover [4], DW
on staggered [6, 5], or mobius DW on the gradient-flowed HISQ [12]), it is
still much larger than the statistical uncertainty and then the mixed action
pion mass can not be used in the analysis. Thus we will just consider the
valence pion mass around the unitary point to extract the low energy
constants.

##  3 Results

In this proceeding, we use the low mode substitution (LMS) to obtain good
signal with reasonable cost [8]. More precisely, we replace the low mode part
of the pion correlation function with small statistics by that from the the
all-to-all propagator using the low lying eigenvectors,

$C_{2}^{\rm
LMS}(t_{f})=\frac{1}{N}\sum_{i}[C_{2}(t_{f},\vec{x}_{i},0;S)-C_{2}(t_{f},\vec{x}_{i},0,S_{\rm
low})]+\frac{1}{L^{3}\times
T}\sum_{t_{0},\vec{x}}C_{2}(t_{f}+t_{0},\vec{x},t_{0};S_{\rm low}),$ (2)

where

$\langle
C_{2}(t_{f},\vec{x},t_{0};S)\rangle=\langle\sum_{\vec{y}}Tr[S^{\dagger}(\vec{y},t_{f};\vec{x},t_{0})S(\vec{y},t_{f};\vec{x},t_{0})]\rangle,$
(3)

$S_{\rm low}$ is the low mode part of the point source (or grid source without
gauge fixing, equivalently) propagator $S$, and $x_{1,...,N}$ are $N$ kinds of
the source positions used in the production to estimate the rest part of
$C_{2}$.

Figure 3: The effective mass of the pion correlation function around the
physical point, on two ensembles at 0.11 fm (left panel) and 0.08 fm (right
panel). The cases with LMS have much better signal and the statistical
fluctuations are significantly suppressed.

As in Fig. 3, the effective decay rates of $C_{2}$ (green dots) show obvious
statistical fluctuation as functions of the source-sink separation $t$ if the
LMS is not applied. But when its low mode part is replaced by the all-to-all
propagators, the fluctuation is highly suppressed and then one can extract the
pion mass with much higher precision. Thus based on the linear interpolation,
we obtained the renormalized quark mass at $\overline{\textrm{MS}}$ 2 GeV with
$m_{\pi}=134.98$ MeV as

$m^{\overline{\textrm{MS}}(2~{}{\rm
GeV})}_{ud}(0.114~{}\mathrm{fm})=3.80(4)(10)(3)~{}{\rm MeV},\
m^{\overline{\textrm{MS}}(2~{}{\rm
GeV})}_{ud}(0.084~{}\mathrm{fm})=3.77(1)(7)(2)~{}{\rm MeV},$ (4)

at two lattice spacings respectively, and three uncertainties are the
statistical error, the truncation error of the perturbative matching from the
RI/MOM scheme to the MS-bar scheme, and the other uncertainty of the
renormalization constants $Z_{S}$ [10]. After a linear $a^{2}$ extrapolation,
we predict the average light quark mass to be

$m^{\overline{\textrm{MS}}(2~{}{\rm GeV})}_{ud}=3.74(4)(5)(5)(3)~{}{\rm MeV},$
(5)

where we take the difference between the values at the smaller lattice spacing
and that in the continuum as the last uncertainty, to estimate the systematic
uncertainty from the continuum extrapolation. Note that the truncation error
of the perturbative matching at different lattice spacing are correlated and
then will be suppressed during the continuum extrapolation. Such a value is
much larger than the current FLAG average, 3.36(4) MeV. It would related to
the RI/MOM renormalization scheme we used, as we found that the SMOM scheme
used by part of the quark mass and condensate calculations would be have
additional systematic uncertainty due to its highly non-trivial $a^{2}p^{2}$
dependence [10].

Figure 4: The pion decay constant without (left panel) and with (right panel)
LMS at $a\sim 0.11$ fm (pruple boxes) and 0.08 fm (green triangles).

For the pion decay constant, one can obtain it from the correlation function
with either the $\bar{\psi}\gamma_{5}\gamma_{4}\psi$ or
$\bar{\psi}\gamma_{5}\psi$ interpolation field,

$\displaystyle(m_{q_{1}}+m_{q_{2}})\langle
0|\bar{\psi}\gamma_{5}\psi|\pi\rangle$ $\displaystyle=M^{2}_{\pi}f_{\pi},$
$\displaystyle Z_{A}\langle 0|\bar{\psi}\gamma_{4}\gamma_{5}\psi|\pi\rangle$
$\displaystyle=M_{\pi}f_{\pi}.$ (6)

Since the latter definition will depend on the value of $Z_{A}$, we will
concentrate on the first definition to extract the pion decay constant. The
result without and with LMS at two lattice spacings are shown in Fig. 4. One
can see that the LMS is very helpful to improve the signal, and the results at
two lattice spacings are consistent with each other. Eventually we obtain the
pion decay constant at the physical point as

$f_{\pi}(0.114~{}\mathrm{fm})=0.1301(3)~{}{\rm GeV},\
f_{\pi}(0.084~{}\mathrm{fm})=0.1304(1)~{}{\rm GeV},$ (7)

and that in the continuum with the linear $a^{2}$ extrapolation is

$f_{\pi}=0.1307(4)(3)~{}{\rm GeV},$ (8)

where the second error estimates the systematic uncertainty from the continuum
extrapolation by taking difference between the value at smaller lattice
spacing and continuum.

Figure 5: Quark mass dependence of the pion mass and decay constant based on
the FLAG values.

The expansions of the pion mass square and decay constant as the function of
the low energy constants $\Sigma$, $F$ and also quark mass $m$ are known based
on $N_{f}=2$ chiral effective theory (Note that $F_{\pi}\equiv
f_{\pi}/\sqrt{2}$),

$\displaystyle M_{\pi}^{2}$
$\displaystyle=M^{2}[1-\frac{1}{2}x(\ln\frac{M^{2}_{\pi,{\rm
phys}}}{M^{2}}+{\ell}_{3})+\mathcal{O}(x^{2})],$ (9) $\displaystyle F_{\pi}$
$\displaystyle=F[1+x(\ln\frac{M^{2}_{\pi,{\rm
phys}}}{M^{2}}+\ell_{4})+\mathcal{O}(x^{2})],$ (10)

where $x=\frac{M^{2}}{(4\pi F)^{2}}$, $M^{2}=\frac{2\Sigma m}{F^{2}}\propto
m$, $M_{\pi,{\rm phys}}=134.98$ MeV, and the lattice averages of the next
leading order (NLO) low energy constants are ${\ell}_{3}=3.07(64)$ and
${\ell}_{4}=4.02(45)$ respectively [2]. Based on the lattice averages of
$\Sigma$ and $F$, the quark mass dependence of the pion mass and decay
constant are shown in Fig. 5.

If we use the FLAG values of ${\ell}_{3,4}$ and our determination of the
physical quark mass and decay constant, we would estimate

$\displaystyle F=87.0(5)(7)~{}{\rm
MeV},\quad\Sigma^{\overline{\textrm{MS}}(2~{}{\rm GeV})}$
$\displaystyle=\big{(}266(3)(1)~{}{\rm MeV}\big{)}^{3}.$ (11)

##  4 Summary

In summary, we calculated the pion masses and decay constants with different
valence quark masses using the overlap fermion, on the 2+1 flavor DW ensembles
with physical pion mass at two lattice spacings. We found that the mixed
action effect is much smaller than the other mixed action setup so far, while
it is still much larger than the statistical uncertainty needed by a high
precision quark mass determination. Thus we only consider the pure valence
pion mass at the unitary point, and use the current FLAG averages of the NLO
low energy constants to determine the LO low energy constants, $\Sigma$ and
$F$. The preliminary estimates are the following:

$\displaystyle m^{\overline{\textrm{MS}}(2~{}{\rm GeV})}_{ud}$
$\displaystyle=3.74(4)(5)(5)(3)~{}{\rm MeV},$
$\displaystyle\Sigma^{\overline{\textrm{MS}}(2~{}{\rm GeV})}$
$\displaystyle=\big{(}266(2)(1)~{}{\rm MeV}\big{)}^{3},$ $\displaystyle
F_{\pi}$ $\displaystyle=92.4(3)(2)~{}{\rm MeV},$ $\displaystyle F$
$\displaystyle=87.0(5)(7)~{}{\rm MeV}.$ (12)

The decay constant we obtain here is consistent with the FLAG value 86.2(5)
MeV. But the chiral condensate $\big{(}266(2)(1)~{}{\rm MeV}\big{)}^{3}$ is
somehow lower than the FLAG average $\big{(}272(5)~{}{\rm MeV}\big{)}^{3}$,
while close to our previous determination from the Dirac spectrum [13],
$\big{(}260(1)(2)~{}{\rm MeV}\big{)}^{3}$. The tension between those two
values is under investigation. If we force the physical quark mass to be the
FLAG average 3.36(4) MeV, then the chiral condensate will be
$\big{(}276(3)~{}{\rm MeV}\big{)}^{3}$ and perfectly consistent with the FLAG
average of the chiral condensate.

In the next step, we would try to combine the results of the other ensembles
to extract the LO and NLO low energy constant simultaneously, it would
suppress the uncertainty of our determination of $F$ and $\Sigma$. At the same
time, the other sources of the systematic uncertainties like the lattice
spacing determination, isospin symmetry breaking effect, partially quenching
effect (the sea pion mass is 4 MeV higher than 135 MeV) and so on, should be
investigated.

## References

  * [1] T. Blum et al.  Domain wall QCD with physical quark masses.  Phys. Rev. D, 93(7):074505, 2016.
  * [2] S. Aoki et al.  FLAG Review 2019: Flavour Lattice Averaging Group (FLAG).  Eur. Phys. J. C, 80(2):113, 2020.
  * [3] Stephen R. Sharpe and Noam Shoresh.  Physical results from unphysical simulations.  Phys. Rev. D, 62:094503, 2000.
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  * [5] C. Aubin, Jack Laiho, and Ruth S. Van de Water.  Discretization effects and the scalar meson correlator in mixed-action lattice simulations.  Phys. Rev. D, 77:114501, 2008.
  * [6] Kostas Orginos and Andre Walker-Loud.  Mixed meson masses with domain-wall valence and staggered sea fermions.  Phys. Rev. D, 77:094505, 2008.
  * [7] M. Lujan, A. Alexandru, Y. Chen, T. Draper, W. Freeman, M. Gong, F. X. Lee, A. Li, K. F. Liu, and N. Mathur.  The $\Delta_{mix}$ parameter in the overlap on domain-wall mixed action.  Phys. Rev. D, 86:014501, 2012.
  * [8] A. Li et al.  Overlap Valence on 2+1 Flavor Domain Wall Fermion Configurations with Deflation and Low-mode Substitution.  Phys. Rev. D, 82:114501, 2010.
  * [9] Ting-Wai Chiu and Sergei V. Zenkin.  On solutions of the Ginsparg-Wilson relation.  Phys. Rev. D, 59:074501, 1999.
  * [10] Fangcheng He and Yi-Bo Yang.  Renormalization constants of overlap quark bilinear operators from RI/MOM and RI/SMOM scheme.  In 38th International Symposium on Lattice Field Theory, 12 2021\.
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# A Generalization of the Erdős-Kac Theorem

Matthew Levy
<EMAIL_ADDRESS>

Joseph Squillace
<EMAIL_ADDRESS>
###### Abstract.

Given a natural number $n$, let $\omega\left(n\right)$ denote the number of
distinct prime factors of $n$, let $Z$ denote a standard normal variable, and
let $P_{n}$ denote the uniform distribution on $\left\\{1,\ldots,n\right\\}$.
The Erdős-Kac Theorem states that if $N\left(n\right)$ is a uniformly
distributed variable on $\\{1,\ldots,n\\}$, then
$\omega\left(N\left(n\right)\right)$ is asymptotically normally distributed as
$n\to\infty$ with both mean and variance equal to $\log\log n$. The
contribution of this paper is a generalization of the Erdős-Kac Theorem to a
larger class of random variables by considering perturbations of the uniform
probability mass $1/n$ in the following sense. Denote by $\mathbb{P}_{n}$ a
probability distribution on $\left\\{1,\ldots,n\right\\}$ given by
$\mathbb{P}_{n}\left(i\right)=1/n+\varepsilon_{i,n}$. We provide sufficient
conditions on $\varepsilon_{i,n}$ so that the number of distinct prime factors
of a $\mathbb{P}_{n}$-distributed random variable is asymptotically normally
distributed, as $n\to\infty$, with both mean and variance equal to $\log\log
n$. Our main result is applied to prove that the number of distinct prime
factors of a positive integer with the Harmonic$\left(n\right)$ distribution
also tends to the normal distribution, as $n\to\infty$. In addition, we
explore sequences of distributions on the natural numbers such that
$\omega(n)$ is normally distributed in the limit. In addition, one of our
theorems and its corollaries generalize a result from the literature involving
the limit of $Zeta\left(s\right)$ distributions as the parameter $s\to 1$.

##  1\. Introduction

Given a natural number $n$, the number of distinct prime factors of $n$ is
denoted $\omega\left(n\right)$. The function $\omega$ may be written as
$\omega\left(n\right)=\sum_{p|n}1$, where the sum is over all prime factors of
$n$. In 1917, Hardy and Ramanujan (p. 270 of [5]) proved that the number of
distinct prime factors of a natural number $n$ is about $\log\log n$. In
particular, they showed that the normal order of $\omega\left(n\right)$ is
$\log\log n$; i.e., for every $\varepsilon>0$, the proportion of the natural
numbers for which the inequalities

$\left(1-\varepsilon\right)\log\log
n\leq\omega\left(n\right)\leq\left(1+\varepsilon\right)\log\log n$

do not hold tends to $0$ as $n\to\infty$–with a typical error of size
$\sqrt{\log\log n}$. Informally speaking, the Erdős-Kac Theorem
generalizes111While the Hardy-Ramanujan theorem provides information about the
average behavior of the number of prime factors of a natural number, the
Erdős-Kac theorem offers a more detailed probabilistic description of their
distribution, by taking into account not just the average number but also the
variability around that average. the Hardy-Ramanujan Theorem by showing that
$\omega\left(n\right)$ is approximately distributed as

$\log\log n+Z\sqrt{\log\log n}$

for large $n$, where $Z$ denotes a standard normal variable. More precisely,
the Erdős-Kac Theorem is the following result (p. 738 of [4]).

######  Theorem 1.

Let $n>1$. Let $P_{n}$ denote the uniform distribution on
$\left\\{1,2,,\ldots,n\right\\}$, and let $Z$ denote a standard normal
variable. As $n\to\infty$,

$P_{n}\left(m\leq n:\omega\left(m\right)-\log\log n\leq x\left(\log\log
n\right)^{1/2}\right)\to\mathbb{P}\left(Z\leq x\right).$

The contribution of this paper is to extend the Erdős-Kac Theorem to a larger
class of random variables on the set
$\left[n\right]\coloneqq\left\\{1,2,\ldots,n\right\\}$ which also have,
asymptotically, $\log\log n+Z\cdot\sqrt{\log\log n}$ many distinct prime
factors.

###  1.1. A Generalization of Erdős-Kac Theorem for $\omega\left(\cdot\right)$

Define a probability mass function (PMF) $\mathbb{P}_{n}$ on $\left[n\right]$
given by

(1) $\mathbb{P}_{n}\left(i\right)=\frac{1}{n}+\varepsilon_{i,n}.$

Due to the axioms of probability, the terms $\varepsilon_{i,n},1\leq i\leq n$,
satisfy the constraints

(2) $\sum_{i=1}^{n}\varepsilon_{i,n}=0;\,\,n\in\mathbb{N}$

and

(3)
$\varepsilon_{i,n}\in\left[-\frac{1}{n},1-\frac{1}{n}\right];\,\,n\in\mathbb{N},1\leq
i\leq n.$

The motivation for defining $\mathbb{P}_{n}$ in terms of the uniform
distribution is due to Durrett’s proof (Theorem 3.4.16 in [3]) of the Erdős-
Kac Theorem. Replacing the uniform distribution $P_{n}$ with the distribution
$\mathbb{P}_{n}$ in Durrett’s proof naturally yields some constraints that the
terms $\varepsilon_{i,n},1\leq i\leq n$, must satisfy in order to conclude
that an integer-valued $\mathbb{P}_{n}$-distributed random variable has about
$\log\log n+Z\sqrt{\log\log n}$ distinct prime factors. Our main result is the
following theorem, where $\left\lfloor\cdot\right\rfloor$ denotes the floor
function.

######  Theorem 2.

(Generalized Erdős-Kac Theorem for $\omega$) Let $Z$ denote a standard normal
variable. Suppose the following statements are true.

  * •

There exists a constant $C\in\mathbb{R}$ such that for all $n>1$ and for all
primes $p$ with $p>n^{1/\log\log n}$,

(4) $\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\varepsilon_{lp,n}\leq\frac{C}{p}.$

  * •

There exists a constant $D\in\mathbb{R}$ such that

(5) $\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\varepsilon_{lp_{1}\cdots p_{k},n}\leq\frac{D}{n}$

for all $n>1$, for each $k$, and for all $k$-tuples
$\left(p_{1},\ldots,p_{k}\right)$ consisting of distinct primes of size at
most $n^{1/\log\log n}$.

  * •

For any prime $p$,

(6) $\lim_{n\to\infty}\sum_{l=1}^{\lfloor n/p\rfloor}\varepsilon_{lp,n}=0.$

Let $\mathbb{P}_{n}^{*}$ denote the PMF obtained by imposing the constraints
$\left(\ref{eq:4}-\ref{eq:6}\right)$ on the PMF $\mathbb{P}_{n}$ given by
$\mathbb{P}_{n}\left(i\right)=\frac{1}{n}+\varepsilon_{i,n}$. As $n\to\infty,$

$\displaystyle\mathbb{P}_{n}^{*}\left(m\leq n:\omega\left(m\right)-\log\log
n\leq x\left(\log\log n\right)^{1/2}\right)$
$\displaystyle\to\mathbb{P}\left(Z\leq x\right).$

######  Remark.

If $\varepsilon_{i,n}=0$ for all $i\leq n$, then $\mathbb{P}_{n}^{*}=P_{n}$
and Theorem 1 is obtained.

###  1.2. Outline

The proof of Theorem 2 is provided in $\S$2; the proof applies the method of
moments and is motivated by Durrett’s proof of the Erdős-Kac Theorem (Theorem
3.4.16 in [3]). Moreover, in $\S$2, the constraints
$\left(\ref{eq:4}-\ref{eq:6}\right)$ are applied to ensure that
$\mathbb{P}_{n}^{*}$ also satisfies Durrett’s method of moment bounds. In
$\S\S$2.1, Theorem 2 is applied to show that the number of distinct prime
factors of a random natural number chosen according to the
Harmonic$\left(n\right)$ distribution is asymptotically normally distributed
with both mean and variance equal to $\log\log n$. In $\S$3, Theorem 2 is used
to prove statements about convex combinations of distributions satisfying
constraints $\left(\ref{eq:4}-\ref{eq:6}\right)$. In $\S$4, we define
conditions that ensure $\omega(X_{j}\left(n\right))$ is asymptotically
normally distributed, with mean and variance both equal to $\log\log n$, for a
sequence of random variables $\left(X_{j}\left(n\right)\right)_{j\geq 1}$ as
$j\to\infty$ and $n\to\infty$. In $\S$5, Theorem 2 is applied to show that the
number of distinct prime factors of a randomly chosen integer according to any
of the following distributions has the same limiting distribution as the case
of a uniform variable:

  * •

Any convex combination of the Harmonic$\left(n\right)$ and
uniform$\left(n\right)$ distributions,

  * •

The $Zeta\left(s\right)$ distribution as $s\to 1$,

  * •

The Logarithmic$\left(s\right)$ distribution as $s\to 1$,

  * •

A geometric power series distribution as $s\to 1$,

  * •

A Logarithmic-Zeta$\left(s,\alpha\right)$ distribution as
$(s,\alpha)\to(1,1)$.

##  2\. Proving Theorem 2

Define $\alpha_{n}\coloneqq n^{1/\log\log n}$.

######  Lemma 1.

As $n\to\infty$

$\left(\sum_{\alpha_{n}<p\leq n}\left(\frac{1}{p}+\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\varepsilon_{lp,n}\right)\right)/\left(\log\log
n\right)^{1/2}\to 0.$

###### Proof.

Given $n$ and any prime $p$ with $p>\alpha_{n}$, we have

$-\frac{\left\lfloor
n/p\right\rfloor}{n}\overset{\left(\ref{eq:3}\right)}{\leq}\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\varepsilon_{lp,n}\overset{\left(\ref{eq:4}\right)}{\leq}\frac{C}{p}.$

Therefore,

(7) $\frac{1}{p}+\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\varepsilon_{lp,n}\in\left[0,\frac{C+1}{p}\right]$

for all $n$. Thus,

$\left(\sum_{\alpha_{n}<p\leq n}\left(\frac{1}{p}+\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\varepsilon_{lp,n}\right)\right)/\left(\log\log
n\right)^{1/2}\to 0$

due to $\left(\ref{eq:7}\right)$ along with the fact that Durrett (p.135 of
[3]) shows

$\left(\sum_{\alpha_{n}<p\leq n}\frac{1}{p}\right)/\left(\log\log
n\right)^{1/2}\to 0.$

∎

The following lemma is proved by Durrett (p. 156 of [3]).

######  Lemma 2.

If $\varepsilon>0$, then $\alpha_{n}\leq n^{\varepsilon}$ for large $n$ and
hence

(8) $\frac{\alpha_{n}^{r}}{n}\to 0$

for all $r<\infty$.

###### Proof of Theorem 2.

Given a natural number $m$ and a prime $p$, define
$\delta_{p}\left(m\right)=1$ if $p$ divides $m$, and $0$ otherwise. Let

$g_{n}\left(m\right)=\sum_{p\leq\alpha_{n}}\delta_{p}\left(m\right)$

denote the number of distinct prime factors of $m$ of size at most
$\alpha_{n}$, and let $\mathbb{E}_{n}$ denote expectation with respect to
$\mathbb{P}_{n}^{*}$. Then

$\displaystyle\mathbb{E}_{n}\left(\sum_{\alpha_{n}<p\leq n}\delta_{p}\right)$
$\displaystyle=\sum_{\alpha_{n}<p\leq
n}\mathbb{P}_{n}^{*}\left(m:\delta_{p}\left(m\right)=1\right)$
$\displaystyle=\sum_{\alpha_{n}<p\leq n}\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\mathbb{P}_{n}^{*}\left(m:m=lp\right)$
$\displaystyle\overset{\left(\ref{eq:1}\right)}{=}\sum_{\alpha_{n}<p\leq
n}\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\left(\frac{1}{n}+\varepsilon_{lp,n}\right)$
$\displaystyle\leq\sum_{\alpha_{n}<p\leq
n}\left(\frac{1}{p}+\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\varepsilon_{lp,n}\right),$

so by Lemma 1 it suffices to prove Theorem 2 for $g_{n}$; i.e., replacing
$\omega\left(m\right)$ with $g_{n}\left(m\right)$ in the statement of Theorem
2 does not affect the limiting distribution.

Consider a sequence $\left(X_{p}\right)_{p\geq 2}$ of independent Bernoulli
random variables with prime-valued indices such that
$\mathbb{P}\left(X_{p}=1\right)=1/p$ and
$\mathbb{P}\left(X_{p}=0\right)=1-1/p$. Note that

$\mathbb{E}\left(\delta_{p}\right)=\frac{\left\lfloor
n/p\right\rfloor}{n}+\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\varepsilon_{lp,n}\overset{\left(6\right)}{\to}1/p$

as $n\to\infty$. Let

$\displaystyle S_{n}$ $\displaystyle\coloneqq$
$\displaystyle\sum_{p\leq\alpha_{n}}X_{p},$ $\displaystyle b_{n}$
$\displaystyle\coloneqq$ $\displaystyle\mathbb{E}\left(S_{n}\right),$
$\displaystyle a_{n}^{2}$ $\displaystyle\coloneqq$
$\displaystyle\text{Var}\left(S_{n}\right).$

By Lemma 1, $b_{n}$ and $a_{n}^{2}$ are both $\log\log n+o\left(\left(\log\log
n\right)^{1/2}\right)$, so it suffices to show

$\mathbb{P}_{n}^{*}\left(m:g_{n}\left(m\right)-b_{n}\leq
xa_{n}\right)\to\mathbb{P}\left(Z\leq x\right).$

An application of Theorem 3.4.10 of [3] shows

$\left(S_{n}-b_{n}\right)/a_{n}\to Z,$

and since $\left|X_{p}\right|\leq 1$, it follows from Durrett’s second proof
of Theorem 3.4.10 [3] that

$\mathbb{E}\left(\left(S_{n}-b_{n}\right)/a_{n}\right)^{r}\to\mathbb{E}\left(Z^{r}\right)$

for all $r$. Using the notation from that proof (and replacing $i_{j}$ by
$p_{j}$) it follows that

$\displaystyle\mathbb{E}\left(S_{n}^{r}\right)$
$\displaystyle=\sum_{k=1}^{r}\sum_{r_{i}}\frac{r!}{r_{1}!\cdots
r_{k}!}\frac{1}{k!}\sum_{p_{j}}\mathbb{E}\left(X_{p_{1}}^{r_{1}}\cdots
X_{p_{k}}^{r_{k}}\right),$

where the sum $\sum_{r_{i}}$ extends over all $k$-tuples of positive integers
for which $r_{1}+\cdots+r_{k}=r$, and $\sum_{p_{j}}$ extends over all
$k$-tuples of distinct primes in $\left[n\right]$. Since
$X_{p}\in\left\\{0,1\right\\}$, the summand in
$\sum_{p_{j}}\mathbb{E}\left(X_{p_{1}}^{r_{1}}\cdots X_{p_{k}}^{r_{k}}\right)$
is

$\mathbb{E}\left(X_{p_{1}}\cdots X_{p_{k}}\right)=\frac{1}{p_{1}\cdots p_{k}}$

by independence of the $X_{p}$’s. Moreover,

$\displaystyle\mathbb{E}_{n}\left(\delta_{p_{1}}\cdots\delta_{p_{k}}\right)$
$\displaystyle\leq\mathbb{P}_{n}\left(m:\delta_{p_{1}}\left(m\right)=\delta_{p_{2}}\left(m\right)\cdots=\delta_{p_{k}}\left(m\right)=1\right)$
$\displaystyle=\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\mathbb{P}_{n}\left(m:m=lp_{1}\cdots p_{k}\right)$
$\displaystyle\overset{\left(1\right)}{=}\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\left(\frac{1}{n}+\varepsilon_{lp_{1}\cdots
p_{k},n}\right)$ $\displaystyle=\frac{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}{n}+\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\varepsilon_{lp_{1}\cdots p_{k},n}$
$\displaystyle\overset{\left(5\right)}{\leq}\frac{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}{n}+\frac{D}{n}.$

The two terms $\mathbb{E}\left(X_{p_{1}}\cdots X_{p_{k}}\right)$ and
$\mathbb{E}_{n}\left(\delta_{p_{1}}\cdots\delta_{p_{k}}\right)$ differ by at
most

$\text{max}\left\\{\frac{1}{p_{1}\cdots
p_{k}}-\frac{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}{n}-\frac{D}{n},\frac{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}{n}+\frac{D}{n}-\frac{1}{p_{1}\cdots
p_{k}}\right\\}\leq\text{max}\left\\{\frac{1-D}{n},\frac{D}{n}\right\\}.$

Therefore, the two $r$th moments differ by

$\displaystyle\left|\mathbb{E}\left(S_{n}^{r}\right)-\mathbb{E}_{n}\left(g_{n}^{r}\right)\right|$
$\displaystyle\leq$
$\displaystyle\sum_{k=1}^{r}\sum_{r_{i}}\frac{r!}{r_{1}!\cdots
r_{k}!}\frac{1}{k!}\sum_{p_{j}}\max\left\\{\frac{1-D}{n},\frac{D}{n}\right\\}.$
$\displaystyle\leq$
$\displaystyle\max\left\\{\frac{1-D}{n},\frac{D}{n}\right\\}\cdot\left(\sum_{p\leq\alpha_{n}}1\right)^{r}$
$\displaystyle\leq$
$\displaystyle\max\left\\{\frac{1-D}{n},\frac{D}{n}\right\\}\cdot\alpha_{n}^{r}$
$\displaystyle\overset{\left(\ref{eq:8}\right)}{\to}$ $\displaystyle 0.$

Using binomial expansions and the inequality above, we see that

$\displaystyle\left|\mathbb{E}\left(\left(\left(S_{n}-b_{n}\right)/a_{n}\right)^{r}\right)-\mathbb{E}_{n}\left(\left(\left(g_{n}-b_{n}\right)/a_{n}\right)^{r}\right)\right|$
$\displaystyle=\left|1/a_{n}^{r}\right|\left|\mathbb{E}\left(\left(S_{n}-b_{n}\right)^{r}\right)-\mathbb{E}_{n}\left(\left(g_{n}-b_{n}\right)^{r}\right)\right|$
$\displaystyle\leq\left|1/a_{n}^{r}\right|\cdot\max\left\\{\frac{1-D}{n},\frac{D}{n}\right\\}\sum_{k=0}^{r}\binom{r}{k}\alpha_{n}^{k}b_{n}^{r-k}$
$\displaystyle=\left|1/a_{n}^{r}\right|\cdot\max\left\\{\frac{1-D}{n},\frac{D}{n}\right\\}\left(\alpha_{n}+b_{n}\right)^{r}.$

Therefore, since $b_{n}\leq\alpha_{n}$, we have

$\left|\mathbb{E}\left(\left(\left(S_{n}-b_{n}\right)/a_{n}\right)^{r}\right)-\mathbb{E}_{n}\left(\left(\left(g_{n}-b_{n}\right)/a_{n}\right)^{r}\right)\right|\overset{\left(\ref{eq:8}\right)}{\to}0$

for all $r$ as well. Since
$\mathbb{E}\left(\left(\left(S_{n}-b_{n}\right)/a_{n}\right)^{r}\right)\to\mathbb{E}\left(Z^{r}\right)$
for all $r$, this completes the proof of Theorem 2. ∎

The following definition is based on Theorem 2.

######  Definition 1.

We refer to distributions satisfying constraints
$\left(\ref{eq:4}-\ref{eq:6}\right)$ as E-K distributions.

###  2.1. The Harmonic Distribution

Now we will apply Theorem 2 to show that the harmonic distributions are E-K
distributions. Given $n\in\mathbb{N},$ consider an integer in $\left[n\right]$
chosen according to the Harmonic$\left(n\right)$ distribution, whose PMF is
given by

$H_{n}\left(i\right)\coloneqq\frac{1}{i\sum_{j=1}^{n}\frac{1}{j}},1\leq i\leq
n.$

If $i\in\left[n\right]$, then equation $\left(\ref{eq:1}\right)$ implies

$\varepsilon_{i,n}=\frac{1}{i\sum_{j=1}^{n}\frac{1}{j}}-\frac{1}{n}.$

Therefore,

$\displaystyle\sum_{l=1}^{\left\lfloor n/p\right\rfloor}\varepsilon_{lp,n}$
$\displaystyle=\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\left(\frac{1}{lp\sum_{i=1}^{n}\frac{1}{i}}-\frac{1}{n}\right)$
$\displaystyle=\frac{\sum_{i=1}^{\left\lfloor
n/p\right\rfloor}\frac{1}{i}}{p\sum_{l=1}^{n}\frac{1}{i}}-\frac{\lfloor
n/p\rfloor}{n}$ $\displaystyle\leq\frac{1}{p}-\frac{\left\lfloor
n/p\right\rfloor}{n}$ $\displaystyle\leq\frac{1}{p},$

so $\left(\ref{eq:4}\right)$ holds with $C=1$. Moreover,

$\displaystyle\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\varepsilon_{lp_{1}\cdots p_{k},n}$
$\displaystyle=\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\left(\frac{1}{lp_{1}\cdots
p_{k}\sum_{i=1}^{n}\frac{1}{i}}-\frac{1}{n}\right)$
$\displaystyle=\frac{\sum_{i=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\frac{1}{i}}{p_{1}\cdots
p_{k}\sum_{i=1}^{n}\frac{1}{i}}-\frac{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}{n}$ $\displaystyle\leq\frac{1}{p_{1}\cdots
p_{k}}-\left(\frac{\frac{n}{p_{1}\cdots p_{k}}-1}{n}\right)$
$\displaystyle=1/n,$

so $\left(\ref{eq:5}\right)$ holds with $D=1$. Finally,

$\displaystyle\lim_{n\to\infty}\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\varepsilon_{lp,n}$
$\displaystyle=\lim_{n\to\infty}\left(\frac{\sum_{i=1}^{\left\lfloor
n/p\right\rfloor}\frac{1}{i}}{p\sum_{l=1}^{n}\frac{1}{i}}-\frac{\lfloor
n/p\rfloor}{n}\right)$
$\displaystyle\sim\frac{1}{p}\frac{\log\left(n/p\right)}{\log n}-\frac{1}{p}$
$\displaystyle\to 1/p-1/p$ $\displaystyle=0,$

so constraint $\left(\ref{eq:6}\right)$ holds. By Theorem 2, this shows that
the number of distinct prime factors of an $H_{n}$-distributed random variable
is asymptotically normally distributed, as $n\to\infty$, with both mean and
variance equal to $\log\log n$.

##  3\. Convex Combinations of E-K Distributions

The following theorem shows that any convex sum of two E-K distributions is
also an E-K distribution.

######  Theorem 3.

(Convex Combinations for Erdős-Kac (CLT)) Let $n>1$ and $0\leq\lambda\leq 1$
be fixed. Suppose $d_{1,n}$ and $d_{2,n}$ are two PMFs on $\left[n\right]$
satisfying the constraints $\left(\ref{eq:4}-\ref{eq:6}\right)$ for all $n>1$.
Then any PMF of the form

(9) $\mathbb{P}^{*}_{n}\left(i\right)=\lambda
d_{1,n}\left(i\right)+\left(1-\lambda\right)d_{2,n}\left(i\right),1\leq i\leq
n$

also satisfies constraints $\left(\ref{eq:4}-\ref{eq:6}\right)$. In
particular,

$\displaystyle\mathbb{P}_{n}^{*}\left(m\leq n:\omega\left(m\right)-\log\log
n\leq x\left(\log\log n\right)^{1/2}\right)$
$\displaystyle\to\mathbb{P}\left(Z\leq x\right)$

as $n\to\infty$.

###### Proof.

Denote

$\varepsilon_{i,n}=d_{1,n}\left(i\right)-1/n,$
$\varepsilon^{\prime}_{i,n}=d_{2,n}\left(i\right)-1/n,$

and

$\hat{\varepsilon}_{i,n}=\mathbb{P}^{*}_{n}\left(i\right)-1/n.$

We have

$\displaystyle\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\hat{\varepsilon}_{lp,n}$
$\displaystyle=\sum_{l=1}^{\left\lfloor n/p\right\rfloor}\left(\lambda
d_{1,n}\left(lp\right)+\left(1-\lambda\right)d_{2,n}\left(lp\right)-\frac{1}{n}\right)$
$\displaystyle\overset{\left(\ref{eq:1}\right)}{=}\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\left(\lambda\left(\varepsilon_{lp,n}+1/n\right)+\left(1-\lambda\right)\left(\varepsilon^{\prime}_{lp,n}+1/n\right)-\frac{1}{n}\right)$
$\displaystyle=\sum_{l=1}^{\lfloor
n/p\rfloor}\left(\lambda\varepsilon_{lp,n}+\left(1-\lambda\right)\varepsilon_{lp,n}^{\prime}\right)$
$\displaystyle=\lambda\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\varepsilon_{lp,n}+\left(1-\lambda\right)\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\varepsilon^{\prime}_{lp,n}$ $\displaystyle\leq\frac{\lambda
C_{1}+\left(1-\lambda\right)C_{2}}{p},$

where the latest inequality is obtained by applying constraint
$\left(\ref{eq:4}\right)$ to both $d_{1,n}$ and $d_{2,n}$; therefore,
$\left(\ref{eq:4}\right)$ holds for the PMF given by $\left(\ref{eq:9}\right)$
with $C=\lambda C_{1}+\left(1-\lambda\right)C_{2}$. Similarly,

$\displaystyle\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\hat{\varepsilon}_{lp_{1}\cdots p_{k},n}$
$\displaystyle=\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\left(\lambda d_{1,n}\left(lp_{1}\cdots
p_{k}\right)+\left(1-\lambda\right)d_{2,n}\left(lp_{1}\cdots
p_{k}\right)-\frac{1}{n}\right)$
$\displaystyle\overset{\left(\ref{eq:1}\right)}{=}\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\left(\lambda\left(\varepsilon_{lp_{1}\cdots
p_{k},n}+1/n\right)+\left(1-\lambda\right)\left(\varepsilon^{\prime}_{lp_{1}\cdots
p_{k},n}+1/n\right)-\frac{1}{n}\right)$
$\displaystyle=\sum_{l=1}^{\lfloor\frac{n}{p_{1}\cdots
p_{k}}\rfloor}\left(\lambda\varepsilon_{lp_{1}\cdots
p_{k},n}+\left(1-\lambda\right)\varepsilon_{lp_{1}\cdots
p_{k},n}^{\prime}\right)$
$\displaystyle=\lambda\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\varepsilon_{lp_{1}\cdots
p_{k},n}+\left(1-\lambda\right)\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\varepsilon^{\prime}_{lp_{1}\cdots p_{k},n}$
$\displaystyle\leq\frac{\lambda D_{1}+\left(1-\lambda\right)D_{2}}{n},$

where the latest inequality is obtained by applying constraint
$\left(\ref{eq:5}\right)$ to both $d_{1,n}$ and $d_{2,n}$; therefore,
$\left(\ref{eq:5}\right)$ holds for the PMF given by $\left(\ref{eq:9}\right)$
with $D=\lambda D_{1}+\left(1-\lambda\right)D_{2}$. Furthermore,

$\displaystyle\lim_{n\to\infty}\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\hat{\varepsilon}_{lp,n}$
$\displaystyle=\lambda\lim_{n\to\infty}\sum_{l=1}^{\lfloor
n/p\rfloor}\varepsilon_{lp,n}+\left(1-\lambda\right)\lim_{n\to\infty}\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\varepsilon_{lp,n}^{\prime}$ $\displaystyle=\lambda\cdot
0+\left(1-\lambda\right)\cdot 0$ $\displaystyle=0,$

where the second equation uses the fact that the distributions $d_{1}$ and
$d_{2}$ satisfy constraint $\left(\ref{eq:6}\right)$. Therefore,
$\left(\ref{eq:6}\right)$ holds for the PMF given by
$\left(\ref{eq:9}\right)$.

∎

By Theorem 3 and mathematical induction, we obtain the following.

######  Corollary 1.

Suppose $d_{1,n},\ldots,d_{k,n}$ are PMFs on $\left[n\right]$ satisfying the
constraints $\left(\ref{eq:4}-\ref{eq:6}\right)$, and suppose
$\left(\lambda_{j}\right)_{j=1}^{k}$ is a sequence of nonnegative numbers
which add to unity. The PMF

$\mathbb{P}^{*}_{n}\left(i\right)=\sum_{j=1}^{k}\lambda_{j}d_{j,n}(i)$

also satisfies constraints $\left(\ref{eq:4}-\ref{eq:6}\right)$. Therefore,

$\displaystyle\mathbb{P}_{n}^{*}\left(m\leq n:\omega\left(m\right)-\log\log
n\leq x\left(\log\log n\right)^{1/2}\right)$
$\displaystyle\to\mathbb{P}\left(Z\leq x\right)$

as $n\to\infty$.

We now present an alternate proof to the remark under Corollary 3.4 in [1];
their proof uses both the Tauberian theorems for the harmonic distribution and
the uniform distribution simultaneously. We present a proof based on Theorems
2 and 3 of this paper.

######  Corollary 2.

(Convex Combinations of Harmonic Distribution and Uniform Distribution) Let
$\lambda\in\left[0,1\right]$ and consider the PMF given by

$\mathbb{P}^{*}_{n}\left(i\right)=\frac{\lambda}{ih_{n}}+\frac{1-\lambda}{n};1\leq
i\leq n,$

where $h_{n}$ is the $n^{th}$ harmonic number given by
$h_{n}\coloneq\sum_{i=1}^{n}1/i$, then

$\mathbb{P}_{n}^{*}\left(m\leq n:\omega\left(m\right)-\log\log n\leq
x\left(\log\log n\right)^{1/2}\right)\to\mathbb{P}\left(Z\leq x\right)$

as $n\to\infty$.

###### Proof.

As noted in the remark after Theorem 2, the uniform distribution satisfies
constraints $\left(\ref{eq:4}-\ref{eq:6}\right)$, and in $\S\S$2.1 we showed
the harmonic distribution satisfies constraints
$\left(\ref{eq:4}-\ref{eq:6}\right)$. By Theorem 3, their convex sums also
satisfy constraints $\left(\ref{eq:4}-\ref{eq:6}\right)$. ∎

The following theorem shows that if given the PMF $1/n+\varepsilon_{i,n}$ of a
sequence of E-K distributions, under additional constraints, the PMF
$1/n-\varepsilon_{i,n}$ is also the PMF of a sequence of E-K distributions.

######  Theorem 4.

(Reflection Theorem) Let $\mathbb{P}_{n}$ satisfy constraints
$\left(\ref{eq:4}\right)$ and $\left(\ref{eq:6}\right)$ for all $n>1$. In
addition, assume $\varepsilon_{i,n}\in\left[-\frac{1}{n},\frac{1}{n}\right]$
for all $i$ and for all $n$, and that for all $k$-tuples of distinct primes
$p_{1},\ldots,p_{k}$, there exists a constant $D\geq 0$ such that

(10) $-\frac{D}{n}\leq\sum_{l=1}^{\lfloor\frac{n}{p_{1}\cdots
p_{k}}\rfloor}\varepsilon_{lp_{1}\cdots p_{k},n}\leq\frac{D}{n}.$

If

$\mathbb{P}_{n}^{*}\left(i\right)\coloneqq\frac{1}{n}-\varepsilon_{i,n},$

then

$\displaystyle\mathbb{P}_{n}^{*}\left(m\leq n:\omega\left(m\right)-\log\log
n\leq x\left(\log\log n\right)^{1/2}\right)$
$\displaystyle\to\mathbb{P}\left(Z\leq x\right)$

as $n\to\infty$.

###### Proof.

Define $\varepsilon_{i,n}^{\prime}\coloneq-\varepsilon_{i,n}$, so

$\displaystyle\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\varepsilon_{lp,n}^{\prime}$
$\displaystyle=-\sum_{l=1}^{\left\lfloor n/p\right\rfloor}\varepsilon_{i,n}$
$\displaystyle\overset{\left(\ref{eq:3}\right)}{\leq}\frac{\lfloor
n/p\rfloor}{n}$ $\displaystyle\leq\frac{1}{p}$

so constraint $\left(\ref{eq:4}\right)$ holds with $C=1$. Similarly,

$\displaystyle\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\varepsilon_{lp_{1}\cdots p_{k},n}^{\prime}$
$\displaystyle=-\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\varepsilon_{lp_{1}\cdots
p_{k},n}\overset{\left(\ref{eq:10}\right)}{\leq}\frac{D}{n},$

so constraint $\left(\ref{eq:5}\right)$ holds. Moreover,

$\displaystyle\lim_{n\to\infty}\sum_{l=1}^{\lfloor
n/p\rfloor}\varepsilon_{lp,n}^{\prime}$
$\displaystyle=-\lim_{n\to\infty}\sum_{l=1}^{\lfloor
n/p\rfloor}\varepsilon_{lp,n}$ $\displaystyle=0,$

where the latest limit is due to the assumption that $\mathbb{P}_{n}$
satisfies constraint $\left(\ref{eq:6}\right)$. ∎

##  4\. A Central Limit Theorem for E-K Distributions on $\mathbb{N}$

We would like to formalize the notion of asymptotic normality for
$\omega(X_{j})$ in the limit of a sequence of random variables
$\left(X_{j}\right)_{j\geq 1}$ on $\mathbb{N}$. Towards that goal we provide
the following definition.

######  Definition 2.

Given an infinite sequence of random variables $X_{1},X_{2},\ldots$ defined on
$\mathbb{N}$, let $X_{j}\left(n\right)$ denote the truncation of $X_{j}$ on
$\left[n\right]$. Define
$\varepsilon_{l,j,n}\coloneqq\mathbb{P}\left(X_{j}\left(n\right)=l\right)-1/n$.
The sequence $X_{1},X_{2},\ldots$ is said to satisfy the uniformity along
primes property if the following holds. There exists constants $C$ and $D$
such that for all $n$ there exists a $d\geq 1$ such that for all $j\geq d$:

  * •

For each prime $p>\alpha_{n}$

$\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\varepsilon_{lp,j,n}\leq\frac{C}{p}.$

  * •

For each $k\geq 1$ and for all $k$-tuples $\left(p_{1},\ldots,p_{k}\right)$
consisting of distinct primes of size at most $\alpha_{n}$

$\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\varepsilon_{lp_{1}\cdots p_{k},j,n}\leq\frac{D}{n}.$

In addition, for any prime $p$,

$\lim_{j\to\infty}\lim_{n\to\infty}\sum_{l=1}^{\left\lfloor
n/p\right\rfloor}\varepsilon_{lp,j,n}=0.$

######  Theorem 5.

Assume $X_{1},X_{2},\ldots$ is an infinite sequence of random variables
defined on $\mathbb{N}$ satisfying the uniformity along primes property. Let
$\mathbb{P}_{j,n}$ be the probability distribution of $X_{j}\left(n\right)$ on
$\left[n\right]$. If $\left(n_{j}\right)_{j\in\mathbb{N}}$ is any sequence in
$\mathbb{N}\setminus\\{1\\}$ tending to $\infty$ such that
$(\ref{eq:4}-\ref{eq:5})$ hold for $X_{j}\left(n_{j}\right)$ for all $j\geq
1$, then

$\mathbb{P}_{j,n_{j}}\left(m\leq n_{j}:\omega\left(m\right)-\log\log n_{j}\leq
x\left(\log\log n_{j}\right)^{1/2}\right)\to\mathbb{P}\left(Z\leq x\right)$

as $j\to\infty$.

###### Proof.

Let’s proceed by contradiction. That is, suppose
$\left(n_{j}\right)_{j\in\mathbb{N}}$ is any sequence in
$\mathbb{N}\setminus\left\\{1\right\\}$ tending to $\infty$ and suppose there
exist a sequence $\left(j_{k}\right)_{k\in\mathbb{N}}$, some
$x_{0}\in\mathbb{R}$, and some $\varepsilon_{0}>0$ such that

$\left|\mathbb{P}_{j_{k},n_{j_{k}}}\left(m\leq
n_{j_{k}}:\omega\left(m\right)-\log\log n_{j_{k}}\leq x_{0}\left(\log\log
n_{j_{k}}\right)^{1/2}\right)-\mathbb{P}\left(Z\leq
x_{0}\right)\right|\geq\varepsilon_{0}$

for all $k\geq 1$. Consider the following PMF on $\left[n\right]$ for all
$n>1$, which satisfies the hypotheses of Theorem 2:

$\mathbb{P}_{n}^{*}\left(i\right)\coloneq\begin{cases}1/n&\text{if
}n_{j_{k}}\not=n\text{ for all }k\geq 1,\\\
\mathbb{P}_{j_{k^{\prime}\left(n\right)},n}\left(i\right)&\text{if
}n_{j_{k^{\prime}\left(n\right)}}=n,\end{cases}$

where $k^{\prime}\left(n\right)$ is the smallest integer satisfying
$n_{j_{k^{\prime}\left(n\right)}}=n$. It is clear that $\mathbb{P}_{n}^{*}$ is
defined for all $n>1$ and satisfies the hypotheses of Theorem 2 due to
Definition 2. Let $\varepsilon=\varepsilon_{0}$, then there is some $d\geq 1$
such that

$\left|\mathbb{P}_{j_{k^{\prime}\left(n\right)},n_{j_{k^{\prime}\left(n\right)}}}\left(m\leq
n_{j_{k^{\prime}\left(n\right)}}:\omega\left(m\right)-\log\log
n_{j_{k^{\prime}\left(n\right)}}\leq x_{0}\left(\log\log
n_{j_{k^{\prime}\left(n\right)}}\right)^{1/2}\right)-\mathbb{P}\left(Z\leq
x_{0}\right)\right|<\varepsilon$

if $n\geq d$, by Theorem 2. The fact that at least one
$k^{\prime}\left(n\right)$ exists such that
$n_{j_{k^{\prime}\left(n\right)}}=n$ when $n\geq d$ is due to the fact that
$n_{j_{k}}\to\infty$ as $k\to\infty$, so there are infinitely many
$k^{\prime}\left(n\right)$ such that $n_{j_{k^{\prime}\left(n\right)}}=n$ when
$n\geq d$. Thus, Theorem 5 holds by contradiction. ∎

###  4.1. Moment Generating Functions and Characteristic Functions

######  Definition 3.

Let $X$ denote a random variable. The moment generating function (MGF) of $X$
is defined as $M_{X}\left(t\right)=\mathbb{E}\left(e^{tX}\right)$, and the
characteristic function (CF) of $X$ is defined as
$\phi_{X}\left(t\right)=\mathbb{E}\left(e^{itX}\right)$.

We have the following two facts about the MGF and the CF that will be used in
the proof of Corollary 3:

  * •

The $k^{th}$ derivative of $M_{X}\left(t\right)$ at $t=0$ is the $k^{th}$
moment of $X$.

  * •

Lèvy’s Continuity Theorem: Convergence in distribution,
$X_{n}\overset{d}{\to}X$, for a sequence of random variables is equivalent to
pointwise convergence, $\phi_{X_{n}}\to\phi_{X}$, of the corresponding CFs on
all of $\mathbb{R}$.

Corollary 3 and Corollary 4 below allow us to state Theorem 7.1 of [2] in a
slightly different way.

######  Corollary 3.

Suppose $X_{1},X_{2},\ldots$ is an infinite sequence of random variables
defined on $\mathbb{N}$ that satisfies the uniformity along primes property.
Let $\left(n_{j}\right)_{j\in\mathbb{N}}$ be any sequence in
$\mathbb{N}\setminus\left\\{1\right\\}$ tending to $\infty$ such that (4 \- 5)
hold for $X_{j}\left(n_{j}\right)$ for all $j\geq 1$. Let $\mathbb{P}_{j}$ be
the probability distribution of $X_{j}$ on $\mathbb{N}$, $\mathbb{P}_{j,n}$ be
the probability distribution of $X_{j}\left(n\right)$ on $\left[n\right]$, and
define
$\mu_{j,n}\coloneqq\mathbb{E}_{j,n}\left(\omega\left(X_{j}\left(n\right)\right)\right)$,
then the following is true:

$\mathbb{P}_{j,n_{j}}\left(m\leq n_{j}:\omega\left(m\right)-\mu_{j,n_{j}}\leq
x\left(\mu_{j,n_{j}}\right)^{1/2}\right)\to\mathbb{P}\left(Z\leq x\right)$

as $j\to\infty$. Additionally, if $\mu_{j}<\infty$ for all $j\geq 1$ and

$\lim_{j\to\infty}\left(\mu_{j}/\left(\log\log
n_{j}\right)^{1/2}-\left(\log\log n_{j}\right)^{1/2}\right)=0,$

then

$\mathbb{P}_{j}\left(m:\omega\left(m\right)-\mu_{j}\leq
x\left(\mu_{j}\right)^{1/2}\right)\to\mathbb{P}\left(Z\leq x\right)$

as $j\to\infty$, where

$\mu_{j}\coloneqq\mathbb{E}_{j}\left(\omega\left(X_{j}\right)\right)=\lim_{n\to\infty}\mu_{j,n}.$

We also have the following asymptotic properties:

$\lim_{j\to\infty}\left(\mu_{j,n_{j}}/\left(\log\log
n_{j}\right)^{1/2}-\left(\log\log n_{j}\right)^{1/2}\right)=0,$
$\lim_{j\to\infty}\left(\mu_{j,n_{j}}^{1/2}-\left(\log\log
n_{j}\right)^{1/2}\right)=0,$

and

$\lim_{j\to\infty}\frac{\mu_{j,n_{j}}}{\log\log
n_{j}}=\lim_{j\to\infty}\frac{\mu_{j}}{\log\log n_{j}}=1.$

###### Proof.

Let $M_{j,n_{j}}$ and $\phi_{j,n_{j}}$ be the MGF and CF of

$\frac{\omega\left(X_{j}\left(n_{j}\right)\right)-\log\log
n_{j}}{\sqrt{\log\log n_{j}}},$

and let $M_{j,\mu}$ and $\phi_{j,\mu}$ be the MGF and CF of

$\frac{\omega\left(X_{j}\left(n_{j}\right)\right)-\mu_{j,n_{j}}}{\sqrt{\mu_{j,n_{j}}}}.$

We have

$M_{j,n_{j}}\left(t\right)=e^{-t\left(\log\log
n_{j}\right)^{1/2}}M_{\omega\left(X_{j}\left(n_{j}\right)\right)}\left(\left(\log\log
n_{j}\right)^{-1/2}t\right)$

and

$M_{j,\mu}\left(t\right)=e^{-t\mu_{j,n_{j}}^{1/2}}M_{\omega\left(X_{j}\left(n_{j}\right)\right)}\left(\left(\mu_{j,n_{j}}\right)^{-1/2}t\right).$

Let’s compute the mean from the MGF $M_{j,n_{j}}\left(t\right)$:

$\displaystyle M_{j,n_{j}}^{\prime}(0)$ $\displaystyle=-\left(\log\log
n_{j}\right)^{1/2}e^{-t\left(\log\log
n_{j}\right)^{1/2}}M_{\omega\left(X_{j}\left(n_{j}\right)\right)}\left(\left(\log\log
n_{j}\right)^{-1/2}t\right)$ $\displaystyle\phantom{=}+e^{-t\left(\log\log
n_{j}\right)^{1/2}}M^{{}^{\prime}}_{\omega\left(X_{j}\left(n_{j}\right)\right)}\left(\left(\log\log
n_{j}\right)^{-1/2}t\right)\left(\log\log n_{j}\right)^{-1/2}\Big{|}_{t=0}$
$\displaystyle=-\left(\log\log n_{j}\right)^{1/2}+\mu_{j,n_{j}}\left(\log\log
n_{j}\right)^{-1/2}$ $\displaystyle\to 0,$

where the last limit holds due to the proof of Theorem 2 (we showed the
moments all approach the moments of the normal distribution) which is used in
the proof of Theorem 5 above; thus,

(11) $\lim_{j\to\infty}\left(\mu_{j,n_{j}}/\left(\log\log
n_{j}\right)^{1/2}-\left(\log\log n_{j}\right)^{1/2}\right)=0$

holds. Therefore,

$\lim_{j\to\infty}\frac{\mu_{j,n_{j}}}{\log\log n_{j}}=1.$

From equation $\left(\ref{eq:11}\right)$ we can conclude

$\lim_{j\to\infty}\left(\mu_{j,n_{j}}^{1/2}-\left(\log\log
n_{j}\right)^{1/2}\right)=0.$

This is because,

$\left|\mu_{j,n_{j}}^{1/2}-\left(\log\log
n_{j}\right)^{1/2}\right|=\left|\frac{\mu_{j,n_{j}}-\left(\log\log
n_{j}\right)}{\mu_{j,n_{j}}^{1/2}+\left(\log\log
n_{j}\right)^{1/2}}\right|\leq\left|\frac{\mu_{j,n_{j}}-\log\log
n_{j}}{\left(\log\log n_{j}\right)^{1/2}}\right|\to 0.$

Now we want to show convergence in distribution using the characteristic
functions. We use the fact that the characteristic function
$\phi_{\omega\left(X_{j}\left(n_{j}\right)\right)}\left(t\right)$ is
continuous at $t=0$. According to properties of continuous functions, we have
the ratios

(12) $\frac{e^{-it\mu_{j,n_{j}}^{1/2}}}{e^{-it\left(\log\log
n_{j}\right)^{1/2}}}=e^{-it\left(\mu_{j,n_{j}}^{1/2}-\left(\log\log
n_{j}\right)^{1/2}\right)}\to e^{0}=1,$

and also

(13)
$\frac{\phi_{\omega\left(X_{j}\left(n_{j}\right)\right)}\left(\left(\mu_{j,n_{j}}\right)^{-1/2}t\right)}{\phi_{\omega\left(X_{j}\left(n_{j}\right)\right)}\left(\left(\log\log
n_{j}\right)^{-1/2}t\right)}\to\frac{\phi_{\omega\left(X_{j}\left(n_{j}\right)\right)}\left(0\right)}{\phi_{\omega\left(X_{j}\left(n_{j}\right)\right)}\left(0\right)}=1$

as $j\to\infty$. This gives us the required convergence of $\phi_{j,\mu}$ at
any $t$; i.e.,

$\displaystyle\phi_{j,\mu}\left(t\right)=e^{-it\mu_{j,n_{j}}^{1/2}}\phi_{\omega\left(X_{j}\left(n_{j}\right)\right)}\left(\left(\mu_{j,n_{j}}\right)^{-1/2}t\right)$
$\displaystyle\overset{\left(\ref{eq:12},\ref{eq:13}\right)}{\sim}e^{-it\left(\log\log
n_{j}\right)^{1/2}}\phi_{\omega\left(X_{j}\left(n_{j}\right)\right)}\left(\left(\log\log
n_{j}\right)^{-1/2}t\right)$ $\displaystyle\to
e^{-t^{2}/2}=\phi_{Z}\left(t\right).$

By Lèvy’s Continuity Theorem,

$\mathbb{P}_{j,n_{j}}\left(m\leq n_{j}:\omega\left(m\right)-\mu_{j,n_{j}}\leq
x\left(\mu_{j,n_{j}}\right)^{1/2}\right)\to\mathbb{P}\left(Z\leq x\right).$

Suppose additionally that $\mu_{j}<\infty$ for all $j\geq 1$ and

$\lim_{j\to\infty}\left(\mu_{j}/\left(\log\log
n_{j}\right)^{1/2}-\left(\log\log n_{j}\right)^{1/2}\right)=0.$

Similarly, let $\phi_{j,\mu_{j}}$ be the characteristic function of

$\frac{\omega\left(X_{j}\right)-\mu_{j}}{\sqrt{\mu_{j}}}.$

Then

$\displaystyle\phi_{j,\mu_{j}}\left(t\right)=e^{-it\mu_{j}^{1/2}}\phi_{\omega\left(X_{j}\right)}\left(\left(\mu_{j}\right)^{-1/2}t\right)$
$\displaystyle\sim e^{-it\left(\log\log
n_{j}\right)^{1/2}}\phi_{\omega\left(X_{j}\left(n_{j}\right)\right)}\left(\left(\log\log
n_{j}\right)^{-1/2}t\right)$ $\displaystyle\to
e^{-t^{2}/2}=\phi_{Z}\left(t\right),$

for any $t\in\mathbb{R}$. Thus, again we have that

$\mathbb{P}_{j}\left(m:\omega\left(m\right)-\mu_{j}\leq
x\left(\mu_{j}\right)^{1/2}\right)\to\mathbb{P}\left(Z\leq x\right),$

by Lèvy’s Continuity Theorem. This completes the proof. ∎

######  Remark.

We can replace $\mu_{j}$ in Corollary 3 with any sequence $a_{j}$ that
satisfies the limit

$\lim_{j\to\infty}\left(a_{j}/\left(\log\log n_{j}\right)^{1/2}-\left(\log\log
n_{j}\right)^{1/2}\right)=0,$

to get a similar result. This is a generalization of Theorem 7.1 in [2] which
states this result for particular sequences of $Zeta\left(s\right)$
distributions as their parameter $s\to 1$. In their result, the mean was
$\sum_{p}p^{-s}$, where the sum is over all prime numbers.

######  Definition 4.

Given an infinite sequence of random variables $X_{1},X_{2},\ldots$ defined on
$\mathbb{N}$, let $X_{j}\left(n\right)$ denote the truncation of $X_{j}$ on
$\left[n\right]$. If $X_{1},X_{2},\ldots$ satisfies the uniformity along
primes property, and $X_{j}\left(n\right)$ satisfies (4 \- 5) for any $n$ and
any $j$, then we say the sequence satisfies the complete uniformity along
primes property.

The following lemma will be applied in Corollary 4.

######  Lemma 3.

If $X_{1},X_{2},\ldots$ has the complete uniformity along primes property and
$\mu_{j}<\infty$ for all $j\geq 1$, then $\mu_{j}\to\infty$ as $j\to\infty$.

###### Proof.

For any sequence $\left(n_{j}\right)_{j\in\mathbb{N}}$ in
$\mathbb{N}\setminus\left\\{1\right\\}$ tending to $\infty$ we have $\log\log
n_{j}\sim\mu_{j,n_{j}}\sim\mu_{j}$ as $j\to\infty$. ∎

######  Corollary 4.

If $X_{1},X_{2},\ldots$ has the complete uniformity along primes property and
$\mu_{j}<\infty$ for all $j\geq 1$, then

$\mathbb{P}_{j}\left(m:\omega\left(m\right)-\mu_{j}\leq
x\left(\mu_{j}\right)^{1/2}\right)\to\mathbb{P}\left(Z\leq x\right)$

as $j\to\infty$.

###### Proof.

Define $n_{j}\coloneqq\max\left\\{\left\lfloor
e^{e^{\mu_{j}}}\right\rfloor,2\right\\}$ so that $n_{j}\to\infty$ by Lemma 3
and the limit

$\lim_{j\to\infty}\left(\mu_{j}/\left(\log\log
n_{j}\right)^{1/2}-\left(\log\log n_{j}\right)^{1/2}\right)=0$

in Corollary 3 holds. ∎

###  4.2. The Zeta Distribution

We show the zeta distribution has the complete uniformity along primes
property as $s\to 1$. Given $s>1$, denote by $Z_{s}$ the $Zeta\left(s\right)$
distribution so that for any $j\in\mathbb{N}$,

$Z_{s}\left(j\right)=\frac{1}{j^{s}\zeta\left(s\right)},$

where

$\zeta\left(s\right)=\sum_{j\geq 1}\frac{1}{j^{s}}$

denotes the Riemann zeta function. Since Theorem 2 involves distributions
defined on $\left[n\right]$, restrict the $Zeta\left(s\right)$ distribution to
$\left[n\right]$ and then normalize by dividing by
$\sum_{i=1}^{n}\frac{1}{i^{s}\zeta\left(s\right)}$; i.e., for
$j\in\left[n\right]$,

$\displaystyle Z_{s,n}\left(j\right)$
$\displaystyle\coloneqq\frac{\frac{1}{j^{s}\zeta\left(s\right)}}{\sum_{i=1}^{n}\frac{1}{i^{s}\zeta\left(s\right)}}$
$\displaystyle=\frac{1}{j^{s}\sum_{i=1}^{n}\frac{1}{i^{s}}};$

and $Z_{s,n}$ is known as the Zipf distribution with parameters $n$ and $s$.
For a Zifp distribution, we have

$\varepsilon_{i,n}\overset{\left(\ref{eq:1}\right)}{=}\frac{1/i^{s}}{\sum_{j=1}^{n}1/j^{s}}-\frac{1}{n}.$

Thus,

$\displaystyle\sum_{l=1}^{\lfloor n/p\rfloor}\varepsilon_{lp,n}$
$\displaystyle=\frac{\sum_{l=1}^{\lfloor
n/p\rfloor}1/\left(lp\right)^{s}}{\sum_{l=1}^{n}1/l^{s}}-\frac{\lfloor
n/p\rfloor}{n}$ $\displaystyle=\frac{1}{p^{s}}\frac{\sum_{l=1}^{\lfloor
n/p\rfloor}1/l^{s}}{\sum_{l=1}^{n}1/l^{s}}-\frac{\lfloor n/p\rfloor}{n}$
$\displaystyle\leq\frac{1}{p^{s}}$ $\displaystyle\leq\frac{1}{p},$

so constraint $\left(\ref{eq:4}\right)$ holds with $C=1$ for any $s$.
Moreover,

$\displaystyle\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\varepsilon_{lp_{1}\cdots p_{k},n}$
$\displaystyle=\frac{\sum_{l=1}^{\lfloor\frac{n}{p_{1}\cdots
p_{k}}\rfloor}1/\left(lp_{1}\cdots
p_{k}\right)^{s}}{\sum_{l=1}^{n}1/l^{s}}-\frac{\lfloor\frac{n}{p_{1}\cdots
p_{k}}\rfloor}{n}$ $\displaystyle=\frac{1}{\left(p_{1}\cdots
p_{k}\right)^{s}}\frac{\sum_{l=1}^{\lfloor
n/p\rfloor}1/l^{s}}{\sum_{l=1}^{n}1/l^{s}}-\frac{1}{p_{1}\cdots
p_{k}}+\frac{1}{n}$ $\displaystyle\leq\frac{1}{\left(p_{1}\cdots
p_{k}\right)^{s}}-\frac{1}{p_{1}\cdots p_{k}}+\frac{1}{n}$
$\displaystyle\leq\frac{1}{n},$

so constraint $\left(\ref{eq:5}\right)$ holds with $D=1$ for any $s$.
Furthermore,

$\displaystyle\lim_{n\to\infty}\sum_{l=1}^{\lfloor
n/p\rfloor}\varepsilon_{lp,n}$
$\displaystyle=\lim_{n\to\infty}\frac{\sum_{l=1}^{\lfloor
n/p\rfloor}1/\left(lp\right)^{s}}{\sum_{l=1}^{n}1/l^{s}}-\frac{1}{p}$
$\displaystyle=\frac{1}{p^{s}}-\frac{1}{p}$ $\displaystyle\overset{s\to
1}{\to}0;$

therefore, $\left(\ref{eq:6}\right)$ holds as $s\to 1$.

######  Corollary 5.

Let $\left(a_{j}\right)_{j\in\mathbb{N}}$ be any sequence of real numbers such
that $0<a_{j}\to\infty$. If $X_{j}$ is a sequence of
$Zeta\left(1+1/a_{j}\right)$-distributed random variables, then

$\mu_{j}=\sum_{p}p^{-\left(1+1/a_{j}\right)}<\infty,$
$\lim_{j\to\infty}\mu_{j}=\infty,$

and

$\mathbb{P}_{j}\left(m:\omega\left(m\right)-\mu_{j}\leq
x\left(\mu_{j}\right)^{1/2}\right)\to\mathbb{P}\left(Z\leq x\right)$

as $j\to\infty$.

This latest limit is the statement of Theorem 7.1 in [2].

###  4.3. Convex Combinations of E-K Distributions on $\mathbb{N}$

We can also take convex combinations of sequences to form new sequences which
satisfy the uniformity along primes property.

######  Corollary 6.

Let $\lambda\in\left[0,1\right]$. Let $X_{1},X_{2},\ldots$ and
$Y_{1},Y_{2},\ldots$ be two sequences of random variables on $\mathbb{N}$ that
both satisfy the complete uniformity along primes property. Then we can define
a new sequence $Z_{1},Z_{2},\ldots$ such that each $Z_{j}$ is the convex
combination of $X_{j}$ and $Y_{j}$. In particular,

$\mathbb{P}_{Z_{j}}\left(i\right)\coloneqq\lambda\mathbb{P}_{X_{j}}\left(i\right)+\left(1-\lambda\right)\mathbb{P}_{Y_{j}}\left(i\right);i\geq
1,j\geq 1.$

Then $Z_{j}$ satisfies the complete uniformity along primes property and thus
Theorem 5 holds for the $Z_{j}$ sequence.

###### Proof.

The proof is similar to that of Theorem 3. ∎

######  Corollary 7.

Apply Corollary 6 to any convex combination of a finite number of sequences

$\left(X_{1,1},X_{1,2},\ldots\right),\ldots,\left(X_{n,1},X_{n,2},\ldots\right)$

that all satisfy the complete uniformity along primes property. Then the
conclusion of Corollary 6 holds for this convex combination.

###### Proof.

Use induction and Corollary 6. ∎

##  5\. Illustrative Examples

In this section, we give examples from the class of distributions
$\mathbb{P}^{*}_{n}$ that satisfy the hypotheses
$\left(\ref{eq:4}-\ref{eq:6}\right)$ and examples of limits of distributions
to which we can apply Theorem 5. It will be shown that the statement of
Theorem 2 holds when $\mathbb{P}_{n}^{*}$ is replaced with either the
Harmonic$\left(n\right)$ distribution or a convex combination of Harmonic and
Uniform Distributions. Then we show that the $Zeta\left(s\right)$ and the
Logarithmic distribution satisfy Theorem 5 as their parameters tend towards
limits. We introduce a $2$-parameter family of distributions
$LZ\left(s,\alpha\right)$ which includes the Logarithmic distribution and Zeta
distribution as special cases (when $\alpha=1$ and $s=1$ respectfully). We
also look at a geometric power series distribution that converges to the
normal distribution on all truncations as $s\to 1$.

###  5.1. The Harmonic Distribution

This was proved in $\S\S$2.1 above.

###  5.2. The Zeta Distribution

This was proved in $\S\S$4.2 above.

###  5.3. The Logarithmic Distribution

Given a real number $s$ with $0<s<1$, a logarithmic distribution with
parameter $s$ is given by

$L_{s}\left(i\right)\coloneq\frac{-1}{\log\left(1-s\right)}\frac{s^{i}}{i};\,\,i\in\mathbb{N}.$

Now we will show that the number of distinct prime factors,
$\omega\left(\cdot\right)$, of a truncated log-distributed variable has the
same central limit theorem as the uniform distribution as $n\to\infty$ and
$s\to 1$. For $i\in\left[n\right]$ we have

$\displaystyle L_{s,n}\left(i\right)$
$\displaystyle\coloneq\frac{\frac{-1}{\log\left(1-s\right)}\frac{s^{i}}{i}}{\sum_{l=1}^{n}\frac{-1}{\log\left(1-s\right)}\frac{s^{l}}{l}}=\frac{\frac{s^{i}}{i}}{\sum_{l=1}^{n}\frac{s^{l}}{l}},$

so

$\displaystyle\varepsilon_{i,n}$
$\displaystyle\overset{\left(\ref{eq:1}\right)}{=}\frac{\frac{s^{i}}{i}}{\sum_{l=1}^{n}\frac{s^{l}}{l}}-\frac{1}{n}.$

Therefore,

$\displaystyle\sum_{l=1}^{\left\lfloor\frac{n}{p}\right\rfloor}\varepsilon_{lp,n}$
$\displaystyle=\sum_{l=1}^{\left\lfloor\frac{n}{p}\right\rfloor}\left(\frac{\frac{\left(s\right)^{lp}}{lp}}{\sum_{l=1}^{n}\frac{s^{l}}{l}}-\frac{1}{n}\right)$
$\displaystyle\leq\frac{1}{p}\frac{\sum_{l=1}^{\left\lfloor\frac{n}{p}\right\rfloor}\frac{\left(s^{p}\right)^{l}}{l}}{\sum_{l=1}^{n}\frac{s^{l}}{l}}-\frac{\lfloor
n/p\rfloor}{n}$ $\displaystyle\leq\frac{1}{p}-\frac{\lfloor n/p\rfloor}{n}$
$\displaystyle\leq\frac{1}{p},$

so $\left(\ref{eq:4}\right)$ holds with $C=1$. Similarly,

$\displaystyle\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\varepsilon_{lp_{1}\cdots p_{k},n}$
$\displaystyle=\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\left(\frac{\frac{\left(s\right)^{lp_{1}\cdots
p_{k}}}{lp_{1}\cdots
p_{k}}}{\sum_{l=1}^{n}\frac{s^{l}}{l}}-\frac{1}{n}\right)$
$\displaystyle\leq\frac{1}{p_{1}\cdots
p_{k}}\frac{\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\frac{\left(s^{p_{1}\cdots
p_{k}}\right)^{l}}{l}}{\sum_{l=1}^{n}\frac{s^{l}}{l}}-\frac{1}{p_{1}\cdots
p_{k}}+\frac{1}{n}$ $\displaystyle\leq\frac{1}{p_{1}\cdots
p_{k}}-\frac{1}{p_{1}\cdots p_{k}}+\frac{1}{n}$ $\displaystyle=\frac{1}{n},$

so $\left(\ref{eq:5}\right)$ holds with $D=1$. Moreover, we have

$\sum_{l=1}^{\lfloor
n/p\rfloor}\varepsilon_{lp,n}\overset{n\to\infty}{\to}\frac{1}{p}\frac{\log\left(1-s^{p}\right)}{\log\left(1-s\right)}-\frac{1}{p}\overset{s\to
1}{\to}0.$

Therefore, $\left(\ref{eq:6}\right)$ holds, so $\omega\left(L_{i,s}\right)$ is
asymptotically distributed as $\mathcal{N}\left(\log\log n,\log\log n\right)$
as $n\to\infty$ and $s\to 1$.

Although Kac’s heuristic for Theorem 1 is based on the asymptotic independence
in the uniform case, we will show that the events $\\{\text{divisible by
}p\\}$ and $\\{\text{divisible by }q\\}$ are not independent for any value of
$s$. Let $A_{p}$ denote the set of all positive integers divisible by $p$,
then

$\displaystyle L_{s}\left(A_{p}\right)$
$\displaystyle=\sum_{l=1}^{\infty}\frac{-1}{\log\left(1-s\right)}\frac{\left(s^{p}\right)^{l}}{pl}$
$\displaystyle=\frac{1}{p}\frac{\log\left(1-s^{p}\right)}{\log\left(1-s\right)}$

and

$\displaystyle L_{s}\left(A_{p}\cap A_{q}\right)$
$\displaystyle=\sum_{l=1}^{\infty}\frac{-1}{\log\left(1-s\right)}\frac{\left(s^{pq}\right)^{l}}{pql}$
$\displaystyle=\frac{1}{pq}\frac{\log\left(1-s^{pq}\right)}{\log\left(1-s\right)}.$

It is worth noting that

$\lim_{s\to 1}L_{s}\left(A_{p}\cap A_{q}\right)=\frac{1}{pq}=\lim_{s\to
1}\left(L_{s}\left(A_{p}\right)L_{s}\left(A_{q}\right)\right),$

as $s\to 1$; so independence is only approached in the limit.

###  5.4. Geometric Power Series Distribution

Let $s\in\left(0,1\right)$. Define:

$\mathbb{P}\left(i\right)\coloneqq\dfrac{1-s}{s}s^{i};\,\,i\in\mathbb{N}$

then this distribution satisfies the hypotheses of Theorem 2 in the limit as
$s\to 1$. As $s\to 1$ it seems that this distribution is converging to the
uniform distribution on any truncation, so the result is to be expected.

Truncating on $\left[n\right]$ leads to

$\varepsilon_{i,n}\overset{\left(\ref{eq:1}\right)}{=}\frac{s^{i}}{\sum_{j=1}^{n}s^{j}}-\frac{1}{n},$

therefore

$\displaystyle\sum_{l=1}^{\lfloor n/p\rfloor}\varepsilon_{lp,n}$
$\displaystyle=\sum_{l=1}^{\lfloor
n/p\rfloor}\left(\frac{s^{i}}{\sum_{j=1}^{n}s^{j}}-\frac{1}{n}\right)$
$\displaystyle=\frac{s-s^{\lfloor n/p\rfloor+1}}{s-s^{n+1}}-\frac{\lfloor
n/p\rfloor}{n}$ $\displaystyle\overset{s\to 1}{\to}\frac{\lfloor
n/p\rfloor}{n}-\frac{\lfloor n/p\rfloor}{n}$ $\displaystyle=0,$

so constraints $\left(\ref{eq:4}-\ref{eq:6}\right)$ hold with $C=D=1$.

###  5.5. Convex Combination of Harmonic and Uniform

Following Cranston and Mountford in [1], let $\lambda\in[0,1]$ and define

$\mathbb{P}_{n}(i)\coloneqq\frac{\lambda}{ih_{n}}+\left(1-\lambda\right)\frac{1}{n};\,\,i\in\mathbb{N}$

where $h_{n}$ is the $n^{\text{th}}$ harmonic number. Then $\mathbb{P}_{n}$
satisfies the hypotheses of Theorem 2. This is proved in Corollary 2 above
using Theorems 2 and 3.

###  5.6. A Logarithmic-Zeta Distribution

There is a $2$-parameter family of power series distributions on $\mathbb{N}$
for $(s,\alpha)\in\mathbb{R}^{2}$ such that $0<s\leq 1$, $\alpha\geq 1$, and
$s=\alpha=1$ is not allowed; the PMF is given by

$LZ_{s,\alpha}\left(i\right)\coloneq\frac{1}{\sum_{j=1}^{\infty}\frac{s^{j}}{j^{\alpha}}}\frac{s^{i}}{i^{\alpha}};\,\,i\in\mathbb{N}.$

Similarly, there is a truncation of $LZ$ given by

$LZ_{s,\alpha,n}\left(i\right)\coloneq\frac{1}{\sum_{j=1}^{n}\frac{s^{j}}{j^{\alpha}}}\frac{s^{i}}{i^{\alpha}};\,\,i\leq
n,$

and a description of the $\varepsilon_{i,n}$ given by

$\varepsilon_{i,n}\overset{\left(\ref{eq:1}\right)}{=}\frac{1}{\sum_{j=1}^{n}\frac{s^{j}}{j^{\alpha}}}\frac{s^{i}}{i^{\alpha}}-\frac{1}{n}.$

It is clear that a similar type of argument shows that Theorem 2 holds for the
truncated logarithmic-zeta distribution as $(s,\alpha)\to(1,1)$; in
particular, we obtain

$\sum_{l=1}^{\lfloor
n/p\rfloor}\varepsilon_{lp,n}\leq\frac{1}{p^{\alpha}}\leq\frac{1}{p}$

and

$\sum_{l=1}^{\lfloor\frac{n}{p_{1}\cdots
p_{k}}\rfloor}\varepsilon_{lp_{1}\cdots
p_{k},n}\leq\frac{1}{p_{1}^{\alpha}\cdots p_{k}^{\alpha}}-\frac{1}{p_{1}\cdots
p_{k}}+\frac{1}{n}\leq\frac{1}{n}.$

Furthermore,

$\lim_{n\to\infty}\sum_{l=1}^{\lfloor
n/p\rfloor}\varepsilon_{lp,n}\overset{s\to
1}{\to}\frac{1}{p^{\alpha}}-\frac{1}{p}\overset{\alpha\to 1}{\to}0.$

###  5.7. Passage to the Limit:
$\lim\limits_{\begin{subarray}{c}\mathllap{s}\to\mathrlap{1}\\\
\mathllap{\alpha}\to\mathrlap{1}\end{subarray}}LZ_{s,\alpha}$

We would like to look at the behavior of $\omega\left(X\right)$ for $X$
distributed as $LZ_{s,\alpha}$ as $\left(s,\alpha\right)\to\left(1,1\right)$.
The motivation for this comes from the fact that as $(s,\alpha)\to(1,1)$ the
truncated distributions resemble the harmonic distribution which we know
behaves similarly to the uniform distribution on $\left[n\right]$ when
$n\to\infty$ according to Theorem 2.

In Theorem 5.5 of [2], they provide the moment generating function for
$\omega\left(X_{\alpha}\right)$ when $X_{\alpha}$ is a
$Zeta\left(\alpha\right)$ distributed random integer with parameter
$\alpha>1$. They go on to prove that

$\hat{\omega}\left(X_{\alpha}\right)=\frac{\omega\left(X_{\alpha}\right)-\sum_{p}p^{-\alpha}}{\sqrt{\sum_{p}p^{-\alpha}}}\overset{d}{\to}Z$

as $\alpha\to 1$ in Theorem 7.1 of [2] by using this moment generating
function. Later, Cranston and Mountford [1] give a new proof of Theorem 1. The
proof uses Theorem 7.1 about zeta distributions to prove Erdős-Kac in a way
that translates over to settings where zeta functions still make sense.

We generalize Peltzer and Cranston’s Theorem 7.1 in [2] in the following way.
As $(s,\alpha)\to\left(1,1\right)$, we show that

$\hat{\omega}\left(X_{s,\alpha}\right)=\frac{\omega\left(X_{s,\alpha}\right)-\mu_{s,\alpha}}{\sqrt{\mu_{s,\alpha}}}\overset{d}{\to}Z,$

where $\mu_{s,\alpha}$ is the mean of $\omega\left(X_{s,\alpha}\right)$ when
$X_{s,\alpha}$ is distributed as $LZ_{s,\alpha}$. In fact, we do not need
independence, nor do we need to compute the mean for any particular $s,\alpha$
in order to conclude this (neither do we need to compute a MGF).

In $\S$4, we proved a stronger statement than the above statement about
$LZ_{s,\alpha}$; we proved that in general

$\hat{\omega}\left(X_{j}\right)=\frac{\omega\left(X_{j}\right)-\mu_{j}}{\sqrt{\mu_{j}}}\overset{d}{\to}Z,$

as long as the truncated variables $X_{j}\left(n\right)$ have the complete
uniformity along primes property. Here we assume that
$\mu_{j}=\mathbb{E}(\omega\left(X_{j}\right))<\infty$. When
$\left(s,\alpha\right)\to\left(1,1\right)$, we recover the above statement
about $LZ_{s,\alpha}$.

###  5.8. A Non-Example: Zeroing at a Set of Primes

Fix $n\in\mathbb{N}$ and let $p\leq n$ denote a prime. Consider the PMF
defined by

$\mathbb{P}_{n,p}\left(i\right)=\begin{cases}\frac{1}{\\#\left(\left[n\right]\setminus
p\mathbb{N}\right)}&i\not\in p\mathbb{N},\\\ 0&i\in p\mathbb{N}.\end{cases}$

We have

$\varepsilon_{i,n}=\begin{cases}\frac{1}{\\#\left(\left[n\right]\setminus
p\mathbb{N}\right)}-1/n&i\not\in p\mathbb{N},\\\ -1/n&i\in
p\mathbb{N}.\end{cases}$

Therefore,

$\displaystyle\lim_{n\to\infty}\sum_{l=1}^{\lfloor
n/p\rfloor}\varepsilon_{lp,n}$ $\displaystyle=\lim_{n\to\infty}-\frac{\lfloor
n/p\rfloor}{n}$ $\displaystyle=-1/p$ $\displaystyle\neq 0.$

Thus, this PMF does not satisfy constraint $\left(\ref{eq:6}\right)$. We
conjecture that the conclusion of Theorem 2 does not hold for this
distribution.

###  5.9. An Erdős-Kac Theorem for Continuous Variables

Consider a continuous uniform random variable $N_{n}$ on the interval
$\left(0,n\right]$. Then $\left\lceil N_{n}\right\rceil$ is a uniform variable
on $\left[n\right]$. Therefore, we have

######  Corollary 8.

Let $Z$ denote a standard normal variable, and let $X_{n}$ be a continuous
random variable on $\left(0,n\right]$. Define the $\varepsilon_{i,n}$
according to the following relation: $\mathbb{P}_{n}\left(\left\lceil
X_{n}\right\rceil=i\right)=\frac{1}{n}+\varepsilon_{i,n}$. If the constraints

  * •

There exists a constant $C$ such that for all $n>1$ and for all primes $p$
with $p>\alpha_{n}$,

$\sum_{l=1}^{\left\lfloor n/p\right\rfloor}\varepsilon_{lp,n}\leq\frac{C}{p}.$

  * •

There exists a constant $D$ such that

$\sum_{l=1}^{\left\lfloor\frac{n}{p_{1}\cdots
p_{k}}\right\rfloor}\varepsilon_{lp_{1}\cdots p_{k},n}\leq\frac{D}{n}$

for all $n>1$ and, for each $k$, all $k$-tuples
$\left(p_{1},\ldots,p_{k}\right)$ consisting of distinct primes of size at
most $\alpha_{n}$, and

  * •

For any prime $p$,

$\lim_{n\to\infty}\sum_{l=1}^{\lfloor n/p\rfloor}\varepsilon_{lp,n}=0$

all hold, then

$\displaystyle\mathbb{P}_{n}\left(t\leq n:\omega\left(\lceil
t\rceil\right)-\log\log n\leq x\left(\log\log n\right)^{1/2}\right)$
$\displaystyle\to\mathbb{P}\left(Z\leq x\right)$

as $n\to\infty$.

##  6\. Conclusion

Theorem 2 generalizes the Erdős-Kac Theorem for $\omega\left(\cdot\right)$ to
distributions other than the uniform distribution, and this theorem was proved
by imposing constraints $\left(\ref{eq:4}-\ref{eq:6}\right)$ on a PMF of the
form $\mathbb{P}\left(i\right)=1/n+\varepsilon_{i,n}$. We showed that the
uniform and harmonic distributions satisfy these constraints; then, we showed
that any convex sum of these PMFs also satisfies the constraints (4 - 6).

The uniformity property provides a natural way to examine asymptotic
properties of truncations of variables with support $\mathbb{N}$. Given an
infinite sequence $X_{1},X_{2},\ldots$ of random variables on $\mathbb{N}$
satisfying uniformity along primes, Theorem 5 showed that for any sequence
with $n_{j}\to\infty$ as $j\to\infty$, the distribution of
$\omega\left(X_{j}\left(n_{j}\right)\right)$ is asymptotically normally
distributed with mean and variance both equal to $\log\log n_{j}$ as long as
$X_{j}\left(n_{j}\right)$ satisfies (4 \- 6).

The definition of complete uniformity along primes allows us to obtain central
limit theorems regardless of how $n\to\infty$, and allows us to make
asymptotic statements involving the mean of $\omega\left(X_{j}\right)$. This
generalized a statement from [2] involving the mean of
$\omega\left(X_{s}\right)$ as $s\to 1$ when $X_{s}$ is a random
$Zeta\left(s\right)$-distributed variable.

Another way to generalize Theorem 2 would be to incorporate it with other
generalizations, e.g., [1, 2, 6, 7]. By incorporating Theorem 2 with these,
further generalizations can be made in which the original setting is not
$\left[n\right]$, the underlying distribution of the random-integer is not
uniform, and $\omega\left(n\right)$ can be replaced with a more general
strongly additive function $f\left(n\right)$.

We also showed the complete uniformity property holds, and thus normality in
the limit, for $Zeta\left(s\right)$ and a number of similar distributions. It
is suspected by the authors, but not known, whether or not the hypotheses
$\left(\ref{eq:4}-\ref{eq:6}\right)$ are necessary and sufficient for the
conclusion of Theorem 2; we conjecture that is the case.

## References

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