Sums of arithmetic functions running on factorials

Jean-Marie De Koninck and William Verreault

###### Abstract

Given an arithmetic function $f:{\mathbb{N}}\to{\mathbb{R}}$, it is customary
to investigate the behavior of the corresponding sum ${\sum_{n\leq N}f(n)}$
for large $N$. Here, for various classical arithmetic functions $f$, including
the number of distinct prime factors function $\omega(n)$, Euler totient’s
function $\phi(n)$, the number of divisors function $d(n)$, the sum of the
divisors function $\sigma(n)$, as well as the middle divisors functions
$\rho_{1}(n)$ and $\rho_{2}(n)$, we investigate the behavior of $f(n!)$ and
their corresponding sums ${\sum_{n\leq N}f(n!)}$. Finally, if $\lambda$ stands
for the Liouville function, according to the Chowla conjecture, ${\sum_{n\leq
N}\lambda(n)\lambda(n+1)=o(N)}$ as $N\to\infty$; here, we show that the
analogue of the Chowla conjecture for factorial arguments is true as we prove
that, as $N\to\infty$, we have ${\sum_{n\leq
N}\lambda(n!)\lambda((n+1)!)=o(N)}$.

##  1 Introduction

It is common to enquire about the average order of an arithmetic function
$f:{\mathbb{N}}\to{\mathbb{R}}$ by first examining the behavior of the sum

$S_{f}(N):=\sum_{n\leq N}f(n)$

for large $N$. On the other hand, one can find the correct order of various
arithmetic functions $f$ at factorial arguments, that is of $f(n!)$, without
first examining the behavior of $\sum_{n\leq N}f(n!)$, but by simply taking a
closer look at the arithmetic properties of $n!$. Yet, in the literature,
there seems to be only a handful of results regarding the study of $f(n!)$ for
classical arithmetic functions $f$. One of the first such functions to have
been closely examined for its exact order at factorial arguments is the number
of positive divisors function $d(n)$. In fact, Ramanujan [29, 30] obtained
interesting estimates for $d(n!)$, and later, Erdős, Graham, Ivić, and
Pomerance [15] refined his findings, whereas more recently, Jakimczuk [25, 26]
obtained estimates for $\sigma(n!)$, where $\sigma(n)$ stands for the sum of
the positive divisors of $n$.

Here we embark on a systematic study of $f(n!)$ and of $\sum_{n\leq N}f(n!)$
for a variety of classical arithmetic functions. Besides the functions $d(n)$
and $\sigma(n)$, these include, among others, the function $\Omega(n)$, which
counts the total number of prime factors of $n$ counting their multiplicity,
the Euler totient function $\phi(n)$, the kernel function
$\gamma(n):=\prod_{p\mid n}p$, the function $\beta(n):=\sum_{p\mid n}p$, as
well as the middle divisors functions

$\rho_{1}(n):=\max\\{d\mid
n:d\leq\sqrt{n}\\}\qquad\mbox{and}\qquad\rho_{2}(n):=\min\\{d\mid
n:d\geq\sqrt{n}\\}.$

Preliminary results and the statements of our main results are given in
Sections 2 to 4. In Section 5, for various arithmetic functions $f$, we
compare the behavior of $f(n!)$ with the maximum of $f(n)$. The proofs of the
main results are laid out in Section 6. We conclude in Section 7 by showing
that the counterpart at factorial arguments of the famous Chowla conjecture is
in fact true.

Note that throughout this text, the letters $c_{i}$ stand for explicit
positive constants, but not necessarily the same at each occurrence.

##  2 The case of additive functions

As a prototypical example, consider the additive function $f(n)=\log n$, for
which $f(n!)=\sum_{k\leq n}\log k=S_{\log}(n)$. As we will see, this is not
unusual.

###  2.1 Small additive functions

We first consider the case of “small” additive functions. Here we have in mind
$\omega(n)$ and $\Omega(n)$, the number of distinct prime factors of $n$ and
the total number of prime factors of $n$, respectively. We also let $\pi(x)$
stand for the number of primes up to $x$.

######  Theorem 1.

We have

$\omega(n!)=\pi(n)\qquad\text{and}\qquad\Omega(n!)=S_{\Omega}(n).$

In particular,

$\omega(n!)\sim\frac{n}{\log n}\qquad\text{and}\qquad\Omega(n!)\sim n\log\log
n\qquad(n\to\infty).$

It is well known that (see for instance formulas (6.11) and (6.12) in [11])

$\displaystyle\sum_{n\leq N}\omega(n)$ $\displaystyle=$ $\displaystyle
N\Big{(}\log\log N+c_{1}+O\Big{(}\frac{1}{\log N}\Big{)}\Big{)},$ (1)
$\displaystyle\sum_{n\leq N}\Omega(n)$ $\displaystyle=$ $\displaystyle
N\Big{(}\log\log N+c_{1}+c_{2}+O\Big{(}\frac{1}{\log N}\Big{)}\Big{)},$

where $c_{1}=\gamma+\sum_{p}\big{(}\log(1-\frac{1}{p})+\frac{1}{p}\big{)}$ and
$c_{2}=\sum_{p}\frac{1}{p(p-1)}$, where $\gamma$ stands for the
Euler–-Mascheroni constant. Similarly, it was proved by De Koninck [5] that

$\displaystyle\sum_{2\leq n\leq N}\frac{1}{\omega(n)}$ $\displaystyle=$
$\displaystyle\frac{N}{\log\log N}\Big{(}1+O\Big{(}\frac{1}{\log\log
N}\Big{)}\Big{)},$ $\displaystyle\sum_{2\leq n\leq N}\frac{1}{\Omega(n)}$
$\displaystyle=$ $\displaystyle\frac{N}{\log\log
N}\Big{(}1+O\Big{(}\frac{1}{\log\log N}\Big{)}\Big{)}.$

In fact, more accurate formulas can be found in Chapter 5 of the book of De
Koninck and Ivić [10]. In addition, De Koninck [6] proved that

$\sum_{2\leq n\leq N}\frac{\Omega(n)}{\omega(n)}=N+c_{2}\frac{N}{\log\log
N}+O\Big{(}\frac{N}{(\log\log N)^{2}}\Big{)}.$

On the other hand, a precise estimate for the sum $\sum_{2\leq n\leq
N}1/\pi(n)$ will prove useful. The study of that sum has a long history, a
survey of which can be found in [7]. The best known estimate is due to Ivić
[23] and can be stated as follows:

(2) $\sum_{2\leq n\leq N}\frac{1}{\pi(n)}=\frac{1}{2}\log^{2}N-\log N-\log\log
N+c_{3}+O\Big{(}\frac{1}{\log N}\Big{)},$

where $c_{3}$ is an absolute constant that is known [3] to satisfy
$6.6840<c_{3}<6.7830$.

Analogous results for partial sums of these small additive functions at
factorial arguments are much easier to obtain. In fact, we get most of the
following results by combining the above estimates with Theorem 1.

######  Theorem 2.

We have

(3) $\displaystyle\sum_{n\leq N}\omega(n!)$ $\displaystyle=$
$\displaystyle\frac{N^{2}}{2\log N}\Big{(}1+O\Big{(}\frac{1}{\log
N}\Big{)}\Big{)},$ (4) $\displaystyle\sum_{n\leq N}\Omega(n!)$
$\displaystyle=$ $\displaystyle\frac{N^{2}}{2}\Big{(}\log\log
N+c_{1}+c_{2}+O\Big{(}\frac{1}{\log N}\Big{)}\Big{)},$ (5)
$\displaystyle\sum_{2\leq n\leq N}\frac{1}{\omega(n!)}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\log^{2}N-\log N-\log\log
N+c_{3}+O\Big{(}\frac{1}{\log N}\Big{)},$ (6) $\displaystyle\sum_{2\leq n\leq
N}\frac{1}{\Omega(n!)}$ $\displaystyle=$ $\displaystyle\frac{\log N}{\log\log
N}\Big{(}1+O\Big{(}\frac{1}{\log\log N}\Big{)}\Big{)},$ (7)
$\displaystyle\sum_{2\leq n\leq N}\frac{\Omega(n!)}{\omega(n!)}$
$\displaystyle=$ $\displaystyle N(\log N\log\log N+(c_{1}+c_{2})\log
N+O(\log\log N)).$

###  2.2 Large additive functions

Next, we consider the “large” additive functions

$\beta(n):=\sum_{p\mid n}p,\qquad B(n):=\sum_{p^{\alpha}\|n}\alpha\,p,\qquad
B_{1}(n):=\sum_{p^{\alpha}\|n}p^{\alpha}.$

For these, the following holds.

######  Theorem 3.

We have

$\beta(n!)=\sum_{p\leq n}p,\qquad B(n!)=S_{B}(n),\qquad\text{and}\qquad
B_{1}(n!)=S_{B_{1}}(n).$

In particular,

$\beta(n!)\sim\frac{n^{2}}{2\log n}\qquad\text{and}\qquad B(n!)\sim
B_{1}(n!)\sim\frac{\pi^{2}}{12}\frac{n^{2}}{\log n}\qquad(n\to\infty).$

Alladi and Erdős [2] proved that

$\sum_{n\leq N}\beta(n)\sim\frac{\pi^{2}}{12}\frac{N^{2}}{\log
N}\qquad(N\to\infty),$

and Ivić [22] established that

$\sum_{n\leq N}B(n)=\sum_{n\leq N}\beta(n)+O(N^{2/3})$

and

$\sum_{n\leq N}B_{1}(n)=\sum_{n\leq N}\beta(n)+O(N^{2/3}),$

from which it follows that

(8) $\sum_{n\leq N}B(n)\sim\frac{\pi^{2}}{12}\frac{N^{2}}{\log N}\quad\mbox{
and }\quad\sum_{n\leq N}B_{1}(n)\sim\frac{\pi^{2}}{12}\frac{N^{2}}{\log
N}\qquad(N\to\infty).$

On the other hand, Ivić [19] proved that

$\sum_{2\leq n\leq N}\frac{1}{\beta(n)}=\frac{N}{e^{\sqrt{2\log N\log\log
N}+O(\sqrt{\log N\log\log\log N})}},$

an estimate that was later slightly improved by Ivić and Pomerance [20].

The factorial counterpart of these results is as follows.

######  Theorem 4.

As $N\to\infty$,

(9) $\displaystyle\sum_{2\leq n\leq N}\beta(n!)$ $\displaystyle=$
$\displaystyle\frac{N^{3}}{6\log N}\Big{(}1+O\Big{(}\frac{1}{\log
N}\Big{)}\Big{)},$ (10) $\displaystyle\sum_{2\leq n\leq N}B(n!)$
$\displaystyle\sim$ $\displaystyle\frac{\pi^{2}}{36}\frac{N^{3}}{\log N},$
(11) $\displaystyle\sum_{2\leq n\leq N}B_{1}(n!)$ $\displaystyle\sim$
$\displaystyle\frac{\pi^{2}}{36}\frac{N^{3}}{\log N}.$

Furthermore, letting $f$ be any of the three functions $\beta$, $B$ and
$B_{1}$, then

$\sum_{n=2}^{\infty}\frac{1}{f(n!)}<\infty.$

##  3 The case of multiplicative functions

###  3.1 The classical sums

Let us first consider the multiplicative function $\gamma$ defined by
$\gamma(1)=1$ and

$\gamma(n):=\prod_{p\mid n}p\qquad(n\geq 2),$

as well as Euler totient’s function $\phi(n)$, which stands for the number of
positive integers up to $n$ that are relatively prime to $n$.

######  Theorem 5.

We have

(12) $\gamma(n!)=\prod_{p\leq n}p\qquad\text{and}\qquad\phi(n!)=n!\prod_{p\leq
n}(1-1/p).$

In particular,

(13) $\gamma(n!)=e^{(1+o(1))n}\qquad\text{and}\qquad\phi(n!)\sim
n!\frac{e^{-\gamma}}{\log n}\qquad(n\to\infty).$

Perhaps the most challenging examples come from multiplicative functions that
are neither completely nor strongly multiplicative, for reasons that will
become apparent in Section 6.1. Of particular interest are the functions
$d(n)$ and $\sigma(n)$, the number of positive divisors of $n$ and the sum of
the positive divisors of $n$, respectively, as well as their higher-order
variants $d_{k}(n)$ and $\sigma_{\kappa}(n)$ for $k\geq 2$ an integer and
$\kappa$ a real number. Here $d_{k}(n)$ is the number of ways of representing
$n$ as a product of $k$ positive integers and $\sigma_{\kappa}(n)$ is the sum
of the $\kappa$-th powers of the positive divisors of $n$, that is,

$d_{k}(n)=\sum_{a_{1}\cdots
a_{k}=n}1\qquad\text{and}\qquad\sigma_{\kappa}(n)=\sum_{d\mid n}d^{\kappa},$

where the $a_{i}$’s run through the positive integers.

Ramanujan [29, 30] studied the behavior of $d(n!)$, conjecturing that for any
$\varepsilon>0$,

$C^{\frac{n}{\log n}(1-\varepsilon)}<d(n!)<C^{\frac{n}{\log
n}(1+\varepsilon)},$

where
$C=(1+1)\sqrt{1+\frac{1}{2}}\sqrt[3]{1+\frac{1}{3}}\sqrt[4]{1+\frac{1}{4}}\cdots\approx
3.51750$, and then proving [29] the more explicit

(14) $d(n!)=C^{\frac{n}{\log n}+o(\frac{n}{\log^{2}n})}\qquad(n\to\infty).$

See [1] for more on this account. Later, Erdős, Graham, Ivić, and Pomerance
[15] obtained an asymptotic series expansion for $\log d(n!)$ by showing that
for any given integer $M\geq 0$,

$d(n!)=\exp\left\\{\frac{n}{\log
n}\sum_{\ell=0}^{M}\frac{c_{\ell}}{\log^{\ell}(n)}+O\Big{(}\frac{n}{\log^{M+2}n}\Big{)}\right\\},$

where $c_{\ell}=\int_{1}^{\infty}\frac{\log(\left\lfloor
t\right\rfloor+1)}{t^{2}}\log^{\ell}t\,dt$. In particular,

(15) $\log d(n!)\sim c_{0}\frac{n}{\log n}\qquad(n\to\infty),$

where $c_{0}$ can be shown to be equal to $\log C$. Indeed, on the one hand,

$c_{0}=\int_{1}^{\infty}\frac{\log(\left\lfloor
t\right\rfloor+1)}{t^{2}}\,dt=\sum_{s=1}^{\infty}\int_{s}^{s+1}\frac{\log(s+1)}{t^{2}}\,dt=\sum_{s=2}^{\infty}\log
s\int_{s-1}^{s}\frac{dt}{t^{2}}=\sum_{s=2}^{\infty}\frac{\log s}{s(s-1)},$

and on the other hand,

$\log
C=\sum_{s=1}^{\infty}\frac{1}{s}\log\Big{(}1+\frac{1}{s}\Big{)}=\sum_{s=1}^{\infty}\frac{1}{s}(\log(s+1)-\log
s)=\sum_{s=2}^{\infty}\Big{(}\frac{1}{s-1}-\frac{1}{s}\Big{)}\log
s=\sum_{s=2}^{\infty}\frac{\log s}{s(s-1)}.$

In fact, as we will see, each $c_{\ell}$ is given by a similar explicit
formula, and furthermore

(16) $c_{\ell}=(\ell+1)!+O((1-2^{-\ell-1})\ell!).$

We shall generalize these results to $d_{k}(n!)$, the number of $k$-divisors
of $n!$.

######  Theorem 6.

Let $k\geq 2$ and $M\geq 0$ be positive integers. Then

$d_{k}(n!)=\exp\left\\{\frac{n}{\log
n}\sum_{\ell=0}^{M}\frac{c^{(k)}_{\ell}}{\log^{\ell}n}+O\Big{(}\frac{n}{\log^{M+2}n}\Big{)}\right\\},$

where

(17)
$c_{\ell}^{(k)}=\ell!\sum_{j=0}^{\ell}\frac{1}{j!}\sum_{s=1}^{\infty}\frac{\log^{j}s}{s}\log\Big{(}1+\frac{k-1}{s}\Big{)}.$

Moreover,

$c_{\ell}^{(k)}=(k-1)(\ell+1)!+O\big{(}(k-1)^{2}(1-2^{-\ell-1})\ell!\big{)}.$

In particular, Theorem 6 implies that

(18) $\log d_{k}(n!)\sim c_{0}^{(k)}\frac{n}{\log n}\qquad(n\to\infty).$

######  Remark 3.1.

Equation (17) allows us to estimate any of the constants $c_{\ell}^{(k)}$ for
fixed $\ell$ and $k$. For instance, the first five values for $k=2$ are
$c_{0}^{(2)}\approx 1.2578$, $c_{1}^{(2)}\approx 2.1139$, $c_{2}^{(2)}\approx
6.1145$, $c_{3}^{(2)}\approx 24.1764$, $c_{4}^{(2)}\approx 120.3601$. It is
readily seen that approximation (16) is rapidly accurate.

In the same vein, Fedorov [16] studied the number of divisors of the central
binomial coefficient $\binom{2n}{n}$ and obtained a similar asymptotic series.
These results are easily extended to the $k$-divisor function.

######  Theorem 7.

Let $k\geq 2$ and $M\geq 0$ be positive integers. Then

(19) $\frac{1}{\log k}\log
d_{k}\Big{(}\binom{2n}{n}\Big{)}=\pi(2n)-\pi(n)+\frac{n}{\log
n}\sum_{\ell=0}^{M}\frac{b_{\ell}}{\log^{\ell}n}+O\Big{(}\frac{n}{\log^{M+2}n}\Big{)},$

where

$b_{\ell}=\sum_{n=1}^{\infty}\int_{n+1/2}^{n+1}\frac{\log^{\ell}t}{t^{2}}\,dt.$

######  Remark 3.2.

Note that the right-hand side of (19) does not depend on $k$. We also deduce
from this equality and the computation $b_{0}=\log 4-1$ that

$\log d_{k}\Big{(}\binom{2n}{n}\Big{)}\sim\log k\log 4\cdot\frac{n}{\log
n}\qquad(n\to\infty).$

It is clear that the same ideas can be used with the multiplicative function
$\rho$ defined as the product of the exponents in the prime factorization of
$n$; that is, $\rho(1)=1$ and, for $n\geq 2$,

$\rho(n):=\rho(p_{1}^{a_{1}}\cdots p_{r}^{a_{r}})=a_{1}\cdots a_{r},$

which in particular yields $\log\rho(n!)\sim\frac{n}{\log n}e_{0}$, where

$e_{0}=\int_{1}^{\infty}\frac{\log\lfloor
t\rfloor}{t^{2}}\,dt=\sum_{s=1}^{\infty}\frac{\log s}{s(s+1)}.$

This asymptotic formula was first obtained in [24] but it was stated in a
slightly different form, much weaker than the full statement here.

######  Theorem 8.

Let $M\geq 0$ be a positive integer. Then

$\rho(n!)=\exp\left\\{\frac{n}{\log
n}\sum_{\ell=0}^{M}\frac{e_{\ell}}{\log^{\ell}n}+O\left(\frac{n}{\log^{M+2}n}\right)\right\\},$

where

$e_{\ell}=\ell!\sum_{j=0}^{\ell}\frac{1}{j!}\sum_{s=1}^{\infty}\frac{\log^{j}(s+1)}{s+1}\log\Big{(}1+\frac{1}{s}\Big{)}.$

As for the sum of the positive divisors of $n!$, it was studied by Jakimczuk
in [25, 26], where he showed that

(20) $\displaystyle\sigma(n!)\sim e^{\gamma}n!\log n\qquad(n\to\infty),$

and, more explicitly, that

$\sigma(n!)=e^{\gamma}\sqrt{2\pi n}\Big{(}\frac{n}{e}\Big{)}^{n}\log
n\Big{(}1+O\Big{(}\frac{1}{\log n}\Big{)}\Big{)}.$

However, the proofs of these results rely on convoluted arguments, which we
simplify and generalize to $\sigma_{\kappa}(n!)$, while improving the error
term.

######  Theorem 9.

For all integers $n\geq 2$,

$\sigma(n!)=n!\,e^{\gamma}\,\log
n\Big{(}1+O\Big{(}\frac{1}{\log^{3}n}\Big{)}\Big{)},$

and for any real $\kappa>1$,

$\sigma_{\kappa}(n!)=(n!)^{\kappa}\,\zeta(\kappa)\Big{(}1+O\Big{(}\frac{\log
n}{n}\Big{)}\Big{)}.$

Here $\zeta(\cdot)$ is the Riemann zeta function

(21) $\zeta(s)=\sum_{n\geq
1}\frac{1}{n^{s}}=\prod_{p}(1-\frac{1}{p^{s}})^{-1}\qquad(\text{Re}(s)>1).$

###  3.2 Sums of multiplicative functions running on factorials

Performing partial summation with the estimate in [11, Problem 6.5], one
obtains

$\sum_{n\leq N}\gamma(n)=(1+o(1))\frac{c}{2}N^{2}\qquad(N\to\infty),$

where $c=\prod_{p}\big{(}1-\frac{1}{p(p+1)}\big{)}$. Also, it is known (see
Theorems 320, 324, and 330 in [18]) that

$\displaystyle\sum_{n\leq N}\phi(n)$ $\displaystyle=$
$\displaystyle\frac{3}{\pi^{2}}N^{2}+O(N\log N),$ $\displaystyle\sum_{n\leq
N}d(n)$ $\displaystyle=$ $\displaystyle N(\log N+2\gamma-1+O(N^{-1/2})),$
$\displaystyle\sum_{n\leq N}\sigma(n)$ $\displaystyle=$
$\displaystyle\frac{\pi^{2}}{12}N^{2}+O(N\log N).$

Moreover, one can prove (see Problem 6.12 and Theorem 6.19 in [11]) that

$\sum_{n\leq N}\log d(n)=\log 2\cdot N\Big{(}\log\log
N+c_{1}+c_{4}+O\Big{(}\frac{1}{\log N}\Big{)}\Big{)},$

where $c_{1}=\gamma+\sum_{p}\big{(}\log(1-\frac{1}{p})+\frac{1}{p}\big{)}$ and
$c_{4}=\sum_{p}\frac{\log(1+1/p)}{p(p-1)}$. For the partial sums of
$\sigma(n)/\phi(n)$ and $\phi(n)/\sigma(n)$, the following holds.

######  Proposition 1.

Given any arbitrarily small number $\delta>0$,

(22) $\sum_{n\leq
N}\frac{\sigma(n)}{\phi(n)}=d_{1}N+O(N^{\frac{1}{2}+\delta})\qquad\mbox{and}\qquad\sum_{n\leq
N}\frac{\phi(n)}{\sigma(n)}=d_{2}N+O(N^{\frac{1}{2}+\delta}),$

where

$\displaystyle d_{1}$ $\displaystyle=$
$\displaystyle\prod_{p}\Big{(}1+\frac{2p^{2}-1}{p(p+1)(p-1)^{2}}\Big{)}\approx
3.61744,$ $\displaystyle d_{2}$ $\displaystyle=$
$\displaystyle\prod_{p}\Big{(}1-\frac{2}{p(p+1)}-\frac{p-1}{p^{2}}\Big{(}\frac{1}{(p+1)(p^{2}+p+1)}+\frac{1}{(p^{2}+p+1)(p^{3}+p^{2}+p+1)}+\cdots\Big{)}\Big{)}$
$\displaystyle\approx$ $\displaystyle 0.45783.$

Although we do not claim that the results in 1 are new, we could not find them
in the literature, so for completeness we shall prove them in Section 6.2.

For sums running on factorials, we instead have the following.

######  Theorem 10.

We have

(23) $\displaystyle\sum_{n\leq N}\gamma(n!)$ $\displaystyle=$ $\displaystyle
e^{N+O\big{(}\frac{N}{\log N}\big{)}},$ (24) $\displaystyle\sum_{n\leq
N}\phi(n!)$ $\displaystyle=$ $\displaystyle e^{-\gamma}\frac{N!}{\log
N}\Big{(}1+O\Big{(}\frac{1}{\log^{3}N}\Big{)}\Big{)},$ (25)
$\displaystyle\sum_{n\leq N}\log d(n!)$ $\displaystyle=$ $\displaystyle
c_{0}\frac{N}{\log N}\Big{(}1+O\Big{(}\frac{1}{\log N}\Big{)}\Big{)},$ (26)
$\displaystyle\sum_{n\leq N}\sigma(n!)$ $\displaystyle=$ $\displaystyle
e^{\gamma}N!\log N\Big{(}1+O\Big{(}\frac{1}{\log^{3}N}\Big{)}\Big{)},$ (27)
$\displaystyle\sum_{n\leq N}\frac{\sigma(n!)}{\phi(n!)}$ $\displaystyle=$
$\displaystyle e^{2\gamma}\left(N\log^{2}N-2N\log
N+2N\right)\Big{(}1+O\Big{(}\frac{1}{\log^{3}N}\Big{)}\Big{)},$ (28)
$\displaystyle\sum_{n\leq N}\frac{\phi(n!)}{\sigma(n!)}$ $\displaystyle=$
$\displaystyle
e^{-2\gamma}\frac{N}{\log^{2}N}\Big{(}1+O\Big{(}\frac{1}{\log^{3}N}\Big{)}\Big{)}.$

##  4 The middle divisors of $n!$

We shall also treat the case of arithmetic functions which are neither
additive nor multiplicative. We are interested in the middle divisors
$\rho_{1}(n)$ and $\rho_{2}(n)$ of a positive integer $n$, which are defined
as

$\rho_{1}(n)=\max\\{d\mid
n:d\leq\sqrt{n}\\}\qquad\mbox{and}\qquad\rho_{2}(n)=\min\\{d\mid
n:d\geq\sqrt{n}\\}.$

In particular,

$\rho_{1}(n)\leq\sqrt{n}\leq\rho_{2}(n)\qquad\mbox{and}\qquad\rho_{1}(n)\rho_{2}(n)=n.$

In 1976, Tenenbaum [31] proved that

(29) $\sum_{n\leq N}\rho_{2}(n)=\dfrac{\pi^{2}}{12}\dfrac{N^{2}}{\log
N}\Big{(}1+O\Big{(}\dfrac{1}{\log N}\Big{)}\Big{)}$

and that, given any $\varepsilon>0$, there exists $N_{0}=N_{0}(\varepsilon)$
such that for all $N\geq N_{0}$,

(30) $\dfrac{N^{3/2}}{(\log N)^{\delta+\varepsilon}}<\sum_{n\leq
N}\rho_{1}(n)\ll\dfrac{N^{3/2}}{(\log N)^{\delta}(\log\log N)^{1/2}},$

where

(31) $\delta=1-\dfrac{1+\log\log 2}{\log 2}\approx 0.086071.$

In 2020, De Koninck and Razafindrasoanaivolala [8] generalised estimate (29)
by establishing that, given a real number $a>0$ and a positive integer $k$,

$\displaystyle\sum_{n\leq N}\rho_{2}(n)^{a}=c_{0}\dfrac{N^{a+1}}{\log
N}+c_{1}\dfrac{N^{a+1}}{\log^{2}N}+\cdots+c_{k-1}\dfrac{N^{a+1}}{\log^{k}N}+O\Big{(}\dfrac{N^{a+1}}{\log^{k+1}N}\Big{)},$

where, for $\ell=0,1,\ldots,k-1$, the constants $c_{\ell}=c_{\ell}(a)$ are
explicitly given. They also proved that, given any integer $k\geq 1$ and any
real number $r>-1$,

$\sum_{n\leq N}\dfrac{\rho_{2}(n)}{\rho_{1}(n)^{r}}=h_{0}\dfrac{N^{2}}{\log
N}+h_{1}\dfrac{N^{2}}{\log^{2}N}+\cdots+h_{k-1}\dfrac{N^{2}}{\log^{k}N}+O\left(\dfrac{N^{2}}{\log^{k+1}N}\right)$

where $h_{0}=\dfrac{\zeta(r+2)}{2}$ and for each $1\leq\ell\leq k-1$,

$h_{\ell}=\left(\dfrac{r+2}{2}\right)c_{\ell}+\sum_{\nu=0}^{\ell-1}\dfrac{rc_{\nu}}{2}\prod_{m=\nu}^{\ell-1}\left(\dfrac{m+1}{2}\right),$

with, for each $\nu=0,1,\ldots,\ell$,

$c_{\nu}=\dfrac{\nu!}{(r+2)^{\nu+1}}\sum_{j=0}^{\nu}\dfrac{(r+2)^{j}(-1)^{j}\zeta^{(j)}(r+2)}{j!}.$

In 2008, Ford [17] considerably improved (30) by showing that

(32) $\displaystyle\sum_{n\leq N}\rho_{1}(n)\asymp\dfrac{N^{3/2}}{(\log
N)^{\delta}(\log\log N)^{3/2}}.$

Interestingly no asymptotic formula has ever been established for the sum
$\sum_{n\leq N}\rho_{1}(n)$.

Recently, De Koninck and Razafindrasoanaivolala [9] established that

(33) $\sum_{\begin{subarray}{c}4\leq n\leq N\\\
n\neq\text{prime}\end{subarray}}\frac{\log\rho_{2}(n)}{\log\rho_{1}(n)}=N\log\log
N+O(N)$

and also that, for all $N$ sufficient large,

(34) $c_{1}\,N<\sum_{2\leq n\leq
N}\frac{\log\rho_{1}(n)}{\log\rho_{2}(n)}<c_{2}\,N,$

where

$\displaystyle c_{1}$ $\displaystyle=$ $\displaystyle 1-\log
2+\int_{1}^{2}\frac{1-\log
u}{u(u+1)}\,du+\int_{3}^{\infty}\frac{u-1}{u+1}\frac{\rho(u-1)}{u}\,du\approx
0.528087,$ $\displaystyle c_{2}$ $\displaystyle=$ $\displaystyle 2-2\log
2\approx 0.613706.$

As we will now see, the behavior of each of the sums

(35) $\sum_{n\leq N}\rho_{1}(n!),\quad\sum_{n\leq
N}\rho_{2}(n!),\quad\sum_{n\leq
N}\dfrac{\rho_{1}(n!)}{\rho_{2}(n!)},\quad\sum_{n\leq
N}\dfrac{\rho_{2}(n!)}{\rho_{1}(n!)},\quad\sum_{2\leq n\leq
N}\frac{\log\rho_{1}(n!)}{\log\rho_{2}(n!)},\quad\sum_{2\leq n\leq
N}\frac{\log\rho_{2}(n!)}{\log\rho_{1}(n!)}$

is easier to study.

######  Theorem 11.

Letting $c$ be the constant appearing below estimate (68), we have

$\displaystyle\sum_{n\leq N}\rho_{1}(n!)$ $\displaystyle=$
$\displaystyle\sqrt{N!}\Big{(}1+O\Big{(}\frac{1}{N^{c}}\Big{)}\Big{)},$
$\displaystyle\sum_{n\leq N}\rho_{2}(n!)$ $\displaystyle=$
$\displaystyle\sqrt{N!}\Big{(}1+O\Big{(}\frac{1}{N^{c}}\Big{)}\Big{)}.$

######  Theorem 12.

Letting $c$ be as above, we have

$\displaystyle\sum_{n\leq N}\frac{\rho_{1}(n!)}{\rho_{2}(n!)}$
$\displaystyle=$ $\displaystyle N+O\left(N^{1-c}\right),$
$\displaystyle\sum_{n\leq N}\frac{\rho_{2}(n!)}{\rho_{1}(n!)}$
$\displaystyle=$ $\displaystyle N+O\left(N^{1-c}\right).$

######  Theorem 13.

We have

$\displaystyle\sum_{2\leq n\leq N}\frac{\log\rho_{1}(n!)}{\log\rho_{2}(n!)}$
$\displaystyle=$ $\displaystyle N+O(1),$ $\displaystyle\sum_{2\leq n\leq
N}\frac{\log\rho_{2}(n!)}{\log\rho_{1}(n!)}$ $\displaystyle=$ $\displaystyle
N+O(1).$

Observe that the main difficulty in finding an estimate for any of the sums
appearing in (35) at arguments $n$ (and not $n!$) is that the values of the
function $\rho_{1}(n)$ (resp. $\rho_{2}(n)$) are spread out over the interval
$[1,\sqrt{n}]$ (resp. $[\sqrt{n},n]$). On the contrary, the values of
$\rho_{1}(n!)$ and $\rho_{2}(n!)$ are all near $\sqrt{n!}$, as is clarified in
the following proposition.

######  Proposition 2.

To any given sufficiently large integer $n$, one can associate two positive
integers $d_{1}<d_{2}$ both dividing $n!$ and satisfying

(36)
$\Big{(}1-\frac{1}{n^{c}}\Big{)}\sqrt{n!}<d_{1}<\sqrt{n!}<d_{2}<\Big{(}1+\frac{2}{n^{c}}\Big{)}\sqrt{n!},$

where $c$ is the constant appearing below estimate (68), thereby implying that

(37)
$\Big{(}1-\frac{1}{n^{c}}\Big{)}\sqrt{n!}<\rho_{1}(n!)<\sqrt{n!}<\rho_{2}(n!)<\Big{(}1+\frac{2}{n^{c}}\Big{)}\sqrt{n!}.$

### Numerical data

From the construction used in Section 6.5 in the proof of Proposition 2 to
create the numbers $d_{1}=d_{1}(n)$ for each $n\geq 6$, it is clear that
$d_{1}\leq\rho_{1}(n)$. Interestingly, one can check that $d_{1}=\rho_{1}(n)$
in the cases $n=7,10,22$. These are perhaps the only ones with that particular
property.

In the particular cases $n=10,20,30,40$, we obtain the following data.  $n$
$n!$ $d_{1}$ $\lfloor\sqrt{n!}\rfloor$ $\lfloor\sqrt{n!}\rfloor/d_{1}$ 10 3
628 800 1890 1904 1.007407407 $2^{8}\cdot 3^{4}\cdot 5^{2}\cdot 7$ $2\cdot
3^{3}\cdot 5\cdot 7$ 20 2 432 902 008 176 640 000 1 558 878 750 1 559 776 268
1.000575746 $2^{18}\cdot 3^{8}\cdot 5^{4}\cdot 7^{2}\cdot 11\cdot 13\cdot
17\cdot 19$ $2\cdot 3^{3}\cdot 5^{4}\cdot 11\cdot 13\cdot 17\cdot 19$ 30
265252859812191058636308480000000 16283616779520000 16286585271694955
1.000182299 $2^{26}\cdot 3^{14}\cdot 5^{7}\cdot 7^{4}\cdot 11^{2}\cdot
13^{2}\cdot 17\cdot 19\cdot 23\cdot 29$ $2^{11}\cdot 3^{10}\cdot 5^{4}\cdot
17\cdot 19\cdot 23\cdot 29$

When $n=40$,

$\displaystyle n!$ $\displaystyle=$ $\displaystyle
815\,915\,283\,247\,897\,734\,345\,611\,269\,596\,115\,894\,272\,000\,000\,000$
$\displaystyle=2^{38}\cdot 3^{18}\cdot 5^{9}\cdot 7^{5}\cdot 11^{3}\cdot
13^{3}\cdot 17^{2}\cdot 19^{2}\cdot 23\cdot 29\cdot 31\cdot 37,$
$\displaystyle d_{1}$ $\displaystyle=$ $\displaystyle
903\,280\,055\,273\,332\,645\,708\,800=2^{10}\cdot 3^{15}\cdot 5^{2}\cdot
7\cdot 11\cdot 13^{3}\cdot 19\cdot 23\cdot 29\cdot 31\cdot 37,$
$\displaystyle\lfloor\sqrt{n!}\rfloor$ $\displaystyle=$ $\displaystyle
903\,280\,290\,523\,322\,408\,635\,610,$
$\displaystyle\lfloor\sqrt{n!}\rfloor/d_{1}$ $\displaystyle=$ $\displaystyle
1.000000260.$

Observe that $\rho_{1}(10!)=1\,890$, $\rho_{1}(20!)=1\,559\,376\,000$,
$\rho_{1}(30!)=16\,286\,248\,192\,500\,000$, and
$\rho_{1}(40!)=903\,280\,055\,273\,332\,645\,708\,800$.

##  5 A comparison with the maximum of arithmetic functions

In Sections 2 and 3, we compared some estimates on partial sums of additive
and multiplicative functions with their factorial counterpart. It is
interesting to compare the behavior of $f(n!)$ with the maximum of $f(n)$.

In this section, we will use the estimate

(38) $\log n!=n\log n-n+O(\log n),$

which follows from the fact that $\sum_{k\leq n}\log k=\int_{1}^{n}\log
t\,dt+O(\log n)$. It is a weak form of Stirling’s formula

(39) $n!\sim\Big{(}\frac{n}{e}\Big{)}^{n}\sqrt{2\pi n}\qquad(n\to\infty).$

Writing $m=n!$, estimate (38) thus implies that, as $n$ tends to infinity,

(40) $\log n\sim\log\log m\qquad\text{and}\qquad n\sim\frac{\log m}{\log\log
m}.$

Sometimes $f(m)$ is very far from the maximum of $f(n)$ over $n\leq x$. For
instance, $\max_{n\leq x}\Omega(n)=\lfloor\log x/\log 2\rfloor,$ with equality
if and only if $k$ is the largest positive integer such that $n=2^{k}\leq x$.
However, by combining Theorem 1 and (40), we get that

$\Omega(m)\sim\log m\frac{\log\log\log m}{\log\log m}\qquad(m\to\infty).$

Other times, $f(m)$ is almost as large as $f(n)$. Indeed, we know that
integers $n\leq x$ have at most $(1+o(1))\frac{\log x}{\log\log x}$ distinct
prime factors, with equality if and only if $n$ is the product of the first
$y$ primes, where $y$ is chosen maximally so that $n=\prod_{p\leq y}p\leq x$.
We thus have $\omega(n)=\pi(y)$, and since $\omega(m)=\pi(n)$, we see that it
is almost of this form, and indeed we have

$\omega(m)\sim\frac{\log m}{(\log\log m)^{2}}.$

There is a similar pattern for $\gamma(n)$. It is clear that $\max_{n\leq
x}\gamma(n)=\prod_{p\leq y}p$ where $y$ is chosen as above. Thus we see that
$\max_{n\leq x}\gamma(n)=\gamma(y!)=x^{1+o(1)}$ as $x\to\infty$, since
$y\sim\log x$. On the other hand,

$\gamma(m)\sim m^{\frac{1+o(1)}{\log\log m}}\qquad(m\to\infty).$

Similarly, for $d(n)$, a classical result of Wigert [33] gives

$\log d(n)\leq\frac{\log 2\log n}{\log\log n}+O\Big{(}\frac{\log n}{(\log\log
n)^{2}}\Big{)},$

with equality if and only if $n$ is the product of the first $r$ primes,
$r\to\infty$, while we see from (15) and (40) that

$\log d(m)\sim c_{0}\frac{\log m}{(\log\log m)^{2}}\qquad(m\to\infty),$

as already observed in [15]. More generally,

$\log d_{k}(n)\leq(\log k+o(1))\frac{\log n}{\log\log n}\qquad(n\to\infty),$

and we get from (18) that

$\log d_{k}(m)\sim c^{(k)}_{0}\frac{\log m}{(\log\log
m)^{2}}\qquad(m\to\infty).$

We shall also observe that $f(m)$ sometimes attains the minimal or maximal
value of $f(n)$. For example, we have

$\phi(m)\sim e^{-\gamma}\frac{m}{\log\log
m}\qquad\text{and}\qquad\sigma(m)\sim e^{\gamma}m\log\log
m\qquad(m\to\infty),$

while classical results (see Theorems 323, 328 in [18]) give

$\liminf_{n\to\infty}\frac{\phi(n)\log\log
n}{n}=e^{-\gamma}\qquad\text{and}\qquad\limsup_{n\to\infty}\frac{\sigma(n)}{n\log\log
n}=e^{\gamma}.$

In other words, $n=m!$ attains a minimal order for $\phi(n)$ and a maximal
order for $\sigma(n)$. Similarly, for any real $\kappa>1$ we have

$\limsup_{n\to\infty}\frac{\sigma_{\kappa}(n)}{n^{\kappa}}=\zeta(\kappa),$

while we know from Theorem 9 that

$\sigma_{\kappa}(m)\sim m^{\kappa}\,\zeta(\kappa)\qquad(m\to\infty).$

Note that the behavior of $\sigma_{\kappa}(n)$ is entirely different for
$0<\kappa<1$.

##  6 Proofs of the main results

Our main tool will be de Polignac’s formula (attributed by L.E. Dickson [14]
to Legendre)

(41) $n!=\prod_{p\leq n}p^{\alpha_{p}(n)},\mbox{ where
}\alpha_{p}(n)=\sum_{i=1}^{\infty}\left\lfloor\frac{n}{p^{i}}\right\rfloor=\sum_{i=1}^{\lfloor\log_{p}n\rfloor}\left\lfloor\frac{n}{p^{i}}\right\rfloor.$

The above formula can be somewhat simplified. Indeed, it is easy to show that

(42)
$\sum_{i=1}^{\infty}\left\lfloor\frac{n}{p^{i}}\right\rfloor=\frac{n-s_{p}(n)}{p-1},$

where $s_{p}(n)$ stands for the sum of the digits of $n$ in base $p$. For a
proof of (42), see Problem 438 in the book of De Koninck and Mercier [12].

We will also rely on classical results, including the following well-known
version of the prime number theorem (see, e.g., [21]):

(43) $\pi(x)=\text{Li}(x)+O(x\exp(-\sqrt{\log x})),$

where $\text{Li}(x)=\int_{2}^{x}\frac{dt}{\log t}\sim\frac{x}{\log x}$ as
$x\to\infty$. We shall at times use the prime number theorem in the simpler
form

(44) $\pi(N)=\frac{N}{\log N}\Big{(}1+O\Big{(}\frac{1}{\log N}\Big{)}\Big{)},$

which can be equivalently stated as

(45) $\theta(N)=N+O\Big{(}\frac{N}{\log N}\Big{)},$

where $\theta$ is the Chebyshev function

(46) $\theta(N):=\sum_{p\leq N}\log p.$

From partial summation and the prime number theorem, one easily deduces the
following asymptotic formula for the sum of the primes up to $N$:

(47) $\sum_{p\leq N}p\sim\frac{N^{2}}{2\log N}\qquad(N\to\infty).$

We will also need Mertens’ theorem, which we use in a strong form due to
Dusart [13]:

(48) $\prod_{p\leq x}\Big{(}1-\frac{1}{p}\Big{)}=\frac{e^{-\gamma}}{\log
x}\Big{(}1+O\Big{(}\frac{1}{\log^{3}x}\Big{)}\Big{)}.$

###  6.1 Proofs of Theorems 1, 3, and 5

Let $f$ be an additive function and $g$ a multiplicative function. Recall that
we say that $f$ is completely additive (resp., $g$ is completely
multiplicative) if $f(nm)=f(n)+f(m)$ (resp., $g(nm)=g(n)g(m)$) for any
positive integers $n$ and $m$. Also, $f$ is strongly additive (resp., $g$ is
strongly multiplicative) if for any prime $p$ and positive integer $a$,
$f(p^{a})=f(p)$ (resp., $g(p^{a})=g(p)$).

Note that $\omega$ and $\beta$ are strongly additive, whereas $\Omega$, $B$,
and $B_{1}$ are completely additive. Theorems 1 and 3 thus follow from the
following more general observation, as well as from the estimates (1), (8),
(44), and (47).

######  Proposition 3.

Let $f$ be a completely additive function and $g$ a strongly additive
function. Then $f(n!)=S_{f}(n)$ and $g(n!)=\sum_{p\leq n}g(p)$.

The multiplicative counterpart to Proposition 3 is as follows.

######  Proposition 4.

Let $f$ be a completely multiplicative function and $g$ a strongly
multiplicative function. Then $f(n!)=\prod_{k\leq n}f(k)$ and
$g(n!)=\prod_{p\leq n}g(p)$.

To prove Theorem 5, first note that both $\gamma(n)$ and $\phi(n)/n$ are
strongly multiplicative, thus proving the identities in (12). Second, use that
$\log\gamma(n!)=\sum_{p\leq n}\log p$, which by the prime number theorem (43)
is $\sim n$ as $n\to\infty$. The last estimate in (13) follows from Mertens’
theorem given through estimate (48).

######  Remark 6.1.

The de Polignac formula (41) implies that for any additive function $f$,

$f(n!)=\sum_{p\leq n}f(p^{\alpha_{p}(n)}),$

so that the assertion $f(n!)=S_{f}(n)$ for completely additive functions $f$
is in fact equivalent to

(49) $\sum_{p\leq n}\alpha_{p}(n)f(p)=\sum_{k\leq n}f(k),$

which is perhaps not so obvious. The analogous result for completely
multiplicative functions says that $\prod_{p\leq
n}f(p)^{\alpha_{p}(n)}=\prod_{k\leq n}f(k)$, which can also be deduced
directly from (49) by taking the $\log$ on both sides.

###  6.2 Proofs of Theorems 2, 4, and 10

Note that (5) is an immediate consequence of (2). Most of the other estimates
in 2 follow at once from standard partial summation techniques.

For instance, using partial summation with Theorem 1 and (44) gives

$\sum_{2\leq n\leq N}\omega(n!)=\sum_{2\leq n\leq N}\frac{n}{\log
n}\Big{(}1+O\Big{(}\frac{1}{\log n}\Big{)}\Big{)}=\frac{N^{2}}{2\log
N}\Big{(}1+O\Big{(}\frac{1}{\log N}\Big{)}\Big{)},$

which proves (3).

Using Theorem 1 and identity (1), we get

$\displaystyle\sum_{2\leq n\leq N}\Omega(n!)$ $\displaystyle=\sum_{2\leq n\leq
N}n\Big{(}\log\log n+c_{1}+c_{2}+O\Big{(}\frac{1}{\log n}\Big{)}\Big{)}$
$\displaystyle=\sum_{2\leq n\leq N}n\log\log n+(c_{1}+c_{2})\sum_{2\leq n\leq
N}n+O\Big{(}\sum_{2\leq n\leq N}\frac{n}{\log n}\Big{)}$ (50)
$\displaystyle=\sum_{2\leq n\leq N}n\log\log
n+(c_{1}+c_{2})\frac{N^{2}}{2}+O\Big{(}\frac{N^{2}}{\log N}\Big{)}.$

By partial summation, the remaining sum is

$\displaystyle\Big{(}\frac{N^{2}}{2}+O(N)\Big{)}\log\log
N-\int_{2}^{N}\frac{\frac{t^{2}}{2}+O(t)}{t\log t}\,dt$
$\displaystyle=\frac{N^{2}}{2}\log\log N+O(N\log\log
N)+O\Big{(}\int_{2}^{N}\frac{t}{\log t}\,dt\Big{)}$ (51)
$\displaystyle=\frac{N^{2}}{2}\log\log N+O\Big{(}\frac{N^{2}}{\log N}\Big{)},$

the last inequality following from integration by parts. Combining (50) and
(51) gives (4).

We proceed in a similar way to prove (6). By Theorem 1,

$\displaystyle\sum_{2\leq n\leq N}\frac{1}{\Omega(n!)}$
$\displaystyle=\sum_{2\leq n\leq N}\frac{1}{n\big{(}\log\log
n+c_{1}+c_{2}+O\big{(}\frac{1}{\log n}\big{)}\big{)}}$
$\displaystyle=\sum_{2\leq n\leq N}\frac{1}{n\log\log
n\big{(}1+O\big{(}\frac{1}{\log\log n}\big{)}\big{)}}$
$\displaystyle=\sum_{2\leq n\leq N}\frac{1}{n\log\log
n}\Big{(}1+O\Big{(}\frac{1}{\log\log n}\Big{)}\Big{)}.$

By partial summation, the main term is

$\sum_{2\leq n\leq N}\frac{1}{n\log\log n}=\frac{\log N+O(1)}{\log\log
N}-\int_{2}^{N}\frac{\log t+O(1)}{t\log t(\log\log t)^{2}}\,dt.$

Integrating by parts, this last integral is

$\displaystyle\ll\int_{2}^{N}\frac{dt}{t(\log\log t)^{2}}=\frac{\log
N}{(\log\log N)^{2}}+O(1)-\int_{2}^{N}\frac{dt}{t(\log\log
t)^{3}}=O\Big{(}\frac{\log N}{(\log\log N)^{2}}\Big{)}.$

The same upper bound holds for the remaining sum in the error term since

$\displaystyle\sum_{2\leq n\leq N}\frac{1}{n(\log\log
n)^{2}}=\int_{2}^{N}\frac{dt}{t(\log\log t)^{2}}+O(1).$

Combining these estimates yields (6).

Finally, combining Theorem 1 with (1) and (44) yields

$\displaystyle\sum_{2\leq n\leq N}\frac{\Omega(n!)}{\omega(n!)}$
$\displaystyle=\sum_{2\leq n\leq N}\frac{\log n\big{(}\log\log
n+c_{1}+c_{2}+O\big{(}\frac{1}{\log n}\big{)}\big{)}}{1+O\big{(}\frac{1}{\log
n}\big{)}}$ $\displaystyle=\sum_{2\leq n\leq N}\log n\Big{(}\log\log
n+c_{1}+c_{2}+O\Big{(}\frac{1}{\log
n}\Big{)}\Big{)}\Big{(}1+O\Big{(}\frac{1}{\log n}\Big{)}\Big{)}$
$\displaystyle=\sum_{2\leq n\leq N}\log n\log\log n+(c_{1}+c_{2})\sum_{2\leq
n\leq N}\log n+O\Big{(}\sum_{2\leq n\leq N}\log\log n\Big{)}$
$\displaystyle=S_{1}+S_{2}+S_{3},$

say. Clearly, $S_{2}=(c_{1}+c_{2})\,N\log N+O(N)$ by (38) and
$S_{3}=O(N\log\log N)$. By partial summation and (38),

$\displaystyle S_{1}$ $\displaystyle=(N\log N+O(N))\log\log
N-\int_{2}^{N}\frac{(t\log t+O(t))}{t\log t}\,dt$ $\displaystyle=(N\log
N+O(N))\log\log N+O(N)$ $\displaystyle=N\log N\log\log N+O(N\log\log N).$

Estimate (7) follows.

Now we prove 4. From Theorem 3, we know that

$\sum_{2\leq n\leq N}\beta(n!)\sim\sum_{2\leq n\leq N}\frac{n^{2}}{2\log
n}\qquad(N\to\infty).$

We can approximate the sum by an integral and integrate by parts to conclude
that

$\sum_{2\leq n\leq N}\frac{n^{2}}{2\log n}\sim\int_{2}^{N}\frac{t^{2}}{2\log
t}\,dt\sim\frac{N^{3}}{6\log N}\qquad(N\to\infty).$

Estimate (9) follows. We proceed similarly to prove (10). Using Theorem 3, (8)
and partial summation, we get

$\sum_{2\leq n\leq N}B(n!)\sim\sum_{2\leq n\leq
N}\frac{\pi^{2}}{12}\frac{N^{2}}{\log
N}\sim\int_{2}^{N}\frac{\pi^{2}}{12}\frac{t^{2}}{\log
t}\,dt\sim\frac{\pi^{2}}{36}\frac{N^{3}}{\log N}\qquad(N\to\infty).$

The exact same proof works for $\sum_{2\leq n\leq N}B_{1}(n!)$. Now letting
$f$ be any of the functions $\beta$, $B$ and $B_{1}$, we have

$\sum_{n=2}^{\infty}\frac{1}{f(n!)}\ll\sum_{n=2}^{\infty}\frac{\log
n}{n^{2}}<\infty.$

This concludes the proof of 4.

Before proceeding with 10, we give a proof of 1.

###### Proof of 1.

First observe that for any real number $s>1$, we have

$\displaystyle\sum_{n=1}^{\infty}\frac{\sigma(n)/\phi(n)}{n^{s}}$
$\displaystyle=$
$\displaystyle\prod_{p}\Big{(}1+\frac{\sigma(p)/\phi(p)}{p^{s}}+\frac{\sigma(p^{2})/\phi(p^{2})}{p^{2s}}+\cdots\Big{)}$
$\displaystyle=$
$\displaystyle\zeta(s)\prod_{p}\Big{(}1-\frac{1}{p^{s}}\Big{)}\Big{(}1+\frac{(p+1)/(p-1)}{p^{s}}+\frac{(p^{2}+p+1)/(p(p-1))}{p^{2s}}+\cdots\Big{)}$
$\displaystyle=$ $\displaystyle\zeta(s)F(s),$

say. It is clear that

$F(s)=\prod_{p}\Big{(}1+\Big{(}\frac{(p+1)/(p-1)}{p^{s}}-\frac{1}{p^{s}}\Big{)}+\Big{(}\frac{(p^{2}+p+1)/(p(p-1))}{p^{2s}}-\frac{(p+1)/(p-1)}{p^{2s}}\Big{)}+\cdots\Big{)}.$

Since we intend to use Wintner’s theorem (see Theorem 6.13 in [11], as well as
its refinement given in Problem 6.3), we need to check that $F(s)$ converges
absolutely at $s=\frac{1}{2}+\delta$ and verify that indeed $F(1)=d_{1}$. One
can easily check that the first of these two conditions is satisfied. To
verify the second condition, observe that

$\displaystyle F(1)$ $\displaystyle=$
$\displaystyle\prod_{p}\Big{(}1+\frac{2}{p(p-1)}+\frac{1}{p^{3}(p-1)}+\frac{1}{p^{5}(p-1)}+\cdots\Big{)}$
$\displaystyle=$
$\displaystyle\prod_{p}\Big{(}1+\frac{2}{p(p-1)}+\frac{1}{p(p+1)(p-1)^{2}}\Big{)}$
$\displaystyle=$
$\displaystyle\prod_{p}\Big{(}1+\frac{2p^{2}-1}{p(p+1)(p-1)^{2}}\Big{)},$

thus establishing our first claim with an application of Wintner’s theorem.

To establish the second claim, one can proceed as above and in the end obtain
that

$\sum_{n=1}^{\infty}\frac{\phi(n)/\sigma(n)}{n^{s}}=\zeta(s)G(s),$

where

$G(1)=\prod_{p}\Big{(}1-\frac{2}{p(p+1)}-\frac{p-1}{p^{2}}\Big{(}\frac{1}{(p+1)(p^{2}+p+1)}+\frac{1}{(p^{2}+p+1)(p^{3}+p^{2}+p+1)}+\cdots\Big{)}\Big{)},$

thus completing the proof of our second assertion. ∎

We now move on to 10. To prove (23), we shall show that

$\sum_{n\leq N}\gamma(n!)=e^{\theta(N)+O(\log N)},$

where $\theta$ is the Chebyshev function defined in (46), since then the
result follows from the prime number theorem in the form (45). Let $r$ be the
unique integer satisfying $p_{r}\leq N<p_{r+1}$. Then, one can easily see that

(52) $\sum_{n\leq N}\gamma(n!)=\sum_{i=1}^{r-1}(p_{i+1}-p_{i})\prod_{p\leq
p_{i}}p+(N-p_{r})\prod_{p\leq p_{r}}p.$

Using Bertrand’s postulate, we have that $p_{i+1}-p_{i}<p_{i}$ and therefore

(53) $\sum_{i=1}^{r-1}(p_{i+1}-p_{i})\prod_{p\leq
p_{i}}p<\sum_{i=1}^{r-1}p_{i}\prod_{p\leq p_{i}}p<\sum_{i=1}^{r-1}\prod_{p\leq
p_{i+1}}p<r\prod_{p\leq p_{r}}p.$

Combining (52) and (53), we may write that

(54) $(N-p_{r})\prod_{p\leq p_{r}}p<\sum_{n\leq N}\gamma(n!)<r\prod_{p\leq
p_{r}}p+(N-p_{r})\prod_{p\leq p_{r}}p=(N-p_{r}+r)\prod_{p\leq p_{r}}p.$

From the definition of $\theta(N)$, we have

(55) $\prod_{p\leq p_{r}}p=\prod_{p\leq N}p=e^{\theta(N)}.$

On the other hand, it is clear that $N-p_{r}<N-p_{r}+r<N$, implying that both
$N-p_{r}$ and $N-p_{r}+r$ are $e^{O(\log N)}$. Combining this observation with
(54) and (55), it is immediate that

$\sum_{n\leq N}\gamma(n!)=e^{\theta(N)}e^{O(\log N)}=e^{\theta(N)+O(\log N)},$

as claimed.

To prove (24), first observe that

$\displaystyle\sum_{2\leq n\leq N}\frac{n!}{\log n}$
$\displaystyle=\frac{N!}{\log N}\Big{(}1+\frac{1}{N}\frac{\log
N}{\log(N-1)}+\frac{1}{N(N-1)}\frac{\log(N-1)}{\log(N-2)}+\cdots\Big{)}$
$\displaystyle=\frac{N!}{\log N}\Big{(}1+O\Big{(}\frac{1}{N}\Big{)}\Big{)},$

and, similarly,

$\sum_{2\leq n\leq
N}\frac{n!}{\log^{4}n}=\frac{N!}{\log^{4}N}\Big{(}1+O\Big{(}\frac{1}{N}\Big{)}\Big{)}.$

Combining these estimates with Theorem 5 and (48), we obtain

$\displaystyle\sum_{2\leq n\leq N}\phi(n!)$ $\displaystyle=\sum_{2\leq n\leq
N}n!\frac{e^{-\gamma}}{\log
n}\Big{(}1+O\Big{(}\frac{1}{\log^{3}n}\Big{)}\Big{)}$
$\displaystyle=e^{-\gamma}\frac{N!}{\log
N}\Big{(}1+O\Big{(}\frac{1}{\log^{3}N}\Big{)}\Big{)}.$

The proof of (26) is similar. On the other hand, (25) follows from partial
summation after using the estimate

$\log d(n!)=c_{0}\frac{n}{\log n}+O\Big{(}\frac{n}{\log^{2}n}\Big{)},$

a consequence of (14).

Using Theorems 5 and 9, we may write that

$\displaystyle\frac{\sigma(n!)}{\phi(n!)}$ $\displaystyle=$
$\displaystyle\frac{\prod_{p\leq
n}\big{(}1-\frac{1}{p^{\alpha_{p}(n)+1}}\big{)}}{\prod_{p\leq
n}\big{(}1-\frac{1}{p}\big{)}^{2}}$ $\displaystyle=$ $\displaystyle
e^{2\gamma}\log^{2}n\Big{(}1+O\Big{(}\frac{1}{\log^{3}n}\Big{)}\Big{)}\Big{(}1+O\Big{(}\frac{\log
n}{n}\Big{)}\Big{)}$ $\displaystyle=$ $\displaystyle
e^{2\gamma}\log^{2}n\Big{(}1+O\Big{(}\frac{1}{\log^{3}n}\Big{)}\Big{)}.$

Summing this last expression over all integers $n\in[2,N]$, we obtain

$\displaystyle\sum_{2\leq n\leq N}\frac{\sigma(n!)}{\phi(n!)}$
$\displaystyle=$ $\displaystyle e^{2\gamma}\sum_{2\leq n\leq
N}\log^{2}n\Big{(}1+O\Big{(}\frac{1}{\log^{3}n}\Big{)}\Big{)}$
$\displaystyle=$ $\displaystyle
e^{2\gamma}\Big{(}1+O\Big{(}\frac{1}{\log^{3}N}\Big{)}\Big{)}\int_{2}^{N}\log^{2}t\,dt$
$\displaystyle=$ $\displaystyle e^{2\gamma}(N\log^{2}N-2N\log
N+2N)\Big{(}1+O\Big{(}\frac{1}{\log^{3}N}\Big{)}\Big{)},$

thus establishing (27).

The proof of (28) follows along the same lines, thus completing the proof of
10.

###  6.3 Proofs of Theorems 6, 7, and 8

Our approach is at first similar to the one used in [15] for each integer
$k=2$. To work with $d_{k}(n!)$, simply recall that
$d_{k}(p^{a})=\binom{k+a-1}{k-1}$ for $k\geq 2$, so that

$d_{k}(n!)=\prod_{p\leq n}\binom{k+\alpha_{p}(n)-1}{k-1}$

by (41). It is more convenient to work with $\log d_{k}(n!)$, splitting the
resulting sum at $p=n^{3/4}$. Since

$\binom{k+\alpha_{p}(n)-1}{k-1}\ll_{k}\alpha_{p}^{k-1}(n)$

and $\alpha_{p}(n)<\frac{n}{p-1}$ (this last inequality following from (42)),
we get

$\sum_{p\leq n^{3/4}}\log\binom{k+\alpha_{p}(n)-1}{k-1}\ll_{k}\sum_{p\leq
n^{3/4}}\log\Big{(}\frac{n}{p-1}\Big{)}\ll_{k}\log n\sum_{p\leq
n^{3/4}}1\ll_{k}n^{3/4},$

where in this last inequality we used (44).

The main contribution to the sum comes from the primes $p\in(n^{3/4},n]$,
where ${\alpha_{p}(n)=\lfloor n/p\rfloor}$. We have

(56) $\sum_{n^{3/4}<p\leq
n}\log\binom{k+\alpha_{p}(n)-1}{k-1}=\int_{n^{3/4}}^{n}\log\binom{k+\left\lfloor\frac{n}{x}\right\rfloor-1}{k-1}\,d\pi(x).$

Writing $\pi(x)=\text{Li}(x)+E(x)$ where $E(x)=O(x\exp(-\sqrt{\log x}))$ by
(43), partial summation on the right-hand side of (56) gives

$I:=\int_{n^{3/4}}^{n}\log\binom{k+\left\lfloor\frac{n}{x}\right\rfloor-1}{k-1}\,\frac{dx}{\log
x}$

as the main term, while the contribution of the error term is

$\int_{n^{3/4}}^{n}\log\binom{k+\left\lfloor\frac{n}{x}\right\rfloor-1}{k-1}\,dE(x)\ll_{k}\log
n\int_{n^{3/4}}^{n}dE(x)\ll_{k}n\log n\exp(-\sqrt{\log n}).$

To estimate the remaining integral $I$, we start with the substitution $t=n/x$
so that

$I=n\int_{1}^{n^{1/4}}\log\binom{k+\left\lfloor
t\right\rfloor-1}{k-1}\frac{dt}{t^{2}\log(n/t)}.$

Since $1\leq t\leq n^{1/4}$, we have that for any fixed integer $M\geq 0$,

$\frac{1}{\log(n/t)}=\frac{1}{\log n\big{(}1-\frac{\log t}{\log
n}\big{)}}=\frac{1}{\log n}\left\\{\sum_{\ell=0}^{M}\Big{(}\frac{\log t}{\log
n}\Big{)}^{\ell}+O\Big{(}\frac{\log t}{\log n}\Big{)}^{M+1}\Big{)}\right\\},$

from which we get

(57) $I=\frac{n}{\log n}\int_{1}^{n^{1/4}}\log\binom{k+\left\lfloor
t\right\rfloor-1}{k-1}\left\\{\sum_{\ell=0}^{M}\Big{(}\frac{\log t}{\log
n}\Big{)}^{\ell}+O\Big{(}\frac{\log t}{\log
n}\Big{)}^{M+1}\Big{)}\right\\}\frac{dt}{t^{2}}.$

Since $\int_{n^{1/4}}^{\infty}\log\binom{k+\left\lfloor
t\right\rfloor-1}{k-1}(\log t)^{\ell}\frac{dt}{t^{2}}$ converges, and
furthermore

$\int_{n^{1/4}}^{\infty}\log\binom{k+\left\lfloor t\right\rfloor-1}{k-1}(\log
t)^{\ell}\frac{dt}{t^{2}}\ll_{k}\int_{n^{1/4}}^{\infty}\frac{\log^{\ell+1}t}{t^{2}}\,dt\ll_{k}\frac{\log^{\ell+1}n}{n^{1/4}}$

for any $0\leq\ell\leq M$, the error term in (57) is

$O\Big{(}\frac{n}{\log^{M+2}n}\int_{1}^{n^{1/4}}\log\binom{k+\left\lfloor
t\right\rfloor-1}{k-1}\log^{M+1}t\,\frac{dt}{t^{2}}\Big{)}=O\Big{(}\frac{n}{\log^{M+2}n}\Big{)},$

while the main term is

$\begin{split}\frac{n}{\log n}\sum_{\ell=0}^{M}&\frac{1}{(\log
n)^{\ell}}\left\\{\int_{1}^{\infty}\log\binom{k+\left\lfloor
t\right\rfloor-1}{k-1}\log^{\ell}t\,\frac{dt}{t^{2}}+O\Big{(}\frac{\log^{\ell+1}n}{n^{1/4}}\Big{)}\right\\}\\\
&=\frac{n}{\log n}\sum_{\ell=0}^{M}\frac{c_{\ell}^{(k)}}{(\log
n)^{\ell}}+O(n^{3/4}),\end{split}$

where

(58) $c_{\ell}^{(k)}=\int_{1}^{\infty}\log\binom{k+\left\lfloor
t\right\rfloor-1}{k-1}\log^{\ell}t\,\frac{dt}{t^{2}}.$

This proves Theorem 6 with the constants given in (58). We note that similar
constants appeared in [8, 28], where the authors worked with integrals
involving fractional parts. Here we use the same approach as the one used in
these papers to show that the constants given in (58) can be expressed as in
(17).

We have

$\displaystyle c_{\ell}^{(k)}=\int_{1}^{\infty}\log\binom{k+\left\lfloor
t\right\rfloor-1}{k-1}\log^{\ell}t\,\frac{dt}{t^{2}}$
$\displaystyle=\sum_{s=1}^{\infty}\int_{s}^{s+1}\log\binom{k+s-1}{k-1}\log^{\ell}t\,\frac{dt}{t^{2}}$
$\displaystyle=\sum_{s=1}^{\infty}\log\binom{k+s-1}{k-1}I_{\ell}(s),$

where

$I_{\ell}(s):=\int_{s}^{s+1}\frac{\log^{\ell}t}{t^{2}}\,dt.$

Integrating by parts, we get

$I_{\ell}(s)=\left.-\frac{\log^{\ell}t}{t}\right|_{s}^{s+1}+\ell\int_{s}^{s+1}\frac{\log^{\ell-1}t}{t^{2}}\,dt=\frac{\log^{\ell}s}{s}-\frac{\log^{\ell}(s+1)}{s+1}+\ell\cdot
I_{\ell-1}(s).$

Repeating $\ell-1$ times, and using that
$I_{0}(s)=\int_{s}^{s+1}t^{-2}\,dt=\frac{1}{s(s+1)}$, we get

$\displaystyle I_{\ell}(s)$
$\displaystyle=\frac{\log^{\ell}s}{s}-\frac{\log^{\ell}(s+1)}{s+1}+\sum_{i=1}^{\ell-1}\ell(\ell-1)\cdots(\ell-i+1)\Big{(}\frac{\log^{\ell-i}s}{s}-\frac{\log^{\ell-i}(s+1)}{s+1}\Big{)}+\frac{\ell!}{s(s+1)}$
$\displaystyle=\sum_{i=0}^{\ell}\frac{\ell!}{(\ell-i)!}\Big{(}\frac{\log^{\ell-i}s}{s}-\frac{\log^{\ell-i}(s+1)}{s+1}\Big{)},$

from which

$\displaystyle c_{\ell}^{(k)}=$
$\displaystyle\sum_{s=1}^{\infty}\log\binom{k+s-1}{k-1}\sum_{i=0}^{\ell}\frac{\ell!}{(\ell-i)!}\Big{(}\frac{\log^{\ell-i}s}{s}-\frac{\log^{\ell-i}(s+1)}{s+1}\Big{)}$
$\displaystyle=\sum_{i=0}^{\ell}\frac{\ell!}{(\ell-i)!}\sum_{s=1}^{\infty}\log\binom{k+s-1}{k-1}\Big{(}\frac{\log^{\ell-i}s}{s}-\frac{\log^{\ell-i}(s+1)}{s+1}\Big{)}$
$\displaystyle=\sum_{i=0}^{\ell}\frac{\ell!}{(\ell-i)!}\sum_{s=1}^{\infty}\Big{(}\log\binom{k+s-1}{k-1}-\log\binom{k+s-2}{k-1}\Big{)}\frac{\log^{\ell-i}s}{s}$
$\displaystyle=\sum_{i=0}^{\ell}\frac{\ell!}{(\ell-i)!}\sum_{s=1}^{\infty}\log\Big{(}\frac{k+s-1}{s}\Big{)}\frac{\log^{\ell-i}s}{s}.$

Summing the terms backwards, we get the desired expression

$c_{\ell}^{(k)}=\ell!\sum_{j=0}^{\ell}\frac{1}{j!}\sum_{s=1}^{\infty}\frac{\log^{j}s}{s}\log\Big{(}1+\frac{k-1}{s}\Big{)}.$

Finally, note that

$\sum_{s=1}^{\infty}\frac{\log^{j}s}{s}\log\Big{(}1+\frac{k-1}{s}\Big{)}=O(1)+(k-1)\sum_{s=1}^{\infty}\frac{\log^{j}s}{s^{2}}+O\Big{(}(k-1)^{2}\sum_{s=1}^{\infty}\frac{\log^{j}s}{s^{3}}\Big{)}.$

Approximating the sum by an integral, we have

$\sum_{s=1}^{\infty}\frac{\log^{j}s}{s^{2}}=\int_{1}^{\infty}\frac{\log^{j}t}{t^{2}}\,dt+O(1)=\int_{0}^{\infty}t^{j}e^{-t}\,dt+O(1)=\Gamma(j+1)+O(1),$

where $\Gamma(j+1)$ is the Gamma function at $j+1$, which equals $j!$. In the
same way,

$\sum_{s=1}^{\infty}\frac{\log^{j}s}{s^{3}}=O\Big{(}\frac{j!}{2^{j+1}}\Big{)},$

so that

$c_{\ell}^{(k)}=\ell!\sum_{j=0}^{\ell}\Big{(}k-1+O\Big{(}\frac{(k-1)^{2}}{2^{j+1}}\Big{)}\Big{)}=(k-1)(\ell+1)!+O\big{(}(k-1)^{2}(1-2^{-\ell-1})\ell!\big{)}.$

This proves the last claim of Theorem 6.

Theorem 8 is proved in the exact same way, but with the simpler constants

$e_{\ell}=\int_{1}^{\infty}\frac{\log\lfloor
t\rfloor\log^{\ell}t}{t^{2}}\,dt\qquad(\ell\geq 0).$

To prove Theorem 7, simply observe that

$\binom{2n}{n}=\prod_{p\leq 2n}p^{\alpha_{p}\binom{2n}{n}},$

where

$\alpha_{p}\binom{2n}{n}=\sum_{j\geq
1}\Big{(}\left\lfloor\frac{2n}{p^{j}}\right\rfloor-2\left\lfloor\frac{n}{p^{j}}\right\rfloor\Big{)}.$

It follows that

$\log d_{k}\Big{(}\binom{2n}{n}\Big{)}=\sum_{p\leq
2n}\log\binom{k+\alpha_{p}\binom{2n}{n}-1}{k-1}.$

If $n<p\leq 2n$, then $\log\binom{k+\alpha_{p}\binom{2n}{n}-1}{k-1}$ reduces
to $\log k$, so that

$\sum_{n<p\leq
2n}\log\binom{k+\alpha_{p}\binom{2n}{n}-1}{k-1}=(\pi(2n)-\pi(n))\log k.$

To deal with the remaining part, use the exact same method as in the proof of
Theorem 6 to obtain that for any integer $M\geq 0$,

$\sum_{p\leq n}\log\binom{k+\alpha_{p}\binom{2n}{n}-1}{k-1}=\frac{n}{\log
n}\sum_{\ell=0}^{M}\frac{b_{\ell}^{(k)}}{\log^{\ell}n}+O\Big{(}\frac{n}{\log^{M+2}n}\Big{)},$

where this time

$b_{\ell}^{(k)}=\int_{1}^{\infty}\log\binom{k+[2t]-2[t]-1}{k-1}\log^{\ell}t\,\frac{dt}{t^{2}}.$

Now observe that the integrand vanishes unless $n+\frac{1}{2}\leq t<n+1$ for
some integer $n\geq 1$, in which case it becomes $t^{-2}\log^{\ell}t\log k.$
The result follows.

###  6.4 Proof of Theorem 9

Recall from Mertens’ theorem (48) that $\prod_{p\leq x}(1-1/p)\sim
e^{-\gamma}/\log x$ as $x\to\infty$. Perhaps surprisingly, the behavior of the
similar product $\prod_{p\leq n}(1-p^{-\alpha_{p}(n)-1})$ is very different,
since it is asymptotic to $1$ as $n\to\infty$.

######  Lemma 1.

For any integer $n\geq 2$,

$\prod_{p\leq
n}\Big{(}1-\frac{1}{p^{\alpha_{p}(n)+1}}\Big{)}=1+O\Big{(}\frac{\log
n}{n}\Big{)}.$

###### Proof.

Write

$A(n)=\prod_{p\leq n}\Big{(}1-\frac{1}{p^{\alpha_{p}(n)+1}}\Big{)}.$

Clearly, $A(n)<1$ for all $n\geq 2$. We will therefore focus our attention on
finding a lower bound for $A(n)$ which is “very close” to 1.

Let $r\geq 3$ be a fixed integer (which will eventually be chosen to be
large). Then we have

$A(n)=\prod_{p\leq
n/r}\Big{(}1-\frac{1}{p^{\alpha_{p}(n)+1}}\Big{)}\cdot\prod_{n/r<p\leq
n}\Big{(}1-\frac{1}{p^{\alpha_{p}(n)+1}}\Big{)}={A^{\prime}}_{r}(n)\cdot{A^{\prime\prime}}_{r}(n),$

say.

Since, for all primes $p\leq n$, we have that $\alpha_{p}(n)\geq 1$, it
follows that $\displaystyle{\frac{1}{p^{\alpha_{p}(n)+1}}\leq\frac{1}{p^{2}}}$
and therefore that

$1-\frac{1}{p^{\alpha_{p}(n)+1}}\geq 1-\frac{1}{p^{2}},$

which means that

$A^{{}^{\prime\prime}}_{r}(n)=\prod_{n/r<p\leq
n}\Big{(}1-\frac{1}{p^{\alpha_{p}(n)+1}}\Big{)}\geq\prod_{n/r<p\leq
n}\Big{(}1-\frac{1}{p^{2}}\Big{)}.$

Since for all $y\in(0,1)$, we have $\log(1-y)>-2y$, we may write that

(59) $A^{{}^{\prime\prime}}_{r}(n)\geq\exp\Big{\\{}\sum_{n/r<p\leq
n}\log(1-1/p^{2})\Big{\\}}>\exp\Big{\\{}-2\sum_{n/r<p\leq
n}\frac{1}{p^{2}}\Big{\\}}.$

Now observe that

(60) $\sum_{n/r<p\leq
n}\frac{1}{p^{2}}<\sum_{p>n/r}\frac{1}{p^{2}}<\int_{n/r}^{\infty}\frac{dt}{t^{2}}=\frac{r}{n}.$

Using (60) in (59), we obtain

(61)
$A^{{}^{\prime\prime}}_{r}(n)>\exp\Big{\\{}-2\frac{r}{n}\Big{\\}}=1+O\left(\frac{r}{n}\right).$

We now move to find a lower bound for $A^{{}^{\prime}}_{r}(n)$. First observe
that for $p\leq n/r$, we have

$\alpha_{p}(n)\geq\left\lfloor\frac{n}{p}\right\rfloor\geq\left\lfloor
r\right\rfloor=r,$

which implies that $\alpha_{p}(n)+1\geq r+1$ and therefore that

(62) $A^{{}^{\prime}}_{r}(n)\geq\prod_{p\leq
n/r}\Big{(}1-\frac{1}{p^{r+1}}\Big{)}>\prod_{p}\Big{(}1-\frac{1}{p^{r+1}}\Big{)}=\frac{1}{\zeta(r+1)}.$

Approximating (21) by an integral, we have for any real $s\geq 3$,

$\zeta(s)\leq
1+\frac{1}{2^{s}}+\int_{2}^{\infty}t^{-s}\,dt=1+\frac{1}{2^{s}}+\frac{1}{2^{s-1}(s-1)}\leq
1+\frac{1}{2^{s}}+\frac{1}{2^{s}}=1+\frac{2}{2^{s}}.$

Using this inequality with $s=r+1$ in (62), we find that

(63)
$A^{{}^{\prime}}_{r}(n)>\frac{1}{\zeta(r+1)}\geq\frac{1}{1+1/2^{r}}>1-\frac{1}{2^{r}}.$

Combining (61) with (63), while choosing
$\displaystyle{r=\left\lfloor\frac{\log n}{\log 2}\right\rfloor}$, we obtain
that

(64) $A(n)=A^{{}^{\prime}}_{r}(n)\cdot
A^{{}^{\prime\prime}}_{r}(n)>\Big{(}1+O\Big{(}\frac{\log
n}{n}\Big{)}\Big{)}\cdot\Big{(}1+O\Big{(}\frac{1}{n}\Big{)}\Big{)}=1+O\Big{(}\frac{\log
n}{n}\Big{)}.$

Combining our first observation to the effect that $A(n)<1$ with (64), we have
thus established that

$A(n)=1+O\Big{(}\frac{\log n}{n}\Big{)}.\qed$

We can now prove Theorem 9. Let $\kappa\geq 1$ and observe that for any prime
power $p^{a}$,

(65)
$\sigma_{\kappa}(p^{a})=1+p^{\kappa}+\cdots+p^{a\kappa}=\frac{p^{(a+1)\kappa}-1}{p^{\kappa}-1}=p^{a\kappa}\frac{1-1/p^{(a+1)\kappa}}{1-1/p^{\kappa}}.$

It follows from equations (41) and (65) that

$\displaystyle\sigma_{\kappa}(n!)=$ $\displaystyle=\prod_{p\leq
n}p^{\alpha_{p}(n)\kappa}\frac{1-1/p^{(\alpha_{p}(n)+1)\kappa}}{1-1/p^{\kappa}}$
$\displaystyle=\frac{(n!)^{\kappa}}{\prod_{p\leq
n}(1-\frac{1}{p^{\kappa}})}\cdot\prod_{p\leq
n}\Big{(}1-\frac{1}{p^{(\alpha_{p}(n)+1)\kappa}}\Big{)}.$

Letting $A_{\kappa}(n)=\prod_{p\leq n}(1-1/p^{(\alpha_{p}(n)+1)\kappa})$, we
see from (48) that

$\sigma(n!)=n!e^{\gamma}\log
n(1+O\Big{(}\frac{1}{\log^{3}n}\Big{)}\Big{)}\cdot A_{1}(n),$

and from (21) that for $\kappa>1$,

$\sigma_{\kappa}(n!)=(n!)^{k}\zeta(k)\cdot A_{\kappa}(n).$

It remains to see that $1>A_{\kappa}(n)>A_{1}(n)=A(n)$ to conclude that the
estimate in Lemma 1 holds for any fixed $\kappa\geq 1$, from which the result
follows.

###  6.5 Proofs of Theorems 11, 12, and 13

Recall that studying sums of the middle divisors is much easier at factorial
arguments because the values of $\rho_{1}(n!)$ and $\rho_{2}(n!)$ are close to
$\sqrt{n!}$, according to 2.

###### Proof of 2.

First of all, it is clear that (37) is an immediate consequence of (36). We
will therefore focus on the proof of (36). We start with the following
construction. First recall that

(66) $n!=\prod_{p\leq n}p^{\alpha_{p}(n)},\quad\mbox{where
}\alpha_{p}(n)=\frac{n-s_{p}(n)}{p-1},$

with $s_{p}(n)$ standing for the sum of the digits of $n$ in base $p$.
Observing that $\alpha_{p}(n)=1$ for all primes $p\in(n/2,n]$, we may write
the factorisation of $n!$ as

(67) $n!=\underbrace{2\cdots 2}_{\alpha_{2}(n)\,\mbox{\tiny
times}}\cdot\underbrace{3\cdots 3}_{\alpha_{3}(n)\,\mbox{\tiny
times}}\cdot\underbrace{5\cdots 5}_{\alpha_{5}(n)\,\mbox{\tiny times}}\cdots
p_{\pi(n/2)+1}\cdots p_{\pi(n)},$

where $p_{i}$ stands for the $i$-th prime, and therefore in the above
$p_{\pi(n)}=P(n!)=P(n)$, the largest prime factor of $n$.

Set $s_{1}=s_{1}(n):=p_{\pi(n/2)+1}\cdots p_{\pi(n)}$ and
$s_{2}=s_{2}(n)=n!/s_{1}$, so that

$n!=\underbrace{2^{\alpha_{2}(n)}\cdot 3^{\alpha_{3}(n)}\cdots
p_{k}^{\alpha_{p_{k}}(n)}}_{s_{2}}\cdot\underbrace{p_{k+1}\cdots
p_{\pi(n)}}_{s_{1}},$

where $k=\pi(n/2)$.

Recall the following result of R. Tijdeman [32].

> Given an infinite sequence $n_{1}<n_{2}<\cdots$ of $B$-friable numbers (or
> $B$-smooth numbers), there exists a positive constant $c(B)$ such that
>
> (68) $n_{i+1}-n_{i}\ll\frac{n_{i}}{\log^{c(B)}n_{i}}.$

It is easily seen that if $B_{1}<B_{2}$, then $c(B_{2})\geq c(B_{1})$.
Therefore, one can assume the existence of a positive number $c$ satisfying
$c\leq c(B)$ for all integers $B\geq 2$. In fact, Langevin [27] later obtained
a lower bound for this constant $c$, a very small one indeed (actually
somewhat smaller than $1/10^{36}$) but nevertheless effective. In the
following, we will therefore assume that $c\in(0,1)$.

For convenience, let us assume that $n\geq 6$ and consider the set $S$ of all
the divisors of $s_{2}$ which are no larger than $\sqrt{n!}/s_{1}$ and let
$r:=\max S$. Finally, let $d_{1}=r\cdot s_{1}$. It is clear that $d_{1}\mid
n!$ and that $d_{1}\leq\sqrt{n!}$.

Since $s_{1}=s_{1}(n)>1$, it follows that $\sqrt{n!}$ is never an integer.
Furthermore, observe that if

$n_{1}<n_{2}<\cdots<n_{k}=:d_{1}<\sqrt{n!}<n_{k+1}=:d_{1}^{+}$

is the list of all the positive divisors of $n!$ smaller than $\sqrt{n!}$ plus
the one located immediately after $\sqrt{n!}$, then it follows from (68) that
for some positive constant $A$,

(69) $n_{k+1}-n_{k}=d_{1}^{+}-d_{1}<A\frac{d_{1}}{(\log d_{1})^{c}}.$

Since the function $t/(\log t)^{c}$ is increasing for all $t\geq e$, it
follows from (69) that

(70) $d_{1}^{+}-d_{1}<A\frac{\sqrt{n!}}{(\log\sqrt{n!})^{c}},$

and therefore that

$d_{1}>d_{1}^{+}-A\frac{\sqrt{n!}}{(\log\sqrt{n!})^{c}}>\sqrt{n!}-A\frac{\sqrt{n!}}{(\log\sqrt{n!})^{c}}.$

Using a weak form of Stirling’s formula, we have
$\displaystyle{\sqrt{n!}>\left(\frac{n}{e}\right)^{n/2}}$, thereby implying
that

(71) $(\log\sqrt{n!})^{c}>\Big{(}\frac{n}{2}(\log n-1)\Big{)}^{c}>n^{c},$

provided $n\geq 21$.

Using (71) in (70), we obtain

(72)
$d_{1}>\sqrt{n!}-\frac{\sqrt{n!}}{n^{c}}=\sqrt{n!}\Big{(}1-\frac{1}{n^{c}}\Big{)},$

thus completing the proof of the inequalities on the left-hand side of (36).
Those on the right-hand side will easily follow by simply setting
$d_{2}:=n!/d_{1}$; indeed, we then have

$d_{2}=\frac{n!}{d_{1}}<\frac{n!}{(1-1/n^{c})\sqrt{n!}}<\sqrt{n!}\Big{(}1+\frac{2}{n^{c}}\Big{)},$

provided $n$ is sufficiently large. ∎

We will also be using the following lemma, whose proof is immediate.

######  Lemma 2.

For any given integer $N\geq 1$, we have

$\displaystyle\sum_{n=1}^{N}n!$ $\displaystyle=$ $\displaystyle
N!\Big{(}1+O\Big{(}\frac{1}{N}\Big{)}\Big{)},$
$\displaystyle\sum_{n=1}^{N}\sqrt{n!}$ $\displaystyle=$
$\displaystyle\sqrt{N!}\Big{(}1+O\Big{(}\frac{1}{\sqrt{N}}\Big{)}\Big{)}.$

We now prove 11. It follows from Proposition 2 that

(73) $\sum_{n\leq N}\rho_{1}(n!)=\sum_{n\leq
N}\sqrt{n!}\Big{(}1+O\Big{(}\frac{1}{n^{c}}\Big{)}\Big{)}=\sum_{n\leq
N}\sqrt{n!}+O\Big{(}\sum_{n\leq N}\frac{\sqrt{n!}}{n^{c}}\Big{)}.$

Observe that

(74) $\displaystyle\sum_{n\leq N}\frac{\sqrt{n!}}{n^{c}}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{N!}}{N^{c}}\Big{(}1+\Big{(}\frac{N}{N-1}\Big{)}^{c}\frac{1}{\sqrt{N}}+\Big{(}\frac{N}{N-2}\Big{)}^{c}\frac{1}{\sqrt{N(N-1)}}+\cdots\Big{)}$
$\displaystyle=$
$\displaystyle\frac{\sqrt{N!}}{N^{c}}\Big{(}1+O\Big{(}\frac{1}{\sqrt{N}}\Big{)}\Big{)}=O\Big{(}\frac{\sqrt{N!}}{N^{c}}\Big{)}.$

Substituting (74) in (73), and using Lemma 2, we get that

$\displaystyle\sum_{n\leq N}\rho_{1}(n!)$ $\displaystyle=$
$\displaystyle\sum_{n\leq N}\sqrt{n!}+O\Big{(}\frac{\sqrt{N!}}{N^{c}}\Big{)}$
$\displaystyle=$
$\displaystyle\sqrt{N!}\Big{(}1+O\Big{(}\frac{1}{\sqrt{N}}\Big{)}\Big{)}+O\Big{(}\frac{\sqrt{N!}}{N^{c}}\Big{)}$
$\displaystyle=$
$\displaystyle\sqrt{N!}\Big{(}1+O\Big{(}\frac{1}{N^{c}}\Big{)}\Big{)},$

thus proving the first estimate of Theorem 11. The second estimate is obtained
through a similar reasoning.

We move on to 12. Using Proposition 2, we obtain

(75) $\sum_{n\leq N}\frac{\rho_{1}(n!)}{\rho_{2}(n!)}=\sum_{n\leq
N}\frac{\sqrt{n!}(1+O(1/n^{c}))}{\sqrt{n!}(1+O(1/n^{c}))}=\sum_{n\leq
N}\Big{(}1+O\Big{(}\frac{1}{n^{c}}\Big{)}\Big{)}=N+O\Big{(}\sum_{n\leq
N}\frac{1}{n^{c}}\Big{)}.$

Approximating this last sum by an integral, we get

$\sum_{n\leq
N}\frac{1}{n^{c}}=\int_{1}^{N}\frac{1}{t^{c}}dt+O(1)=\frac{N^{1-c}}{1-c}+O(1)=O(N^{1-c}).$

Using this last estimate in (75), we obtain that

$\sum_{n\leq N}\frac{\rho_{1}(n!)}{\rho_{2}(n!)}=N+O(N^{1-c}),$

thus establishing the first estimate of Theorem 12. The second estimate can be
proved in a similar manner.

To prove 13, we use Proposition 2 and Stirling’s formula (39) to obtain

$\displaystyle\sum_{2\leq n\leq N}\frac{\log\rho_{1}(n!)}{\log\rho_{2}(n!)}$
$\displaystyle=$ $\displaystyle\sum_{2\leq n\leq
N}\frac{\log(\sqrt{n!}(1+O(1/n^{c})))}{\log(\sqrt{n!}(1+O(1/n^{c})))}=\sum_{2\leq
n\leq
N}\frac{\log\sqrt{n!}+\log(1+O(1/n^{c}))}{\log\sqrt{n!}+\log(1+O(1/n^{c}))}$
$\displaystyle=$ $\displaystyle\sum_{2\leq n\leq
N}\frac{\log\sqrt{n!}+O(1/n^{c})}{\log\sqrt{n!}+O(1/n^{c})}=\sum_{2\leq n\leq
N}\frac{1+O\Big{(}\frac{1}{n^{c}\log\sqrt{n!}}\Big{)}}{1+O\Big{(}\frac{1}{n^{c}\log\sqrt{n!}}\Big{)}}$
$\displaystyle=$ $\displaystyle\sum_{2\leq n\leq
N}\Big{(}1+O\Big{(}\frac{1}{n^{c+1}\log n}\Big{)}\Big{)}=N+O\Big{(}\sum_{2\leq
n\leq N}\frac{1}{n^{c+1}\log n}\Big{)}$ $\displaystyle=$ $\displaystyle
N+O(1),$

since the series $\displaystyle{\sum_{n=2}^{\infty}\frac{1}{n^{c+1}\log n}}$
converges. This establishes the first estimate in Theorem 13. The second
estimate is proved similarly.

##  7 An analogue of Chowla’s conjecture for factorial arguments

We may consider many other questions regarding arithmetic functions at
factorial arguments. We note that the authors of [15] showed that

$\frac{d(n!)}{d((n-1)!)}=1+\frac{P(n)}{n}+O\Big{(}\frac{1}{n^{1/2}}\Big{)},$

where $P(n)$ is the largest prime factor of $n$, as in Section 6.5. The same
ratio is easier to understand for most other arithmetic functions. For
instance, observe that if $f$ is completely additive, then

$\frac{f(n!)}{f((n-1)!)}=1+\frac{f(n)}{S_{f}(n-1)},$

while if $f$ is strongly additive,

$\frac{f(n!)}{f((n-1)!)}=\begin{cases}1+\frac{f(n)}{\sum_{p\leq
n-1}f(p)}&\text{if }n\text{ is prime},\\\ 1&\text{otherwise}.\end{cases}$

In particular, the additive functions $\omega$, $\Omega$, and $\beta$ are such
that the ratio of their consecutive values at factorial arguments is $\sim 1$
as $n\to\infty$, but $B$ and $B_{1}$ are not, which can be seen by picking $n$
to be a product of primes.

On the other hand, for the multiplicative functions $\gamma$ and $\phi$, the
results are more interesting, since for any completely multiplicative function
$f$,

$\frac{f(n!)}{f((n-1)!)}=f(n),$

and for any strongly multiplicative function $f$,

$\frac{f(n!)}{f((n-1)!)}=\begin{cases}f(n)&\text{if }n\text{ is prime},\\\
1&\text{otherwise}.\end{cases}$

Using that $\gamma(n)$ and $\frac{\phi(n)}{n}$ are strongly multiplicative, we
get

$\frac{\gamma(n!)}{\gamma((n-1)!)}=\begin{cases}n&\text{if }n\text{ is
prime},\\\ 1&\text{otherwise},\end{cases}$

and

$\frac{\phi(n!)}{\phi((n-1)!}=\begin{cases}n-1&\text{if }n\text{ is prime},\\\
n&\text{otherwise}.\end{cases}$

These types of considerations are key in the following argument.

Let $\lambda(n):=(-1)^{\Omega(n)}$ stand for the Liouville function. A famous
conjecture due to Chowla [4] in its simplest form can be stated as follows.

> Conjecture (Chowla) As $x\to\infty$,
>
> $\sum_{n\leq x}\lambda(n)\lambda(n+1)=o(x).$

Interestingly, the analogous form of Chowla’s conjecture at factorial
arguments is true. Indeed, we can prove the following result.

######  Theorem 14.

As $x\to\infty$,

$\sum_{n\leq x}\lambda(n!)\lambda((n+1)!)=o(x).$

###### Proof.

We will show that

(76) $\sum_{2\leq n\leq x}\lambda((n-1)!)\lambda(n!)=o(x)\qquad(x\to\infty).$

Clearly,

$\lambda((n-1)!)\lambda(n!)=(-1)^{\sum_{k\leq n-1}\Omega(k)+\sum_{k\leq
n}\Omega(k)}=(-1)^{2\sum_{k\leq
n-1}\Omega(k)+\Omega(n)}=(-1)^{\Omega(n)}=\lambda(n).$

Therefore, in order to prove (76), we only need to prove that

(77) $\sum_{2\leq n\leq x}\lambda(n)=o(x)\qquad(x\to\infty).$

Since one can easily show that (77) holds if and only if $\sum_{2\leq n\leq
x}\mu(n)=o(x)$ as $x\to\infty$, a result which is equivalent to the prime
number theorem, the proof of Theorem 14 is complete. ∎

######  Remark 7.1.

By a similar reasoning, one easily sees that Chowla’s conjecture is in fact
equivalent to the statement

$\sum_{n\leq x}\lambda((n-1)!)\lambda((n+1)!)=o(x)\qquad(x\to\infty),$

a sort of symmetric version of 14.

######  Remark 7.2.

It is clear that one can adapt the proof of Theorem 14 to prove that if
$(a_{n})_{n\geq 1}$ is any sequence of positive integers (not necessarily
monotonic), then

(78) $\sum_{n\leq x}\lambda(a_{n})\lambda(na_{n})=o(x)\qquad(x\to\infty).$

Hence, by choosing $a_{n}=(n-1)!$ in (78), we obtain (76). On the other hand,
by choosing $a_{n}=n$ and $a_{n}=P(n)$, we have

$\sum_{n\leq x}\lambda(n)\lambda(n^{2})=o(x)\qquad(x\to\infty)$

and

$\sum_{n\leq x}\lambda(P(n))\lambda(nP(n))=o(x)\qquad(x\to\infty).$

Moreover, according to a more general version of Chowla’s conjecture, given
any positive integer $k$,

(79) $\sum_{n\leq
x}\lambda(n)\lambda(n+1)\cdots\lambda(n+k-1)=o(x)\qquad(x\to\infty).$

It is clear that one can adapt the proof of Theorem 14 to prove that if $k$ is
an even integer, then the factorial version of (79) holds.

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Département de mathématiques et de statistique, Université Laval
Québec, QC, G1V 0A6, Canada
<EMAIL_ADDRESS>
Département de mathématiques et de statistique, Université Laval
Québec, QC, G1V 0A6, Canada
<EMAIL_ADDRESS>


# An intermediate phase between jammed and un-jammed amorphous solids

Yuliang Jin1, Itamar Procaccia2 and Tuhin Samanta2 1Institute of Theoretical
Physics, Chinese Academy of Sciences,Beijing 100190, China, 2Dept. of Chemical
Physics, The Weizmann Institute of Science, Rehovot 76100, Israel

###### Abstract

A significant amount of attention was dedicated in recent years to the
phenomenon of jamming of athermal amorphous solids by increasing the volume
fraction of the microscopic constituents. At a critical value of the volume
fraction, pressure shoots up from zero to finite values with a host of
critical exponents discovered and discussed. In this letter, we advance
evidence for the existence of a second transition, within the jammed state of
two-dimensional granular systems, that separates two phases characterized by
different mechanical screening regimes. Explicitly, highly packed systems are
quasi-elastic with quadrupole-screening, and more loosely jammed systems
exhibit anomalous mechanics with dipole screening. Evidence is given for a
clear transition between these two regimes, reminiscent of the intermediate
hexatic phase of crystal melting in two-dimensional crystals.

The concepts of rigidity and jamming in granular systems are topics of intense
studies for the community of amorphous solids [1, 2, 3]. For athermal hard
sphere systems, the transition from zero to infinite pressure upon jamming, is
clearly defined in terms of a critical isostatic number of contacts $Z^{*}$
[4, 5, 6]. This mathematically clean phenomenology is getting somewhat blurred
when temperature, or friction, are included in the relevant physics of
amorphous solids. Notably, even the packing fraction in which jamming is
observed is not well defined, since it can depend on details of preparation or
compressing protocols [7].

The difficulty in identifying a sharp transition between solids and liquids is
not particular to amorphous solids. In fact, it is well known that between
perfect crystals in two dimensions and their fluid phase there exist an
intervening hexatic phase in which the material exhibits intermediate
properties between solid and liquid [8]. Motivated by the celebrated theory of
the hexatic phase transition we have recently investigated the anomalous
mechanics that appears between elastic amorphous solids and their fluid
analogs [9, 10, 11, 12, 13]. In this Letter we present evidence that the
transition between an elastic phase and the novel (rigid) phase in which
elasticity is anomalous, is in fact sharp, at least in two dimensions. We
emphasize that this transition is different from the jamming or rigidity
transition that has been studied in the literature [1, 2]. The different
phases are characterized by different responses to non-uniform strains, such
that the resulting displacement field is sensitive to different mechanical
screening mechanisms, which are either quadrupolar or dipolar elastic charges.

The concept of screening in modern physics arises naturally in the
electrostatics of continuous media, where microscopic mechanisms are available
to damp the externally imposed electric field. A famous example of non-
electric screening is the Kosterlitz–Thouless (KT) transition [14], where
vortex-pairs dissociate to produce a coulomb gas phase where monopoles
(unbound vortices) are available to screen external fields, similar to the
Debye-screening in the electrostatic analog. In the field of mechanics, the
nucleation of structural defect in response to external loads, is the basis
for theories of crystal melting [15, 16], where solids are supplemented with
screening quadrupoles (dislocation pairs), hexatics with screening dipoles
(unbound dislocations) and liquids with screening monopoles (unbound
disclinations).

While thermal melting and the hexatic phase transition necessitate finite-
temperature statistical mechanics, recent research on a variety of athermal
systems pointed out the existence of similar transition in cellular tissue
models [17] and vibrated granular matter [18, 19, 20]. These observations were
fundamentally based on structural characteristics, including short and quasi-
long range translational and orientational order, observed via correlation
functions.

In contrast to these thermal or mechanically agitated examples of hexatic
phases, we present here theoretical and numerical indications for the
existence of a transition to a dipoles-screened phase within the jammed regime
of athermal amorphous granular systems. This finding is based on recent
research, in which it was discovered that the prevalence of plastic events in
amorphous solids results in screening phenomena that are akin, but richer and
different, to screening effects in electrostatics [9, 10, 11, 12, 13]. Plastic
events, which are typically quadrupoles in the displacement field, can act as
screening charges. It was shown that when the density of plastic quadrupoles
is low, their effect is limited to renormalizing the elastic moduli, but the
structure of (linear) elasticity theory remains intact. This is analogous to
dipole screening in dielectrics. On the other hand, when the nucleation cost
of quadrupoles drops, the quadrupoles density becomes high, and the nucleation
of effective dipoles defined by the gradients of their density, cannot be
neglected. The presence of effective dipoles changes the analytic form of the
response to strains, in ways that are in fundamental clash with standard
elasticity theory. It was concluded that one needs to consider a new theory,
and this emergent theory was confirmed by comparing its predictions to results
of extensive experiments and simulations [9, 10, 11, 12, 13]. While dipole
screening was observed in both two and three dimensions, in this letter we
focus on two dimensional systems in which one can demonstrate a clear
transition as stated above.

Having realized that gradients of quadrupole density act as effective dipoles,
it became evident that the vast majority of experiments and numerical
simulations that study the mechanics of amorphous solids are not using the
best strain protocols. Indeed, in most studies researchers employ simple and
pure shear, or tensile compression or extension. To expose the unusual and
interesting mechanical properties of amorphous solids it is advisable to
employ non-uniform strains. To this aim we have employed, in both experiments
[10] and simulations [9, 10, 12], a circular geometry in two dimensions and a
spherical one in three dimensions [13]. In the former case we then inflate a
central disk, to observe the resulting radial component of the displacement
field $d_{r}$. Theoretically we expect that in a phase where only quadrupole
screening is dominant, an inflation $r_{\rm in}\to r_{\rm in}+d_{0}$ of a disk
of initial Radius $r_{\rm in}$ in a system of outer radius $r_{\rm out}$ will
result in radial displacement

$\displaystyle
d_{r}(r)=d_{0}\frac{r_{\text{in}}\left(r^{2}-r_{\text{out}}^{2}\right)}{r\left(r_{\text{in}}^{2}-r_{\text{out}}^{2}\right)},\quad\text{in
two dimensions}\ .$ (1)

On the other hand, in a phase that is governed by dipole screening we
theoretically expect the same radial component of the displacement field to
obey

$d_{r}(r)=d_{0}\frac{Y_{1}(r\,\kappa)J_{1}(r_{\text{out}}\kappa)-J_{1}(r\,\kappa)Y_{1}(r_{\text{out}}\kappa)}{Y_{1}(r_{\text{in}}\kappa)J_{1}(r_{\text{out}}\kappa)-J_{1}(r_{\text{in}}\kappa)Y_{1}(r_{\text{out}}\kappa)}\
.$ (2)

Here $J_{1}$ and $Y_{1}$ are the circular Bessel functions of the first and
second kind respectively. The parameter $\kappa$ has the dimension of inverse
length and is referred to as the screening parameter. When $\kappa\to 0$ the
expression (2) tends to Eq. (1).

Figure 1: Maps of the magnitude of the displacement field after inflation of
25% in the radius of the inner disk, $N=80000$. Upper panel: high pressure
$P=29.35$. Lower panel: low pressure $P=0.394$

.

Obviously, the difference between Eqs. (1) and (2) is striking. The first
exhibits a monotonous decay of an outward displacement until it vanishes at
the outer boundary, whereas the latter allows oscillations, and even negative
(inward pointing) displacements although the imposed inflation points
outwards. The main question that we raise here is whether there exists a clear
transition, as a function of an intensive parameter in a given athermal
amorphous system, separating material phases in which the mechanical response
tends to jump from Eq. (1) to Eq. (2) with a finite value of $\kappa$. We show
next that in 2-dimensions the answer is affirmative, the intensive parameter
for a granular jammed system is the pressure, and the transition is indeed
clear.

To demonstrate the transition we investigate frictionless assemblies of small
disks that are at mechanical equilibrium, prepared with a desired target
pressure $P$ and confined in a circular two-dimensional area with a fixed
outer circular wall. Open source codes (LAMMPS [21]) are used to perform the
simulations. Every simulation begins with a configuration of $N=80000$ bi-
disperse disks of mass $m=1$, placed randomly in a circular area with a radius
$r_{\text{out}}=172$ in dimensionless units. Half of the small spheres have a
radius $R_{1}=0.45$ and the other half a radius $R_{2}=0.65$. One larger disk
is not placed randomly, but rather fixed to the center of coordinates. To
reach a desired pressure we begin with a chosen packing fraction and the
system is relaxed to mechanical equilibrium by solving Newton’s second law of
motion with damping. This process is carried out until the desired target
pressure is reached and forces are minimized to values smaller than $10^{-6}$.
The normal contact force is Hertzian with force constant $k_{n}=2\times
10^{5}$, following the Discrete Element Method of Ref. [22]. The tangential
contact force is zero as the system is frictionless.

After achieving a mechanically stable configuration at a target pressure, we
inflate the central disk by 25%. The displacement field is denoted
${\bm{d}}(r,\theta)$ and the radial component is obtained as an angle average,
$d_{r}(r)\equiv(2\pi)^{-1}\oint_{0}^{2\pi}{\bm{d}}\cdot\hat{r}d\theta$ where
$\hat{r}\equiv{\bm{r}}/r$. The displacement field exhibits qualitatively
different appearance at high and low pressures as exemplified in Fig. 1. At
high pressures the displacement field is centered around the inflated disk as
is expected from Eq. (1). In contrast, at low pressure the displacement field
is spread out throughout the system, in correspondence with Eq. (2). A
quantitative comparison is provided by plotting the radial component
$d_{r}(r)$, cf. Fig. 2.

Figure 2: Green dots: the radial component of the displacement field that
corresponds to the data in Fig. 1, averaged over the angles. Dashed line is
the fit to theory. Upper panel: high pressure $P=29.35$. Lower panel: low
pressure $P=0.394$, $\kappa=0.023$

While the upper panel shows the typical decay of an elastic solution, the
lower panel presents negative radial displacement that result from screening
and the Bessel functions in Eq. (2).

The simulations indicate a clear transition from quasi-elastic to anomalous
response. The best way to demonstrate the transition is to measure the
screening parameter $\kappa$ as a function of the pressure. In Fig. 3 we
present the measured screening parameter as a function of $\ln(P^{-1})$.

Figure 3: The screening parameter $\kappa$ as a function of the logarithm of
the inverse pressure. A transition between material phases with quasi-elastic
response and with anomalous response is clearly observed.

The screening parameter was measured in two independent ways. In one we simply
fitted the measured radial component of the displacement field to Eq. (2), see
for example Fig. 2 lower panel. The second method relied on a direct
measurement of the presence of dipoles. This will connect the observed
transition to the well known hexatic phase transition which appears in two-
dimensional melting. It should be stressed that in the present context the
existence of dipoles, or in fact of a dipole density
${\bm{{\mathcal{P}}}}(r,\theta)$, does not refer to the material structure,
but rather to the presence of dipoles in the displacement field. To this aim
we refer to the theory presented in Ref. [11]. It was shown that the dipole
density can be measured directly by the following line integral:

$\oint_{\partial\Omega}{\bm{{\mathcal{P}}}}(r,\theta)\cdot\mathbf{n}\,\text{dl}=\oint_{\partial\Omega}(\nabla(\nabla\cdot{\bm{d}}))\cdot\mathbf{n}\,\text{dl}\
.$ (3)

The line integral can be taken around any closed loop. Due to obvious
conservation laws we expect that when the loop encircles the whole system, the
net dipole included should be zero, whereas with a loop enclosing any part of
the system the integral will not vanish only when there is dipole density in
the enclosed area $\Omega$. In the case of our circular systems with radial
symmetry both sides of this equation can be evaluated analytically. The final
result is [11]

$\oint_{\partial\Omega}{\bm{{\mathcal{P}}}}(r,\theta)\cdot\mathbf{n}\,\text{dl}=-\kappa^{2}\oint_{\partial\Omega}{\bm{d}}(r,\theta)\cdot\mathbf{n}\,\text{dl}\
.$ (4)

In Fig. 4 we present the function ${\mathcal{P}}(r)$ computed as a function of
the radius $r$ of the loop integral for our system with $N=80000$ disks.

Figure 4: The dipole density included in a radius of size r, as a function of
r, for the system with $N=80000$, for different pressures. Upper panel:
calculation using the simulation data. Lower panel: calculation based on the
analytic formulae (4) with $d(r)$ taken from Eqs. (1) or (2) (with the
measured value of $\kappa$).

The function ${\mathcal{P}}(r)$ is computed in two ways. In the upper panel
the simulation data was used according to Eq. (4), whereas in the lower panel
Eq. (4) was applied to the analytic formulae Eqs. (1) or (2) (with the
measured value of $\kappa$). At pressures with quasi-elastic response,
${\mathcal{P}}(r)$ vanishes for every value of $r$. In the anomalous regime
${\mathcal{P}}(r)$ is not zero, in very good agreement between the two methods
of computation, showing that the transition is tightly associated with the
appearance of dipole densities in the displacement field at these conditions.
Finally, the value of the screening parameter $\kappa$ was determined by
taking the ratio of the two integrals in Eq. (4) , giving us $\kappa^{2}$.

The values of the screening parameter $\kappa$ as a function of inverse
pressure is shown in linear-log scales in Fig. 3. The values shown were
obtained as an average between the two methods of measurement. For pressure
$P\geq 3.5\pm 0.3$ the response is quasi-elastic with $\kappa=0$. For pressure
$P\leq 3.5\pm 0.3$ the response is anomalous. The scatter in the values of
$\kappa$ in the anomalous regime is typical to the considerable sample-to-
sample fluctuations in the values of the screening parameter. We should note
that once the screening parameter differs from zero it appears quite
independent of pressure.

To understand the transition and the apparent constancy of $\kappa$ as a
function of pressure, we need to determine when we can expect an avalanche of
plastic events that can span a region of size $\kappa^{-1}$. Start with
estimating the maximal size of a blob that can become unstable and go through
a plastic deformation. In our amorphous configurations of disks interacting
via Hertzian forces. the pressure depends on excess coordination number
$\Delta Z\equiv Z-Z^{*}$ according to [3, 23]

$p\sim(\phi-\phi_{J})^{3/2}\sim\Delta Z^{3}\ ,$ (5)

where $Z^{*}=4$ is the coordination number at jamming. When $\Delta Z=0$,
breaking any contact will render the system unstable. On the other hand, when
$\Delta Z>0$ one can afford breaking more than one bond, in fact one can break
a whole circumference of bonds of length $\ell^{d-1}$ as long as [24],

$Z\ell^{d-1}=\Delta Z\ell^{d}\ .$ (6)

If the region is smaller than this $\ell$, it is unstable to such breaking,
and if larger, the region is always stable and rigid. We then interpret this
length as the maximal blob size that can participate in an avalanche of
plastic events. This length scale depends on pressure like

$\ell\sim\frac{Z}{\Delta Z}\sim p^{-1/3}\ .$ (7)

This is represented by the black line in Fig. 5.

Figure 5: A sketch of the expected transition in the observed value of the
screening parameter $\kappa$

.

Next the question is what is the pressure dependence of $\kappa$. From Ref.
[13] one reads

$\frac{\mu_{1}^{2}}{\mu_{2}\left(\lambda+2\mu\right)}=\kappa^{2}\ .$ (8)

Here $\mu_{1}$ and $\mu_{2}$ are new moduli associated with the quadrupole and
dipole terms in the energy function. The combination of Lame’ coefficients
$(\lambda+2\mu)=B+G$, where $B$ and $G$ are the bulk and shear moduli
respectively. This combination is dominated by
$B\sim(\phi-\phi_{J})^{\alpha-2}$ because $G\sim(\phi-\phi_{J})^{\alpha-3/2}$
with $\alpha=5/2$. The coefficients $\mu_{1}$ and $\mu_{2}$ are second
derivatives of the energy with respect to strain, either directly or through
the quadrupole field $Q$. Thus they are both expected to scale like $B$.
Therefore $\kappa$ should be independent of pressure. However, $\kappa$ can
exist only when a blob of the order of $\kappa^{-1}$ exceeds the scale $\ell$.
Accordingly we predict that the observed value of $\kappa$ will be zero for
high pressures and constant for small pressures, with a jump when
$\kappa\approx\ell^{-1}$. This is the red line in Fig. 5. Here we propose that
the sketch presented in Fig. 5 rationalizes the numerical results shown in
Fig. 3. We should note however that the pressure where the transition is
observed can depend on the magnitude of the inflation at the central disk and
on the microscopic properties of the amorphous material.

Acknowledgments: IP thanks Michael Moshe for very useful discussions. This
work has been supported in part by the the joint grant between the Israel
Science Foundation and the National Science Foundation of China, and by the
Minerva Foundation, Munich, Germany.

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