TITLE: Is there a difference between topological defects and topological soliton?
QUESTION [4 upvotes]: Is there a difference between topological defects and topological soliton? Or are these objects the same thing? I ask this because it very common find some papers whose the authors itself refer, for example, the domain wall's, monopoles, cosmic string, etc, as topological defects, while in others, the same objects are called of topological solitons. As the authors aren't specific about this  It seems that defects and topological solitons are the same things. But will be this idea conceptually correct?

REPLY [1 votes]: Yes, there is. In static situations, both are strictly the same thing. But, when analyze the dynamic of these structures, we see that they are differents.


TITLE: Doubt in proof of theorem 8.20 of Apostol Modular functions and Dirichlet series in number theory
QUESTION [3 upvotes]: I am self studying number theory from Tom M Apostol Dirichlet Series and Modular functions in number theory and I have a doubt in theorem 8.20 of the book.
I am attaching images of relevant results.

The Theorem--

My doubt is in last  paragraph  of proof how by integral analog of Landau theorem function on left is analytic in half plane $\sigma$ > 1/2 .

Landau theorem -

What I could deduce using Landau theorem--> The function on right is analytic. So, the function on left converges for $ \sigma$ >1 , which is not similar to result deduced by Apostol.
Can somebody please help to make deduction given in the book .

REPLY [1 votes]: The first part is to assume that  for some $\Re(s) > 1$, $L_n(s)=\sum_{n\le x}\lambda(n)n^{-s}= 0$, then take a sequence $t_k$ such that for $p\le x$, $p^{-s-it_k}\to  -p^{-s}$ which implies that $n^{-s-it_k}\to\lambda(n)n^{-s}$ and hence $\zeta_n(z+s+it_k)\to L_n(z+s)$ uniformly around $z=0$ which means that for $k$ large enough $\zeta_n(s+it_k+z)$ has a zero near $z=0$.
The last part is that $\forall x > X,\sum_{n\le x} \lambda(n)n^{-1}\ge 0$ implies $F(s)=\int_X^\infty (\sum_{n\le x}\lambda(n)n^{-1})x^{-s}dx$ has a singularity at its abscissa of convergence $\sigma$.
Proof : if $F$ is analytic then $F(\sigma-\epsilon) = \sum_{k\ge 0}\frac{\epsilon^k}{k!} \int_X^\infty (\sum_{n\le x}\lambda(n)n^{-1})x^{-\sigma} (\log x)^k dx$ if it is finite then we can invert $\sum,\int$ obtaining that $\int_X^\infty (\sum_{n\le x}\lambda(n)n^{-1})x^{-\sigma+\epsilon}dx$ is finite.
The analyticity of the LHS shows $\sigma=1/2$.

I don't know about the converse, assuming the RH what can we say about the zeros of $\zeta_n$ and the sign of $\sum_{n\le x}\lambda(n)n^{-1}$ ?


TITLE: Velocity field arrows along null clines as well as outside null clines
QUESTION [0 upvotes]: For Question 8 (as well as in general), I don't understand how to sketch velocity field arrows along the null clines as well as outside the null clines. For this question the f1 null cline would be when y-x=0 so y=x and the f2 null cline would be when y+x=0 so y=-x. I understand that along null cline 1 the velocity field arrows will be pointing upwards or downwards and along null cline 2 velocity field arrows will be pointing left or right but how do we determine whether they will be upwards or downwards or left or right? Also how would we determine the velocity field arrows outside the null clines?
Any help would be much appreciated.

REPLY [0 votes]: Well, even if you've asked to sketch vector field somewhere, you still can (and sometimes should or even must) do some calculations that will give you a rigorous understanding of vector field behaviour.
You know the equations of null clines, so you could directly substitute coordinates of point from null cline to vector field and check signs of its components. For example, take the $f_1$ null cline: each point of it has coordinates $(x, x), \; x \in \mathbb{R}$. Then just plug $(x, x)$ into the $(y-x) {\bf e_1}  + (y+x) {\bf e_2}$ and you'll obtain $2x \cdot {\bf e_2}$.
So this gives you an exact formula which describes how vector field behaves along null cline $f_1$.
Now if you want to check vector field outside null clines, the easiest way to do this is (again!) to plug some point. Null clines define borders of domains where components have constant and definite sign. When you sketch null clines, the plane $\mathbb{R}^2$ will be split into several domains and you'll just have to check only one point from each domain. This would be enough to sketch direction of vector field everywhere.
Hope this will help.


TITLE: Intuitively predict the direction of the acceleration vector of a single parameter vector valued function
QUESTION [2 upvotes]: The velocity vector of a Vector valued function (or position function), (for example $f(t)= \lt x(t),y(t) \gt$)
 seems to always be tangent to the curve of motion, its pretty intuitive why that is.
However intuition about what the acceleration vector $f ^{\prime \prime}(t)$ looks like or where does it point is hard to find.
Can I somehow predict or get an intuition on what the acceleration vector might be or where does it point to? how?.

REPLY [1 votes]: The acceleration vector is not uniquely determined by the curve, it depends on the way it is parametrized.  However, when the curve is parametrized by it's arclength, the acceleration vector of the curve is perpendicular to the velocity vector in 2 dimensions. In higher dimensions, this is no longer true.
The arclength of a curve is given by $s=\int_{t_1}^{t_2}dt\sqrt{x'(t)^2+y'(t)^2}$, where $t$ is any given parametrization. If the curve is given by $r(t)=(x(t),y(t))$ and it's arclength parametrization is $r(s)=(X(s), Y(s))$ then
$$\hat{T}(s)=\Big(\frac{dX}{ds}, \frac{dY}{ds}\Big)=\frac{dr}{ds}=\frac{dr}{dt}\frac{dt}{ds}=\Big(\frac{x'(t)}{\sqrt{x'(t)^2+y'(t)^2}},\frac{y'(t)}{\sqrt{x'(t)^2+y'(t)^2}}\Big)$$
and we conclude that
$$||\hat{T}(s)||=1$$
Now are allowed to write $\frac{dX}{ds}=\cos\theta(s),\frac{dY}{ds}=\sin\theta(s)$
and the acceleration of the curve can be written as
$$a(s)=\frac{dT(s)}{ds}=\theta'(s)(-\sin\theta, \cos\theta)=\frac{d\theta}{ds}\hat{n}$$
and it's easy to show that $T\cdot \hat{n}=0$ and thus the acceleration is perpendicular to the velocity in this case.
If the curve is parametrized by any other parameter, the direction of the acceleration is not perpendicular to the tangent vector anymore, since we can write $\frac{dx}{dt}=||v(t)||\cos\theta(t), \frac{dy}{dt}=||v(t)||\sin\theta(t)$ and thus
$$a(t)=\frac{dv}{dt}=\frac{d||v||}{dt}(\cos\theta(t), \sin\theta(t))+||v||\frac{d\theta}{dt}(-\sin\theta(t), \cos\theta(t))=\frac{d||v||}{dt}\hat{T}+||v||\frac{d\theta}{dt}\hat{n}$$
which shows that the acceleration now has a component in the tangent direction as well, which is zero if $t=s$.
EDIT: In Newtonian physics, the trajectory of the particle is determined by the equation of motion $m\frac{d^2r(t)}{dt^2}=F$. In this case it is clear that this equation determines the direction of the acceleration of the particle, and the acceleration is not necessarily the same as the geometrical acceleration of the curve. In other words, the time parameter $t$ has a physical meaning and it's relationship to the arclength $s=s(t)$ is determined by the equations of motion and it's not chosen freely, like in differential geometry. One has to solve the equations of motion to be able to tell the relationship between the curve and the body's acceleration. However, the last equation written above for $a(t)$ is correct, and it represents the decomposition of the particle's acceleration to tangential and normal components.


TITLE: Question about special orthogonal Lie group construction
QUESTION [3 upvotes]: Working through homework and I run into this problem:

Suppose the Lie group $SO^{+}(2,2)$ is presented as the group of all transformations in its associated space. How do you determine whether a given 4x4 transformation matrix is an element of the group?

I'm slightly familiar with the $SO^{+}(1,3)$ Lie group, but have almost no experience with group theory as a whole. So far, I seem to remember the conditions require orthogonality between coordinates, preservation of the orientation (of subgroups?), and elements with a determinant of 1, but still feel like I'm missing something. I expect given a matrix representation of the form:
$\left( \begin{array}{ccc}
a & b & 0 & 0\\
c & d & 0 & 0\\
0 & 0 & e & f\\
0 & 0 & g & h\end{array} \right) $
Do I need to ensure the determinants of the components
$\left( \begin{array}{ccc}
a & b \\
c & d \\
\end{array} \right) $
and $\left( \begin{array}{ccc}
e & f \\
g & h \\
\end{array} \right) $
are related  in a certain way to prevent orientation shifts (such as reflections), or do I only need to pay attention to the total determinant of a transformation?

REPLY [2 votes]: For $SO(2,2)$, You should check that it preserves the quadratic form $-x^2-y^2+z^2+w^2$, and that the matrix has determinant one. I guess the $+$ means that the transformation preserves both orientations, meaning that in you example both smaller determinants have to be equal to one.


text:
\begin{document}

\title[An analog of perfect numbers]{An analog of perfect numbers involving the unitary totient function}
\author[Tomohiro Yamada]{Tomohiro Yamada*}
\address{\llap{*\,}Center for Japanese language and culture\acr
                   Osaka University\acr
                   562-8558\acr
                   8-1-1, Aomatanihigashi, Minoo, Osaka\acr
                   JAPAN}
\email{tyamada1093@gmail.com}

\subjclass[2010]{Primary 11A05, 11N36; Secondary 11A25, 11A51.}
\keywords{Unitary divisor; totient; Lehmer's problem; sieve methods; arithmetic functions}

\begin{abstract}
We shall give some results for an integer divisible by its unitary totient.
\end{abstract}

\maketitle

\section{Introduction}\label{intro}

In 1932, Lehmer \cite{Leh1} asked whether there exists any composite numbers $n$ such that
the Euler totient $\vph(N)$ divides $N-1$ or not.
Obviously, if $N$ is prime, then $\vph(N)=N-1$ and vice versa.
Lehmer's problem is still unsolved.

On the other hand, it is easy to show that $\vph(N)$ divides $N$
if and only if $N$ is of the form $2^m 3^n$ (see, for example, p.p. 196--197 of \cite{Sie}).
Such an integer can be considered as a totient version of a multiperfect number,
i.e., an integer dividing its sum of divisors.
However, it is not known whether there exists any odd multiperfect numbers,
nor whether there exist infinitely many even multiperfect numbers.

An divisor $d$ of $N$ is called a unitary divisors of $N$ if $\gcd(d, N/d)=1$,
denoted by $d\mid\mid N$.
Subbarao \cite{Sub} considered the problem analogous to Lehmer's problem
involving $\vph^*$, the unitary analogue of $\vph$.
So $\vph^*$ is defined by
\begin{equation}
\vph^*(N)=\prod_{p^e\mid\mid N}(p^e-1),
\end{equation}
where the product is over all prime powers unitarily dividing $N$.
We call the value $\vph^*(N)$ the {\it unitary totient} of an integer $N$.
Subbarao conjectured that $\vph^*(N)$ divides $N-1$ if and only if $N$ is a prime power.
This conjecture is still unsolved.
However, Subbarao and Siva Rama Prasad \cite{SS}
showed that $N$ must have at least eleven distinct prime factors if $N$ is not a prime power
and $\vph^*(N)$ divides $N-1$.

Now the problem naturally arises when $\vph^*(N)$ divides $N$.
We can easily find some instances (see also \url{https://oeis.org/A319481}):
\begin{equation}\label{eq01}
\begin{split}
N_1 & =1, N_2=2, N_3=2\cdot 3, N_4=2^2\cdot 3,\\
N_5 & =2^3\cdot 3\cdot 7,\\
N_6 & =2^4\cdot 3\cdot 5,\\
N_7 & =2^5\cdot 3\cdot 5\cdot 31,\\
N_8 & =2^8\cdot 3\cdot 5\cdot 17,\\
N_9 & =2^{11}\cdot 3\cdot 5\cdot 11^2\cdot 23\cdot 89,\\
N_{10} & =2^{16}\cdot 3\cdot 5\cdot 17\cdot 257,\\
N_{11} & =2^{17}\cdot 3\cdot 5\cdot 17\cdot 257\cdot 131071,\\
N_{12} & =2^{32}\cdot 3\cdot 5\cdot 17\cdot 257\cdot 65537.
\end{split}
\end{equation}
All of these examples satisfy $N=2\vph^*(N)$ except $N=1=\vph^*(1)$ and $N=6=3\vph^*(6)$.

Such integers were implicitly referred in \cite{FJ} as possible orders of groups with
perfect order subsets, groups $G$ with the number of elements in each order subset dividing $\abs{G}$.
Integers equal to twice of its unitary totient had been introduced in the OEIS in 1999
by Yasutoshi Kohmoto\\ \url{https://oeis.org/A030163}.

At the JANT meetings held at Kyoto in the spring of 2009, the author showed some basic results
for integers with this property.
Here the author showed that $N$ must be divisible by one of the first five (i.e. all known)
Fermat primes and have a prime power factor $>10^8$.
Later Ford, Konyagin and Florian Luca \cite{FKL} studied properties of such integers
and showed that such integers $N$ must be divisible by $3$ and $N/\vph^*(N)\leq 85$.

This paper is a revised version of the unpublished note presented at the JANT meetings
and contains some new results which the author recently proved.

We would like to state main results.

\begin{thm}\label{thm2}
If $N$ is an integer divisible by $\vph^*(N)$ other than given in \eqref{eq01},
then $\omega(N)\geq 8$, where $\omega(N)$ denotes the number of distinct prime factors of $N$.
\end{thm}

\begin{thm}\label{thm3}
If $N$ is none of the twelve instances given above, then $N$ must be divisible by an odd prime factor
at least $10^5$ and an odd prime power at least $10^8$.
\end{thm}

\begin{thm}\label{thm5}
There exist only finitely many integers $N$ divisible by $\vph^*(N)$
which are products of consecutive primes.
\end{thm}

Theorems \ref{thm2} and \ref{thm3} of our results are proved by elementary means
but require a fair amount of computation.
Theorem \ref{thm5} are proved using sieve methods combined with
the analytic theory of the distribution of primes in arithmetic progressions.

\section{Basic properties}
From now on, we assume that $N$ is an integer satisfying $\varphi^*(N)\mid N$.

We set $h(n)=n/\vph^*(n)$ for a positive integer $n$.
Hence, $N$ is $\vph^*$-multiperfect if and only if $h(N)$ is an integer.
We factor $N=2^e p_1^{e_1} p_2^{e_2} \cdots p_r^{e_r}$ and put $P_i=p_i^{e_i}$,
where $p_1<p_2<\cdots<p_r$ are the odd prime factors of $N$.
We write $M_k=2^e P_1 P_2 \cdots P_k$ for $k=1, 2, \ldots, r$ and $M_0=2^e$.
Moreover, let $Q_1<Q_2<\cdots<Q_m$ be the odd prime power factors of $N$.
It is easy to see that $Q_k\geq p_k$ for every $k$.
Indeed, among $Q_1, Q_2, \ldots, Q_k$ is some $Q_j=p_t^{e_t}$ with $t\geq k$.
We note that $P_i$ and $Q_i$ are not necessarily equal.
For example, if $N=2^e\cdot 3^2\cdot 5\cdot 11\cdots$, then $P_1=3^2$ and $P_2=5$
while $Q_1=5$ and $Q_2=3^2$.

\begin{lem}\label{lm11}
We have the following properties:
\begin{itemize}
\item[a)] If $N=2^e$, then $e=1$.
\item[b)] For each $i=1, 2, \ldots, r$, $p_i^{e_i}-1$ divides $N$.
In particular, $p_i-1$ must divide $N$.
\item[c)] If $m$ is a squarefree integer dividing $N$, then $\vph(m)=\vph^*(m)$ must divide $N$.
\item[d)] If $2^e\mid\mid N$, then $N$ has at most $e$ distinct odd prime factors.
More exactly, if $2^f$ divides $h(N)$ and $2^{f_i}\mid (p_i-1)$ for each odd prime factor
$p_i (i=1, 2, \ldots, r)$ of $N$, then we have $f+\sum_{i=1}^r f_i\leq e$.
\end{itemize}
\end{lem}

\begin{proof}
\begin{itemize}
\item[a)] $\varphi^*(N)=(2^e-1)\mid 2^e=N$ and therefore we must have $2^e=2$.
\item[b)] $(p_i-1)\mid(p_i^{e_i}-1)\mid\varphi^*(N)\mid N$.
\item[c)] Factor $m=\prod_j q_j$ and set $f_j$ to be the exponent of $q_j$ dividing $N$.
Since $\vph^*(q_j)=(q_j-1)\mid (q_j^{f_j}-1)=\vph^*(q_j^{f_j})$ for each $j$,
we have $\vph^*(m)\mid \vph^*(\prod_j q_j^{f_j})\mid \vph^*(N)\mid N$.
\item[d)]
If $2^f$ divides $h(N)$ and $2^{f_i}\mid (p_i-1)$ for each odd prime factor
$p_i (i=1, 2, \ldots, m)$ of $N$, then
$2^{\sum_{i=1}^r f_i}\mid (p_1-1)\cdots(p_r-1)\mid \vph^*(N)$
and therefore $2^{f+\sum_{i=1}^r f_i}\mid h(N)\vph^*(N)=N$.
\end{itemize}
\end{proof}

It immediately follows from d) that $1$ is the only odd integer
divisible by its unitary totient.

The property b) makes no reference to the exponent of $p$ dividing $N$.
This is the key tool of our study.
To express the latter part of the property b) in other words,
if $p$ does not divide $N$, then no prime $\equiv 1\pmod{p}$ divides $N$.
In particular, the smallest odd prime factor of $N$ must be a Fermat prime,
a prime of the form $2^{2^k}+1$.

The property c) is a generalization of the latter part of the property b).
We write $\prod_{i=1}^k q_i^{f_i} \rightarrow \prod_{j=1}^l r_j^{g_j}$
if $\prod_{j=1}^l r_j^{g_j}$ divides $\prod_{i=1}^k (q_i^{f_i}-1)$
for two sets of distinct primes $q_i$'s and $r_j$'s.
Now the property c) can be generalized further.
\begin{lem}\label{lm12}
Let $m_i (i=1, 2, \ldots, k)$ be squarefree integers,
$p, q$ be primes and $f\geq 0, g>0$ be arbitrary integers.
If $q^{gl}\mid\mid N$ for some integer $l>0$ and
$q^g\rightarrow m_1\rightarrow m_2\rightarrow \cdots \rightarrow m_k\rightarrow p^f$,
then $p^f\mid N$.
\end{lem}
\begin{proof}
We see that $m_1\mid (q^g-1)\mid (q^{gl}-1)=\vph^*(q^{gl})\mid \vph^*(N)\mid N$.
Since $m_1$ is squarefree, we have $m_2\mid \vph^*(m_1)\mid \vph^*(N)\mid N$
by the property c).
Inductively, each $m_i (i=1, 2, \ldots , k)$ divides $N$.
Hence, $p^f\mid\vph^*(m_k)\mid \vph^*(N)\mid N$.
\end{proof}

Now we introduce two well-known lemmas concerning prime factors
of the $n$-th cyclotomic polynomial, which we denote by $\Phi_n(X)$.
Lemma \ref{lm21} follows from Theorems 94 and 95 in Nagell \cite{Nag}.
Lemma \ref{lm22} has been proved by Bang \cite{Ban} and rediscovered by many authors
such as Zsigmondy \cite{Zsi}, Dickson \cite{Dic} and Kanold \cite{Ka1, Ka2}.

\begin{lem}\label{lm21}
Let $p, q$ be distinct primes with $q\neq 2$ and $n$ be a positive integer.
Then, $q$ divides $p^c-1$ if and only if the multiplicative order of $q$ modulo $p$
divides $n$.
\end{lem}

\begin{lem}\label{lm22}
If $a$ is an integer greater than $1$, then $a^n-1$ has
a prime factor which does not divide $a^k-1$ for any $k<n$,
unless $(a, n)=(2, 1), (2, 6)$ or $n=2$ and $a+1$ is a power of $2$.
Moreover, such a prime factor must be congruent to $1$ modulo $n$.
\end{lem}

Since $2^6-1=3^2\cdot 7$, we see that $a^k-1$ always has a prime factor congruent to
$1$ modulo $k$ unless $(a, k)=(2, 1)$.  Thus the following result holds.

\begin{lem}\label{lm13}
If $a$ is prime, $k$ is a positive integer and $a^k\mid\mid N$, then $k\mid N$.
\end{lem}

\begin{proof}
We may assume that $k>1$.
Since $a$ is prime, $a^k-1=\vph^*(a^k)$ divides $N$ by b) of Lemma \ref{lm11}.
By Lemma \ref{lm22}, $a^k-1$ has at least one prime factor $p\equiv 1\pmod{k}$,
which divides $N$.
Using b) of Lemma \ref{lm11} again, we see that $k$ divides $N$.
\end{proof}

Concerning the number of distinct prime factors of $N$,
it is not difficult to see the following result.

\begin{lem}\label{lma1}
If $N\neq 1, 2, 6, 12, 168, 240$,
then $2^5\mid N$ and $\omega(N)\geq 5$.
\end{lem}
\begin{proof}
If $\omega(N)=1$, then $N=2^e$.
The only possibility is $e=1$ since, if $e>1$, then $h(N)<2/\vph^*(2)=2$.

If $\omega(N)=2$, then $N=2^{e_0} p_1^{e_1}$.
If $h(N)\geq 3$, then we must have $e_0=1$ and $p_1^{e_1}=3$
since, otherwise, $h(N)<2\times (3/2)=3$.
If $h(N)=2$, then $e_0\geq 2$ by Lemma \ref{lm11}.
Now the only possibility is $e_0=2$ and $p_1^{e_1}=3$
since, otherwise, $h(N)<(4/3)\times (3/2)=2$.

If $N=2^2 p_1^{e_1} p_2^{e_2}$,
then $p_1\equiv p_2\equiv 3\pmod{4}$ and $h(N)$ must be odd by Lemma \ref{lm11},
implying that $h(N)\geq 3$.
But $h(N)\leq (4/3)(3/2)(7/6)=7/3<3$, which is a contradiction.

If $N=2^3 p_1^{e_1} p_2^{e_2} p_3^{e_3}$,
then $h(N)\geq 3$ and $p_1\equiv p_2\equiv p_3\equiv 3\pmod{4}$ by Lemma \ref{lm11}.
But $h(N)\leq (8/7)(3/2)(7/6)(11/10)=11/5<3$.

If $N=2^4 p_1^{e_1} p_2^{e_2} p_3^{e_3} p_4^{e_4}$,
then we must have $h(N)\geq 3$ and $p_i\equiv 3\pmod{4} (i=1, 2, 3, 4)$ by Lemma \ref{lm11}
but $h(N)\leq (16/15)(3/2)(7/6)(11/10)(19/18)<3$.

If $N=2^3 p_1^{e_1} p_2^{e_2}$,
then we must have $h(N)=2$ since $h(N)\leq (8/7)(3/2)(5/4)=15/7<3$.
Hence, $p_1\equiv p_2\equiv 3\pmod{4}$.
The only possibility is $(p_1^{e_1}, p_2^{e_2})=(3, 7)$
since, otherwise, $h(N)<(8/7)(3/2)(7/6)=2$.

If $N=2^e p_1^{e_1} p_2^{e_2}$ with $e\geq 4$,
then we must have $(e, p_1^{e_1}, p_2^{e_2})=(4, 3, 5)$
since, otherwise, $h(N)<(16/15)(3/2)(5/4)=2$.

If $N=2^4 p_1^{e_1} p_2^{e_2} p_3^{e_3}$,
then $h(N)=2$ since $h(N)\leq (16/15)(3/2)(5/4)(7/6)=7/3<3$.
Now we must have $p_1\equiv p_2\equiv p_3\equiv 3\pmod{4}$ by Lemma \ref{lm11}.
we must have $(p_1^{e_1}, p_2^{e_2})=(3, 7)$
and therefore $p_3^{e_3}=15$, which is inadmissible.

Now, we see that, unless $N=1, 2, 6, 12, 168, 240$, we must have $2^5\mid N$ and $\omega(N)\geq 4$.
\end{proof}

In general, there exist only finitely many positive integers $N$ divisible by $\vph^*(N)$
with a given number of distinct prime factors.

\begin{lem}\label{lm23}
$N\leq 2^{2^{r+1}}-2^{2^r}$.
\end{lem}
\begin{proof}
It is implicit in Lemma 2 of \cite{Coo} that, for any positive integers $a, b, k$ and $1<n_1<n_2<\cdots <n_k$
satisfying
\begin{equation}
\prod_{i=1}^k\left(1-\frac{1}{n_i}\right)\leq\frac{a}{b}<\prod_{i=1}^{k-1}\left(1-\frac{1}{n_i}\right),
\end{equation}
the inequality
\begin{equation}\label{eq20}
a\prod_{i=1}^k n_i\leq (a+1)^{2^k}-(a+1)^{2^{k-1}}
\end{equation}
must hold.
An explicit proof is given by \cite{Nie} with \eqref{eq20} replaced by $a\prod_{i=1}^k n_i<(a+1)^{2^k}$.
A slight change enables to prove \eqref{eq20}.
Indeed, observing that $an_1\leq a(a+1)=(a+1)^2-(a+1)$ for $k=1$, we can prove \eqref{eq20}
by induction of $r$ proceeding as the proof of Lemma 1 of \cite{Nie}.
Now the lemma immediately follows applying this with $a=1, b=N/\vph^*(N), k=r+1$ and $n_1=2^e, n_{i+1}=P_i$.
\end{proof}

This gives the following upper bound for the $P_i$'s.

\begin{lem}\label{lma0}
For any $i=1, 2, \ldots, r$, we have $P_i(P_i-1)\mid N$
and $P_i\leq 2^{2^{r-1}}$.
\end{lem}
\begin{proof}
The former statement immediately follows from b. of Lemma \ref{lm11}.
By Lemma \ref{lm23}, we have $P_i(P_i-1)\leq N\leq 2^{2^{r+1}}-2^{2^r}$,
which immediately gives the latter statement.
\end{proof}

Moreover, we can see that, for a given $M_k$, the next prime factor $p_{k+1}$ must be bounded.

\begin{lem}\label{lm29}
For each $k=0, 2, \ldots, r-1$, we have
\begin{equation}\label{eq291}
\left(\frac{p_{k+1}}{p_{k+1}-1}\right)^{r-k}\geq \prod_{i=k+1}^r \frac{p_i}{p_i-1}\geq \frac{h(N)}{h(M_k)}.
\end{equation}
Furthermore, $p_{k+1}-1$ must divide $M_k/\gcd(M_k, \vph^*(M_k))$.
\end{lem}
\begin{proof}
\eqref{eq291} immediately follows by observing that
\begin{equation}
\prod_{i=k+1}^m \frac{p_i}{p_i-1}\geq h(p_{k+1}^{e_{k+1}}\cdots p_r^{e_r})=\frac{h(N)}{h(M_k)}.
\end{equation}
We can easily see that $L_k=\gcd(M_k, \vph^*(M_k))(p_{k+1}-1)$ divides $N$
and any prime factor of $L_k$ must be smaller than $p_{k+1}$.
Hence, $L_k$ divides $M_k$ and the latter statement holds.
\end{proof}

\section{The number of distinct prime factors of $N$}

In Lemma \ref{lma1}, we already proved that $\omega(N)\geq 5$.
In this section, we shall prove Theorem \ref{thm2}.

Assume that $N\neq 1, 2, 6, 12, 168, 240$ is an integer such that $\omega(N)\leq 7$ and $\vph^*(N)$ divides $N$.
If $h(N)\geq 3$, then we must have $\omega(N)\geq 8$ since, otherwise,
\begin{equation}
h(N)<h(2^5\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17)=\frac{32}{31}\cdot\frac{3}{2}\cdot\frac{5}{4}\cdot\frac{7}{6}\cdot\frac{11}{10}\cdot\frac{13}{12}\cdot\frac{17}{16}<3.
\end{equation}
Hence, in the remaining part of this section, we may assume that,
$N\neq 1, 2, 6, 12, 168, 240$ is an integer such that $\omega(N)\leq 7$ and $N=2\vph^*(N)$.

We would like to show that $N$ must be one of the instances given in \eqref{eq01}.
This would be proved via several lemmas.

\begin{lem}\label{lmb1}
$3\mid N$.
\end{lem}
\begin{proof}
Although this is proved in \cite{FKL}, we would like to give a proof
since the proof is much simpler given that $\omega(N)\leq 7$.
Assume that $3$ does not divide $N$.
By Lemma \ref{lma1}, we must have $\omega(N)\geq 5$.
If $\omega(N)=5$, then $h(N)\leq h(2^5\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17)<2$
and, if $\omega(N)\geq 6$, then $h(N)\leq h(2^6\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19)<2$.
Hence, we have a contradiction and therefore $3$ must divide $N$.
\end{proof}

\begin{lem}\label{lmb2}
$3\mid\mid N$.
\end{lem}
\begin{proof}
From the previous lemma, we know that $3\mid N$.
Assume that $3^2\mid N$.
Observing that
$h(2^7\cdot 3^2\cdot 5\cdot 7\cdot 17\cdot 19\cdot 23)<2$,
we must have $P_2=5, P_3=7$ and $P_4\leq 13$.

If $P_4=11$, then $P_5=13$ or $17$ by Lemma \ref{lm29}.
If $(P_4, P_5)=(11, 13)$, then $2^{11}\mid N$ by Lemma \ref{lm11}
and therefore $2047/1028\leq h(3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot P_6)<2$.
Hence, we must have $43<P_6<47$, which is impossible.
If $(P_4, P_5)=(11, 17)$, then $2^{13}\mid N$ and $23<P_6<29$, a contradiction
(we cannot have $P_6=25=p_2^2$ or $27 =p_1^3$).

If $P_4=13$, then $P_5=17, 2^{14}\mid N$ and $P_6<19$, a contradiction.
Hence, $3\mid\mid N$.
\end{proof}

\begin{lem}\label{lmb3}
$5\mid N$.
\end{lem}
\begin{proof}
We know that $P_1=3$ by the previous lemma.
If $p_2>7$, then $h(N)\leq h(2^7\cdot 3\cdot 13\cdot 17\cdot 23\cdot 29\cdot 47)<2$, which is a contradiction.

Now we assume that $p_2=7$.
We observe that $\gcd(p_i-1, 15)=1$ for $i\geq 3$, which implies that neither $11$ nor $13$ divide $N$.
Hence, we see that $p_3\geq 17, p_4\geq 29, p_5\geq 59$ and $p_6\geq 113$.
If $p_3=17$, then $h(N)\leq h(2^9\cdot 3\cdot 7\cdot 17\cdot 29\cdot 59\cdot 113)<2$.
If $p_3>17$, then $p_3\geq 29, p_4\geq 59, p_5\geq 113$ and
$h(N)\leq h(2^6\cdot 3\cdot 5\cdot 29\cdot 59\cdot 113\cdot 127)<2$.
Thus we see that $p_2=7$ is impossible.
Hence, we must have $p_2=5$.
\end{proof}

\begin{lem}\label{lmb4}
$5\mid\mid N$.
\end{lem}
\begin{proof}
We know that $p_1=3, e_1=1$ and $p_2=5$.
We shall show that $e_2$ cannot be divisible by any prime $\leq 29$ unless $e_2\geq 9$ is a power of three.

{\it Case I.} If $e_2$ is even, then $N$ has no prime factor $\equiv 1\pmod{3}$
since otherwise $3^2\mid (5^2-1)(p_i-1)\mid \vph^*(N)\mid N$ for some $i$, contrary to $3\mid\mid N$.
Moreover, $2^9\mid 2(3-1)(5^2-1)(p_3-1)(p_4-1)(p_5-1)(p_6-1)\mid N$.
Hence, we must have
$h(N)\leq h(2^9\cdot 3\cdot 5^2\cdot 11\cdot 17\cdot 23\cdot 29)<2$.

{\it Case II.} If $e_2=3$, then Lemma \ref{lm29} gives that $p_3\leq 11$
since $2^8$ must divide $N$ and $h(2^8\cdot 3\cdot 5^3\cdot 13\cdot 17\cdot 19\cdot 23)<2$.

If $p_3=7$, then we must have $p_4, p_5, p_6 \equiv 2\pmod{3}$ as above.
Since $h(2^8\cdot 3\cdot 5^3\cdot 7^2\cdot 11\cdot 17\cdot 23)<2$
and $h(2^8\cdot 3\cdot 5^3\cdot 7\cdot 23\cdot 29\cdot 41)<2$,
we must have $e_3=1$ and $p_4\leq 17$.
In other words, we must have $p_4=11$ or $p_4=17$.

If $p_4=11$, then, we must have $e_4=1$
since $h(2^8\cdot 3\cdot 5^3\cdot 7\cdot 11^2\cdot 17\cdot 23)<2$.
Furthermore, by Lemma \ref{lm29}, $p_5\leq 73$ and $p_5-1=2^m u$ with $u$ dividing $5^3\cdot 7\cdot 11$.
Hence, $p_5=17, 23, 29, 41$ or $71$.
If $e_5\geq 3$, then $p_5^{e_5}-1$ has always a prime factor $\geq 47$.  Hence, $p_6\geq 47$ and $h(N)<2$.
If $e_5$ is even, then we must have $3\mid p_5^{e_5}-1$ and $3^2\mid (7-1)(p_5^{e_5}-1)\mid N$ while $3\mid\mid N$,
which is a contradiction.
Hence, we must have $e_5=1$ and, $p_5=41$ or $71$ since otherwise $h(N)\geq h(3\cdot 5^3\cdot 7\cdot 11\cdot 29)>2$.

If $p_5=41$, then Lemma \ref{lm29} gives that $p_6^{e_6}=257$ or $p_6^{e_6}=281$,
leading to a contradiction observing that
$h(2^9\cdot 3\cdot 5^3\cdot 7\cdot 11\cdot 41\cdot 281)>2>h(2^{10}\cdot 3\cdot 5^3\cdot 7\cdot 11\cdot 41\cdot 257)$.
If $p_5=71$, then Lemma \ref{lm29} gives that $p_6^{e_6}=89$ or $p_6^{e_6}=113$,
which is impossible since $h(2^8\cdot 3\cdot 5^3\cdot 7\cdot 11\cdot 71\cdot 89)<2$.
Thus we cannot have $p_4=11$.

If $p_4=17$, then, proceeding as above, we must have $e_4=1$ and $p_5^{e_5}=29$.
It follows from Lemma \ref{lm29} that no prime is appropriate for $p_6$.
Thus we cannot have $p_4=17$ and therefore $p_3=7$.

If $p_3=11$, then $p_4^{e_4}=17$ but Lemma \ref{lm29} yields that
there exists no appropriate prime for $p_5$.

{\it Case III.} If $5$ divides $e_2$, then $11\cdot 71$ divides $N$.
We must have $p_3^{e_3}=7$ since otherwise $h(N)\leq h(2^8\cdot 3\cdot 5^5\cdot 11\cdot 13\cdot 17\cdot 71)<2$.
$p_4^{e_4}=11$.
If $p_5\geq 53$, then $h(N)\leq h(2^8\cdot 3\cdot 5^5\cdot 7\cdot 11\cdot 53\cdot 71)<2$.
If $p_5\leq 47$, then, by Lemma \ref{lm29}, we must have $p_5=17, 23, 29$ or $41$.
If $e_5\geq 2$, then $h(N)\leq h(2^8\cdot 3\cdot 5^5\cdot 7\cdot 11\cdot 17^2\cdot 71)<2$.
If $e_5=1$, then $h(N)>h(\cdot 3\cdot 7\cdot 11\cdot 41\cdot 71)>2$.
Both cases lead to contradictions and therefore $e_2$ cannot be divisible by $5$.

{\it The other cases.} If $7$ divides $e_2$, then $3^2$ must divide $N$ since $5^7\rightarrow 19531\rightarrow 3^2$, which contradicts to Lemma \ref{lmb2}.
Similarly, it follows from
$5^{11}\rightarrow 12207031\rightarrow 521\rightarrow 13$ and $13\cdot 12207031\rightarrow 3^2$,
$5^{13}\rightarrow 305175781\rightarrow 3^2$,
$5^{17}\rightarrow 409\cdot 466344409\rightarrow 3^2$,
$5^{19}\rightarrow 191\rightarrow 19\rightarrow 3$ and $5^{19}\rightarrow 6271\rightarrow 3$
and
$5^{23}\rightarrow 332207361361\rightarrow 3^2$
that, if $11, 13, 17, 19$ or $23$ divides $e_2$, then $3^2$ must divide $N$,
contrary to Lemma \ref{lmb2} again.

Now, if $e_2$ is a power of three, then $3^2\mid e_2$, $19\mid (5^9-1)\mid N$ and $3^2\mid (19-1)\mid N$,
contrary to Lemma \ref{lmb2}.
Thus we see that $e_2$ must have a prime factor $\geq 29$ if $5^2\mid N$.
However, Lemma \ref{lma0} immediately gives that $5^{e_2}\leq 2^{64}$.
Hence, we must have $e_2=1$.
\end{proof}

\begin{lem}\label{lmb5}
If $p=7, 11$ or $13$, then $p^2\mid\mid N$.
If a prime $17\leq p<200$ divides $N$, then $e\leq 2$ and $p\in \{23, 31, 47, 67, 83, 137\}$
or $e=1$ and $p\in \{17, 29, 41, 53, 59, 61, 71, 89, 97, 103, 107, 113, 131, 139, 167, 179, 191, 193, 197\}$.
Moreover, if $p=233, 241, 257, 409, 641, 769, 1021, 1361$ or $1637$ divides $N$, then $p\mid\mid N$.
\end{lem}

\begin{proof}
We shall only give a proof for $p=7$.
Assume that $7\mid N$ and let $e$ be its exponent.
We observe that $7^2$ must divide $N$ since $h(105)>2$.

It follows from $7^3\rightarrow 3^2$, $7^7\rightarrow 4733\rightarrow 3$,
$7^{11}\rightarrow 1123\rightarrow 3$, $7^{17}\rightarrow 2767631689\rightarrow 3^2$
and $7^{19}\rightarrow 419\rightarrow 19\rightarrow 3^2$
together with $7\rightarrow 3$
that, if $e_2$ is divisible by one of $3, 7, 11, 17$ and $19$, then
$3^2$ must divide $N$, contrary to Lemma \ref{lmb2}.

Similarly, if $e_2$ is divisible by $4, 5$ or $13$, then
$5^2$ must divide $N$ since $7^4\rightarrow 5^2$, $7^5\rightarrow 2801\rightarrow 5^2$
and $7^{13}\rightarrow 16148168401\rightarrow 5^2$,
contrary to Lemma \ref{lmb4}.

Since $7^{23}>2^{64}$, Lemma \ref{lma0} yields that we must have $e=2$.
The proof for the other primes is similar to the case $p=7$.
\end{proof}

\begin{lem}\label{lmb6}
$7$ cannot divide $N$.  In other words, $p_3\geq 11$.
\end{lem}
\begin{proof}
Assume that $7$ divides $N$.

We already know that $7^2\mid\mid N$ from the previous lemma.

We must have $11\leq p_4<71$.
since $h(2^{10}\cdot 3\cdot 5\cdot 7^2\cdot 71\cdot 73\cdot 77)<2$.
Hence, $p_4=11, 17, 29$ or $41$.
By the previous lemma, we must have $p_4^{e_4}=11^2, 17, 29$ or $41$.
If $p_4^{e_4}=11^2$, then $3^2\mid (7^2-1)(11^2-1)\mid N$, which contradicts Lemma \ref{lmb2}.
If $p_4^{e_4}=17$, then $h(N)>h(3\cdot 5\cdot 7^2\cdot 17)>2$, a contradiction again.

If $p_4^{e_4}=29$, then, with the aid of Lemma \ref{lm29}, we must have $p_5=41, 59, 71, 101, 113$, $e_5\geq 2$ or $p_5=197, 233$.
By Lemma \ref{lmb5}, we must have $P_5=197$ or $233$.
If $P_5=197$, then Lemma \ref{lm29} yields that $p_6=281$.
However this is impossible since
$h(2^{12}\cdot 3\cdot 5\cdot 7^2\cdot 29\cdot 197\cdot 281)>2>h(2^{13}\cdot 3\cdot 5\cdot 7^2\cdot 29\cdot 197\cdot 281)$.
If $P_5=233$, then there exists no appropriate prime for $p_6$ in view of Lemma \ref{lm29}.

If $p_4^{e_4}=41$, then, combining Lemmas \ref{lm29} and \ref{lmb5}, we must have $P_5=83$.
But, now there exists no appropriate prime for $p_6$ in view of Lemma \ref{lm29}.
\end{proof}

\begin{lem}\label{lmb7}
$N=2^{11}\cdot 3\cdot 5\cdot 11^2\cdot 23\cdot 89$ or $p_3^{e_3}\in \{17, 31, 31^2, 41, 61\}$.
\end{lem}
\begin{proof}
Now we know that $p_3\in \{11, 13, 17, 31, 41, 61\}$.
Assume that $p_3\leq 13$.
Since $h(3\cdot 5\cdot 13)=195/96>2$, we must have $e_3\geq 2$.
By Lemma \ref{lmb5}, we must have $p_3^{e_3}=11^2$.

Now Lemma \ref{lm29} implies that
$p_4=2^{m_4} 11^{f_4}+1$ and $p_4<53$.  The only possibility is $p_4=23$.
Lemma \ref{lmb5} gives that $e_4\leq 2$.
If $e_4=2$, then using Lemma \ref{lm29} again, $p_5<41$ and $p_5=2^{m_5} 11^{f_5} 23^{g_5}+1$, which is impossible.
If $e_4=1$, then Lemma \ref{lm29} gives that $p_5=89$.
If $e=11$, then $N=2^{11}\cdot 3\cdot 5\cdot 11^2\cdot 23\cdot 89$ satisfies $h(N)=2$.
If $e\geq 12$, then we see that $2^{11}<p_6^{e_6}<2^{12}$ and $p_6^{e_6}-1$ divides $2^e\cdot 23\cdot 89$.
However, there exists no such prime $p_6$.
\end{proof}

\begin{lem}\label{lmb8}
We can never have $p_3=31, 41$ or $61$.
\end{lem}

\begin{proof}
If $p_3=31$, then Lemma \ref{lm29} implies that
$p_4<257$ and $p_4$ must be of the form $2^f\cdot 31^g$, which is impossible.
If $P_3=41$, then Lemma \ref{lm29} implies that $p_4<83$ and $p_4$ must be of the form $2^{f_1}\cdot 3^{f_2}\cdot 41^{f_3}$
with $f_2\leq 1$, which is again impossible.
Similarly, $P_3=61$ is impossible.
\end{proof}

Now the remaining case is the case $P_1=3, P_2=5$ and $P_3=17$.

\begin{lem}\label{lmb9}
If $P_1=3, P_2=5, P_3=17, P_4=257$, then
$N=N_{10}, N_{11}$ or $N_{12}$.
\end{lem}
\begin{proof}
If $2^{16}\mid\mid N$, then $N=N_{10} M$ for some integer $M$ coprime to $N_{10}$.
Since $h(N_{10})=2$, we must have $M=1$ and $N=N_{10}$.

If $2^{17}\mid N$, then
$h(2^{e_0}\cdot 3\cdot 5\cdot 17\cdot 257)\leq 2(1-1/131071)$.

If $2^{17}\mid\mid N$, then
$65537\leq p_5\leq Q_5\leq 262143$
and $p_5-1$ must divide $2\cdot 3\cdot 5\cdot 17\cdot 257$.
The only possibility is $p_5=131071$.
If $e_5>1$, then $p_5<p_6\leq Q_5\leq 262143$ and
$p_6-1$ must divide $2\cdot 131071$, which is impossible.
Hence, we have $e_5=1$ and $N=N_{11} M$ for some integer $M$ coprime to $N_{11}$.
Since $h(N_{11})=2$, we must have $M=1$ and $N=N_{11}$.

If $2^{18}\mid N$, then $65537\leq p_5\leq Q_5<174761$
and $p_5-1$ must divide $2^{e-16}\cdot 3\cdot 5\cdot 17\cdot 257$.
Hence, $p_5$ must be $65537, 82241, 87041, 98689, 131071$ or $163841$.
If $p_5=163841$, then $P_6<200000$ and $P_6-1$ divides $2^e\cdot 3\cdot 17\cdot 257$, which is impossible.
If $p_5=131071$, then $P_6<262144$ and $P_6-1$ divides $2^e\cdot 131071$, which is impossible.
If $p_5=98689$, then $2^{24}\mid N$, $P_6<200000$ and $P_6-1$ divides $2^e\cdot 5\cdot 17\cdot 98689$, which is impossible.
If $p_5=87041$, then $2^{27}\mid N$, $P_6<266000$ and $P_6-1$ divides $2^e\cdot 3\cdot 257\cdot 87041$, which is impossible.
If $p_5=82241$, then $2^{23}\mid N$, $P_6<340000$ and $P_6-1$ divides $2^e\cdot 3\cdot 17\cdot 87041$.
Hence, we must have $P_6=328961$, which is impossible since $h(2^{24} P_1 P_2 P_3 P_4 P_5 P_6)>2>h(2^{25} P_1 P_2 P_3 P_4 P_5 P_6)$.

If $p_5=65537$, then $2^{32}\mid N$.
If $p_5=65537$ and $2^{33}\mid N$, then
$2^{32}\leq P_6<2^{33}-1$ and $2^e-1$ must divide $3\cdot 5\cdot 17\cdot 257\cdot 65537\cdot P_6$.
We must have $e=61$ and $P_6=2^{61}-1$.
However, this is inappropriate since $P_6-1$ does not divide $2^{61} \cdot 3\cdot 5\cdot 17\cdot 257\cdot 65537$.
Hence, $2^{33}$ cannot divide $N$ and $N$ must be $N_{12}$.
We note that, if $F_5=2^{32}+1$ were prime, then $e=64$ and $P_6=2^{32}+1$ would suffice.
\end{proof}

\begin{lem}\label{lmb10}
If $P_4\neq 257$, then we must have $P_4\in \{409, 641, 769, 1021\}$.
\end{lem}
\begin{proof}
By Lemma \ref{lm29}, we must have $p_4\in \{31, 41, 61, 97, 103, 137, 193, 241,\\
257, 409, 641, 769, 1021\}$.

If $p_4\leq 257$ and $P_4\neq 257$, then, we must have $e_4>1$ since $h(3\cdot 5\cdot 17\cdot 251)>2$.
It immediately follows from Lemma \ref{lmb5} that
$p_4=31$ or $137$ and $e_4=2$.
In view of Lemma \ref{lm29}, there exists no prime appropriate for $p_5$.

If $p_4>257$, then Lemma \ref{lmb5} immediately gives that $e_4=1$
and $P_4\in \{409, 641, 769, 1021\}$.
\end{proof}

\begin{lem}\label{lmb11}
We cannot have $P_1=3, P_2=5, P_3=17$ and $p_4>257$.
\end{lem}
\begin{proof}
By the previous lemma, $p_4=409, 641, 769$ or $1021$
and, by Lemma \ref{lmb5}, we must have $e_4=1$.
If $p_4\geq 769$, then, by Lemma \ref{lm29}, we must have $p_5\geq 1021$,
which is impossible since $2^{11}$ must divide $N$ and
$h(N)\leq h(2^{11}\cdot 3\cdot 5\cdot 17\cdot 769\cdot 1021\cdot 1031)<2$.
If $p_4=409$, then, $2^{13}$ must divide $N$ and,
by Lemma \ref{lm29}, we must have $409<p_5<1500$
and $p_5-1$ must divide $2^e\cdot 5\cdot 409$.
However, there exists no such prime.
If $p_4=641$, then, by Lemma \ref{lm29}, we must have $p_5=769$.
Observing that $2^{24}$ must divide $N$ and using Lemma \ref{lm29} again,
we find that $p_6<1000$ and $p_6-1$ divides $2^e\cdot 769$,
which is impossible.
Hence, no prime $>257$ is appropriate for $p_4$.
\end{proof}

Now the theorem immediately follows combining Lemmas \ref{lmb7}-\ref{lmb11}.

\section{The largest prime (power) factor of $N$}

In this section, we prove Theorem \ref{thm3}.
We write $P(n)$ for the largest prime divisor of $n$.

We begin by proving the former part of Theorem \ref{thm3}.
Firstly, we need to find all primes $p$ and integers $k$ such that $p<10^5$ and $P(p^k-1)<10^5$.
This work can be done using the method of Goto and Ohno \cite{GO},
who showed that an odd perfect number must have a prime factor $>10^8$.
For the use in the proof of the latter part of Theorem \ref{thm3},
we shall also determine all exponents $k$ for which $P(2^k-1)<10^8$.

\begin{lem}\label{lm51}
If $P(2^k-1)<10^5$, then $k\leq 16$ or $k\in\{18, 20, 21, 22, 24, 25,\\
26, 28, 29, 30, 32, 36, 40, 42, 44, 45, 48, 50, 52, 60, 84\}$.
Moreover, if $P(2^k-1)<10^8$, then $k\leq 30$ or
\[\begin{split}
k\in\{
& 32, 33, 34, 35, 36, 38, 39, 40, 42, 43, 44, 45, 46, 47, 48, 50, 51,\\
& 52, 53, 54, 55, 56, 57, 58, 60, 63, 64, 66, 68, 70, 72, 75, 76, 78,\\
& 81, 84, 90, 92, 96, 100, 102, 105, 108, 110, 132, 140, 156, 180, 210\}.
\end{split}\]
\end{lem}

\begin{rem}
Recently, Stewart \cite{Ste} showed that, given two integers $b>a>1$, we have
$P(a^k-b^k)/k>\exp(\log k/(104\log\log k))$ for $k$ sufficiently large in terms of $\omega(ab)$.
We note that it is easy to show that the largest prime {\it power} divisor of $a^k-b^k$
is $>Ck\vph(k)\log a/\log (k\log a)$ for some absolute constant $C>0$.
\end{rem}

\begin{proof}
Here we shall prove the latter statement.
The latter yields the former after checking each exponent.

We may assume that $k>30$.
Let $\Phi_k(x)=\prod_\zeta (x-\zeta)$ denote the $k$-th cyclotomic polynomial,
where $\zeta$ runs over all primitive $k$-th roots of the unity.
Observing that each primitive $k$-th root of unity appears together with its conjugate
and $(2-\zeta)(2-\bar\zeta)\geq 3$, we have $\Phi_k(2)\geq 3^{\vph(k)/2}$ for any $k$.

By Lemma \ref{lm21}, any prime factor of $\Phi_k(2)$ must be congruent to $1\pmod{k}$.
Moreover, Lehmer\cite{Leh2} shows that if $p<10^8$ and $p^2\mid\Phi_k(2)$, then $p=1093$ or $p=3511$.
Hence, if $P(2^k-1)<10^8$, then the product of primes below $10^8$
congruent to $1\pmod{k}$ must be $\geq 3^{\vph(k)/2}/(1093\cdot 3511)$ and therefore
\begin{equation}
\frac{3^{\vph(k)/2}}{1093\cdot 3511}\leq \prod_{i<10^5/k} (ik+1)<(10^5)^{10^5/k}.
\end{equation}
This yields that
\begin{equation}\label{eq51}
k\varphi(k) \leq 2\cdot 10^8\cdot \log(10^8)/\log 3+\log(1093\cdot 3511).
\end{equation}
If $k\geq 510510$, then $\vph(k)\geq 92160$.  If $144000\leq k<510510$, then $\vph(k)/k\geq 5760/30030$.
In neither case, \eqref{eq51} can hold.

Now we show that $k\leq 500$.  This can be done by computing the product of primes below $10^8$
congruent to $1\pmod{k}$ for each $500<k\leq 144000$.
Indeed, we confirmed that, for each $500<k\leq 144000$,
\begin{equation}
\sum_{p\equiv 1\pmod{k}, p\mid \Phi_k(2)}p<\frac{\vph(k)\log 3}{2}-\log(1093\cdot 3511).
\end{equation}

Finally, for each $k\leq 500$, we factored $2^k-1$ and found all $k$'s satisfying $P(2^k-1)<10^8$,
which are given in the lemma.
\end{proof}

Nextly, we shall determined all prime powers $p^k$ with $3\leq p<10^5$ and $P(p^k-1)<10^5$.

\begin{lem}\label{lm52}
All prime powers $p^k$ such that $3\leq p<10^5$, $k\in \{7, 8, 9, 10, 12,\\ 20\}$ and $P(p^k-1)<10^5$
are given in Table \ref{tbl}.
If $P(3^k-1)<10^5$, then $k\leq 54$.
If $P(5^k-1)<10^5$, then $k\leq 30$.
Furthermore, if $7\leq p<10^5$, $k>10$, $k\neq 12, 20$ and $P(p^k-1)<10^5$, then
$(p, k)=(7, 14), (7, 24), (11, 21), (11, 24),\\ (13, 11), (67, 16)$.
\end{lem}

\begin{table}\label{tbl}
\caption{All prime powers $p^k$ such that $3\leq p<10^5$, $k\in \{7, 8, 9, 10, 12, 20\}$ and $P(p^k-1)<10^5$}\begin{center}
\begin{small}
\begin{tabularx}{\linewidth}{| c | X |}
 \hline
$k$ & $p$ \\
 \hline
$7$ & $2, 3, 5, 7, 11, 19, 59, 79, 269, 359$ \\
$8$ & $3, 5, 7, 11, 13, 17, 19, 31, 37, 41, 43, 47, 59, 67, 79, 83, 107, 127, 137, 149$,
$223, 227, 233, 239, 263, 269, 271, 359, 389, 401, 499, 563, 571, 587, 617, 773$,
$809, 823, 881, 971, 1061, 1091, 1201, 1213, 1319, 1487, 1579, 1637, 1657, 1669$,
$1783, 1907, 2351, 2383, 2399, 2677, 2741, 3109, 3163, 3373, 3631, 3847, 3851$,
$4877, 5167, 6451, 7237, 7699, 8081, 9239, 9397, 9733, 10099, 10181, 10691$,
$11483, 12721, 14051, 14149, 15427, 16067, 16607, 16987, 18979, 19531, 20129$,
$25253, 25633, 27073, 35837, 37783, 41893, 42391, 46327, 46889, 47041, 49253$,
$53831, 57173, 58013, 60101, 62497, 65951, 66541, 69457, 75931, 82241, 82261$,
$84229, 87721, 88339, 88819, 91499, 92333, 95917, 99523$ \\
$9$ & $2, 3, 5, 7, 19, 29, 31, 37, 43, 53, 379, 1019, 63599$ \\
$10$ & $3, 5, 7, 11, 13, 17, 19, 31, 53, 67, 113, 197, 421, 569$ \\
$12$ & $3, 5, 7, 11, 13, 17, 19, 23, 29, 41, 47, 53, 73, 79, 89, 97, 101, 103, 113$,
$137, 139, 197, 251, 271, 307, 367, 389, 397, 401, 421, 467, 479, 487, 907, 1013$,
$1319, 1451, 1627, 1697, 3083, 4027, 22051, 30977, 52889$ \\
$20$ & $3, 5, 7, 13, 17$ \\
 \hline
\end{tabularx}
\end{small}
\end{center}
\end{table}

\begin{proof}
Such prime powers $p^k$ with $k$ \textit{odd prime} can be taken from the table of
Goto and Ohno, who determined all such prime powers with $p<10^8$ and $P(p^k-1)<10^8$.
Their table shows that, if $p, r$ are odd primes such that $r>5, p<10^5$ and $P(p^r-1)<10^5$, then
$(p, r)=(13, 11)$, $p=3$ and $r=11$ or $17$ or
$r=7$ and $p=3, 5, 7, 11, 19, 59, 79, 269$ or $359$.
Furthermore, for any prime $p\geq 3$ and $r_1, r_2\geq 5$, we have $P(p^{r_1 r_2}-1)>10^5$.
We note that, there exist exactly $125$ odd primes $p<10^5$ such that $P(p^5-1)<10^5$.

We confirmed that $P(p^9-1)<10^5$ only for $p=3, 5, 7, 19, 29, 31, 37, 43, 53,\\ 379, 1019$ or $63599$,
among which, $P(p^{27}-1)<10^5$ only for $p=3, 5$ and $P(p^{81}-1)<10^5$ holds for no odd prime $p<10^5$.
Moreover, we confirmed that $P(p^8-1)$ for exactly $116$ odd primes $p<10^5$
and $P(p^{16}-1)$ only for $p=3, 5$ or $67$ among primes below $10^5$.
Moreover, $P(p^{32}-1)>10^5$ for any prime $p<10^5$.

If $p\geq 3, r\geq 7$ and $P(p^{3r}-1)<10^5$, then $(p, r)=(11, 7)$.
If $p$ is an odd prime and $P(p^{15}-1)<10^5$, then $p=3, 5$ and $P(p^{45}-1)>10^5$.
Now we know all prime powers $p^k$ with $k$ odd for which $P(p^k-1)<10^5$.

Finally, we checked each prime $p<10^5$ and exponent $k=2^f r$
with $f\leq 4$ and $r\geq 3$ prime or $r=9, 15, 21, 27$ such that $P(p^r-1)<10^5$.
We found that $p=3, r=14, 18, 24, 30, 54$, $p=5, r=14, 27, 30$, $p=7, r=14, 24$, $p=11, r=21, 24$,
$(p, r)=(13, 11), (17, 24), (67, 16)$ or $(p, r)$ must be in Table \ref{tbl}.
This completes the proof.
\end{proof}

Now our computer search starting from $2^k$ for each $k$ given in Lemma \ref{lm51},
yields that the eleven instances $N_1, N_2, \ldots, N_{10}$ and $N_{12}$
given in the introduction are all ones with $P(N)<10^5$.

Our algorithm can be explained in the following way:
\begin{enumerate}
\item Begin with $p_0=2, e_0=e, i=0, m_i=0$;
\item We factor $p_{m_i}^{e_{m_i}}-1$; For each prime factor $q$ of $p_{m_i}^{e_{m_i}}-1$,
we put $p_j=q$ for some $j$ with $p_j=0$ and relate $p_j$ to $p_{m_i}$ if there is no $j$ with $p_j=q$;
Moreover, we let $f_{m_i, j}$ be the exponent of $p_j$ dividing $p_{m_i}^{e_{m_i}}-1$;
\item For each unmarked $p_j$, we factor $p_j-1$ and mark $p_j$;  For each prime factor $q$ of $p_j-1$,
we put $p_h=q$ for some $h$ with $p_h=0$ and relate $p_h$ to $p_{m_i}$ if there is no $h$ with $p_h=q$;
Moreover, we let $f_{j, h}$ be the exponent of $p_h$ dividing $p_j-1$;
Until all $p_j$'s are marked, we repeat this step;
\item If there is some $j$ with $\sum_h f_{j, h}>e_j>0$, then jump to Step 4A;  Otherwise,
if there is some $j$ with $p_j>0$ and $e_j=0$, jump to Step 4B;  Otherwise jump to Step 4C;
\item[(4A)] We clear all $p_j$'s, all $f_{j, h}$'s for all prime $p_j$ related to $p_{m_i}$;
Jump to Step 5;
\item[(4B)] If there is some $j$ with $e_j=0$, then we set $m_{i+1}$ to be one of such $j$('s);
We set $e_{m_{i+1}}=1$;  We set $i=i+1$;  Jump to Step 2;
\item[(4C)] We output $N=\prod p_j^{e_j}$ where $j$ runs over all $j$ with $p_j\neq 0$;
Jump to Step 5;
\item[(5)] If $p_{m_i}>2$, then jump to Step 5A;  Otherwise, jump to Step 5B;
\item[(5A)] We set $e_{m_i}=e_{m_i}+1$;  If $p_{m_i}^{e_{m_i}}<10^8$, then jump to Step 2;  Otherwise,
We clear $p_{m_i}, e_{m_i}$;  We clear $m_i$;  We set $i=i-1$;  Jump to Step 2;
\item[(5B)] We set $e_{m_i}$ to the next member of the current value of $e_{m_i}$ in the set given in
Lemma \ref{lm52};  If there is no more member, then we terminate;
\end{enumerate}

We illustrate our proof in the case $e=32$.

If $2^{32}\mid\mid N$, then $3\cdot 5\cdot 17\cdot 257\cdot 65537=2^{32}-1$ divides $N$.
In particular, $3$ must divide $N$.
By Lemma \ref{lm52}, we must have $3^k\mid\mid N$ with $1\leq k\leq 12, 14\leq k\leq 18$ or $k\in\{20, 22, 24, 27, 28, 30, 34, 48, 54\}$.

If $k$ is even, then we must have $2^{33}\mid\vph^*(3^k\cdot 5\cdot 17\cdot 257\cdot 65537)\mid N$,
contrary to the assumption that $2^{32}\mid\mid N$.
If $3$ divides $k$, then $13\mid (3^3-1)\mid N$ and
$2^{33}\mid\vph^*(3\cdot 5\cdot 13\cdot 17\cdot 257\cdot 65537)\mid N$, a contradiction again.

If $5$ divides $k$, then $11^2\mid (3^5-1)\mid N$ and $5^2\mid \vph^*(2^{32}\cdot 11)\mid N$.
Hence, we must have $5^l\mid\mid N$ with $2\leq l\leq 10$, $l=12, 14, 15, 16, 18, 20$ or $30$.
If $l$ is even, then $2^{33}\mid\vph^*(3\cdot 5^2\cdot 17\cdot 257\cdot 65537)\mid N$,
a contradiction.  If $3$ divides $l$, then $31\mid (5^3-1)$ divides $N$ and
$2^{33}\mid \vph^*(3\cdot 5\cdot 11\cdot 17\cdot 31\cdot 257\cdot 65537)\mid N$,
a contradiction.  $l=5, 7$ lead to a similar contradiction and
$5$ can never divide $k$.
$p\mid k$ with $p=7, 11, 13, 17$ also lead to a similar contradiction
and we must have $k=1$.

Proceeding similarly, we must have $3\cdot 5\cdot 17\cdot 257\cdot 65537\mid\mid N$
and $N=2^{32}\cdot 5\cdot 17\cdot 257\cdot 65537\mid\mid N$.

Our procedure worked for the other exponents given in Lemma \ref{lm51}
and yielded only twelve instances given in the Introduction.
Some exponents required several hours.
This proves the former part of Theorem \ref{thm3}.

Now we shall prove that $N$ must be divisible by a prime power $>10^8$.
$e$ must belong to the set given in Lemma \ref{lm51}.
As in the previous section, we show that $2^e\mid\mid N$ cannot occur
unless $N$ is one of twelve instances given above
for each $e$ in this set.

For example, we show that $e\neq 210$ in the following way:

$2^{210}-1=3^2\cdot 7^2\cdot 11\cdot 31\cdot 43\cdot 71\cdot 127\cdot 151\cdot 211\cdot 281\cdot 331\cdot 337\cdot 5419\cdot 29191\cdot 86171\cdot 106681\cdot 122921\cdot 152041\cdot 664441\cdot 1564921$ must divide $N$.
Hence, $(3-1)(7-1)(11-1)\ldots(1564921-1)=2^{35}\cdot 3^{20}\cdot 5^{15}\cdot 7^{15}\cdot 11\cdot 23\cdot 43\cdot 113\cdot 127\cdot 139\cdot 181\cdot 439\cdot 1231$ must divide $N$.
Thus $3^{20}>10^8$ must divide $N$, which is contradiction.

Our procedure terminated for all exponents given in Lemma \ref{lm51}
and yielded only twelve instances given in the Introduction.
This completes the proof of Theorem \ref{thm3}.

\section{A product of consecutive primes}

The purpose of this section is to prove the Theorem \ref{thm5}.

In this section, $\sum^\prime$ denotes the sum over primes in a given range.
We need a Brun-Titchmarsh type theorem in the following form.

\begin{lem}\label{lm71}
During the statement and the proof of this lemma, let $c$ denote an effectively computable absolute constant which may take different value at each occurrence.
Moreover, let $Q_1=\log^{9/2} x$ and $Q=x^{1/2}/\log^{9/2} x$.
Then, we have
\begin{equation}
\sideset{}{^\prime}\sum_{q<Q} \max_{y\leq x} \abs{\pi(y; q, 1)-\frac{\pi(y)}{\vph(q)}}<\frac{cx}{\log^2 x}
\end{equation}
for $x>0$.
\end{lem}

\begin{proof}
By Theorem 1.2 of \cite{AH}, we have
\begin{equation}
\begin{split}
& \sum_{q\leq Q}\frac{q}{\vph(q)}\sideset{}{^*}\sum_{\chi\pmod{q}} \max_{y\leq x}\abs{\psi(y, \chi)} \\
& <c_0(4x+2x^\frac{1}{2} Q^2+6x^\frac{2}{3} Q^\frac{3}{2}+5x^\frac{5}{6} Q)\log^\frac{7}{2} x,
\end{split}
\end{equation}
where $c_0=48.83236\cdots$ and $\sum^*_{\chi\pmod{q}}$ denotes
the sum over all primitive characters $\chi\pmod{q}$.

Dividing the sum into intervals of the form $(2^k Q_1, 2^{k+1} Q_1]$
$(k=0, 1, \ldots,\\ \floor{\log(Q/Q_1)/\log 2}$, we have
\begin{equation}
\begin{split}
& \sum_{Q_1<q\leq Q}\frac{1}{\vph(q)}\sideset{}{^*}\sum_{\chi\pmod{q}} \max_{y\leq x}\abs{\psi(y, \chi)} \\
& <c\left(\frac{x}{Q_1}+x^\frac{1}{2} Q+x^\frac{2}{3} Q^\frac{1}{2}+5x^\frac{5}{6}\log x\right)\log^\frac{7}{2} x.
\end{split}
\end{equation}
Limiting $q$ in the sum to primes, we obtain
\begin{equation}
\begin{split}
& \sideset{}{^\prime}\sum_{Q_1<q\leq Q} \max_{y\leq x}\abs{\psi(y; q, 1)-\frac{\psi(y)}{\vph(q)}-1} \\
& <c\left(\frac{x}{Q_1}+x^\frac{1}{2} Q+x^\frac{2}{3} Q^\frac{1}{2}+5x^\frac{5}{6}\log x\right)\log^\frac{7}{2} x
\end{split}
\end{equation}
and, recalling that $Q_1=\log^{9/2} x$ and $Q=x^{1/2}/\log^{9/2} x$,
\begin{equation}\label{eq71}
\sideset{}{^\prime}\sum_{Q_1<q\leq Q} \max_{y\leq x}\abs{\psi(y; q, 1)-\frac{\psi(y)}{\vph(q)}}
<c\frac{x}{\log x}.
\end{equation}

It is implicit in the proof of Theorem 4 of Chen and Wang \cite{ChW} that
for a Dirichlet character $\chi$ modulo $q\leq \log^{9/2} x$ and $x\geq \exp(\exp(9.7))$, we have
\begin{equation}\label{eq72}
\abs{\psi(x, \chi)-E_0 x}\leq \frac{0.022x}{\log^{7.5} x}+E_0 \log x+E_1\frac{x^\beta}{\beta},
\end{equation}
where $E_0=1$ if $\chi$ is principal and $E_0=0$ otherwise
and $\beta$ denotes a real zero of $L(s, \chi)$ greater than $1-0.1077/\log q$ and $E_1=1$ if it exists
and $E_1=0$ otherwise (For more general results, see the author's recent paper \cite{Ymd}).
Moreover, $E_1=1$ occurs for at most one character among all Dirichlet characters modulo $k$.

From Kadiri \cite{Kad}, we know that, there exists at most one modulus $q_0\leq Q_1$
such that a Dirichlet $L$-function $L(s, \chi)$ has a real zero $s=\beta>1-1/4.0904\log Q_1$ for some character $\chi\pmod{q_0}$.
We shall call the modulus $q_0$ exceptional if it exists and other moduli $\leq Q_1$ nonexceptional.
Now, it immediately follows from \eqref{eq72} that, if a prime $q\leq Q_1$ is nonexceptional, then
\begin{equation}
\max_{y\leq x}\abs{\psi(y; q, 1)-\frac{\psi(y)}{\vph(q)}}<\frac{cx}{\log^{7.5} x}+2x^{1-1/4.0904\log Q_1}<\frac{c_1 x}{\log^{7.5} x}
\end{equation}
for another effectively computable constant $c_1$.
Using Theorem 3 of \cite{LW} stating that $\beta\leq 1-\pi/0.4923q_0^{1/2}\log^2 q_0$,
we have $x^\beta/q_0<cx(\log\log x)^4/\log^2 x$ and therefore
\begin{equation}
\max_{y\leq x}\abs{\psi(y; q_0, 1)-\frac{\psi(y)}{\vph(q_0)}}<\frac{cx}{\log^{7.5} x}+\frac{c^\prime x(\log\log x)^4}{\log^2 x}<\frac{c_2 x}{\log x}
\end{equation}
for some effectively computable constants $c^\prime$ and $c_2$.
Hence, we obtain
\begin{equation}
\sideset{}{^\prime}\sum_{q\leq Q_1} \max_{y\leq x}\abs{\psi(y; q, 1)-\frac{\psi(y)}{\vph(q)}}<\frac{cx}{\log x}.
\end{equation}
Combining with \eqref{eq71}, we have
\begin{equation}\label{eq71a}
\sideset{}{^\prime}\sum_{q<Q} \max_{y\leq x} \abs{\psi(y; q, 1)-\frac{\psi(y)}{\vph(q)}}<\frac{cx}{\log x}.
\end{equation}

Let
\begin{equation}
\Pi(y; q, a)=\sum_{n\leq y, n\equiv a\pmod{q}}\frac{\Lm(n)}{\log n}=\sum_{p^k\leq y, p^k\equiv a\pmod{q}}\frac{1}{k}
\end{equation}
and put $x_1=x/\log^2 x$.
Moreover, we write $E_f(y; q, a)$ for the error term $\abs{f(y; q, a)-f(y)/\vph(q)}$ for
arithmetic functions $f=\pi, \Pi, \theta, \psi$.
Using partial summation, we have
\begin{equation}\label{eq71b}
\begin{split}
E_\Pi(x, q, 1)-& E_\Pi(x_1, q, 1) \\
= & \frac{E_\psi(x, q, 1)}{\log x}-\frac{E_\psi(x_1, q, 1)}{\log x_1}-\int_{x_1}^x \frac{E_\psi(y, q, 1)}{y\log^2 y}dy \\
\leq & \frac{\abs{E_\psi(x, q, 1)}}{\log x}+\frac{\abs{E_\psi(x_1, q, 1)}}{\log x_1}+\frac{\max_{y\leq x}\abs{E_\psi(y, q, 1)}}{\log x_1} \\
\leq & \frac{2\max_{y\leq x}\abs{E_\psi(y, q, 1)}}{\log x_1}+\frac{cx_1}{\log x_1}.
\end{split}
\end{equation}
In Section 8 of \cite{AH}, it is shown that
\begin{equation}\label{eq71c}
\abs{E_\Pi(x_1; q, a)-E_\pi(x_1; q, a)}\leq 2x_1^\frac{1}{2}\leq 2x^\frac{1}{2}
\end{equation}
for $y\leq x$.

Finally, observing that $Q\leq x_1^{1/2}$, the Brun-Titchmarsh inequality given in \cite{MV} gives
\begin{equation}\label{eq71d}
\sideset{}{^\prime}\sum_{q\leq Q}\abs{E_{\pi}(x_1, q, 1)}\leq \sideset{}{^\prime}\sum_{q\leq Q}\frac{4x_1}{q\log x_1}\leq \frac{c_3 x_1\log\log x}{\log x_1}\leq \frac{cx\log\log x}{\log^3 x_1},
\end{equation}
where $c_3$ denotes some effectively computable constant.

Now the Lemma follows from \eqref{eq71a}, \eqref{eq71b}, \eqref{eq71c} and \eqref{eq71d}.
\end{proof}

We shall prove Theorem \ref{thm5}.
Let $0<\alpha_1, \alpha_2<1/2$ be two constants.
Hereafter, $c$ and $c^\prime$ denote some effectively computable constants
which may take different value at each occurrence and depend only on $\alpha_1$ and $\alpha_2$.
Assume that $\vph^*(N)$ divides $N$ and $N$ is divisible by exactly all primes below $x$.
Moreover, we may assume that $x$ is so large that $\abs{\pi(x)-x/\log x}<0.001x/\log x$.
For any prime $q$, $q^{\pi(x; q, 1)}\mid \prod_{p\leq x, p\equiv 1\pmod{q}} (p-1)\mid \vph^*(N)\mid N$.
Hence,
\begin{equation}
\sideset{}{^\prime}\prod_{q\leq x}\left(\frac{q^{\pi(x; q, 1)}}{q^{\pi(x; q, 1)}-1}\right)\geq \prod_{q^f\mid\mid N}\left(\frac{q^f}{q^f-1}\right)=h(N)\geq 2
\end{equation}
and therefore
\begin{equation}\label{eq70}
\begin{split}
\sideset{}{^\prime}\sum_{q\leq x}\frac{1}{q^{\pi(x; q, 1)}}
> & \sideset{}{^\prime}\sum_{q\leq x}\log\left(\frac{q^{\pi(x; q, 1)}}{q^{\pi(x; q, 1)}-1}\right)-\sideset{}{^\prime}\sum_{q\leq x}\frac{1}{q^{2\pi(x; q, 1)}} \\
> & \log 2-\frac{c}{\log x},
\end{split}
\end{equation}
where the last inequality follows observing that $q^{2\pi(x; q, 1)}>c\log^2 x$ for $q<\log x$ by \eqref{eq72}.

Let $U$ be the set of primes $q$ such that $\pi(x; q, 1)<0.001x/(q\log x)$.
We let
\begin{equation}
L=\sum_{p\leq x} \log (p-1)=\sum_{\substack{p, q\leq x,\\ p, q\text{: prime},\\ q\mid (p-1)}} \log q=\sideset{}{^\prime}\sum_{q<x/2} \pi(x; q, 1)\log q
\end{equation}
and divide $L$ into
\begin{equation}
L_1=\sideset{}{^\prime}\sum_{q\leq x^{1/2-\alpha_1}} \pi(x; q, 1)\log q,
\end{equation}
\begin{equation}
L_{2, 1}=\sum_{x^{1/2-\alpha_1}\leq q\leq x^{1-\alpha_2}, q\in U}\pi(x; q, 1)\log q,
\end{equation}
\begin{equation}
L_{2, 2}=\sum_{x^{1/2-\alpha_1}\leq q\leq x^{1-\alpha_2}, q\not\in U}\pi(x; q, 1)\log q
\end{equation}
and
\begin{equation}
L_3=\sideset{}{^\prime}\sum_{x^{1-\alpha_2}\leq q<x/2} \pi(x; q, 1)\log q.
\end{equation}

We can easily see that
\begin{equation}\label{eq731}
L=\sum_{p\leq x} \log (p-1)>\theta(x)-\sum_{p\leq x}\frac{2}{p}>0.999x.
\end{equation}

Since $x^{\alpha_1}>\log^5 x$, we can apply Lemma \ref{lm71} to obtain
\begin{equation}\label{eq732}
\begin{split}
L_1\leq & \pi(x)\sideset{}{^\prime}\sum_{q\leq Q} \frac{\log q}{q-1}+(\alpha_1\log x)\sideset{}{^\prime}\sum_{q\leq Q} \abs{\pi(x; q, 1)-\frac{\pi(x)}{q-1}} \\
\leq & 1.001\pi(x)\log Q+\frac{c\alpha_1 x}{\log x} \\
\leq & 0.501x.
\end{split}
\end{equation}

Using the inequality $\abs{\sum_{q<z}^\prime (\log q)/q-E}<1/2\log z$ with $E=-1.33258\cdots$ for $z\geq 319$ (see Theorem 6 of \cite{RS}), we have
\begin{equation}\label{eq733}
L_{2, 1}\leq \frac{0.001x}{\log x}\sideset{}{^\prime}\sum_{x^{1/2-\alpha_1}\leq q\leq x^{1-\alpha_2}}\frac{\log q}{q}\leq 0.001x
\end{equation}
for sufficiently large $x$.

Let $S(y, a)$ be the number of integers $q\leq y$ such that both $q$ and $aq+1$ are prime.
We observe that
\begin{equation}\label{eq75}
\sideset{}{^\prime}\sum_{x^{1-\alpha_2}\leq q<x/2} \pi(x; q, 1)=\sum_{\substack{x^{1-\alpha_2}\leq q<x/2, q\text{: prime},\\ p=aq+1\leq x, p\text{: prime}}}1
\leq \sum_{2\leq a\leq x^{\alpha_2}}S(x/a, a).
\end{equation}
Proceeding as in Subsection 2.3.2 of \cite{Gre} immediately yields that
\begin{equation}
S(x/a, a)<\frac{c\psi(a)(x/a)}{a\log^2 (x/a)}<\frac{c\psi(a)x}{(1-\alpha_2)^2 a^2 \log^2 x}
\end{equation}
for each $a$, where we note that $c$ does not depend on $a$ and $\psi(a)=a\prod_{p\mid a}(1+1/p)$
now denotes the Dedekind $\psi$-function, not the second Chebyshev function.
Thus the last sum in \eqref{eq75} is at most
\begin{equation}
\frac{cx}{(1-\alpha_2)^2 \log^2 x}\sum_{2\leq a\leq x^{\alpha_2}}\frac{\psi(a)}{a^2}
<\frac{(c+0.0001) c^\prime \alpha_2 x}{(1-\alpha_2)^2 \log x}
\end{equation}
and therefore
\begin{equation}\label{eq734}
L_3<\frac{(c+0.0001)c^\prime \alpha_2 x}{(1-\alpha_2)^2}.
\end{equation}

By \eqref{eq731}, \eqref{eq732}, \eqref{eq733} and \eqref{eq734},
there exists an absolute and effectively computable constant $\delta>0$ such that,
if $\alpha_2<\delta$, then
\begin{equation}\label{eq76}
L_{2, 2}>L-(L_1+L_{2, 1}+L_3)>0.497x
\end{equation}
for sufficiently large $x$.

On the other hand, using the Brun-Titchmarsh theorem again,
\begin{equation}
\begin{split}
L_{2, 2}< & \sum_{\substack{q\not\in U,\\ x^{1/2-\alpha_1}\leq q\leq x^{1-\alpha_2}}}\frac{2x\log q}{(q-1)\log (x/q)} \\
< & \frac{2.0001(1-\alpha_2)x}{\alpha_2}\sum_{\substack{q\not\in U,\\ x^{1/2-\alpha_1}\leq q\leq x^{1-\alpha_2}}}\frac{1}{q}
\end{split}
\end{equation}
and
\begin{equation}\label{eq77}
\begin{split}
\sum_{\substack{q\not\in U,\\ x^{1/2-\alpha_1}\leq q\leq x^{1-\alpha_2}}}\frac{1}{q}>\frac{0.248\alpha_2}{1-\alpha_2}.
\end{split}
\end{equation}

Since $\abs{E_\pi(x; q, 1)}>0.998x/(q\log x)$ for any prime $q<x^{1/2-\alpha_1}$ in $U$,
Lemma \ref{lm71} gives
\begin{equation}
\sum_{q\leq x^{1/2-\alpha_1}, q\in U}\frac{1}{q}<\frac{1.01\log x}{x}\sum_{q\leq x^{1/2-\alpha_1}, q\in U}E_\pi(x; q, 1)<\frac{c}{\log x}.
\end{equation}
Thus \eqref{eq70} gives
\begin{equation}
\begin{split}
& \sum_{\substack{q\in U,\\ x^{1/2-\alpha_1}\leq q\leq x^{1-\alpha_2}}}\frac{1}{q}\geq \sum_{\substack{q\in U,\\ x^{1/2-\alpha_1}\leq q\leq x^{1-\alpha_2}}}\frac{1}{q^{\pi(x; q, 1)}} \\
> & \log 2-\frac{c}{\log x}-\sideset{}{^\prime}\sum_{x^{1-\alpha_2}<q<x} \frac{1}{q}-\sum_{\substack{q\not\in U,\\ q\leq x^{1-\alpha_2}}}\frac{1}{q^{0.001x/(q\log x)}} \\
> & \log 2+\log(1-\alpha_2)-\frac{c^\prime}{\log x}.
\end{split}
\end{equation}
Hence,
\begin{equation}
\begin{split}
\sum_{\substack{q\not\in U,\\ x^{1/2-\alpha_1}\leq q\leq x^{1-\alpha_2}}}\frac{1}{q}
< & \log\frac{1-\alpha_2}{\frac{1}{2}-\alpha_1}+\frac{c}{\log x}-\sum_{\substack{q\in U,\\ x^{1/2-\alpha_1}\leq q\leq x^{1-\alpha_2}}}\frac{1}{q} \\
< & \log\frac{1-\alpha_2}{\frac{1}{2}-\alpha_1}-\log 2-\log (1-\alpha_2)+\frac{c}{\log x} \\
= & \log\frac{1}{1-2\alpha_1}+\frac{c}{\log x}.
\end{split}
\end{equation}
Now \eqref{eq77} implies
\begin{equation}
\frac{0.248\alpha_2}{1-\alpha_2}<\log\frac{1}{1-2\alpha_1}+\frac{c}{\log x}.
\end{equation}

Taking $0<\alpha_1<1$ so that
\begin{equation}
\frac{(1-\alpha_2)x}{\alpha_2}\log\frac{1}{1-2\alpha_1}<0.248,
\end{equation}
we have $c/\log x>c^\prime$ for sufficiently large $x$, which implies $x<c$.
This proves Theorem \ref{thm5}.

\section{Acknowledgement}

The author is gratefully thankful for the referee's helpful comments and
informations on some literatures including \cite{FKL}.

config: arxiv


TITLE: The non-empty intersection of two open discs contains an open disc.
QUESTION [2 upvotes]: Is the following argument correct?
Let $D_1$ and $D_2$ be any open discs in $\mathbf{R}^2$ with $D_1\cap D_2\neq\varnothing$. If $(a,b)$ is any point in $D_1\cap D_2$, show that ther exists an open disc $D_{(a,b)}$ with center $(a,b)$ such that $D_{(a,b)}\subset D_1\cap D_2$.

Proof. Let $D_1,D_2$ denote two arbitrary discs in $\mathbf{R}^2$ having $\alpha = (a_1,b_1)$ and $\beta = (a_2,b_2)$ as there centers
  and radii $r_1$ and $r_2$ respectively, that is \begin{align*} D_1 =
 \{(x,y)\in\mathbf{R}^2:(x-a_1)^2+(y-b_1)^2<r_1^2\}\\ D_2 =
 \{(x,y)\in\mathbf{R}^2:(x-a_2)^2+(y-b_2)^2<r_2^2\} \end{align*} Let
  $\gamma = (a,b)\in D_1\cap D_2$. We define the disc $D_{(a,b)}$
  similar to $D_1,D_2$ but having center $(a,b)$ and radius $r =
 \min\{\frac{r_1-d_1}{8},\frac{r_2-d_2}{8}\}$, where $d_1$ and $d_2$
  are defined as follows
  \begin{align*} d(\gamma,(\alpha,\gamma)) = d_1
 = \sqrt{(a-a_1)^2+(b-b_1)^2}\\ d(\gamma,(\beta,\gamma)) = d_2 = \sqrt{(a-a_2)^2+(b-b_2)^2}
\end{align*} Now let $(x,y)$ be an
  arbitrary point inside the disc $D_{(a,b)}$, appealing to the triangle
  inequality then yields
  \begin{align*} \sqrt{(x-a_1)^2+(y-b_1)^2}
 &=d(\alpha,(x,y))\leq d(\alpha,\gamma)+d(\gamma,(x,y))\\
 &=d_1+r<d_1+\frac{r_1-d_1}{8}=\frac{r_1}{8}<r_1
\end{align*}
\begin{align*} \sqrt{(x-a_2)^2+(y-b_2)^2} &= d(\beta,(x,y))\leq
 d(\beta,\gamma)+d(\gamma,(x,y))\\
 &=d_2+r<d_2+\frac{r_2-d_2}{8}=\frac{r_2}{8}<r_2 \end{align*} Since our
  choice of $(x,y)$ was arbitrary, we have $D_{(a,b)}\subset D_1\cap D_2$.
$\blacksquare$

REPLY [1 votes]: The proof is correct and mostly good, there are only minor mistakes and typos.
You write
$$
d_1+r<d_1+\frac{r_1-d_1}{8}=\frac{r_1}{8}<r_1
$$
and here the equality is false.
But this can be easily fixed, for example (using $d_1< r_1$)
$$
d_1+r<d_1+\frac{r_1-d_1}{8}=\frac{r_1}{4}<r_1 .
$$
The same issue exist with $d_2,r_2$ below.
In my opinion it would be nicer if you choose
$$
 r= \min\{r_1-d_1,r_2-d_2\},
$$
then your proof would still work,
but your choice is not wrong!
Small typos: You probably mean to write
$d(\gamma,\alpha)$ or $d(\gamma,(a_1,b_1))$ instead of
$d(\gamma,(\alpha,\gamma))$ and
$d(\gamma,\beta)$ or $d(\gamma,(a_2,b_2))$ instead of
$d(\gamma,(\beta,\gamma))$.


TITLE: How to write $\zeta(s)^3$ as a function of the generalized Mobius function?
QUESTION [1 upvotes]: Popovici (1963) (see link), created a way to extend the Mobius function, $\mu(n)$, to the complex plane.
The Mobius $\mu(n)$ function is such that:
$\frac{1}{\zeta{(s)}}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}$
With Popovici's extension, $\mu_k(n)$, we get a generalization:
$\frac{1}{\zeta{(s)^k}}=\sum_{n=1}^{\infty}\frac{\mu_k(n)}{n^s}$, where $μ_k=μ∗...∗μ$ taken $k$ times on a Dirichlet convolution is Popovici’s function.
My question is..., how about $\zeta{(s)}^3$? Does his generalization of $\mu(n)$ work in this case?
And what the function $\mu_{-3}(n)$ would be in this case?
Popovici's generalized Mobius function

REPLY [1 votes]: Popovici's generalised Möbius function $\mu_k$ is multiplicative, so is defined by its values on $p^a$ with $p\in\Bbb P,\,a\in\Bbb Z_{\ge 0}$ viz. $$\mu_k(p^a)=(-1)^a\binom{k}{a}=\frac{(-1)^a\prod_{j=0}^{a-1}(k-j)}{a!}.$$The last expression can be evaluated for $k=-3$, viz.$$\mu_{-3}(p^a)=\frac{(a+2)!}{a!2}=\frac{(a+1)(a+2)}{2}.$$Indeed, the binomial theorem implies $$(1-p^{-s})^{-3}=\sum_{a=0}^\infty\frac{(a+1)(a+2)}{2}p^{-as},$$to which we can apply $\prod_{p\in\Bbb P}$ to prove $(\zeta(s))^3=\sum_{n\ge 1}\mu_{-3}(n)n^{-s}$.


TITLE: Find the number of different lines containing $8$ points
QUESTION [2 upvotes]: A $10\times 10\times 10$ grid of points consists of all points in space of the form $(i,j,k)$ where $i,j,k$ are integers between $1$ and $10$ inclusive. Find the number of distinct lines containing exactly $8$ of these lattice points.

I don't understand why the following solution considers all possible lines. Can someone prove this?

Consider a pair of opposite faces of the cube. There are $4$ lines $l_i,1\leq i\leq 4$ with $8$ collinear points on the top face (for example one line contains the points $(i,i-2, 10)$ for $3\leq i\leq 10$). For each of these lines, draw a rectangular plane containing one of the $l_i$'s that is perpendicular to the top face. Each rectangular plane will contain $16$ lines containing $8$ points, $10$ of which are parallel to the top face and $6$ of which run diagonally through the rectangular plane. We can repeat this for each of the $3$ pairs of opposite faces to get $3\cdot 16\cdot 4=192$ possibilities.
The only possibilities that are overcounted are those lines passing through the diagonals of rectangular planes. There are $4$ rectangular planes perpendicular to one pair of opposite faces whose intersection with the top face of the cube contains $8$ points, so $4\cdot 6$ lines were overcounted, giving a total of $192-24 = 168.$

Also, which lines are overcounted exactly? I think one of the diagonals mentioned by the solution contains the points $(i,i-2, 11-i)$ for $3\leq i\leq 10$.

Source: https://artofproblemsolving.com/wiki/index.php/2017_AIME_II_Problems/Problem_14 (refer to solution 2)

REPLY [0 votes]: I don't understand why the following solution considers all possible lines. Can someone prove this?

If a line containing exactly $8$ lattice points is parallel to a face of the cube, then the line is parallel to a line containing exactly $8$ lattice points on the face, so every such line is included in the lines written in the given solution.
If a line containing exactly $8$ lattice points is not parallel to the faces, then see case 1 in solution 1 saying

$Case \textrm{ } 1:$ The lines are not parallel to the facesA line through the point $(a,b,c)$ must contain $(a \pm 1, b \pm 1, c \pm 1)$ on it as well, as otherwise, the line would not pass through more than 5 points. This corresponds to the 4 diagonals of the cube.

This explains that every such line is on a plane which is perpendicular to a face of the cube, and so every such line is included in the lines written in the given solution.

Also, which lines are overcounted exactly? I think one of the diagonals mentioned by the solution contains the points $(i,i-2, 11-i)$ for $3\leq i\leq 10$.

For each rectangular plane, there are four lines passing through a corner of the rectangular plane, and there are two lines which do not pass through any corner. The lines passing through a corner are overcounted, and the lines which do not pass through any corner are not overcounted. Therefore, there are $\dfrac{3\times 4\times 4}{2}=24$ overcounted lines.
The following $24$ lines are overcounted :

The line containing $(i,i-2, 11-i)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $x-y-2=0$ and on the plane $y+z-9=0$.

The line containing $(i,i-2, 13-i)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $x-y-2=0$ and on the plane $x+z-13=0$.

The line containing $(i,i-2, i)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $x-y-2=0$ and on the plane $y-z+2=0$.

The line containing $(i,i-2, i-2)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $x-y-2=0$ and on the plane $x-z-2=0$.

The line containing $(i,9-i,11-i)\ (1\leq i\leq 8)$ is overcounted since the line is both on the plane $x+y-9=0$ and on the plane $y-z+2=0$.

The line containing $(i,9-i,9-i)\ (1\leq i\leq 8)$ is overcounted since the line is both on the plane $x+y-9=0$ and on the plane $x+z-9=0$.

The line containing $(i,9-i,i+2)\ (1\leq i\leq 8)$ is overcounted since the line is both on the plane $x+y-9=0$ and on the plane $x-z+2=0$.

The line containing $(i,9-i,i)\ (1\leq i\leq 8)$ is overcounted since the line is both on the plane $x+y-9=0$ and on the plane $y+z-9=0$.

The line containing $(i,i+2,11-i)\ (1\leq i\leq 8)$ is overcounted since the line is both on the plane $x-y+2=0$ and on the plane $y+z-13=0$.

The line containing $(i,i+2,9-i)\ (1\leq i\leq 8)$ is overcounted since the line is both on the plane $x-y+2=0$ and on the plane $x+z-9=0$.

The line containing $(i,i+2,i+2)\ (1\leq i\leq 8)$ is overcounted since the line is both on the plane $x-y+2=0$ and on the plane $x-z+2=0$.

The line containing $(i,i+2,i)\ (1\leq i\leq 8)$ is overcounted since the line is both on the plane $x-y+2=0$ and on the plane $y-z-2=0$.

The line containing $(i,13-i,13-i)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $x+y-13=0$ and on the plane $x+z-13=0$.

The line containing $(i,13-i,11-i)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $x+y-13=0$ and on the plane $y-z-2=0$.

The line containing $(i,13-i,i)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $x+y-13=0$ and on the plane $y+z-13=0$.

The line containing $(i,13-i,i-2)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $x+y-13=0$ and on the plane $x-z-2=0$.

The line containing $(i-2,13-i,11-i)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $y-z-2=0$ and on the plane $x+z-9=0$.

The line containing $(i,i,i-2)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $y-z-2=0$ and on the plane $x-z-2=0$.

The line containing $(i,11-i,i-2)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $y+z-9=0$ and on the plane $x-z-2=0$.

The line containing $(i-2,i-2,11-i)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $y+z-9=0$ and on the plane $x+z-9=0$.

The line containing $(i,11-i,13-i)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $y-z+2=0$ and on the plane $x+z-13=0$.

The line containing $(i-2,i-2,i)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $y-z+2=0$ and on the plane $x-z+2=0$.

The line containing $(i-2,13-i,i)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $y+z-13=0$ and on the plane $x-z+2=0$.

The line containing $(i,i,13-i)\ (3\leq i\leq 10)$ is overcounted since the line is both on the plane $y+z-13=0$ and on the plane $x+z-13=0$.


TITLE: Why does mini quadcopter doesn't maintain altitude?
QUESTION [4 upvotes]: I have a mini quadcopter controlled by remote controller. I have set the motors (e.g 50% thrust) it will remain at altitude at 10 cm~ but once I increase to 80% and then follow by decrease to 50% I should remain at a certain altitude (e.g 2 meters) but it will then drop back down to 10cm~. My question is, why does it drop back? If 50% of the motors can lift the quadcoter, can't It do this at any altitude (I do understand air gets thinner above certain altitude but we talking only a few meters off ground)
From what I know, motor output is Normal force which is pushing it up which is 'constant' and gravity is force pulling it down which is also constant.

REPLY [6 votes]: Great question, I've seen this happen before but never thought about why before.
The reason is Ground effect.
50% thrust isn't enough to hover in free space, but when it gets close to the ground then there is increased lift which acts as to stabilize to altitude.
The pressure is slightly higher under the wing (rotor) when its near the ground which mean extra lift.


TITLE: Why are energy and mass the same?
QUESTION [1 upvotes]: I am aware that energy mass are inter-convertible using the famous $$E=mc^2$$
But why is it that energy and mass are basically the same thing that takes different forms?

I am looking for a theoretical answer rather than answer based on formula.
I am also changing matter to mass as pointed out in the comments.

REPLY [0 votes]: One possible way to look at it is that we tend (anthropomorphism) to regard processes as being fundamentally different from objects and to class physical phenomena into these two categories (objects, processes), as if these two categories were inherently mutually exclusive.
Quantum mechanics teaches us that the particle/wave duality is erroneous; electrons, photons have both characteristics of a particle and a wave.
Another example of physical phenomena which appear at first to be objects but prove to behave in a more dynamic fashion on closer look is given by the example of protons or neutrons.
Most of the mass of the proton (or neutrons) does not result from the sum of the masses of the quarks but from the average kinetic energy of these particles. So basically, most of the mass of the p/n is kinetic energy; therefore, most of the mass of nuclei is kinetic energy, and by extension, most of the mass of atoms (or matter more generally) is kinetic energy.
https://www.forbes.com/sites/startswithabang/2016/08/03/where-does-the-mass-of-a-proton-come-from/?sh=791273ea2e1d
Mass, therefore, is essentially energy; stated otherwise, mass is just an anthropomorphism, a macroscopic designation, just a word for a more complex, dynamic phenomenon.
To put it another way, one couldn't convert mass into energy and conversely, if at some level these two facets of matter were not one and the same thing.


TITLE: Probability that exists at least an edge in the configuration model
QUESTION [6 upvotes]: In this period, I am studying some topics on random networks to understand the modularity optimization used in community detection. In particular, I am trying to understand a model called configuration model, which represents a random graph with a given degree sequence.
You can find some descriptions on the topic here:
http://tuvalu.santafe.edu/~aaronc/courses/5352/fall2013/csci5352_2013_L11.pdf
http://homepage.cs.uiowa.edu/~sriram/196/spring12/lectureNotes/Lecture11.pdf
Now, my problem is that I don't understand why the probability $p_{ij}$ that exists at least an edge between two vertices $i$ and $j$ is evaluated as $p_{ij} = \frac{k_i k_j}{2m}$. I suppose that they are evaluating the probability as the edges were independent. Am i right? Are really the edges independent or is it supposed that they are in a large graph? I don't understand that very well because I have noticed that in some graphs the given probability is not a value between 0 and 1.
It's confusing the fact that I have read on some articles that they call it as the "expected number of edges between two vertex" and on other ones that they call it as the "probability that exists at least an edge between two vertex", too. Is it the same thing?
Could you help me understand the given probability?
Thank you.

REPLY [2 votes]: There is no independence here.  The model assumes that the $2m$ "edge stubs" from the vertices are matched at random: you pick a pair of unconnected stubs, connect them, and repeat until all are connected.  If vertex $i$ has $k_i$ stubs and a different vertex $j$ has $k_j$ stubs, there are $k_i k_j$ pairs of stubs, one for $i$ and one for $j$.
The probability that a particular pair $(stub_1, stub_2)$ are connected is $1/(2m-1)$, since
any of the $2m-1$ stubs other than $stub_1$ has equal probability of ending up connected to $stub_1$.  By linearity of expected value, the expected number of connections between vertices $i$ and $j$ is then $k_i k_j/(2m-1)$.  If $m$ is very large, you might approximate this as $k_i k_j/(2m)$.
The probabilty that there is at least one connection is less than the expected number of connections, if it is possible to have more than one connection.
Under appropriate conditions, the probability of having more than one connection will be small, in which case you can use $k_i k_j//(2m-1)$ as
an approximation to the probability of at least one connection.


TITLE: Implications of weak convergence in Lp e change of measure spaces
QUESTION [3 upvotes]: In the paper "A uniform algebra of analytic functions" of Carne-Cole-and-Gamelin. The Theorem 6.4 says:
Theorem 6.4: Let $\mu$ be a positive measure on some measure space, let $2\leq p<\infty$, and let $N$ be a integer satisfying $N\geq p$. Let $\{f_j\}$ be a sequence in $L_p(\mu)$ such that $f_j\longrightarrow 0$ weakly, and $F(f_j)\longrightarrow 0$ for every $F\in\mathcal{P}_N(L_p(\mu))$. Then $\|f_j\|_p\longrightarrow 0$.
Here, $\mathcal{P}_N(L_p(\mu))$ is the space of N-homogeneous polynomials.
Proof: Since the $f_j$'s are all carried by some $\sigma$-finite set, we can assume that the measure $\mu$ is $\sigma$-finite. Replacing $\mu$ by a mutually absolutely continuous measure, we can assume that $\mu$ is a probability measure. $\cdots$
My questions. Someone can help me understand this change of a arbitrary measure space to a probability measure space? Why the functions are carried by a $\sigma$-finite set?
Thanks for any hint.

REPLY [1 votes]: Let $A_{j,n} = \{ |f_j| > 1/n\}$.  Since $f_j \in L^p$, each $A_{j,n}$ must have finite measure.  So if we let $A = \bigcup_{j,n} A_{j,n}$, then $A$ is $\sigma$-finite.  And every $f_j$ vanishes outside $A$.  This is what they mean by "all $f_j$'s are carried by some $\sigma$-finite set."
I'll leave the part about changing to a probability measure for someone else.


TITLE: Given $\alpha$ , $\beta$ respectively the fifth and the fourth non-real roots of unity then the value of
QUESTION [0 upvotes]: Given $\alpha$ , $\beta$ respectively the fifth and the fourth non-real roots of unity then the value of $\left(1+\alpha \right)\left(1+\beta \right)\left(1+\alpha ^2\right)\left(1+\beta ^2\right)\left(1+\alpha ^3\right)\left(1+\beta ^3\right)\left(1+\alpha ^4\right)$ will be ?
After seeing this question on first sight i though it would be an easy one as $x^4 -1 = 0$ and i will then apply $(x-1)(x-\alpha)...(x-\alpha ^{n-1}))$ but there is + sign between expression so i thought multiplying and dividing $(1-\alpha)$ and $(1-\beta)$ would help but the $(1+\alpha^{3})$ term is making trouble for me .
Options are (A) $0$ (B) $1$ (C) $(1+\alpha+\alpha^{2})(1-\beta^{2})$ (D) $(1+\alpha)(1+\beta+\beta^{2})$

REPLY [1 votes]: Looking at the $\beta$ terms alone, multiplying the first two brackets gives $$(1+\beta)(1+\beta^2)=1+\beta+\beta^2+\beta^3=0$$ so the answer is A


TITLE: Sum of two independent random variables converges in distribution
QUESTION [10 upvotes]: Show that if $X_n$ and $Y_n$ are independent random variables for $1 \le n \le \infty$, $X_n \Rightarrow X_{\infty}$, and $Y_n \Rightarrow Y_{\infty}$, then $X_n + Y_n \Rightarrow X_{\infty} + Y_{\infty}$. Where $\Rightarrow$ means converge weakly or converge in distribution.
Any idea is appreciated.

REPLY [9 votes]: I think I got it:
Since $X_n \Rightarrow X_{\infty}$ and $Y_n \Rightarrow Y_{\infty}$, we have $\varphi_{X_n}(t) \to \varphi_{X_\infty}(t)$ for all t, and $\varphi_{Y_n}(t) \to \varphi_{Y_\infty}(t)$ for all $t$ by the continuity theorem. And thus $$ \varphi_{X_n + Y_n} (t) = \varphi_{X_n} (t) \varphi_{Y_n}(t) \to \varphi_{X_\infty}(t)\varphi_{Y_\infty}(t) = \varphi_{X_\infty + Y_\infty}(t)$$ for all $t$. Since both $\varphi_{X_\infty} (t)$ and $\varphi_{Y_\infty}(t)$ are continuous at $0$, $\varphi_{X_\infty + Y_\infty} (t)$ is continuous at $0$. Therefore by continuity theorem we have $X_n + Y_n \Rightarrow X_\infty + Y_\infty$.


TITLE: What are the cosets of this presentation?
QUESTION [1 upvotes]: I'm reading a book on algebra, and they give a presentation for $S_3$, with 6 elements $\{1, x, x^2, y, x y, x^2y\}$ as $$x^3 = 1,\quad y^2 = 1,\quad y x=x^2y$$
Now later in the book, there is a subgroup $H = \{1, x y\}$.  So they give the left cosets of $H$ in this group as
$$\{1, x y\} = H = xyH$$
$$\{x, x^2 y\} = xH = x^2yH$$
$$\{x^2 y\} = x^2H = yH$$
The last set of equations is what concerns me.  I think it should be
$$\{x^2, y\} = x^2H = yH$$  Am I correct?  If not, what am I missing?

REPLY [1 votes]: As confirmed by Tyler in the comments, you are correct. What appears in the book is missing a comma.


TITLE: Eigenvalues for discrete second derivative with periodic boundary conditions
QUESTION [2 upvotes]: This wikipedia page https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors_of_the_second_derivative for the discrete second derivative ivative says that the eigenvalue problem $v_{k+1}-2v_k+v_{k-1}=\lambda v_k$ in the case of periodic boundary conditions $v_0=v_n$ ($k=1,\ldots,n$) has eigenvalues $$
\lambda_j=\begin{cases}-4\sin^2\left(\frac{\mathrm{\pi}(j-1)}{2n}\right) & j~\text{odd},\\-4\sin^2\left(\frac{\mathrm{\pi}j}{2n}\right)&j~\text{even}.\end{cases}
$$
How is this derived, or is there a reference anywhere? I wrote down the corresponding equations as:
\begin{align*}
-(2+\lambda_1)v_1+v_2+v_N &= 0, \\
v_{j-1}-(2+\lambda_j)v_j + v_{j+1} &= 0,\quad j=2,\ldots,N-1, \\
v_1+v_{N-1}-(2+\lambda_N)v_N &= 0.
\end{align*}
but how to actually solve this system of difference equations is not so clear to me.

REPLY [1 votes]: Great question!
Actually, solving the eigenvalue for circular boundary condition is surprisingly tractable. Consider the matrix
$$
\begin{bmatrix}
-2 & 1 & 0 &0 &...&1\\
1 & -2 & 1 & 0&... &0\\
0&1 & -2  & 1&... &0\\
&&&&...\\
0&0 & 0  & 0&... &1\\
1&0 & 0  & 0&... &-2\\
\end{bmatrix}
$$
It's a circulant matrix, which could be diagonalized by a Discrete Fourier Transform matrix. Generally, if
$$
C=\begin{bmatrix}
c_0 & c_{n-1} & c_{n-2} & ... & c_2 & c_1\\
c_1 & c_{0} & c_{n-1} & ...& c_3 & c_2\\
&&...&&\\
c_{n-2} & c_{n-3} & c_{n-4} & ...& c_0& c_{n-1}\\
c_{n-1} & c_{n-2} & c_{n-3} & ... & c_1&c_0\\
\end{bmatrix}
$$
Then, with $\omega = e^{2\pi i/n}$
$$
v_j=[1,\omega^j,\omega^{2j},...\omega^{(n-1)j}]^T\\
\lambda_j = c_0+c_1 \omega^j+c_2 \omega^{2j}+ ... + c_{n-1}\omega^{(n-1)j}
$$
In this case, $c_0=-2,c_1=1,c_{n-1}=1$, then we can directly compute each eigenvalue $j=0,1,...n-1$
$$
\lambda_j = -2+\omega^j+\omega^{-j}
$$
Then use the basic formula for $\cos$ double angle.
$$
\lambda_j = -2+e^{2j\pi i/n}+e^{-2j\pi i/n}\\
=-2+2\cos(\frac{2\pi j}{n})\\
=-4\sin^2(\frac{\pi j}{n})
$$

Note that this looks different from your answer, but they are the same set of eigenvalues. Notice that
$$
\sin^2(\frac{\pi j}{n})=\sin^2(\frac{\pi (n-j)}{n})  \text{ for } j=1,2...n-1
$$
So any $\lambda_j,j\neq0$ has multiplicity 2, $\lambda_0=0$ has multiplicity 1; if $n$ is even, $\lambda_{n/2}=-4$ has multiplicity 1 too.
So if you sort and re-oder the eigenvalues, they will be the same as your answer.


TITLE: sub and super-levelset regularity for Sobolev functions
QUESTION [4 upvotes]: I'm wondering if there are known results about the "regularity" (in some sense to be determined) of sub and super levelsets of Sobolev functions $u\in W^{1,p}(\mathbb{R}^d)$. More precisely:
Assume $u\in W^{1,p}(\mathbb{R}^d)$, fix a constant $M\in \mathbb{R}$, and let
$$
E_M:=\{x\in \mathbb{R}^d,\hspace{1cm}u(x)>M\}\subset \mathbb{R}^d
$$
be the corresponding super-levelset of $u$. What can we say about the regularity of $E_M$? I actually have no precise idea of what "regularity" should mean here, but I am not talking about regularity of the boundary $\partial E_M$ as in the usual context of PDE's...
I think a natural question to ask is for example: how can one guarantee that $E_M$ is open (for a suitable representative of $u$), or at least that $E_M$ has nonempty interior? For example if $p$ is large enough we know by Sobolev inequality that $u$ is continuous so $E_M$ is open. When the gradient regularity deteriorates, how "ugly" could $E_M$ be? While still being measurable $E_M$ could be quite pathological, for example a Smith-Volterra-Cantor set with positive measure but empty interior (well, maybe not, this is actually my question!)
Also, what if we replace $u\in W^{1,p}$ by a weaker difference-quotient-type condition
$$
u\in L^p\quad\text{and}\quad \sup\limits_{0\neq h\in R^d}\frac{|\tau_h u-u|_{L^p(\mathbb{R}^d)}}{|h|^{\alpha}}<\infty \text{ for some }\alpha\in(0,1),
$$
where $\tau_h u(.)=u(.-h)$ denotes the usual shift???
Thank you in advance for your reference suggestions, input, and comments.

REPLY [3 votes]: One positive answer is that this set is $p$-quasi-open, see some resource about capacity theory, e.g., here: https://math.stackexchange.com/questions/48776/capacity-theory-beginner-resources.


TITLE: Absolutely integrable implies almost surely finite
QUESTION [0 upvotes]: If $X$ is absolutely integrable, i.e., $\mathbb {E}\vert X \vert<\infty$, I want to show that $X$ is almost surely finite, i.e., $\mathbb {P}(\vert X\vert<\infty)=1$. For now, I have the following:
\begin{align*}
\mathbb {E}\vert X\vert=\int_{0}^{\infty}\mathbb {P}(\vert X\vert\geq u)\, du
\end{align*}
I wonder how to argue based on this.

REPLY [1 votes]: Suppose that $P(|X| = \infty) = c > 0$. Thus for all $u \ge 0$ we have  $P(|X| > u) \ge P(|X| = \infty) = c$. Thus $\infty > E|X| = \int_{0}^{\infty} P(|X|\ge u)du \ge \int_{0}^{\infty} c du =\infty$. It's contradiction.