\beta\ge
\gamma$ be numbers in $(0,1)$. Let $y,~z$ and $\Omega$ be functions
of $x$ of growth
$$
y=\exp\left((\log x)^{\alpha+o(1)}\right),\quad z=\exp\left((\log
x)^{\beta+o(1)}\right),\quad \Omega=(\log x)^{\gamma+o(1)}
$$
as $x\to\infty$. We shall make these functions more precise later.
We split the set $\mathfrak{B}(x)$ into six subsets as follows:
\begin{eqnarray*}
\label{eq:Bees} \mathfrak{B}_1 & = & \{n\in \mathfrak{B}(x):P\le
y\};\\
\mathfrak{B}_2 & = & \{n\in \mathfrak{B}(x)\backslash
\mathfrak{B}_1:p^2\mid n~{\rm for~some~prime}~p\ge y\};\\
\mathfrak{B}_3 & = & \{n\in \mathfrak{B}(x)\backslash
(\mathfrak{B}_1\cup \mathfrak{B}_2):\Omega(n)\ge \Omega\};\\
\mathfrak{B}_4 & = & \{n\in \mathfrak{B}(x)\backslash (\cup_{j=1}^3
\mathfrak{B}_j):Q\le z\};\\
\mathfrak{B}_5 & = & \{n\in \#\mathfrak{B}(x)\backslash
(\cup_{j=1}^4
\mathfrak{B}_j):t(Q)\ge Q^{1/3}~{\text{\rm and}}~\Omega(t(Q))\le \Omega\};\\
\mathfrak{B}_6 & = & \mathfrak{B}(x)\backslash (\cup_{j=1}^5
\mathfrak{B}_j).
\end{eqnarray*}
We now bound the cardinalities of each of the sets $\mathfrak{B}_i$
for $i=1,\ldots,6$.
{\bf The set $\mathfrak{B}_1$}. Numbers in $\mathfrak{B}_1$ are
called {\it $y$-smooth}. Well-known results concerning the number of
$y$-smooth numbers $n\le x$ (see \cite{CEP}) show that in our range
the estimate
$$
\#\mathfrak{B}_1=xu^{-u+o(u)}
$$
holds as $u\to\infty$, where $u=\log x/\log y$ ($=(\log
x)^{1-\alpha+o(1)}$ as $x\to\infty$). Thus,
\begin{equation}
\label{eq:B1} \#\mathfrak{B}_1\le x\exp(-(1+o(1))u\log u)\qquad
{\text{\rm as}}~x\to\infty.
\end{equation}
{\bf The set $\mathfrak{B}_2$.} If $p\in [y,x^{1/2}]$ is a fixed
prime, then there are $\lfloor x/p^2\rfloor\le x/p^2$ positive
integers $n\le x$ divisible by $p^2$. Summing up over all the
possible values for $p$, we get that
\begin{equation}
\label{eq:B2} \#\mathfrak{B}_2\le \sum_{y\le p\le
x^{1/2}}\frac{x}{p^2}\le x\sum_{y\le m} \frac{1}{m^2}\le
x\int_{y-1}^{\infty} \frac{dt}{t}\ll \frac{x}{y}.
\end{equation}
{\bf The set $\mathfrak{B}_3$.} Lemma 13 in \cite{LuPo} shows that
uniformly for all integers $k\ge 1$ we have
\begin{equation}
\label{eq:LuPo}
\sum_{\substack{n\le x\\ \Omega(n)\ge k}}1\ll
\frac{kx\log x}{2^k}.
\end{equation}
Applying this with $k=\lfloor \Omega \rfloor$, we get that
\begin{equation}
\label{eq:B3} \#\mathfrak{B}_3\le x\exp\left(-(\log
2+o(1))\Omega\right)\qquad {\text{\rm as}}~x\to\infty.
\end{equation}
From now on until the end of the argument, we consider $n\in
\mathfrak{B}(x)\backslash (\cup_{j=1}^3\mathfrak{B}_j)$. Write $n=Pm$
and observe that $P(m)
z^{1/3}$.
{\bf Case 1.} $Qt(Q)\le x/m$. Let us write $\mathfrak{B}_{5,1}$ for
the subset of $\mathfrak{B}_5$ formed by such numbers $n$. In this
instance, the second term in equation \eqref{eq:possforP} dominates
and the number of possibilities for $P$ when $m$ and $Q$ are fixed
is
$$
\le \frac{2x}{mQt(Q)}.
$$
Fix $d=t(Q)$ and sum up the above bound over all primes $Q$ such
that $t(Q)=d$. Since $Q\equiv 1\pmod d$, it follows that $Q=1+d\ell$
for some positive integer $\ell \le x\Omega/(md)x/m$. We write $\mathfrak{B}_{5,2}$ for the
subset of $\mathfrak{B}_5$ consisting of these numbers. We write
also $\beta(n)=P+\beta(m)=Q\delta$, where $1\le \deltax/m$, it follows that $P$ (hence, $n$) is uniquely
determined by $Q$. We now fix both $m$ and $\delta$ and observe that
$Q\le x\Omega/(m\delta)$. Note also that
$$
P=\beta(n)-\beta(m)=Q\delta-\beta(m)\equiv \delta-\beta(m)\pmod {d}.
$$
Since also $Pm\equiv 1\pmod d$, we get that $d$ divides
$m(\beta(m)-\delta)+1$. This last number is not zero since $m\ge 2$
(if it were zero, then $m(\delta-\beta(m))=1$, which is impossible
for $m\ge 2$). Since $\delta\ge 1$, the size of this number is
\begin{eqnarray*}
|m(\beta(m)-\delta)+1| & < & \max\{m\delta, m P(m)\Omega(m)\}\\
& < & \max\{xm\Omega, m^2\Omega\}\\
& < & xm\Omega<\frac{x^2\Omega}{y} \Omega$. Let
$\mathfrak{B}_{6,1}$ and $\mathfrak{B}_{6,2}$ be the subsets of
$\mathfrak{B}_6$ such that the first and second inequality holds,
respectively.
We first deal with $\mathfrak{B}_{6,1}$. Let ${\mathcal Q}$ be the
set of primes such that $t(Q)\Omega$.
Fix $m$ and $Q\Omega$ and using bound \eqref{eq:sumlargeOmega} for
$t=x\Omega$ and $k=\lfloor \Omega\rfloor$, we get that
\begin{equation}
\label{eq:B62} \#\mathfrak{B}_{6,2}\ll \frac{x(\log
x)^2\Omega^2}{2^{\Omega}}=x\exp(-(\log 2+o(1))\Omega)
\end{equation}
as $x\to\infty$. From estimates \eqref{eq:B61} and \eqref{eq:B62},
we get
\begin{equation}
\label{eq:B6} \#\mathfrak{B}_6\le
\frac{x}{z^{1/3+o(1)}}+\frac{x}{\exp{((\log 2+o(1))\Omega)}}
\end{equation}
as $x\to\infty$.
Comparing bounds \eqref{eq:B1}, \eqref{eq:B2}, \eqref{eq:B3},
\eqref{eq:B4}, \eqref{eq:B5} and \eqref{eq:B6}, we see that the
optimal bounds are obtained when the parameters $y,~z$ and $\Omega$
are chosen such that
$$
\Omega \log 2=\frac{\log z}{3}-(\log\log x+\log 3)\Omega=v\log
v=u\log u.
$$
This gives
\begin{eqnarray*}
\log y & = & (c_1+o(1))(\log x)^{2/3}(\log\log x)^{2/3},\\
\log z & = & (c_2+o(1))(\log x)^{1/3}(\log\log x)^{4/3},\\
\Omega & = & (c_3+o(1))(\log x)^{1/3}(\log\log x)^{1/3}
\end{eqnarray*}
as $x\to\infty$, where $c_1=(\log 2)^{-1/3},~c_2=(\log
2)^{-2/3},~c_3=c_2/3$. Note that this agrees with the conventions we
made at the beginning on $y,~z,~\Omega$ with
$\alpha=2/3,~\beta=\gamma=1/3$. Therefore we have just shown that
\begin{equation}
\label{eq:upB} \#\mathfrak{B}(x)\le x\exp\left(-(c_4+o(1))(\log
x\log\log x)^{1/3}\right)
\end{equation}
as $x\to\infty$, where $c_4=(\log 2)^{1/3}/3$. This finishes the
proof of Theorem \ref{thm:1}.
\medskip
\noindent {\bf Remark.} Notice that as a byproduct of our effort we conclude
immediately, by partial summation, that
$$
\sum_{n\in \mathfrak{B}}\frac{1}{n}<\infty.
$$
\section{Proof of Theorem \ref{thm:2}}
We let $p$ be a large prime and put $M=2^p-1$. Let $k$ be a positive
integer such that $k=o(p)$ holds as $p\to\infty$. Choose positive
integers $a\ge 3$ and $b$ even such that $a+b=k$ and $a-b=1$.
Clearly, $a=(k+1)/2$ and $b=(k-1)/2$, and in order for $a$ and $b$
to be integers with $b$ even we must have $k\equiv 1\pmod 4$. From now on, we
work under this assumption. Let ${\mathcal I}=\lfloor M/(2p^2),
M/p^2\rfloor$. We choose $a-3$ distinct primes in ${\mathcal I}$
which are $1$ modulo $p$ and $b$ distinct primes in ${\mathcal I}$
which are $-1$ modulo $p$. By the Siegel-Walfisz Theorem (note that
that the inequality $p<2\log(M/(2p^2))$ holds for large enough
values of $p$ so we may apply the Siegel-Walfisz Theorem to estimate
the number of primes congruent to either $1$ or $-1$ modulo $p$ in
${\mathcal I}$), it follows that the inequality
$$
U_{\eta}=\pi(M/p^2;p,\eta)-\pi(M/(2p^2);p,\eta)\ge \frac{M}{3p^3\log
M}\qquad {\text{\rm for}}~\eta\in \{\pm 1\}
$$
holds for large values of $p$. Recall that $\pi(x;v,u)$ means the
number of primes $p\le x$ congruent to $u$ modulo $v$. The number of
choices of pairs of primes as above is
\begin{eqnarray}
\label{eq:UV} & = & \binom{U_1}{a-3}\binom{U_{-1}}{b}\ge
\left(\frac{U_1}{a-3}\right)^{a-3}\left(\frac{U_1}{b}\right)^b\nonumber\\
& \ge &
\left(\frac{U_1}{a}\right)^{a-3}\left(\frac{U_{-1}}{b}\right)^{b}\left(1+O\left(\frac{kp^3}{M}\right)
\right)^k\nonumber\\
& \gg & \frac{M^{k-3}}{(3(\log 2)kp^4)^{k-3}}\gg
\frac{M^{k-3}}{p^{5k-15}}
\end{eqnarray}
since $3(\log 2)k2$ for large values
of $p$), it follows that $N$ is even. Furthermore, $N\equiv
a-3-b\pmod p\equiv -2\pmod p$. Thus, $M-N>M-M/p$ is a large odd
number which is congruent to
$$
2^p-1-N\equiv 2-1-(-2)\equiv 3\pmod p.
$$
By a Theorem of Ayoub \cite{Ay} (see also \cite{Ba}), it follows
that for large $p$ the number $M-N$ can be written as
$$
M-N=r_1+r_2+r_3,
$$
where $r_12} \left(1+\frac{1}{(\ell-1)^3}\right).
$$
Observe that $C_{M-N}\gg 1$. In what follows, we will see that at
least half of the above representations will have $r_1>M/p^2$.
Indeed, assume that this is not so. Then $r_1\le M/p^2$ is a prime
congruent to $1$ modulo $p$ and $r_2\le M$ is a prime congruent to
$1$ modulo $p$, and once $r_1$ and $r_2$ are chosen then $r_3$ is
fixed by the equation $r_3=M-N-r_1-r_2$. The number of such pairs
$(r_1,r_2)$ is
$$
\le \pi(M/p^2;p,1)\pi(M;p,1)\ll \frac{M^2}{p^4(\log M)^2}\ll
\frac{M^2}{p^3(\log M)^3}
$$
and the above upper bound is of a smaller order of magnitude then
the function appearing at \eqref{eq:Ay}. Thus, for large $p$, there
are
\begin{equation}
\label{eq:r} \gg \frac{M^2}{p^2(\log M)^3}\gg \frac{M^2}{p^5}
\end{equation}
such representations where $r_1>M/p^2$. From now on we work with
such representations. Now observe that
$\{p_1,\ldots,p_{a-3},q_1,\ldots,q_b,r_1,r_2,r_3\}$ are distinct
primes because $r_1>p_{a-3}$. Let $n$ be the product of the above
$a+b=k$ primes. Then
$$
\beta(n)=\sum_{i=1}^{a-3} p_i+\sum_{j=1}^b q_j+r_1+r_2+r_3=2^p-1
$$
and $n\equiv 1\cdot (-1)^b\equiv 1\pmod p$, therefore $p\mid n-1$,
so $\beta(n)\mid 2^p-1\mid 2^{n-1}-1$. Thus, $n\in \mathfrak{B}$.
The size of $n$ is
$$
n