 K\mu(A)\mu(B),
	\end{align}  
	for some constant $K > 1$.
	\end{enumerate}
	Then $T$ is mixing along $(s_n)$.  
\end{lem}
\noindent Let us denote by  $\{p_n\}$ the sequence of primes, that is, $p_n$ is the $n$th prime and $\omega(n)$  the number of prime factors of $n$. We thus have as a direct consequence 
the following corollary.
\begin{cor}If $\{s_n\}$ is $\{n^k\}$, $k \geq 1$, or $\{p_n^\ell\}$, $\ell \geq 1$, or $\{\omega(n)\}$ and $T$ satisfy \eqref{Orneq:1}. Then $T$ is mixing along $(s_n)$.
\end{cor}
\begin{proof}By the well-know results, for any irrational number $\alpha$, the sequence $\{s_n \alpha\}$ is dense.
\end{proof}
Let us give the proof of Lemma \ref{GO}.
\begin{proof}[Proof of Lemma \ref{GO}]We first prove that $T$ is weakly mixing.  Note that
	since every power of $T$ is ergodic, $T$ can not admit
	eigenvalues which are roots of unity. Suppose $T$ admits
	eigenvalues of the form $e^{2\pi i a}$ for some irrational $a$ and $f$ its associated eigenfunction with $|f|=1$.
	Then, for any $c \in \mathbb C$, obviously  the set $\{x\;:\; f(x)=c\}$ has Lebesgue measure zero. Let $x \in [0,1)$ be such that for every arc $C$ containing $x$, $f^{-1}\big(C\big)$ has positive Lebesgue measure. Since $T$ is totally ergodic, the Lebesgue measure of $f^{-1}\big(C\big)$ 
	($m\big(f^{-1}\big(C\big)\big)$) can be made as small as we please by choosing $C$ small enough. Now, by the assumption \eqref{Ass1}, there is a subsequence $s_{n_j}$ such that $s_{n_j}.a \longrightarrow 1 \textrm{~~mod~} 1.$. Therefore,
	\begin{align*}
	m\Big(T^{-s_{n_k}}f^{-1}\big(C\big) \cap f^{-1}\big(C\big)\Big)&=
	m\Big(f^{-1}\big(e^{-2\pi i s_{n_k}.a}C\cap C\big)\Big)\\
	&  \tend{k} {+\infty}m\Big(f^{-1}\big(C\big)\Big)
		\end{align*}
	But, by our assumption \eqref{Ass2}, we have
	\begin{align*}
	\limsup m\Big(T^{-s_{n_k}}f^{-1}\big(C\big) \cap f^{-1}\big(C\big)\Big) \leq K \Bigg(m\Big(f^{-1}\big(C\big)\Big)\Bigg)^2. 
	\end{align*}
	If $m\Big(f^{-1}\big(C\big)\Big)$ is chosen small enough, we arrive to contradiction, proving that $T$ is weak mixing. To prove that $T$ is
	mixing along $(s_i)$. let us denote by $\Delta_n$ the graph joining given by $T^n$, by which we 
	mean the measure on $X\times X$ defined by
	$\Delta_n(A\times B) = \mu(T^nA\cap B)$, for any two sets
	$A, B \in {\mathcal{B}}$. Since the set of joining is compact in the weak 
	topology of joinings $ \displaystyle (i.e.~\rho_n \rightarrow \rho 
	~{\rm {iff~}} \rho_n(A \times B) \rightarrow \rho(A \times B),~~{\rm {for~each~rectangle~}}(A\times B))$, we can
	find a sequence of graph joining $(\Delta_{s_{n_j}})_{j=1}^\infty$
	converging to some joining, say,  $\lambda$. Whence, by our assumption \eqref{Ass2}, we get
	$$\lambda(A\times B) \leq K (\mu\otimes \mu) (A\times B).$$
	Let us define a joining $\psi$ by
	$$\mu \otimes \mu = \frac{1}{K}\lambda + (1 - \frac{1}{K}) \psi.$$
	But $\mu\otimes\mu$ is ergodic, so an extreme point in the space
	of invariant measures, whence $\lambda = \psi = \mu\otimes \mu$,
	so that for all $A, B \in \mathcal{B}$, $\mu(T^{s_n}A\cap B)
	\rightarrow \mu(A)\mu(B)$, and $T$ is mixing along $(s_n)$. This completes the
	proof of the Lemma.
\end{proof}

\subsection{ M-Towers by the cutting and independent staking method.}
Following \cite[section 2]{Fb}, an $n$-column is an ordered set of intervals $C=(I_1,\cdots,I_n)$, all of the same length. The width of $C$ is defined by $w(C)=|I_1|$, and the height of $C$ by $h(C)=n$. The base of $C$ is $b(C)=I_1$ and the top is $t(C)=I_n$. Associated with $C$ is the map $T_C$ which maps $I_{i}$ linearly to $I_{i+1}$, $1 \leq i \leq n-1$. So, $T_C$ is defined on $C\setminus I_n$. We thus have the following figure.
\vskip 1.0cm
\begin{center}
\begin{tikzpicture}
\draw (0,0) -- (5,0)node[below]{$C_1$};
\draw (0,1) -- (5,1) node[below]{$C_2$} ;
\draw[->,>=latex] (2,0)--(2,1);
\draw (0,2) -- (5,2)  node[below]{$C_3$};
\draw[->,>=latex] (2,1)--(2,2) node[above]{$\vdots$};
 \draw (0,3) -- (5,3)node[below]{$C_n$};
  \draw[->,>=latex] (3,2)--(3,3) ;
\end{tikzpicture}
\end{center}
A tower $G$ is an ordered set of disjoint columns $C_1,C_2, \cdots C_l$. The top of $G$ is
$t(G)=\bigcap_{j=1}^{l}t(C_j)$ and its base is  $b(G)=\bigcap_{j=1}^{l}b(C_j)$.  The width of $G$ is
 $w(G)=|t(G)|=|b(G)|$. The transformation $T_G$ consists of $T_{C_j},$ $1 \leq j \leq l$. A level in column in $C_j$ is considered as a level in $G$. $G$ can be seen also as union of all levels in it. Thus, $T_G$ is defined on $G\setminus t(G).$   


\section{Proof of Theorem \ref{Nat}} \label{Proof-II} We will gives a construction  based on Friedman's construction.
It will be carried out by induction.\\

We start by fixing an $M$-towers $T_{G}$ as  support of mixing time. Therefore, for any $\epsilon>0$, there is a positive integer $N(G,\epsilon)$ such that for any $n >N(G,\epsilon)$, we have
\begin{align}\label{M1}
	\Big|\mu(T_{G}^{n^2}I \cap J)-\frac{\mu(I) \mu(J)}{\mu(G)}\Big|<\epsilon,
\end{align}
where $I,J$ are levels in $G.$  Whence, by construction, we can choose $k=k(G,\epsilon,N)$ so large that $T$ extends $T_k$. We thus have
\begin{align}\label{M2}
\Big|\mu(T_{G}^{n^2}I \cap J)-\frac{\mu(I) \mu(J)}{\mu(G)}\Big|<\epsilon,\; \; N(G,\epsilon) \leq n^2 \leq N
