 $h_1,h_2,h_3$.
\end{lemma}

Apply these lemmas to
$$Q=\frac{1}{f^{(0)}\cdots f^{(n-1)}}Q_1^{(n)},$$
and we have the conclusion.

\begin{prp}Define $f^{(n)}$, $Q_1^{(n)}$ and $Q_2^{(n)}$ by $\eqref{Q1Q2}$. Then $E_6(s=n+1)$ $(n=1,2,\dots)$ factors as $Q_1^{(n)}\circ Q_2^{(n)}$. For a basis $\{h_1,h_2,h_3\}$ of the solution space of $Q_1^{(n)}$, define
  the first-order operators $\{L_1,L_2,L_3\}$ as in Lemma $\ref{LLL}$. Then
  $$Q_1^{(n)}=f^{(0)}\cdots f^{(n-1)}L_1\circ L_2\circ L_3.$$
  Though these three operators  $L_1,L_2$ and $L_3$ are not uniquely determined,
  they are controlled by Lemma $\ref{LLLhhh}$.
\end{prp}

\begin{remark}
  The three operators $L_1,L_2$ and $L_3$ have apparent singularities not only at the roots and the poles of $f^{(0)}\cdots f^{(n-1)}$ but also at the points depending on the choice of the basis $\{h_1,h_2,h_3\}$. We do not know how to control them.
\end{remark}



\section{$S\!E_6$}\label{SE6}
\secttoc
$S\!E_6$ is, by definition, a codimension-two specialization of the exponents of $E_6$:
\begin{equation}e_3-2e_2+e_1=e_4-2e_5+e_6=e_7-2e_8+e_9.\label{E6restEar}
\end{equation}
This equation has many nice properties  such as integral representation of solutions and shift operators that do not come from those of $E_6$. The study of our seventeen equations started from this equation.

\subsection{Definition of $S\!E_6$ as a middle convolution of $S\!E_3$}\label{SE6MC}
The Dotsenko-Fateev operator (\cite{DF}) is an operator of order 3 defined in Section \ref{SE3} as
$$S\!E_3=D\!F(a,b,c,g)=(x-1)^2 x^2\partial^3+D\!F_1\partial^2+D\!F_2\partial+D\!F_3,\quad \partial:=d/dx.$$
Any solution admits the integral representation:
$$\int\omega(x),\quad \omega(x):=\prod_{i=1,2} t_i^a(t_i-1)^b(t_i-x)^c\cdot (t_1-t_2)^g\ dt_1\wedge dt_2.$$
To a solution $v(x)=\int\omega$ of the  Dotsenko-Fateev equation, we multiply $x^{p}(x-1)^{q}$:
$$v(x)\ \to\  v_{pq}(x)=x^{p}(x-1)^{q}v(x),$$
and apply the Riemann-Liouville transformation with parameter $r+1$:
$$v_{pq}(x)\ \to\ \int v_{pq}(t)(t-x)^{r}\, dt,$$
and we get the function of $x$ defined by the integral
$$\int \prod_{i=1,2} t_i^a(t_i-1)^b(t_i-t_3)^c\cdot (t_1-t_2)^g\cdot t_3^{p}(t_3-1)^{q}(t_3-x)^{r}\ dt_1\wedge dt_2\wedge dt_3.$$
Note that the integrand is invariant under the change $t_l\leftrightarrow t_2$. It is well-known that a function defined by such an integral satisfy a Fuchsian differential equation of order the number of bounded chambers cut out by the factors of the integrand.

\subsubsection{Number of bounded chambers and the order of the equation}\label{SE6MCbdd}
Consider in the real $(t_1,t_2,t_3)$-space the arrangement of 
ten planes:
$$ \prod_{i=1,2,3}t_i(t_i-1)\cdot \prod_{1\le i<j\le3}(t_i-t_j)\cdot(t_3-x)=0.$$
The number of bounded chambers cut out by this arrangement is twelve, in fact,
for simplicity we assume $x>1$, then we find eight tetrahedra:
$$\begin{array}{lll}
&0<t_i<t_j<t_k<1\qquad &\{i,j,k\}=\{1,2,3\},\\ [1mm]
&1<t_i<t_j<t_3<x &\{i,j\}=\{1,2\}\end{array}$$
and four triangular prisms:
$$\begin{array}{llll}
&0<t_i<t_j<1,\quad &1<t_3<x\qquad &\{i,j\}=\{1,2\},\\ [1mm]
  &1<t_i<t_3<x,\quad &0<t_j<1\qquad &\{i,j\}=\{1,2\}.\end{array}$$
Though there are twelve bounded chambers, the number of bounded chambers modulo the change $t_1\leftrightarrow t_2$ is six. So the function defined by the integral satisfy an equation of order six; let us call this equation $E\!ar=E\!ar(a,b,c,g,p,q,r)$ for the moment. It is not practical to find $E\!ar$ by differentiating the integral. We make use of a middle convolution.

\subsubsection{$E\!ar$ is $S\!E_6$}\label{SE6MCEar}
When $v$ is a solution of $S\!E_3=x^2(x-1)^2\partial^3+\cdots$, the equation satisfied by $v_{pq}=x^{p}(x-1)^{q}v(x)$ can be written in $(x,\partial)$-form as
  $$x^3(x-1)^3\partial^3+\cdots$$
Since this operator has a term $x^6\partial^3$, to get a $(z,\partial)$-form, we have to multiply $\partial^3$, at least.  
After multiplying $\partial^3$ from the left, we find that we can write the equation in $(z,\partial)$-form, which is a linear combination of 
$z^i\circ\partial^j\ (z=x\partial).$
 We then apply the middle convolution (\S \ref{GenAdd})  with parameter $r+1$, that is, we replace $z$ by $z-r-1$.
 We eventually get an operator
 $$E\!ar=x^3(x-1)^3\partial^6+y_5\partial^5+y_4\partial^4+y_3\partial^3+y_2\partial^2+y_1\partial+y_0,$$
\[\def1.1} \begin{array}{ll{1.2}\setlength\arraycolsep{3pt}
\begin{array}{lcl}
y_{5} &=& x^2(x-1)^2({\rm polynomial\ of\ degree\ 1\ in\ }x),\\           
y_{4} &=& x(x-1)({\rm polynomial\ of\ degree\ 2\ in\ }x),\\           
y_{3} &=& ({\rm polynomial\ of\ degree\ 3\ in\ }x),\\           
y_{2} &=& (r-2)({\rm polynomial\ of\ degree\ 2\ in\ }x),\\         
y_{1} &=& (r-1)(r-2)({\rm polynomial\ of\ degree\ 1\ in\ }x),\\
y_{0} &=& r(r-1)(r-2)(2c+r+1+q+p)\\
&&\  \times (a+2+2c+q+g+r+b+p)(2a+3+2c+q+g+r+2b+p),
\end{array}
\]
and find

\begin{thm}\label{etoar} The equation $E\!ar$ turns out to be a specialization $S\!E_6$ of the equation $E_6$:
  \[\def1.1} \begin{array}{ll{1.1}\setlength\arraycolsep{2pt}