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\def\nn{\nonumber}
\def\thft{{}_3F_2\!\[\begin{matrix}                     -n,\alpha,\beta \\
                    \gamma ,\delta \end{matrix}; 1\]}
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\begin{document}

\newbox\Adr
\setbox\Adr\vbox{
\centerline{\small\uppercase{C. Krattenthaler$^\dagger{}^\ddagger$\ \ 
and\ \ K. Srinivasa Rao}}
\vskip12pt
\centerline{Institut f\"ur Mathematik der Universit\"at Wien,}
\centerline{Strudlhofgasse 4, A-1090 Wien, Austria.}
\centerline{E-mail: {\tt\footnotesize kratt@ap.univie.ac.at}}
\centerline{WWW: \footnotesize\tt http://www.mat.univie.ac.at/\~{}kratt}
\centerline{and}
\centerline{Institute of Mathematical Sciences, Chennai-600 113, India}
\centerline{E-mail: {\tt\footnotesize rao@imsc.ernet.in}}
}
\title[Automatic Generation of Hypergeometric identities]
{\large Automatic Generation of Hypergeometric identities
        by the Beta Integral Method}
\author[C. Krattenthaler and K. Srinivasa Rao]
{\box\Adr}
\thanks{$^\dagger$Research partially supported by the Austrian
Science Foundation FWF, grant P12094-MAT, and by EC's IHRP Programme,
grant HPRN-CT-2001-00272.\newline
\hbox to\hsize{
\kern10pt $^\ddagger$Current address: Institut Girard Desargues,
Universit\'e Claude Bernard Lyon-I,
21, avenue Claude}\newline
\hbox{Bernard,
F-69622 Villeurbanne Cedex, France. E-mail: \tt kratt@euler.univ-lyon1.fr}}
%\email{}
%\dedicatory{}
%\date{}
%\thanks{}
\subjclass[2000]{Primary 33C20;
 Secondary 33C05 33C70}
\keywords{beta integral,
generalized hypergeometric series, Kamp\'e de F\'eriet
functions}

\begin{abstract}
In this article hypergeometric identities (or transformations)
for 
$_{p+1}F_{p}$-series and for Kamp\'e de F\'eriet series of 
unit arguments are derived systematically from known 
transformations of hypergeometric series and products of 
hypergeometric series, respectively, using the beta integral 
method in an automated manner, based on
the {\sl Mathematica} package HYP. As a result we obtain some known and some 
identities which seem to not have been recorded before in literature.
\end{abstract}

\maketitle

\section{Introduction}

Euler's beta integral evaluation
\begin{equation} 
\int_0^1 z^{\alpha -1} (1-z)^{\beta -1} dz  = 
 \frac{\G(\alpha )\,\G(\beta )}{\G(\alpha +\beta )}     
\label{eq2.2}
\end{equation}
provided $\Re(\alpha )>0$ and $\Re(\beta )>0$, is at the heart of many
identities in the theory of hypergeometric series. An example is 
the well known integral representation
(see e.g.\ \cite[(11), Sec.~3.6]{LukeAA})
of a hypergeometric series
\begin{multline} \label{e4}
{}_{p+1}F_p\!\[\begin{matrix} \alpha,\alpha_1,\dots ,\alpha_p\\
                \gamma,\beta_1,\dots ,\beta_{p-1} \end{matrix}; t\]
\\=
\frac {\G(\gamma)} {\G(\alpha)\G(\gamma-\alpha)}
\int _{0} ^{1} z^{\alpha-1}(1-z)^{\gamma-\alpha-1}
{}_pF_{p-1}\!\[\begin{matrix} \alpha_1,\dots ,\alpha_p\\
                \beta_1,\dots ,\beta_{p-1} \end{matrix}; zt\] dz.
\end{multline}
Here we use the standard hypergeometric notation
\begin{equation*}  % \label{eq:hyper}
{}_r F_s\!\left[\begin{matrix} a_1,\dots,a_r\\ b_1,\dots,b_s\end{matrix}; 
z\right]=\sum _{k=0} ^{\infty}\frac {\po{a_1}{k}\cdots\po{a_r}{k}}
{k!\,\po{b_1}{k}\cdots\po{b_s}{k}} z^k\ ,
\end{equation*}
where the Pochhammer symbol $(\alpha)_k$ is defined by
$(\alpha)_k  =  \alpha(\alpha+1)(\alpha+2)\cdots (\alpha+k-1)$,
$k>0$, $(\alpha)_0  = 1$. (In order to prove \eqref{e4}, one would
interchange integration and summation on the right-hand side, and
then use \eqref{eq2.2} to evaluate the integral inside the
summation.)

\medskip

In this paper we propose the exploitation of (what we call) the ``{\it
beta integral method}," a method of deriving ``new" hypergeometric
identities from ``old" ones, using the beta integral evaluation,
which is folklore in the hypergeometric
literature, although appearances can be
only found sporadically (see, for example, \cite[Ch.~3, Ex.~5, 14 and
16]{AnARAA}). It seems that it was never exploited systematically,
probably because of the effort it takes to do all the computations.
However, with the help of a computer, these computations become
completely painless. This is what we propose here: the {\it completely
automatic} application of the beta integral method.

\medskip

To convey the idea, we briefly recall an early occurrence of
this method (but, very likely, not the first) 
in \cite[Sec.~3]{WeEr}. 
We start with the well known  
transformation formula \cite[(1.8.10), with $a,b,c$ replaced by 
$-n,\alpha,\gamma$, respectively]{SlatAC}:
\begin{equation} 
{}_2F_1\!\[\begin{matrix} -n,\alpha\\
                \gamma \end{matrix}; z\] =
                \frac {(\gamma-\alpha)_n}{(\gamma)_n}
                {}_2F_1\!\[\begin{matrix} -n,\alpha \\
                \alpha-n-\gamma+1\end{matrix}; 1-z\],
\label{eq2.3}
\end{equation}
multiply both sides by $z^{\beta-1}(1-z)^{\delta-\beta-1}$, 
integrate both sides with respect to $z$, $0\le z\le 1$,%
\footnote{In fact, we have to temporarily restrict the
parameters $\beta$ and $\delta$ to $\Re (\beta)>0$ and $\Re(\delta-\beta)>0$,
because otherwise the beta integral \eqref{eq2.2} would not converge. 
However, these
restrictions can in the end be removed by analytic continuation. We
shall, if necessary, make similar temporary assumptions without
mentioning in subsequent derivations.} 
interchange integration and summation on both sides, then use
\eqref{eq2.2}, and finally convert the result back to hypergeometric
notation,
to get the transformation formula: 
\begin{equation} 
\thft = \frac{(\gamma-\alpha)_n}{(\gamma)_n}
{}_3F_2\!\[\begin{matrix}                -n,\alpha,\delta-\beta\\
               1+\alpha-\gamma-n,\delta 
\end{matrix}; 1\] ,
\label{eq2.4}
\end{equation}
where $n$ is a nonnegative integer.
In this case we obtain an already known transformation formula,
namely one of Thomae's $_3F_2$-transformation formulas.
(The formula \eqref{eq2.4} can for example be extracted
from Tables II$_A$ and II$_B$ ($Fp(0;4,5)=Fn(4;0,1)$) 
in Bailey's tract \cite{BailAA} which summarizes and groups
the equivalent numerator and denominator functions obtained by Thomae
\cite{ThoeAB} in the notation of Whipple \cite{WhipAA}.)

\medskip

However, 
not only can this method be applied to a large variety of identities,
its application can be {\it completely automatized\/} (as we already
announced) with the help of the
{\sl Mathematica} program HYP
created by one of us \cite{KratBF}. Thus, this method is similar in
spirit to the idea of ``dual identities" in ``WZ-theory" \cite{WiZeAA}
(though it is less sophisticated), 
and the idea of ``parameter augmentation" in
\cite{ChLiAA,ChLiAB} (which, however, introduces just one additional
parameter instead of two as in the ``beta integral method"). 
Also there, the idea is to start with a known
identity, and then apply a certain procedure to
automatically produce a (possibly) new identity. (In WZ-theory, the
procedure consists in finding the ``WZ-pair" which is behind the
original identity, and then use summation over the ``other" variable
to obtain a different identity. ``Parameter augmentation" applies a
certain ($q$-)differential operator to a known identity.)

\medskip

The algorithm has also been applied to 
transformation formulae for 
products of hypergeometric functions known in literature. As a result
formulae for the Kamp\'e de F\'eriet series
of unit arguments, which transform them into single-sum hypergeometric
series, are obtained. Again, some of the
results obtained are known ones but some of the results obtained
seem to be new and interesting.
 
\medskip

In Section~2, the algorithm and its implementation is presented. 
A section of
the {\sl Mathematica} session is reproduced to give a feel for the
methodology adopted. 
In Section~3, we list the results if we apply our algorithm
to known hypergeometric transformation formulae between single-sum
series.
In Section~4, we list a few of the results for Kamp\'e de 
F\'eriet that we obtain when we apply our machinery to identities 
involving products of hypergeometric series. Concluding
remarks are made in Section~5.
 
\section{The algorithm}

Let us recall the basic steps of the algorithm that produced
\eqref{eq2.4} from \eqref{eq2.3}:
\begin{itemize}
\item [(i)] convert the hypergeometric series on both sides of a
given transformation into sums;
\item [(ii)] multiply both sides of the equation by the factors
$z^{\beta-1}(1-z)^{\delta-\beta-1}$;
\item[(iii)] integrate term-by-term with respect to $z$ for 
$0\leq z\leq 1$;
\item [(iv)] interchange integration and summation;
\item [(v)] use the beta integral to evaluate the integrals inside
the summations;
\item[(vi)] convert the sums back into hypergeometric notation.
\end{itemize}

\medskip

We give below (see {\tt In[2]} in the {\sl Mathematica} session below) 
an implementation of this algorithm in {\sl Mathematica}.
There, a function {\tt T} is defined, which is applied to some
equation. To briefly explain the code: Step~(i) is performed 
in the first line ({\tt
X=Ers[Equ,FSUM,$\{$1,2$\}$]}), where the variable {\tt X} is set to the
equation {\tt Equ} where on both sides the hypergeometric series is
converted into a sum (which is achieved by the application of {\tt
FSUM}). Then, in the next line, both sides of the equation are
multiplied by $z^{A-1}(1-z)^{B-A-1}$, thus performing Step~(ii). The
subsequent line has the purpose of bringing everything inside the
summations (which is achieved by {\tt SUMSammle}). Then
Steps~(iii)-(vi) are performed, in the following way. First, the
variable {\tt Y} is set to the integral over $z$, $0\le z\le 1$, of
the {\it summand\/} (represented by {\tt X[[1,1]]}) of the series on
the left hand side of the equation. Subsequently, the result is
summed over $k$, and then converted to hypergeometric notation (the
latter being achieved by {\tt SUMF}). Then the same is done for the
right hand side, and the result is stored in the variable {\tt X}.
Finally, in the last line ({\tt Y==X}) the results on both sides are
equated.

\medskip

To illustrate the use of this algorithm, we adjoin a segment of 
the {\sl Mathematica} output 
where the algorithm is applied to the transformation formula
\cite[(1.3.15)]{SlatAC}. 
Within HYP, the transformation formula is input as Tgl2103 (see {\tt
In[3]}). Then the algorithm is invoked, by applying the function {\tt T}
to the transformation formula. In the
interactive mode the questions raised by {\sl Mathematica} have to 
be answered appropriately. 
The result is displayed in {\tt
Out[4]}. 
\MATH
\goodbreakpoint%
Mathematica 2.2 for DOS 387
Copyright 1988-93 Wolfram Research, Inc.
\goodbreakpoint%
In[1]:= \MATHkleiner \MATHkleiner hyp.m
\goodbreakpoint%
In[2]:= T[Equ\MATHtief ]:=
            Module[%
\MATHlbrace X,Y%
\MATHrbrace ,
              X=Ers[Equ,FSUM,%
\MATHlbrace 1,2%
\MATHrbrace ];
              Mal[z\MATHhoch (A-1)*(1-z)\MATHhoch (B-A-1)];
              X=(Gleichung/.SUMSammle);
              Y=Integrate[X[[1,1]],%
\MATHlbrace z,0,1%
\MATHrbrace ];
              Y=(SUM[Y,%
\MATHlbrace k,0,Infinity%
\MATHrbrace ]/.SUMF);
              X=Integrate[X[[2,1]],%
\MATHlbrace z,0,1%
\MATHrbrace ];
              X=(SUM[X,%
\MATHlbrace k,0,Infinity%
\MATHrbrace ]/.SUMF);
              Y==X
             ]
\goodbreakpoint%
In[3]:= Tgl2103
Do you want to set values for the equation? [y|n]: n
\goodbreakpoint%
           %
\MATHluEck          %
\MATHruEck                          %
\MATHluEck                    %
\MATHruEck 
           %
\MATHvStrich  a, b    %
\MATHvStrich            -a - b + c    %
\MATHvStrich  -a + c, -b + c    %
\MATHvStrich 
Out[3]=  F %
\MATHvStrich      ; z %
\MATHvStrich  == (1 - z)            F %
\MATHvStrich                ; z %
\MATHvStrich 
        2 1%
\MATHvStrich   c      %
\MATHvStrich                       2 1%
\MATHvStrich        c           %
\MATHvStrich 
           %
\MATHloEck          %
\MATHroEck                          %
\MATHloEck                    %
\MATHroEck 
\goodbreakpoint%
In[4]:= T[\%]
Is -a a nonnegative integer?
[y|n]:  n
Is -b a nonnegative integer?
[y|n]:  n
A hypergeometric series is converted into a sum.
Enter a variable for the summation index: k
Is a - c a nonnegative integer?
[y|n]:  n
Is b - c a nonnegative integer?
[y|n]:  n
A hypergeometric series is converted into a sum.
Enter a variable for the summation index: k
\goodbreakpoint%
           %
\MATHluEck             %
\MATHruEck 
           %
\MATHvStrich  A, a, b    %
\MATHvStrich 
         F %
\MATHvStrich         ; 1 %
\MATHvStrich  \MATHGamma (A) \MATHGamma (-A + B)
        3 2%
\MATHvStrich   B, c      %
\MATHvStrich 
           %
\MATHloEck             %
\MATHroEck 
Out[4]= -------------------------------- == 
                      \MATHGamma (B)
\goodbreakpoint
 
        %
\MATHluEck                       %
\MATHruEck 
        %
\MATHvStrich  A, -a + c, -b + c    %
\MATHvStrich 
      F %
\MATHvStrich                   ; 1 %
\MATHvStrich  \MATHGamma (A) \MATHGamma (-a - A - b + B + c)
     3 2%
\MATHvStrich  -a - b + B + c, c    %
\MATHvStrich 
        %
\MATHloEck                       %
\MATHroEck 
\MATHgroesser     ------------------------------------------------------
                       \MATHGamma (-a - b + B + c)

\endMATH

Notice that $\G(A)$ is occurring on both sides of the output
equation. Simplification of the result can be done either by hand or
using the various commands available in HYP. After the following 
replacements, in succession: $c\rightarrow e$, $A\rightarrow c$, 
$B\rightarrow d$, the final result obtained is:
\begin{equation} 
{}_3F_2\!\[\begin{matrix} a,b,c\\
        d,e \end{matrix}; 1\] = \frac {\G(d)\,\G(s)} {\G(-c+d) \,\G(s+c)}
{}_3F_2\!\[\begin{matrix} c,-a+e,-b+e\\
        s+c,d \end{matrix}; 1\]
\label{e7}
\end{equation}
where $s=d+e-a-b-c$, is the parameter excess. This identity 
belongs to the set of ten non-terminating ${}_3F_2[1]$ series
(see \cite[Ex.~7, p.~98]{BailAA}
and \cite[(III)]{Sri1}). 
Within HYP, this output can then be converted into 
any of the commonly used forms of 
\TeX\ (\LaTeX, \AmS-\LaTeX, \AmSTeX\ or Plain-\TeX). In the case of this 
article, conversion into \AmS-\LaTeX\ code was applied, 
which is reproduced here.


\section{``New" single sum hypergeometric identities from old ones}

In this section,
we apply the algorithm of the previous section systematically 
to known hypergeometric series transformations. Whenever we have been
able to trace the obtained identity in the literature, we provide the
respective reference. In any case, we have already seen two such
examples: If we start with the transformation formula \eqref{eq2.3}, 
then the algorithm results in \eqref{eq2.4},
and if we start with the transformation formula displayed as {\tt
Out[3]} of the {\sl Mathematica} session, then the algorithm results
in \eqref{e7}.
%
%A[2]:=
%\ni {\bf 1.} The transformation formula \cite[(1.3.15)]{SlatAC}
%\begin{equation}
%{} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { d}\end{matrix} ;
%   {\displaystyle z}\right ]  = 
%  {{\left( 1 - z \right) }^{-a - b + d}}\,
%   {} _{2} F _{1} \!\left [ \begin{matrix} { -a + d, -b + d}\\ {
%    d}\end{matrix} ; {\displaystyle z}\right ] 
%\label{eq3.1}
%\end{equation}
%results in 
%A[3]:=
%\begin{equation}
%{} _{3} F _{2} \!\left [ \begin{matrix} { a, b, c}\\ { d, e}\end{matrix} ;
%   {\displaystyle 1}\right ]  = 
%\frac {\G( e)\,\G(  d + e-a - b - c)} {\G(  e-c)\,\G(  d 
%+    e-a - b)}
%  {} _{3} F _{2} \!\left [ \begin{matrix} { c, d-a,   d-b}\\ { d, 
%    d + e-a - b}\end{matrix} ; {\displaystyle 1}\right ] \,
%\label{eq3.2}
%\end{equation}
%which belongs to the set of 10 non-terminating ${}_3F_2[1]$
%transformations (\cite[(III)]{Sri1}). 
The transformation formula \cite[(1.7.1.3)]{SlatAC},
$$
{} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { c}\end{matrix} ; {\displaystyle
   z}\right ]  = {{{\left( 1 - z \right) }^{-a}}}
 {{{} _{2} F _{1} \!\left [ \begin{matrix} { a, c-b}\\ {
      c}\end{matrix} ; {\displaystyle -{\frac z {1 - z}}}\right ] }
    },
$$
will also result in one of the 18 terminating ${}_3F_2[1]$ transformations 
(see \cite[(IX), p.~91]{Sri2}), when one of the numerator 
parameters is a negative integer.

\medskip
\ni {\bf 1.} If we start with the quadratic transformation formula 
%A[10]:=
\cite[(3.2)]{RaVeAA}
\begin{equation}
{} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { 1 + a - b}\end{matrix} ;
   {\displaystyle -z}\right ]  = 
  { {{{\left( 1 - z \right) }^{-a}}}}
{} _{2} F _{1} \!\left [ \begin{matrix} { {\frac a 2}, {\frac 1 2} + {\frac a
      2}}\\ { 1 + a - b}\end{matrix} ; {\displaystyle {-\frac {4\,z} {{{\left(
      1 - z \right) }^2}}}}\right ]
\label{eq3.3}
\end{equation}
and assume that $a$ is a nonpositive integer (so that the
$_2F_1$-series terminate), then we obtain
%A[11]:=
\begin{multline}
{} _{3} F _{2} \!\left [ \begin{matrix} { a, b, d}\\ { 1 + a - b,
   e}\end{matrix} ; {\displaystyle -1}\right ]  \\
= 
   \frac {\G( e)\,\G(  e-a - d)} {\G(  e-a)\,\G( e-d)}
  {} _{4} F _{3} \!\left [ \begin{matrix} { {\frac 1 2} + {\frac a 2}, {\frac a
    2}, d, 1 + a - e}\\ { 1 + a - b, {\frac 1 2} + {\frac a 2} + {\frac d 2} -
    {\frac e 2}, 1 + {\frac a 2} + {\frac d 2} - {\frac e 2}}\end{matrix} ;
    {\displaystyle 1}\right ] .
\label{eq3.4}
\end{multline}
This transformation relating a nearly-poised 
${}_3F_2[-1]$ to a ${}_4F_3[1]$ is
valid provided $a$ is a nonpositive integer. However, by a standard
polynomial trick, it can be seen that it is also true if
$a$ is arbitrary but $d$ is a nonpositive integer: let $d$ by a fixed
nonpositive integer. By multiplying both sides of \eqref{eq3.4} by 
$$(e-a)_d\,(1+a-b)_d\,(1+a+d-e)_{2d},$$
both sides become polynomials in $a$ of degree at most $5d$. These
two polynomials agree for all nonpositive $a$, since we
know that \eqref{eq3.4} is true for nonpositive $a$. These are
infinitely many values of $a$, whence the polynomials must be identical.

\medskip
\ni {\bf 2.} If we start with the transformation formula 
%A[12]:=
\cite[(5.10)]{RaVeAA}
\begin{equation}
{} _{2} F _{1} \!\left [ \begin{matrix} { a, {\frac 1 2} + a}\\ { {\frac 1 2} +
   b}\end{matrix} ; {\displaystyle {z^2}}\right ]  = 
  {
    {{{\left( 1 - z \right) }^{-2\,a}}}}
{} _{2} F _{1} \!\left [ \begin{matrix} { 2\,a, b}\\ { 2\,b}\end{matrix} ;
      {\displaystyle {\frac {2\,z} {z-1}}}\right ]
\label{eq3.5}
\end{equation}
and assume that $a$ is a nonpositive integer, then we obtain
%A[13]:=
\begin{equation}
  {} _{4} F _{3} \!\left [ \begin{matrix} { a, {\frac 1 2} + a, {\frac 1 2} +
    {\frac d 2}, {\frac d 2}}\\ { {\frac 1 2} + b, {\frac 1 2} + {\frac e 2},
    {\frac e 2}}\end{matrix} ; {\displaystyle 1}\right ] =
   \frac {\G( e)\,\G( e-2\,a - d )} {\G(  e-2\,a)\,\G( e-d)} 
{} _{3} F _{2} \!\left [ \begin{matrix} { 2\,a, b, d}\\ { 2\,b, 1 + 2\,a + d -
   e}\end{matrix} ; {\displaystyle 2}\right ]  .
\label{eq3.6}
\end{equation}
This identity is true provided $a$ is a nonpositive integer or $d$ is
a nonpositive integer, the latter because of the same arguments as
sketched in item 1.
%which after the replacements $2a\rightarrow a, d\rightarrow -n,
%1+a-e-n\rightarrow d$ and further simplification becomes:
%%A[14]:=
%\begin{equation}
%{} _{3} F _{2} \!\left [ \begin{matrix} { a, b, -n}\\ { 2\,b, d}\end{matrix} ;
%   {\displaystyle 2}\right ]  = 
%  {\frac {      ({ \textstyle -a + d}) _{n} } {({ \textstyle d}) _{n} }}
%{} _{4} F _{3} \!\left [ \begin{matrix} { {\frac 1 2} + {\frac a 2}, {\frac a
%       2}, {\frac 1 2} - {\frac n 2}, -{\frac {n} 2}}\\ { {\frac 1 2} + b, 1 +
%       {\frac a 2} - {\frac d 2} - {\frac n 2}, {\frac 1 2} + {\frac a 2} - {\frac d
%       2} - {\frac n 2}}\end{matrix} ; {\displaystyle 1}\right ],
%\label{eq3.7}
%\end{equation}
It can be found in the literature as a special case of a more general
transformation for basic hypergeometric series
(see \cite[Ex.~3.4, $q\rightarrow 1$, reversed]{GaRaAA}; it is
available as {\tt T3235} within HYP, see \cite{KratAS}).

\medskip
\ni {\bf 3.} If we start with the quadratic transformation formula of Gau{\ss} 
%A[17]:=
\cite[Ex.~4.(iii), p.~97, with $\alpha\to a/2$, $\beta\to b/2$, 
$x\to z$]{BailAA}
\begin{equation}
{} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { {\frac 1 2} + {\frac a 2} +
   {\frac b 2}}\end{matrix} ; {\displaystyle z}\right ]  = 
  {} _{2} F _{1} \!\left [ \begin{matrix} { {\frac a 2}, {\frac b 2}}\\ {
   {\frac 1 2} + {\frac a 2} + {\frac b 2}}\end{matrix} ; {\displaystyle
   4\,\left( 1 - z \right) \,z}\right ] 
\label{eq3.8}
\end{equation}
and assume that $a$ is a nonpositive integer, then we obtain
%A[18]:=
\begin{equation}
{} _{3} F _{2} \!\left [ \begin{matrix} { a, b, d}\\ { {\frac 1 2} + {\frac a 2}
   + {\frac b 2}, e}\end{matrix} ; {\displaystyle 1}\right ]  = 
  {} _{4} F _{3} \!\left [ \begin{matrix} { {\frac a 2}, {\frac b 2}, d, -d +
   e}\\ { {\frac 1 2} + {\frac a 2} + {\frac b 2}, {\frac 1 2} + {\frac e 2},
   {\frac e 2}}\end{matrix} ; {\displaystyle 1}\right ] .
\label{eq3.9}
\end{equation}
This identity is true provided 
both hypergeometric series terminate.
It is the main theorem in \cite{SinVAA}
(actually, there even a $q$-analogue of \eqref{eq3.9} can be found;
see also  \cite[(3.10.13); Appendix (III.21)]{GaRaAA}).
This same result \eqref{eq3.9} will be obtained  
when $z\to 1-z$ in \eqref{eq3.8}.

\medskip
\ni {\bf 4.} If we start with the quadratic transformation formula 
%A[19]:=
\cite[(5.12), reversed]{RaVeAA}
\begin{equation}
{} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { 2\,b}\end{matrix} ;
   {\displaystyle z}\right ]  = 
  {    {{{\left( 1 - z \right) }^{-{a/ 2}}}}}
{} _{2} F _{1} \!\left [ \begin{matrix} { {\frac a 2}, -\frac {a}
{2}+b }\\ { {\frac {1 } 2}+b}\end{matrix} ; {\displaystyle {\frac {{z^2}}
      {4\,\left( z-1 \right) }}}\right ]
\label{eq3.10}
\end{equation}
and assume that $a$ is an even nonpositive integer, then we obtain
%A[20]:=
\begin{equation}
{} _{3} F _{2} \!\left [ \begin{matrix} { a, b, d}\\ { 2\,b, e}\end{matrix} ;
   {\displaystyle 1}\right ]  = 
   \frac {\G(e)\,\G( e-{\frac {a} 2} - d)} {\G( e-{\frac {a} 2})\,
\G(    e-d)}
  {} _{4} F _{3} \!\left [ \begin{matrix} { {\frac a 2}, -{\frac {a} 2} + b,
    {\frac 1 2} + {\frac d 2}, {\frac d 2}}\\ { {\frac 1 2} + b, 1 + {\frac a 2} +
    d - e, -{\frac {a} 2} + e}\end{matrix} ; {\displaystyle
1}\right],
\label{eq3.11}
\end{equation}
provided $a$ is an even nonpositive integer or $d$ is any nonpositive integer.
This identity can also be obtained in a different (but more
complicated) way: In the transformation formula listed as {\tt T4391}
in HYP (see \cite{KratAS}; it is Eq.~(3.5.7) from \cite{GaRaAA} with
$q\to1$) let $e\to\infty$. Then on the right-hand
side the second term vanishes, while the first is a very-well-poised
$_7F_6$-series to which Whipple's $_7F_6$-to-$_4F_3$ transformation 
(see \cite[(2.4.1.1)]{SlatAC}) can be applied. The result is the
above transformation formula \eqref{eq3.11}.

\medskip
\ni {\bf 5.} The transformation formula 
%A[21]:=
\cite[(3.31), reversed]{RaVeAA}
\begin{equation}
{} _{2} F _{1} \!\left [ \begin{matrix} { a, 1 - a}\\ { c}\end{matrix} ;
   {\displaystyle z}\right ]  = 
  {{\left( 1 - z \right) }^{c-1}}\,
   {} _{2} F _{1} \!\left [ \begin{matrix} { -\frac {a} {2}+\frac {c}
{2}, -\frac {1} {2}+\frac {a} {2}+\frac {c} {2}}\\ 
{ c}\end{matrix} ; {\displaystyle 4\,\left( 1 - z \right)
    \,z}\right ] 
\label{eq3.12}
\end{equation}
results in 
%A[22]:=
\begin{multline}
{} _{3} F _{2} \!\left [ \begin{matrix} { 1 - a, a, d}\\ { c, e}\end{matrix} ;
   {\displaystyle 1}\right ]  \\
= 
   \frac {\G( e)\,\G( c - d + e-1)} {\G( c + e-1)\, \G(e-d)}
  {} _{4} F _{3} \!\left [ \begin{matrix} { -{\frac {a} 2} + {\frac c 2},
    -{\frac 1 2} + {\frac a 2} + {\frac c 2}, d, c - d + e-1}\\ { c, -{\frac 1
    2} + {\frac c 2} + {\frac e 2}, {\frac c 2} + {\frac e 2}}\end{matrix} ;
    {\displaystyle 1}\right ],
\label{eq3.13}
\end{multline}
provided both hypergeometric series terminate.

\medskip
\ni {\bf 6.} If we start with the transformation formula 
\cite[(4.10), with $x\to -z$]{BailAB}
%A[36]:=
\begin{equation}
{} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { 2\,b}\end{matrix} ; {\displaystyle
   {-\frac {4\,z} {{{\left( 1 - z \right) }^2}}}}\right ]  = 
  {{\left( 1 - z \right) }^{2\,a}}\,
   {} _{2} F _{1} \!\left [ \begin{matrix} { a, {\frac 1 2} + a - b}\\ { {\frac 1 2} +
    b}\end{matrix} ; {\displaystyle {z^2}}\right ],
\label{eq3.26}
\end{equation}
which is a combination of \cite[(5.10)]{RaVeAA} and 
\cite[(6.2), reversed]{RaVeAA}, 
and assume that $a$ is a nonpositive integer, then we obtain
%A[37]:=
\begin{multline}
{} _{4} F _{3} \!\left [ \begin{matrix} { a, b, d, e}\\ { 2\,b, {\frac d 2} + {\frac e
   2}, {\frac 1 2} + {\frac d 2} + {\frac e 2}}\end{matrix} ; {\displaystyle
   1}\right ]  \\= 
   \frac {\G( 1 - e)\,\G( 1 + 2\,a - d - e)} {\G( 1 + 2\,a - e)\,\G( 1 - d -
    e)}
{} _{4} F _{3} \!\left [ \begin{matrix} { a, {\frac 1 2} + a - b,
    {\frac 1 2} + {\frac d 2}, {\frac d 2}}\\ { {\frac 1 2} + b, {\frac 1 2} + a -
    {\frac e 2}, 1 + a - {\frac e 2}}\end{matrix} ; {\displaystyle
1}\right],
\label{eq3.27}
\end{multline}
provided $a$ or $d$ is a nonpositive integer.

\medskip
\ni {\bf 7.} If we start with the transformation formula (\cite[(4.22), with $\alpha\to
a$, $\beta\to b$, $x\to z$]{BailAB} and 
\cite[(5.12), with $z\to {z}/(2-z)$]{RaVeAA}):
%A[40]:=
\begin{equation}
{} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { {\frac 1 2} + a + b}\end{matrix} ;
   {\displaystyle {\frac {{z^2}} {4\,\left(  z-1 \right) }}}\right ]  = 
  {{\left( 1 - z \right) }^a}\,{} _{2} F _{1} \!\left [ \begin{matrix} { 2\,a, a +
    b}\\ { 2\,a + 2\,b}\end{matrix} ; {\displaystyle z}\right ] 
\label{eq3.28}
\end{equation}
and assume that $a$ is a nonpositive integer, then we obtain
%A[41]:=
\begin{equation}
{} _{4} F _{3} \!\left [ \begin{matrix} { a, b, {\frac 1 2} + {\frac d 2}, {\frac d
   2}}\\ { {\frac 1 2} + a + b, 1 + d - e, e}\end{matrix} ; {\displaystyle
   1}\right ]  = 
   \frac {\G( e)\,\G( a - d + e)} {\G( a + e)\,\G( e-d)}
{} _{3} F _{2} \!\left [ \begin{matrix} { 2\,a, a + b, d}\\ { 2\,a +
    2\,b, a + e}\end{matrix} ; {\displaystyle 1}\right ],
\label{eq3.29}
\end{equation}
provided $a$ or $d$ is a nonpositive integer.

\medskip
\ni {\bf 8.} The transformation formula \cite[(3.31), with $z\to 
{4z}/(1-z)$, $a\to a+c+\frac {1} {2}$, $c\to b$]{RaVeAA}:
%A[42]:=
\begin{equation}
{} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { {\frac 1 2} + a + b}\end{matrix} ;
   {\displaystyle 4\,\left( 1 - z \right) \,z}\right ]  = 
  {{\left( 1 - z \right) }^{{\frac 1 2} - a - b}}\,
   {} _{2} F _{1} \!\left [ \begin{matrix} { {\frac 1 2} + a - b, {\frac 1 2} - a +
    b}\\ { {\frac 1 2} + a + b}\end{matrix} ; {\displaystyle z}\right ] 
\label{eq3.30}
\end{equation}
and assume that $a$ is a nonpositive integer, then we obtain
%A[43]:=
\begin{multline}
{} _{4} F _{3} \!\left [ \begin{matrix} { a, b, d, e-d}\\ { {\frac 1 2} + a + b,
   {\frac 1 2} + {\frac e 2}, {\frac e 2}}\end{matrix} ; {\displaystyle 1}\right ] 
\\    = 
   \frac {\G(e)\,\G( {\frac 1 2} - a - b - d + e)} {\G( {\frac 1 2} - a - b +
    e)\,\G( e-d)}
{} _{3} F _{2} \!\left [ \begin{matrix} { {\frac 1 2} + a - b, {\frac 1 2} - a +
    b, d}\\ { {\frac 1 2} + a + b, {\frac 1 2} - a - b + e}\end{matrix} ;
    {\displaystyle 1}\right ],
\label{eq3.31}
\end{multline}
provided both hypergeometric series terminate.
This transformation can also be obtained by combining the
$_4F_3$-to-$_3F_2$ transformation from \cite{SinVAA} 
that occurred already 
in item~3 with the $_3F_2$-transformation \eqref{e7}.

\medskip
\ni {\bf 9.} If we start with the transformation formula 
%A[25]:=
\cite[(3.4.8), $q\to 1$, reversed]{GaRaAA}
\begin{equation}
\left( 1 + z \right) \,{} _{2} F _{1} \!\left [ \begin{matrix} { a, {\frac 1 2} + a}\\
    { b}\end{matrix} ; {\displaystyle {-\frac {4\,z} {{{\left( 1 - z \right)
    }^2}}}}\right ]  = {{\left( 1 - z \right) }^{2\,a}}\,
   {} _{3} F _{2} \!\left [ \begin{matrix} { 2\,a-1, {\frac 1 2} + a, 2\,a - b}\\ {
    -{\frac 1 2} + a, b}\end{matrix} ; {\displaystyle -z}\right ] 
\label{eq3.16}
\end{equation}
and assume that $a$ is a nonpositive integer, then we obtain
%A[26]:=
\begin{multline}
\left( d + e \right) \,{} _{4} F _{3} \!\left [ \begin{matrix} { a, {\frac 1 2}
    + a, d, -e}\\ { b, {\frac 1 2} + {\frac d 2} - {\frac e 2}, 1 + {\frac d 2} -
    {\frac e 2}}\end{matrix} ; {\displaystyle 1}\right ]  \\
= 
\frac {\G( 1 + e)\,\G( 2\,a - d + e)} {\G( 2\,a +
    e)\,\G( e-d)}
  {} _{4} F _{3} \!\left [ \begin{matrix} { 2\,a-1, {\frac 1 2} + a,  2\,a -
    b, d}\\ { -{\frac 1 2} + a, b, 2\,a + e}\end{matrix} ; {\displaystyle
    -1}\right ], 
\label{eq3.17}
\end{multline}
provided $a$ or $d$ is a nonpositive integer.
(In fact, when applying the beta integral method,
we have to deal with a sum of two series on the left-hand side,
which generates the factor $(d+e)$ in the result.)

\medskip
\ni {\bf 10.} If we start with the transformation formula 
%A[28]:=
\cite[p.~97, Ex.~4.(iv), with $x\to z$]{BailAA}
\begin{equation}
{} _{3} F _{2} \!\left [ \begin{matrix} { {\frac 1 2}a, {\frac 1 2}+{\frac a 2}, 1 + a
   - b - c}\\ { 1 + a - b, 1 + a - c}\end{matrix} ; {\displaystyle
   {-\frac {4\,z} {{{\left( 1 - z \right) }^2}}}}\right ]  = 
  {{\left( 1 - z \right) }^a}{} _{3} F _{2} \!\left [ \begin{matrix} { a, b,
    c}\\ { 1 + a - b, 1 + a - c}\end{matrix} ; {\displaystyle z}\right ] 
\label{eq3.18}
\end{equation}
and assume that $a$ is a nonpositive integer, then we obtain
%A[29]:=
\begin{multline}
{} _{5} F _{4} \!\left [ \begin{matrix} { {\frac 1 2} + {\frac a 2}, {\frac a 2},
   1 + a - b - c, d, 1 - e}\\ { 1 + a - b, 1 + a - c, {\frac 1 2} + {\frac d 2}
   - {\frac e 2}, 1 + {\frac d 2} - {\frac e 2}}\end{matrix} ; {\displaystyle
   1}\right ]  \\
=   \frac {\G( e)\,\G( a - d + e)} {\G( a + e)\,\G(e-d)}
 {} _{4} F _{3} \!\left [ \begin{matrix} { a, b, c, d}\\ { 1 +
    a - b, 1 + a - c, a + e}\end{matrix} ; {\displaystyle 1}\right ],
\label{eq3.19}
\end{multline}
provided $a$ or $d$ is a nonpositive integer.
This is a known transformation between a nearly-poised ${}_4F_3[1]$ series 
and a Saalsch\"utzian ${_5F_4[1]}$ series (\cite[(2.4.2.3)]{SlatAC};
to see this do the replacements
$a\to f$, $b\to 1+f-h$, $c\to h-a$, $e\to g-f$, in \eqref{eq3.19}). 
\medskip

\ni {\bf 11.} If we start with the transformation formula 
%A[32]:=
\cite[p.~97, Ex.~6, with $b\to 1+a-b$, $c\to 1+a-c$, $x\to z$]{BailAA}
\begin{multline}
\left( 1 + z \right) \,{} _{3} F _{2} \!\left [ \begin{matrix} { {\frac 1 2} + {\frac a
    2}, 1 + {\frac a 2}, - a + b + c-1}\\ { b, c}\end{matrix} ; {\displaystyle
    {-\frac {4\,z} {{{\left( z-1 \right) }^2}}}}\right ]  \\
= 
  {{\left( 1 - z \right) }^{1 + a}}\,
   {} _{4} F _{3} \!\left [ \begin{matrix} { a, 1 + {\frac a 2}, 1 + a - b, 1 + a -
    c}\\ { {\frac a 2}, b, c}\end{matrix} ; {\displaystyle z}\right ] 
\label{eq3.20}
\end{multline}
and assume that $a$ is a nonpositive integer, then we obtain
%A[33]:=
\begin{multline}
{} _{5} F _{4} \!\left [ \begin{matrix} { {\frac 1 2} + {\frac a 2}, 1 + {\frac a 2}, 
   - a + b + c-1, d, -e}\\ { b, c, {\frac 1 2} + {\frac d 2} - {\frac e 2}, 1 +
   {\frac d 2} - {\frac e 2}}\end{matrix} ; {\displaystyle 1}\right ]  \\
= 
  {\frac { 1} {d + e}}
\frac {\G( 1 + e)\,\G( 1 + a - d + e)} {\G( 1 + a + e)\,\G(
        e-d)}
{} _{5} F _{4} \!\left [ \begin{matrix} { a, 1 + {\frac a 2}, 1 + a - b, 1 + a -
       c, d}\\ { {\frac a 2}, b, c, 1 + a + e}\end{matrix} ; {\displaystyle
       1}\right ],
\label{eq3.21}
\end{multline}
provided $a$ or $d$ is a nonpositive integer.
On replacing $b\to 1+a-b, c\to 1+a-c, d\to -m, e\to -1-a+w$, this
identity corresponds to \cite[(4.5.2)]{BailAA}.
(The parenthetical remark in item~9 applies also here. However, in the
result, we
moved the factor $(d+e)$, which is generated by the sum of two series on
the left-hand side of \eqref{eq3.20}, to the right-hand side.)

\medskip
\ni {\bf 12.} If we start with the transformation formula 
\cite[(5.8)]{AndSt}:
%A[45]:=
\begin{multline}
\left( 1 - {\frac z 2} \right) \,
   {} _{4} F _{3} \!\left [ \begin{matrix} { 1 + a, a - 2\,b, 1 - a + 2\,b, {\frac 4
    3} + {\frac a 3}}\\ { {\frac 3 2} + b, 1 + a - b, {\frac 1 3} + {\frac a
    3}}\end{matrix} ; {\displaystyle {-\frac {{z^2}} {4\,\left( 1 - z \right)
    }}}\right ]  \\= \left( 1 - 2\,z \right) \,
   {{\left( 1 - z \right) }^{1 + a}}\,
   {} _{4} F _{3} \!\left [ \begin{matrix} { 1 + a, 1 + b, {\frac 1 2} + a - b,
    {\frac 5 3} + {\frac {2\,a} 3}}\\ { 2 + 2\,b, 1 + 2\,a - 2\,b, {\frac 2 3} +
    {\frac {2\,a} 3}}\end{matrix} ; {\displaystyle 4\,\left( 1 - z \right)
    \,z}\right ].
\label{eq3.32} 
\end{multline}
and assume that $a$ is a negative integer, then we obtain
%A[46]:=
\begin{multline}
{} _{6} F _{5} \!\left [ \begin{matrix} { {\frac 4 3} + {\frac a 3}, 1 + a, a - 2\,b, 1
   - a + 2\,b, {\frac 1 2} + {\frac d 2}, {\frac d 2}}\\ { {\frac 1 3} + {\frac a
   3}, 1 + a - b, {\frac 3 2} + b, 1 + d - e, 1 + e}\end{matrix} ; {\displaystyle
   1}\right ]  \\= {\frac {2\,\left( 1 + a - 2\,d + e \right) \,
       } {2\,e-d}}\,
\frac {\G( 1 + e)\,\G( 1 + a - d + e)} {\G( 2 + a + e)\,
       \G(e-d)}\kern2cm\\
\times
{} _{6} F _{5} \!\left [ \begin{matrix} { {\frac 5 3} + {\frac {2\,a} 3}, 1 + a,
       {\frac 1 2} + a - b, 1 + b, d, 1 + a - d + e}\\ { {\frac 2 3} +
       {\frac {2\,a} 3}, 1 + 2\,a - 2\,b, 2 + 2\,b, 1 + {\frac a 2} + {\frac e
       2}, {\frac 3 2} + {\frac a 2} + {\frac e 2}}\end{matrix} ; {\displaystyle
       1}\right ],
\label{eq3.33}
\end{multline}
provided $a$ is a negative integer or $d$ is a nonpositive integer.
(Again, the parenthetical remark of item~9 applies, this time on both
sides. The factors generated appear in the first term on the
right-hand side.)

\medskip
\ni {\bf 13.} If we start with the cubic transformation formula 
\cite[(4.05), with 
$\rho_1\to b$,
$\rho_2\to 3a-b+{\frac 3 2}$, $x\to z$]{BailAB}
%A[34]:=
\begin{equation}
{
    {{{\left( 1 - z \right) }^{-3\,a}}}} 
{} _{3} F _{2} \!\left [ \begin{matrix} { a, {\frac 1 3} + a, {\frac 2 3} + a}\\ { b,
      {\frac 3 2} + 3\,a - b}\end{matrix} ; {\displaystyle {-\frac {27\,z}
      {4\,{{\left( 1 - z \right) }^3}}}}\right ]= 
  {} _{3} F _{2} \!\left [ \begin{matrix} { 3\,a, - 3\,a + 2\,b-1, 2 + 3\,a -
   2\,b}\\ { b, {\frac 3 2} + 3\,a - b}\end{matrix} ; {\displaystyle {\frac z
   4}}\right ] 
\label{eq3.22}
\end{equation}
and assume that $a$ is a nonpositive integer, then we obtain
%A[35]:=
\begin{multline}
{} _{6} F _{5} \!\left [ \begin{matrix} { a, {\frac 1 3} + a, {\frac 2 3} + a, d,
   {\frac 1 2} + {\frac {3\,a} 2} - {\frac e 2}, 1 + {\frac {3\,a} 2} - {\frac e
   2}}\\ { {\frac 3 2} + 3\,a - b, b, {\frac 1 3} + a + {\frac d 3} - {\frac e 3},
   {\frac 2 3} + a + {\frac d 3} - {\frac e 3}, 1 + a + {\frac d 3} - {\frac e
   3}}\end{matrix} ; {\displaystyle 1}\right ]  \\= 
\frac {\G( -3\,a + e)\,\G( e-d)} {\G( e)\,\G( -3\,a - d +
    e)}
  {} _{4} F _{3} \!\left [ \begin{matrix} { 3\,a, 2 + 3\,a - 2\,b, - 3\,a +
2\,b-1,
    d}\\ { {\frac 3 2} + 3\,a - b, b, e}\end{matrix} ; {\displaystyle {\frac 1
    4}}\right ],
\label{eq3.23} 
\end{multline}
provided $a$ or $d$ is a nonpositive integer. This is
an unusual identity featuring a transformation between a 
${}_6F_5[1]$-series and a ${}_4F_3[\frac {1} {4}]$-series.
\medskip

\ni {\bf 14.} If we start with the second cubic transformation formula of Bailey
\cite[(4.06), with 
$\rho_1\to b$, $\rho_2\to \frac {3} {2}+3a-b$, $x\to z$]{BailAB}:
%A[38]:=
\begin{equation}
{} _{3} F _{2} \!\left [ \begin{matrix} { a, {\frac 1 3} + a, {\frac 2 3} + a}\\ { b,
   {\frac 3 2} + 3\,a - b}\end{matrix} ; {\displaystyle {\frac {27\,{z^2}}
   {4\,{{\left( 1 - z \right) }^3}}}}\right ]  = 
  {{\left( 1 - z \right) }^{3\,a}}\,
   {} _{3} F _{2} \!\left [ \begin{matrix} { 3\,a, -{\frac 1 2} + b, 1 + 3\,a - b}\\ {
    2\,b-1, 2 + 6\,a - 2\,b}\end{matrix} ; {\displaystyle 4\,z}\right ] 
\label{eq3.24}
\end{equation}
and assume that $a$ is a nonpositive integer, then we obtain
%A[39]:=
\begin{multline}
{} _{6} F _{5} \!\left [ \begin{matrix} { a, {\frac 1 3} + a, {\frac 2 3} + a, {\frac 1
   2} + {\frac d 2}, {\frac d 2}, e}\\ { {\frac 3 2} + 3\,a - b, b, {\frac d 3} +
   {\frac e 3}, {\frac 1 3} + {\frac d 3} + {\frac e 3}, {\frac 2 3} + {\frac d 3} +
   {\frac e 3}}\end{matrix} ; {\displaystyle 1}\right ]  \\= 
\frac {\G(1 - e)\,\G( 1 + 3\,a - d - e)} {\G( 1 + 3\,a -
    e)\,\G( 1 - d - e)}
  {} _{4} F _{3} \!\left [ \begin{matrix} { 3\,a, 1 + 3\,a - b, -{\frac 1 2} + b, d}\\
    { 2 + 6\,a - 2\,b, 2\,b-1, 1 + 3\,a - e}\end{matrix} ; {\displaystyle
    4}\right ],
\label{eq3.25} 
\end{multline}
provided $a$ or $d$ is a nonpositive integer.

\medskip
\ni {\bf 15.} If we start with the transformation formula \cite[Entry~4 of
Ramanujan, Ch.~11, p.~50]{BcBAA}:
%A[47]:=
\begin{equation}
{} _{2} F _{1} \!\left[ \begin{matrix} { {\frac a 2}, {\frac 1 2} + {\frac a 2}}\\ {
   {\frac 1 2} + b}\end{matrix} ; {\displaystyle {-\frac {4\,z} {{{\left( 1 - z
   \right) }^2}}}}\right ]  = {{\left( 1 - z \right) }^a}\,
   {} _{2} F _{1} \!\left [ \begin{matrix} { a, {\frac 1 2} + a - b}\\ { {\frac 1 2} +
    b}\end{matrix} ; {\displaystyle -z}\right ] 
\label{eq3.34}
\end{equation}
and assume that $a$ is a nonpositive integer, then we obtain
%A[48]:=
\begin{equation}
{} _{4} F _{3} \!\left [ \begin{matrix} { {\frac 1 2} + {\frac a 2}, {\frac a 2}, d,
   e}\\ { {\frac 1 2} + b, {\frac d 2} + {\frac e 2}, {\frac 1 2} + {\frac d 2} +
   {\frac e 2}}\end{matrix} ; {\displaystyle 1}\right ]  = 
   \frac {\G(1 - e)\,\G( 1 + a - d - e)} {\G( 1 + a - e)\,\G( 1 - d -
    e)}
  {} _{3} F _{2} \!\left [ \begin{matrix} { a, {\frac 1 2} + a - b, d}\\ { {\frac 1 2}
    + b, 1 + a - e}\end{matrix} ; {\displaystyle -1}\right ],
\label{eq3.35}
\end{equation}
provided $a$ or $d$ is a nonpositive integer.

\medskip

\ni {\bf 16.} The transformation formula \cite[(1.8.10), with $c$ replaced by 
$1+a+b-c$ and  $z$ replaced by $1-z$]{SlatAC}
expressing the Gau{\ss} solution valid for $|z|<1$, in terms of 
the Gau{\ss} functions valid for $|z-1|<1$ is 
%A[3]:=
\begin{multline}
{} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { c}\end{matrix} ;
   {\displaystyle z}\right ]  = 
  {{\left( 1 - z \right) }^{c-a-b}}\,
    \frac {\G( c)\,\G( a + b - c)} {\G( a)\,\G( b)}
    {} _{2} F _{1} \!\left [ \begin{matrix} { -b + c, -a + c}\\ { 1 - a - b +
     c}\end{matrix} ; {\displaystyle 1 - z}\right ] 
\\
 + 
    \frac {\G(c)\,\G( c -a - b)} { \G(c -b)\,\G( c -a)}
   {} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { 1 + a + b -
     c}\end{matrix} ; {\displaystyle 1 - z}\right ] .
\label{eq3.36}
\end{multline}
If our algorithm is applied, it results in
%A[4]:=
\begin{multline}
{} _{3} F _{2} \!\left [ \begin{matrix} { a, b, c}\\ { d, e}\end{matrix} ;
   {\displaystyle 1}\right ]  = 
    \frac {\G(d)\,\G( d -a - b)} {\G( d -a)\,\G( d-b)}
  {} _{3} F _{2} \!\left [ \begin{matrix} { a, b, -c + e}\\ { 1 + a + b - d,
     e}\end{matrix} ; {\displaystyle 1}\right ] 
\\
 +     \frac {\G(a + b - d)\,\G( d)\,\G( e)\,\G( d + e -a - b - c)}
{\G( a)\,\G( b)\,\G(
      e-c)\,\G(d + e -a - b)}
{} _{3} F _{2} \!\left [ \begin{matrix} {  d-a, d-b, d + e-a -
     b - c}\\ { 1  + d- a - b, d + e-a - b}\end{matrix} ;
     {\displaystyle 1}\right ] 
\label{eq3.37}
\end{multline}
This three-term $_3F_2$-transformation formula can for example be
found in \cite[(4.3.4.2)]{SlatAC}.

\medskip

\ni {\bf 17.} The transformation formula 
\cite[Entry~21 of Ramanujan]{BcBAA}
%A[6]:=
\begin{multline}
{} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { {\frac 1 2} + {\frac a 2} +
   {\frac b 2}}\end{matrix} ; {\displaystyle {\frac {1 - z} 2}}\right ]  = 
  z\,\frac {\G( -{\frac 1 2})\,\G( {\frac 1 2} +
     {\frac a 2} + {\frac b 2})} {\G( {\frac a 2})\,\G( {\frac b 2})}
{} _{2} F _{1} \!\left [ \begin{matrix} { {\frac 1 2} + {\frac a 2},
     {\frac 1 2} + {\frac b 2}}\\ { {\frac 3 2}}\end{matrix} ; {\displaystyle
     {z^2}}\right ] \\
 +     \frac {\G({\frac 1 2})\,\G( {\frac 1 2} + {\frac a 2} +
     {\frac b 2})} {\G( {\frac 1 2} + {\frac a 2})\,\G( {\frac 1 2} + {\frac b
2})}
   {} _{2} F _{1} \!\left [ \begin{matrix} { {\frac a 2}, {\frac b 2}}\\ {
     {\frac 1 2}}\end{matrix} ; {\displaystyle {z^2}}\right ] 
\label{eq3.38}
\end{multline}
results in 
%A[7]:=
\begin{multline}
{} _{3} F _{2} \!\left [ \begin{matrix} { a, b, e-d}\\ { {\frac 1 2} +
   {\frac a 2} + {\frac b 2}, e}\end{matrix} ; {\displaystyle {\frac 1 2}}\right
   ]  = 
    \frac {\G({\frac 1 2})\,\G( {\frac 1 2} + {\frac a 2} +
     {\frac b 2})} {\G( {\frac 1 2} + {\frac a 2})\,\G( {\frac 1 2} + {\frac b
2})}
{} _{4} F _{3} \!\left [ \begin{matrix} { {\frac a 2}, {\frac b 2},
     {\frac 1 2} + {\frac d 2}, {\frac d 2}}\\ { {\frac 1 2}, {\frac 1 2} + {\frac e
     2}, {\frac e 2}}\end{matrix} ; {\displaystyle 1}\right ] 
 \\
+     \frac {\G( -{\frac 1 2})\,\G( {\frac 1 2} + {\frac a 2} +
     {\frac b 2})\,\G( 1 + d)\,\G( e)} {\G( {\frac a 2})\,\G( {\frac b
2})\,\G( d)\,\G( 1 + e)}
{} _{4} F _{3} \!\left [ \begin{matrix} { {\frac 1 2} + {\frac a
     2}, {\frac 1 2} + {\frac b 2}, {\frac 1 2} + {\frac d 2}, 1 + {\frac d 2}}\\ {
     {\frac 3 2}, {\frac 1 2} + {\frac e 2}, 1 + {\frac e 2}}\end{matrix} ;
     {\displaystyle 1}\right ].
\label{eq3.39}
\end{multline}

\section{Products of hypergeometric series and identities for Kamp\'e
de F\'eriet series}

\arraycolsep2pt

There exist many relations connecting products of hypergeometric
series, including the ones called the theorems of Cayley and Orr
\cite[Sec.~10.1]{BailAA}. In this section we apply our machinery to
identities of the form 
\begin{equation} \label{eq:prod-ser}
\text {``product of two hypergeometric series = hypergeometric
series."}
\end{equation}
As before, on the right hand side we will obtain a new hypergeometric
series. However, we will see that
on the left hand side we obtain Kamp\'e de F\'eriet series. These are
hypergeometric double series. The standard notation, which will be
used in the sequel, for these is as follows:
$${} F{}^{A:B;B'} _{C:D;D'} \!\left [ \begin{array}{cccccc} 
(a)&:&(b)&;&(b')&;\\
   (c)&:&(d)&;&(d')&;\end{array} \quad
   {\displaystyle z_1,z_2} \right ]=
\sum _{m,n\ge0} ^{}\frac {\prod _{j=1} ^{A}(a_j)_{m+n}\,
\prod _{j=1} ^{B}(b_j)_{m}\,\prod _{j=1} ^{B'}(b'_j)_{n}} {\prod _{j=1}
^{C}(c_j)_{m+n}\,
\prod _{j=1} ^{D}(d_j)_{m}\,\prod _{j=1} ^{D'}(d'_j)_{n}}
\frac {z_1^m z_2^n} {m!\,n!}.$$
We report below the interesting results
we obtain from the identities of the form \eqref{eq:prod-ser} 
found in \cite[Sec.~2.5]{SlatAC}.

\medskip

\ni {\bf 1.} ``The" standard theorem of the type \eqref{eq:prod-ser}
is the celebrated Clausen's theorem. 
It concerns the square of a Gau{\ss} series and it is 
%A[1]:=
\begin{equation}
{{{} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { {\frac 1 2} + a +
      b}\end{matrix} ; {\displaystyle z}\right ] }^2} = 
  {} _{3} F _{2} \!\left [ \begin{matrix} { 2\,a, 2\,b, a + b}\\ { 2\,a +
   2\,b, {\frac 1 2} + a + b}\end{matrix} ; {\displaystyle z}\right].
\label{eq4.8}
\end{equation}
The {\sl Mathematica} program given in Section 2 above has been
modified to take care of cases like this and it generated the
following result from \eqref{eq4.8}:
%A[2]:=
\begin{equation}
{} F{}^{1:2;2} _{1:1;1} \!\left [ \begin{array}{cccccc} d&:&a, b&;&a, b&;\\
   e&:&{\frac 1 2} + a + b&;&{\frac 1 2} + a + b&;\end{array} \quad
   {\displaystyle x,x} \right ]  = 
  {} _{4} F _{3} \!\left [ \begin{matrix} { 2\,a, 2\,b, a + b, d}\\ { 
   {\frac 1 2} + a + b, 2\,a + 2\,b, e}\end{matrix} ; {\displaystyle x}\right ]
,
\label{eq4.9}
\end{equation}
(The reader should note that the above result is obtained by first
replacing $z$ by $xz$ in \eqref{eq4.8} and subsequently applying our
method.)

\medskip

%A[3]:=
%\begin{equation}
%{} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { -{\frac 1 2} + a +
%    b}\end{matrix} ; {\displaystyle z}\right ] \,
%   {} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { {\frac 1 2} + a +
%    b}\end{matrix} ; {\displaystyle z}\right ]  = 
%  {} _{3} F _{2} \!\left [ \begin{matrix} { 2\,a, 2\,b, a + b}\\ { 2\,a +
%   2\,b-1, {\frac 1 2} + a + b}\end{matrix} ; {\displaystyle z}\right ] 
%\label{eq4.}
%\end{equation}
%A[4]:=
%\begin{equation}
%{} F{}^{1:2;2} _{1:1;1} \!\left [ \begin{matrix} d&:&a, b&;&a, b&;\\
%   e&:&-{\frac 1 2} + a + b&;&{\frac 1 2} + a + b&;\end{matrix} \quad
%   {\displaystyle x,x} \right ]  = 
%  {} _{4} F _{3} \!\left [ \begin{matrix} { 2\,a, 2\,b, a + b, d}\\ { {\frac 1
%   2} + a + b, 2\,a + 2\,b-1, e}\end{matrix} ; {\displaystyle x}\right ] 
%\label{eq4.}
%\end{equation}

%\medskip
%\ni {\bf .} 
%A[5]:=
%\begin{equation}
%{} _{2} F _{1} \!\left [ \begin{matrix} { a, b-1}\\ { -{\frac 1 2} + a +
%    b}\end{matrix} ; {\displaystyle z}\right ] \,
%   {} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { -{\frac 1 2} + a +
%    b}\end{matrix} ; {\displaystyle z}\right ]  = 
%  {} _{3} F _{2} \!\left [ \begin{matrix} { 2\,a, 2\,b-1, a + b-1}\\ {
%   -2 + 2\,a + 2\,b, -{\frac 1 2} + a + b}\end{matrix} ; {\displaystyle
%   z}\right ] 
%\label{eq4.}
%\end{equation}
%A[6]:=
%\begin{equation}
%{} F{}^{1:2;2} _{1:1;1} \!\left [ \begin{matrix} d&:&a, b-1&;&a, b&;\\
%   e&:&-{\frac 1 2} + a + b&;&-{\frac 1 2} + a + b&;\end{matrix} \quad
%   {\displaystyle x,x} \right ]  = 
%  {} _{4} F _{3} \!\left [ \begin{matrix} { 2\,a, a + b-1, 2\,b-1, d}\\
%   { -{\frac 1 2} + a + b, -2 + 2\,a + 2\,b, e}\end{matrix} ; {\displaystyle
%   x}\right ] 
%\label{eq4.}
%\end{equation}

\medskip
\ni {\bf 2.} The transformation which follows from Theorem~VII of 
Bailey's \cite[(2.5.31)]{SlatAC}
%A[7]:=
\begin{equation}
{} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { 1 + a + b - c}\end{matrix}
    ; {\displaystyle z}\right ] \,
   {} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { c}\end{matrix} ;
    {\displaystyle z}\right ]  = 
  {} _{4} F _{3} \!\left [ \begin{matrix} { a, b, {\frac a 2} + {\frac b 2},
   {\frac 1 2} + {\frac a 2} + {\frac b 2}}\\ { a + b, c, 1 + a + b -
   c}\end{matrix} ; {\displaystyle 4\,\left( 1 - z \right) \,z}\right ] 
\label{eq4.10}
\end{equation}
results in 
%A[8]:=
\begin{equation}
{} F{}^{1:2;2} _{1:1;1} \!\left [ \begin{matrix} d&:&a, b&;&a, b&;\\ e&:&1 + a
   + b - c&;&c&;\end{matrix} \quad {\displaystyle 1,1} \right ]  = 
  {} _{6} F _{5} \!\left [ \begin{matrix} { a, {\frac a 2} + {\frac b 2},
   {\frac 1 2} + {\frac a 2} + {\frac b 2}, b, d, e-d}\\ { a + b, 1 + a + b -
   c, c, {\frac 1 2} + {\frac e 2}, {\frac e 2}}\end{matrix} ; {\displaystyle
   1}\right ]. 
\label{eq4.11}
\end{equation}

\medskip
\ni {\bf 3.} If we start with 
the transformation which follows from Theorem~VIII of 
Bailey's \cite[(2.5.32)]{SlatAC} 
%A[9]:=
\begin{equation}
{} _{2} F _{1} \!\left [ \begin{matrix} { a, b}\\ { c}\end{matrix} ;
    {\displaystyle z}\right ] \,
   {} _{2} F _{1} \!\left [ \begin{matrix} { a, -b + c}\\ { c}\end{matrix} ;
    {\displaystyle z}\right ]  = 
  {
    {{{\left( 1 - z \right) }^{-a}}}}\,
{} _{4} F _{3} \!\left [ \begin{matrix} { a, b, -a + c, -b + c}\\ { c,
      {\frac c 2}, {\frac 1 2} + {\frac c 2}}\end{matrix} ; {\displaystyle
      -{\frac {{z^2}} {4\,\left( 1 - z \right) }}}\right ]
\label{eq4.12}
\end{equation}
and assume that $a$ or $b$ is a nonpositive integer, then we obtain
%A[10]:=
\begin{multline}
{} F{}^{1:2;2} _{1:1;1} \!\left [ \begin{matrix} d&:&a, b&;&a, -b + c&;\\
   e&:&c&;&c&;\end{matrix} \quad {\displaystyle 1,1} \right ]  \\
=    \frac {\G( e)\,\G( e -a - d)} {\G( e-a)\,\G( e-d)}
  {} _{6} F _{5} \!\left [ \begin{matrix} { a, b, -a + c, -b + c, {\frac 1 2} +
    {\frac d 2}, {\frac d 2}}\\ { {\frac 1 2} + {\frac c 2}, {\frac c 2}, c, 1 + a
    + d - e, -a + e}\end{matrix} ; {\displaystyle 1}\right ] ,
\label{eq4.13}
\end{multline}
provided $a$, $b$, or $d$ is a nonpositive integer.

\medskip
%\ni {\bf .} 
%A[11]:=
%\begin{equation}
%{} _{1} F _{1} \!\left [ \begin{matrix} { a}\\ { c}\end{matrix} ;
%    {\displaystyle -z}\right ] \,
%   {} _{1} F _{1} \!\left [ \begin{matrix} { a}\\ { c}\end{matrix} ;
%    {\displaystyle z}\right ]  = 
%  {} _{2} F _{3} \!\left [ \begin{matrix} { a, -a + c}\\ { c, {\frac c 2},
%   {\frac 1 2} + {\frac c 2}}\end{matrix} ; {\displaystyle {\frac {{z^2}}
%   4}}\right ] 
%\label{eq4.}
%\end{equation}
%A[12]:=
%\begin{equation}
%{} F{}^{1:1;1} _{1:1;1} \!\left [ \begin{matrix} d&:&a&;&a&;\\
%   e&:&c&;&c&;\end{matrix} \quad {\displaystyle -x,x} \right ]  = 
%  {} _{4} F _{5} \!\left [ \begin{matrix} { a, {\frac d 2}, {\frac 1 2} +
%   {\frac d 2}, -a + c}\\ { {\frac 1 2} + {\frac e 2}, {\frac e 2}, {\frac 1 2} +
%   {\frac c 2}, {\frac c 2}, c}\end{matrix} ; {\displaystyle {\frac {x^2} 4}}\right ]
%\label{eq4.}
%\end{equation}

%\medskip
%\ni {\bf .} 
%A[13]:=
%\begin{equation}
%{} _{0} F _{1} \!\left [ \begin{matrix} { -}\\ { b}\end{matrix} ;
%    {\displaystyle z}\right ] \,
%   {} _{0} F _{1} \!\left [ \begin{matrix} { -}\\ { c}\end{matrix} ;
%    {\displaystyle z}\right ]  = 
%  {} _{2} F _{3} \!\left [ \begin{matrix} { -{\frac 1 2} + {\frac b 2} + {\frac c
%   2}, {\frac b 2} + {\frac c 2}}\\ { b + c-1, c, b}\end{matrix} ;
%   {\displaystyle 4\,z}\right ] 
%\label{eq4.}
%\end{equation}
%A[14]:=
%\begin{equation}
%{} F{}^{1:0;0} _{1:1;1} \!\left [ \begin{matrix} d&:&-&;&-&;\\
%   e&:&a&;&b&;\end{matrix} \quad {\displaystyle x,x} \right ]  = 
%  {} _{3} F _{4} \!\left [ \begin{matrix} { -{\frac 1 2} + {\frac a 2} + {\frac b
%   2}, {\frac a 2} + {\frac b 2}, d}\\ { a, b, a + b-1, e}\end{matrix} ;
%   {\displaystyle 4x}\right ] 
%\label{eq4.}
%\end{equation}

%\medskip
%\ni {\bf .} 
%A[15]:=
%\begin{equation}
%{} _{0} F _{2} \!\left [ \begin{matrix} { -}\\ { a, b}\end{matrix} ;
%    {\displaystyle -z}\right ] \,
%   {} _{0} F _{2} \!\left [ \begin{matrix} { -}\\ { a, b}\end{matrix} ;
%    {\displaystyle z}\right ]  = 
%  {} _{3} F _{8} \!\left [ \begin{matrix} { -{\frac 1 3} + {\frac a 3} + {\frac b
%   3}, {\frac a 3} + {\frac b 3}, {\frac 1 3} + {\frac a 3} + {\frac b 3}}\\ {
%   -{\frac 1 2} + {\frac a 2} + {\frac b 2}, {\frac a 2} + {\frac b 2}, a, b,
%   {\frac a 2}, {\frac 1 2} + {\frac a 2}, {\frac b 2}, {\frac 1 2} + {\frac b
%   2}}\end{matrix} ; {\displaystyle -{\frac {27\,{z^2}} {64}}}\right ] 
%\label{eq4.}
%\end{equation}
%A[16]:=
%\begin{multline}
%{} F{}^{1:0;0} _{1:2;2} \!\left [ \begin{matrix} d&:&-&;&-&;\\ e&:&a,
%b&;&a,%
%   b&;\end{matrix} \quad {\displaystyle -x,x} \right ]  \\
%= 
%  {} _{5} F _{10} \!\left [ \begin{matrix} { -{\frac 1 3} + {\frac a 3} +
%   {\frac b 3}, {\frac a 3} + {\frac b 3}, {\frac 1 3} + {\frac a 3} + {\frac b 3},
%   {\frac 1 2} + {\frac d 2}, {\frac d 2}}\\ { {\frac 1 2} + {\frac a 2}, {\frac a
%   2}, a, {\frac 1 2} + {\frac b 2}, -{\frac 1 2} + {\frac a 2} + {\frac b 2},
%   {\frac a 2} + {\frac b 2}, {\frac b 2}, b, {\frac 1 2} + {\frac e 2}, {\frac e
%   2}}\end{matrix} ; {\displaystyle -{\frac {27} {64}}}x^2\right ] 
%\label{eq4.}
%\end{multline}

%\medskip
%\ni {\bf .} 
%A[17]:=
%\begin{equation}
%{} _{0} F _{1} \!\left [ \begin{matrix} { -}\\ { a}\end{matrix} ;
%    {\displaystyle -z}\right ] \,
%   {} _{0} F _{1} \!\left [ \begin{matrix} { -}\\ { a}\end{matrix} ;
%    {\displaystyle z}\right ]  = 
%  {} _{0} F _{3} \!\left [ \begin{matrix} { -}\\ { a, {\frac a 2}, {\frac 1 2} +
%   {\frac a 2}}\end{matrix} ; {\displaystyle -{\frac {{z^2}} 4}}\right ] 
%\label{eq4.}
%\end{equation}
%A[18]:=
%\begin{equation}
%{} F{}^{1:0;0} _{1:1;1} \!\left [ \begin{matrix} d&:&-&;&-&;\\
%   e&:&a&;&a&;\end{matrix} \quad {\displaystyle -x,x} \right ]  = 
%  {} _{2} F _{5} \!\left [ \begin{matrix} { {\frac 1 2} + {\frac d 2}, {\frac d
%   2}}\\ { {\frac 1 2} + {\frac a 2}, {\frac a 2}, a, {\frac 1 2} + {\frac e 2},
%   {\frac e 2}}\end{matrix} ; {\displaystyle -{\frac {x^2} 4}}\right ] 
%\label{eq4.}
%\end{equation}

%\medskip
%\ni {\bf .} 
%A[19]:=
%\begin{equation}
%{} _{2} F _{0} \!\left [ \begin{matrix} { a, b}\\ { -}\end{matrix} ;
%    {\displaystyle -z}\right ] \,
%   {} _{2} F _{0} \!\left [ \begin{matrix} { a, b}\\ { -}\end{matrix} ;
%    {\displaystyle z}\right ]  = 
%  {} _{4} F _{1} \!\left [ \begin{matrix} { a, b, {\frac a 2} + {\frac b 2},
%   {\frac 1 2} + {\frac a 2} + {\frac b 2}}\\ { a + b}\end{matrix} ;
%   {\displaystyle 4\,{z^2}}\right ] 
%\label{eq4.}
%\end{equation}
%A[20]:=
%\begin{equation}
%{} F{}^{1:2;2} _{1:0;0} \!\left [ \begin{matrix} d&:&a, b&;&a, b&;\\
%   e&:&-&;&-&;\end{matrix} \quad {\displaystyle -x,x} \right ]  = 
%  {} _{6} F _{3} \!\left [ \begin{matrix} { a, {\frac a 2} + {\frac b 2},
%   {\frac 1 2} + {\frac a 2} + {\frac b 2}, b, {\frac 1 2} + {\frac d 2}, {\frac d
%   2}}\\ { a + b, {\frac 1 2} + {\frac e 2}, {\frac e 2}}\end{matrix} ;
%   {\displaystyle 4x^2}\right ] 
%\label{eq4.}
%\end{equation}

\section{Concluding Remarks}

We have considered in this article a method, which we called the 
{\it beta integral 
method}, to obtain transformations and summation theorems from known
transformations of hypergeometric and products of hypergeometric series.
While being itself folklore,
this method was automated with the help of programs written in
{\sl Mathematica}
using the software package HYP developed by one of us \cite{KratBF}.
We obtained some known identities (which we displayed only
partially), and some which we were not able to trace in the
literature, among them some very interesting ones, such as
\eqref{eq3.13}, \eqref{eq3.33} \eqref{eq3.23}, or \eqref{eq3.25}. 

\medskip
Clearly, variations (that can be again automated) may also be
considered. For example, instead of using the beta integral
evaluation, we may replace $z$ by $zt$ at the beginning and then use
the Euler integral representation \eqref{e4} (with $p=1$) of a
$_2F_1[t]$-series. This yields identities which equate single sums to
sums of $_2F_1$-series (aside from some explicit multiplicative
factors). Another direction would be to use other extensions of the
beta integral evaluation, most notably $q$-analogues thereof. However,
it seems that, in particular in the case of $q$-analogues, the
repertory of identities which can be used as a starting point, is very
limited.

\medskip

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