\def\AndrAF{1}
\def\BailAB{2}
\def\BailAC{3}
\def\BresAA{4}
\def\BresAB{5}
\def\CarlAM{6}
\def\EgorAA{7}
\def\GaspAB{8}
\def\GaRaAB{9}
\def\GaRaAA{10}
\def\GeStAB{11}
\def\GOulAE{12}
\def\GOulAG{13}
\def\GOulAI{14}
\def\GOulAJ{15}
\def\GOHsAA{16}
\def\GrKPAA{17}
\def\KratAC{18}
\def\RahmAC{19}
\def\RahmAA{20}
\def\RiorAA{21}
\def\SinDAA{22}
\def\SlatAC{23}
\def\VeJaAD{24}

\def\Z{{\Bbb Z}}
\def\al{\alpha}
\def\be{\beta}
\def\ga{\gamma}
\def\Ga{\Gamma}
\def\de{\delta}
\def\om{\omega}
\def\[{\left[}
\def\]{\right]}

\NoBlackBoxes
%\input format


\def\poq#1#2{(#1;q)_#2}
\def\po#1#2{(#1)_#2}

\topmatter 
\title A NEW MATRIX INVERSE
\endtitle 
\author C.~Krattenthaler
\endauthor 
\affil 
Institut f\"ur Mathematik der Universit\"at Wien,\\
Strudlhofgasse 4, A-1090 Wien, Austria.
\endaffil 
\address{Institut f\"ur Mathematik der Universit\"at Wien,
Strudlhofgasse 4, A-1090 Wien, Austria.}\endaddress
\subjclass{Primary 15A09, 33D20, 33C20;
 Secondary 05A10, 05A19, 05A30, 11B65, 33C70}\endsubjclass
\keywords{matrix inversion, inverse relations.}\endkeywords
\abstract{We compute the inverse of a specific infinite-dimensional
matrix, thus unifying a number of previous matrix inversions.
Our inversion theorem is applied to derive a number of summation
formulas of hypergeometric type.}\endabstract
\endtopmatter
\document 

\subheading{1. Introduction}
Let $F = (f_{nk})_{n,k\in \Z}$ ($\Z$ denotes the set of integers) be
an infinite-dimensional lower-triangular matrix; i.e. $f_{nk}=0 $
unless $n\ge k$. The matrix $(f_{kl}^{-1})_{k,l\in\Z}$ is the inverse
matrix of $F$ if and only if
$$\sum_{n\ge k\ge l}^{}f_{nk}f_{kl}^{-1}=\de_{nl}$$
for all $n,l\in\Z$. Such matrix inversions are very important in many
fields of combinatorics and special functions. For example, when
dealing with combinatorial sums, application of the so-called
``inverse relations" (see (4.1) and (4.2)), 
which base on matrix inversion, helps to
simplify problems, or yields new identities. Riordan dedicated two
chapters of his book \cite{\RiorAA} to inverse relations and its
applications. Riordans inverse relations were classified and given a
unified method of proof by Egorychev \cite{\EgorAA, ch.3}. Studying a
specific class of inverse relations in a series of papers
\cite{\GOulAE--\GOulAJ}, Gould and Hsu \cite{\GOHsAA} finally discovered a very
general matrix inversion, which is equivalent to:

{\it If}
$$A_{nk}=\frac{\prod_{j=k}^{n-1}(a_j+kb_j)}{(n-k)!}\tag 1.1)(1$$ 
{\it then}
$$A_{kl}^{-1}=(-1)^{k-l} \frac{a_l+lb_l} {a_k+kb_k}
\frac{\prod_{j=l+1}^{k}(a_j+{kb}_j)}{(k-l)!}      \ .\tag 1.1)(2$$
(By convenience, products of the form $\prod _{j=u} ^{u-1}$ are
defined to be equal to 1, while for $u>v-1$ a product $\prod _{j=u}
^{v-1}$ by definition is equal to 0.) The inverse pair
(1.1) contains a lot of inverse relations of Gould--type (cf\. \cite
{\RiorAA, pp.~50--51}) and Abel--type (cf\. \cite{\RiorAA, p.~99}).
Carlitz \cite{\CarlAM} found a $q$-analogue of (1.1) which is equivalent
to:

{\it If}
$$B_{nk}=q^{\binom{n-k}2} \frac{\prod_{j=k}^{n-1} (a_j+q^kb_j)}
{(q;q)_{n-k}} 
\tag 1.2)(1$$
{\it then}
$$B_{kl}^{-1}= (-1)^{k-l}\frac{a_l+q^lb_l} {a_k+q^kb_k}
\frac{\prod_{j=l+1}^{k}(a_j+q^kb_j)}{(q;q)_{k-l}}\ ,\tag 1.2)(2$$ 
where $(a;q)_m = (1-a)(1-aq)\cdots(1-aq^{m-1})$. 
Special cases of (1.2) are of great
significance in $q$-series theory. Andrews \cite {\AndrAF} showed
that the Bailey transform \cite {{\BailAB, \BailAC}}, used to prove identities of
Rogers-Ramanujan type, is equivalent to the case $a_j=1$ and
$b_j=-bq^j$ of (1.2). Gessel and Stanton \cite {\GeStAB} used several
specializations of the $a_j=1$, $b_j=-bp^j$ case of (1.2) to derive
a number of basic hypergeometric summations and transformations, and,
once again, identities of Rogers--Ramanujan type.

In \cite {\BresAA} Bressoud considers finite forms of Rogers--Ramanujan
identities. The transform which he uses to prove them is equivalent
to the matrix inversion \cite {\BresAB}:

{\it If}
$$C_{nk}= \frac{(1-aq^{2k}) (b;q)_{n+k}\, (ba^{-1};q)_{n-k}
(ba^{-1})^k} {(1-a)(aq;q)_{n+k}\,(q;q)_{n-k}} \tag 1.3)(1$$ 
{\it then}
$$C _{kl} ^{-1}=\frac {(1-bq^{2l})(a;q)_{k+l}\,(ab^{-1};q)_{k-l}\,(ab^{-1})^l}
{(1-b)(bq;q)_{k+l}\,(q;q)_{k-l}}\ .\tag 1.3)(2$$

The purpose of this paper is to give the following generalization of
(1.1)--(1.3), which will be proved in section~2:
\proclaim{Theorem}Let $(a_i)_{i\in\Z}$, $(b_i)_{i\in\Z}$ and
$(c_i)_{i\in\Z}$ be arbitrary sequences such that $c_i\ne c_j$ if
$i\ne j$.

If$$f_{nk}=\frac {\prod _{j=k} ^{n-1}(a_j+c_kb_j)} {\prod _{j=k+1}
^{n}(c_j-c_k)}\tag 1.4)(1$$
then
$$f_{kl}^{-1}=\frac {a_l+c_lb_l} {a_k+c_kb_k}\frac {\prod
_{j=l+1}^{k}(a_j+c_kb_j)} {\prod _{j=l} ^{k-1}(c_j-c_k)}\ .\tag
1.4)(2$$
\endproclaim
In fact, (1.1) is the special case $c_k=k$, (1.2) is equivalent to
the case $c_k=q^k$, and (1.3) is equivalent to the case
$c_k=q^{-k}+aq^k$, $a_j=aq^{-j-1}+b^2q^{j-1}$, $b_j=-b/q$. Moreover,
replacing $c_k$ by $(1+ac_k^2)/c_k$, $a_j$ by $(a+b_j^2)/c_{j+1}$ and
$b_j$ by $-b_j/c_{j+1}$ in (1.4), we get after some simplification

\proclaim{Corollary} With the assumption of the Theorem, $a\ne
(c_jc_k)^{-1}$ for all $j,k\in\Z$,
if
$$g_{nk}=\frac {\prod _{j=k} ^{n-1}(1-c_kb_j)\prod _{j=k}
^{n-1}(b_j-ac_k)} {\prod _{j=k+1} ^{n}(1-ac_kc_j)\prod _{j=k+1}
^{n}(c_j-c_k)}\tag 1.5)(1$$
then
$$g _{kl} ^{-1}=\frac {(1-c_lb_l)(b_l-ac_l)} {(1-c_kb_k)(b_k-ac_k)}\frac {\prod
_{j=l+1} ^{k}(1-c_kb_j)\prod _{j=l+1} ^{k}(b_j-ac_k)} {\prod _{j=l}
^{k-1}(1-ac_kc_j)\prod _{j=l} ^{k-1}(c_j-c_k)}\ .\tag 1.5)(2$$
\endproclaim          
For $c_k=q^k$ the cases $b_j=bq^j$ and $b_j=bq^{-j}$ are equivalent
to (1.3). In this sense the form (1.5) of (1.4) is the natural
generalization of Bressoud's matrix inverse (1.3), such as (1.4) is
the natural generalization of Gould's and Hsu's (1.1) and Carlitz's
(1.2).

A special case of (1.5), namely (4.3), which involves
rising $q$-factorials with two different bases, has been crucial
in papers by
Gasper and Rahman \cite{{\GaspAB, \RahmAC}}.
They used inversion together with an indefinite bibasic sum to derive
numerous beautiful bibasic,
cubic, and quartic summation formulas for basic
hypergeometric series (see section~3 for ``hypergeometric" definitions). 
(They also extended this method to obtain bibasic,
cubic, and quartic transformation formulas \cite{\GaRaAB, \RahmAA;
\GaRaAA, sec.~3.6}.) 
In section~4 we add two more applications of
this inversion. We derive a 
 summation formula (identity (4.8)) for series of 
hypergeometric type and as a by-product of the second application we
obtain a basic hypergeometric transformation formula (identity (4.12)). 
The former contains an infinite
family of summation formulas for very well-poised
hypergeometric series. Besides, we use the opportunity to clearly
demonstrate that what Gasper and Rahman do in \cite{{\GaspAB, \RahmAC}} 
is indeed inversion,
though in disguise. This fact does not seem to be as accepted as it
should be. (Their extension in \cite{\GaRaAB, \RahmAA; \GaRaAA,
%corrected, C. Krattenthaler
sec.~3.6} has an explanation in terms of ``partial inversion".)
%corrected, C. Krattenthaler

Finally, in section~5 we apply other special cases of (1.4) to obtain
curious identities ((5.5), (5.9), (5.12)) which involve cubic, quartic,
and quintic analogues of rising factorials, respectively.


\subheading{2. Proof of Theorem} In \cite{\KratAC} the author gave a
method for solving Lagrange inversion problems. These are closely
connected with the problem of inverting lower-triangular matrices.
For convenience, by a formal Laurent series (fLs) we 
always mean  a series of the form $\sum _{i\ge m} \al_iz^i$, for some
$m\in\Z$. Given the fLs $a(z)$ and $b(z)$ we introduce the bilinear form
$\langle \, , \rangle$ by 
$$\langle a(z),b(z)\rangle=\langle z^0\rangle a(z)\cdot b(z)\ , $$
where $\langle z^0\rangle c(z)$ denotes the coefficient of $z^0$ in
$c(z)$.
Given any linear operator $L$ acting on fLs, $L\sp*$ denotes the
adjoint of $L$ with respect to $\langle \, , \rangle$; i.e. $\langle La(z),
b(z)\rangle = \langle a(z), L\sp* b(z)\rangle$ for all fLs $a(z)$ and
$b(z)$. What we need is the following special case of \cite{\KratAC,
Theorem 1}.
\proclaim{Lemma} Let $F=(f_{nk})_{n,k\in\Z}$ be an infinite
lower-triangular matrix with $f_{kk}\ne0$ for all $k\in\Z$. For
$k\in\Z$, the formal Laurent series $f_k(z)$ and $\tilde f_k(z)$ are defined
by $f_k(z)=\sum_{n\ge k} f_{nk}z^n$ and $\tilde f_k(z)= \sum _{l\le k}
 f_{kl}^{-1}z^{-l}$, where $(f_{kl}^{-1})_{k,l\in\Z}$ is the uniquely
determined inverse matrix of $F$. Suppose that for $k\in\Z$ holds 
$$Uf_k(z)=c_k\cdot Vf_k(z)\ , \tag 2.1$$
where $U,V$ are linear operators acting on formal Laurent series, $U$
being bijective, and $(c_k)_{k\in\Z}$ is a sequence of constants. If
$h_k(z)$ is a solution of
$$U^* h_k(z)=c_k\cdot V^* h_k(z)\ , \tag  2.2$$
with $h_k(z)\not\equiv 0$ for all $k\in\Z$, then
$$\tilde f_k(z)=\frac{1}{\langle f_k(z),V^* h_k(z)\rangle} V^* h_k(z)\
. \tag 2.3$$
\endproclaim

In order to prove (1.4), we set $f_k(z)=\sum_{n\ge k}f_{nk}z^n$ with
$f_{nk}$ of (1.4)(1). Obviously, for $n\ge k$
$$(c_n-c_k)f_{nk}=(a_{n-1}+c_kb_{n-1})f_{n-1,k} \ . \tag 2.4$$
If we define the linear operators $\Cal A, \Cal B, \Cal C$ by $\Cal A
z^k=a_kz^k, \Cal Bz^k=b_kz^k$ and $\Cal C z^k=c_kz^k$, (2.4) may be
rewritten in the form 
$$(\Cal C-z\Cal A) f_k(z) = c_k(1+z\Cal B)f_k(z)\ , \tag 2.5$$
valid for all $k\in\Z$. (2.5) is a system of equations of type (2.1)
with $U=\Cal C-z\Cal A$ and $V=1+z\Cal B$. The dual equations (2.2) for the
auxiliary fLs $h_k(z)$ in this case read 
$$(\Cal C^*-\Cal A^*z) h_k(z)=c_k(1+\Cal B^*z) h_k(z) \ .\tag
2.6$$
Because of $\Cal A^*z^{-k}=a_kz^{-k}$ etc., by comparing the
coefficients of $z^{-l}$ in (2.6) we obtain
$$(c_l-c_k) h_{kl}=(a_l+c_kb_l) h_{k,l+1} \ .$$ 
If we set $h_{kk}=1$, we get
$$h_{kl}=\frac {\prod _{j=l} ^{k-1}(a_j+c_kb_j)}
{\prod_{j=l}^{k-1}(c_j-c_k)} \ . $$
Taking into account (2.3), here we have $\tilde f_k(z)=(1+\Cal B^*z)h_k(z)$.
Hence again, comparing coefficients of $z^{-l}$ leads to 
$$f_{kl}^{-1}=h_{kl}+b_lh_{k,l+1}
=\frac {a_l+c_lb_l} {a_k+c_kb_k}\frac {\prod _{j=l+1}
^{k}(a_j+c_kb_j)} {\prod _{j=l} ^{k-1}(c_j-c_k)} \ ,$$
which is exactly (1.4)(2).                                                 


\subhead 3. Hypergeometric and basic hypergeometric notation\endsubhead
All the notation and terminology is adopted from \cite{\GaRaAA, pp.~1--6}.
The {\it (generalized) hypergeometric series} is defined by
$${}_r F_s\!\left[\matrix a_1,\dots,a_r\\ b_1,\dots,b_s\endmatrix; 
z\right]=\sum _{n=0} ^{\infty}\frac {\po{a_1}{n}\cdots\po{a_r}{n}}
{n!\,\po{b_1}{n}\cdots\po{b_s}{n}} z^n\ ,$$
where the rising factorial
$(a)_n$ is given by $(a)_n:=a(a+1)\cdots(a+n-1)$,
$n\ge1$, $(a)_0:=1$. The gamma function can be used to extend
rising factorials by defining 
$(a)_\be=\lim _{\ga\to a}\Ga(\ga+\be) / \Ga(\ga)$, $\be$ arbitrary
(cf\. \cite{\GrKPAA, p.~211f}). 
A hypergeometric series $_{r+1}F_r$ is called {\it very well-poised} if
$a_i+b_i=1+a_0$ for $i=1,2,\dots,r$,
and among the parameters $a_i$ occurs $1+a_0/2$. We
use the standard abbreviation for very well-poised hypergeometric
series,
$$_{r+1}V_r(a_0;a_2,a_3,\dots,a_r;z):={}_{r+1}F_r\!\[\matrix
a_0,1+a_0/2,a_2,a_3,\dots,a_r\\
a_0/2,1+a_0-a_2,1+a_0-a_3,\dots,1+a_0-a_r\endmatrix; z\].$$
We shall also use the compact
Gasper-Rahman notation
$$ {(a_1,a_2,\dots,a_r)_n} :=(a_1)_n\,(a_2)_n\,\cdots\,(a_r)_n$$
and 
$$\Ga\bmatrix a_1,a_2,\dots,a_r\\b_1,b_2,\dots,b_s\endbmatrix:=
\frac {\Ga(a_1)\,\Ga(a_2)\,\cdots\,\Ga(a_r)} 
{\Ga(b_1)\,\Ga(b_2)\,\cdots\,\Ga(b_s)}\ .$$

Given a (fixed) complex number $q$ with $|q|<1$,
the {\it basic hypergeometric series\/} is defined by
$${}_r\phi_s\!\left[\matrix a_1,\dots,a_r\\ b_1,\dots,b_s\endmatrix; q,
z\right]=\sum _{n=0} ^{\infty}\frac {\poq{a_1}{n}\cdots\poq{a_r}{n}}
{\poq{q}{n}\poq{b_1}{n}\cdots\poq{b_s}{n}}\left((-1)^nq^{\binom
n2}\right)^{s-r+1}z^n\ ,$$
where, as before, the rising $q$-factorial $(a;q)_n$ is given by 
$(a;q)_n:=(1-a)(1-aq)\cdots(1-aq^{n-1})$, $n\ge1$, $(a)_0:=1$. The
infinite $q$-factorial $(a;q)_\infty:=\prod _{i=0} ^{\infty}(1-aq^{i})$
can be used to extend (finite) $q$-factorials by defining $(a;q)_\be:=
(a;q)_\infty/(aq^\be;q)_\infty$, $\be$ arbitrary. 
A basic hypergeometric series $_{r+1}\phi_r$ is called {\it very
well-poised\/} if $a_ib_i=qa_0$ for $i=1,2,\dots,r$,
and among the parameters $a_i$ occur both $q\sqrt{a_0}$ and
$-q\sqrt{a_0}$. We
use the standard abbreviation for very well-poised basic hypergeometric
series,
$$_{r+1}W_r(a_0;a_3,a_4,\dots,a_r;z):={}_{r+1}\phi_r\!\[\matrix
a_0,q\sqrt{a_0},-q\sqrt{a_0},a_3,a_4,\dots,a_r\\
\sqrt{a_0},-\sqrt{a_0},qa_0/a_2,qa_0/a_3,\dots,qa_0/a_r\endmatrix; z\].$$
We use 
short notations in the basic hypergeometric context which are
analogous to the hypergeometric ones.


All identities in our paper are subject to suitable conditions on the
parameters such that the involved hypergeometric or basic
hypergeometric series converge. We shall not state these conditions
for each identity. The reader should consult \cite{\GaRaAA,
pp.~4--5}.




\subheading{4. Hypergeometric summation formulas from inversion}
There is a standard technique for deriving new summation formulas
from known ones by using inverse matrices (cf\. \cite{{\AndrAF, \GeStAB, \RiorAA}}). If
$(f_{nk})$ and $(f_{kl}^{-1})$ are lower triangular matrices being
inverses of each other, then of course the following is true:
$$\sum _{k=0} ^{n}f_{nk}a_k=b_n\quad \text {\it if and only if}\quad \sum
_{l=0} ^{k}f_{kl}^{-1}b_l=a_k.\tag4.1$$
If one side in (4.1) is known then the other produces another
summation formula. There is also a less used dual version, which in
Riordan's book \cite{\RiorAA} is called ``rotated inversion". It
reads
$$\sum _{n=k} ^{\infty}f_{nk}B_n=A_k\quad \text {\it if and only if}\quad \sum
_{k=l} ^{\infty}f_{kl}^{-1}A_k=B_l,\tag4.2$$
subject to suitable convergence conditions. Again, if one side in
(4.2) is known then the other produces a possibly new identity.

What Gasper and Rahman do in \cite{\GaspAB, \RahmAC}
is rotated inversion, though in disguise. Recall that the rotated
inversion (4.2) bases on the orthogonal relation $\sum _{k=l}
^{n}f_{nk}f^{-1}_{kl}=\de_{nl}$. ($\de_{nl}$ is the usual Kronecker
delta.) Typically, Gasper and Rahman start with such an orthogonality
relation, but with $l=0$, i.e\. $\sum _{k=0}
^{n}f_{nk}f^{-1}_{k0}=\de_{n0}$. (As inverse pair
$(f_{nk})$, $(f_{kl}^{-1})$ they choose special cases of (4.3), with
the roles of $f_{nk}$ and $f_{kl}^{-1}$ interchanged.) Then they
multiply both sides by $B_n$, say, and sum over all $n$. This
gives $\sum _{n=0} ^{\infty}B_n\sum _{k=0}
^{n}f_{nk}f_{k0}^{-1}=B_0$. Next the sums are interchanged to give $\sum
_{k=0} ^{\infty}f_{k0}^{-1}\sum _{n=k} ^{\infty}f_{nk}B_n=B_0$. The
inner sum in all cases is simplified by means of Bailey's nonterminating
extension of Jackson's $_8\phi_7$-sum, to $A_k$ say. Thus the
identity $\sum _{k=0} ^{\infty}f_{k0}^{-1}A_k=B_0$ is proved. Of course,
the reader will have observed that this is exactly the proof of the
left-to-right implication of (4.2), for $l=0$. Hence, why not start
with $l$ generic? The answer is that there is no problem of doing
the same with arbitrary $l$. But in the end it turns out that the
resulting identity is not more general. In fact, a simple parameter
substitution eliminates all occurences of $l$, and one is left with
the identity for $l=0$. This $l=0$ phenomenon will be exemplified in
the following application of the inversion (1.5). 

Before we start, another digression appears to be of interest. Gasper and
Rahman's indefinite bibasic sum \cite{\GaRaAA, (3.6.13)}, which is so
important in their derivations, comes from a telescoping argument
which is based on a mysterious factorization \cite{\GaRaAA, (3.6.10)}
of a difference of two four-term products into a four-term product.
This factorization loses some of its mystery if one observes that it is just
equivalent to the $n=1$ case of Jackson's $_8\phi_7$-sum
\cite{\GaRaAA, (2.6.2)} (replace $a$ by $adp^kq^{k-1}$, $b$ by
$bp^k$, $c$ by $cq^k$, $d$ by $ap^k$ in Jackson's sum with $n=1$).


In this section we
use the inverse matrices of (1.5) with the choices $c_k=q^k$ and
$b_j=bp^j$. This inversion can be written in the form,

{\it if}
$$f_{nk}=\frac {(bp^kq^k;p)_{n-k}\,(bp^kq^{-k}/a;p)_{n-k}}
{(aq^{2k+1};q)_{n-k}\,(q;q)_{n-k}}\ ,\tag4.3)(1$$
{\it then}
$$f_{kl}^{-1}=p^{l-k}\frac {(1-bp^lq^l)(1-ap^{-l}q^l/b)} 
{(1-bp^kq^k)(1-ap^{-k}q^k/b)}
\frac {(bp^kq^k;p^{-1})_{k-l}\,(bp^kq^{-k}/a;p^{-1})_{k-l}}
{(aq^{2k-1};q^{-1})_{k-l}\,(q^{-1};q^{-1})_{k-l}}\ .\tag4.3)(2$$


If in (4.3) we replace $a$ by $q^B$, $b$ by $q^{B-C}$, $q$ by $p^A$, 
and let $p\to1$, we obtain the inverse pair

{\it If}
$$f_{nk}=
\frac {(AB-AC+(1+A)k)_{n-k} (-AC+(1-A)k)_{n-k}}
{(B+2k+1)_{n-k} (n-k)!}\ ,\tag4.4)(1$$
{\it then}
$$\multline
f_{kl}^{-1}={{\left( - A C   +(1-A) l \right)  
     \left( A B - A C + (1+A) l \right)  
}\over{\left( - A C   + (1-A) k \right)  
     \left( A B - A C + (1+A) k \right) }}\\
{{     ({ \textstyle - A B   + A C -(1+A) k }) _{k - l}   
     ({ \textstyle A C -(1-A) k }) _{k - l} }\over
   {  
     ({ \textstyle 1 - B - 2 k}) _{k - l}\,k! }}
\ .
\endmultline\tag4.4)(2$$
For $B_n$ we choose
$$B_n=\frac {(D)_n} {(-B+AB-2AC+D)_n}.\tag4.5$$
Now we form the sum $A_k=\sum _{n=k} ^{\infty}f_{nk}B_n$. In 
hypergeometric notation this is
$$A_k={{({ \textstyle D}) _{k}}\over 
{({ \textstyle -B + A B - 2 A C + D}) _{k} }}
{{{} _{3} F _{2} \!\left [ \matrix { D + k, - A C   + k - A k,
      A B - A C + k + A k}\\ { -B + A B - 2 A C + D + k, 1 + B +
      2 k}\endmatrix ; {\displaystyle 1}\right ]   }
    }\ .
$$
To the $_3F_2$ we apply a nonterminating 
extension of the Pfaff--Saalsch\"utz
summation \cite{\SlatAC, (2.4.4.4)}
$$\multline
{} _{3} F _{2} \!\left [ \matrix { a, b, c}\\ { d, 1 + a + b + c -
   d}\endmatrix ; {\displaystyle 1}\right ]  = 
  \Gamma \left [ \matrix 1 + a - d, 1 + b - d, 1 + c - d, 1 + a + b + c - d\\
    1 - d, 1 + b + c - d, 1 + a + c - d, 1 + a + b - d\endmatrix \right ]
\\- 
    \Gamma \left [ \matrix  d-1, 1 + a - d, 1 + b - d, 1 + c - d, 1 + a + b
     + c - d\\ 1 - d, a, b, c, 2 + a + b + c - 2 d\endmatrix \right]
\\
   {} _{3} F _{2} \!\left [ \matrix { 1 + a - d, 1 + b - d, 1 + c - d}\\ { 2 -
     d, 2 + a + b + c - 2 d}\endmatrix ; {\displaystyle 1}\right ]  ,
\endmultline\tag4.6$$
thus obtaining
$$\align A_k&=
\Gamma \left [ \matrix 1 + B - A B + 2 A C, 1 + B - A B + A C - D - A k,
   1 + B + A C - D + A k\\ 1 +
   B - A B + 2 A C - D - k, 1 + B - D + k, 1 + B + A C + k + A k\endmatrix \right ]
\tag4.7\\
&\hskip2cm
\Gamma \bmatrix  1 + B + 2 k, -B + A B - 2 A C + D, D + k\\
 1 + B -
   A B + A C + k - A k, D, -B + A B - 2 A C + D + k\endbmatrix
\\
&\hskip-15pt 
-\Gamma \left [ \matrix -1 - B + A B - 2 A C + D + k, 1 + B - A B +
    2 A C, 1 + B - A B + A C - D - A k\\ 1 + B - A B + 2 A C - D - k, -
    A C   + k - A k, A B - A C + k + A k\endmatrix \right ]\\
&\hskip2cm
\Gamma\bmatrix  1 + B + A C - D + A k, 1 + B +
    2 k, -B + A B - 2 A C + D\\
    2 + 2 B - A B +
    2 A C - D + k, D, -B + A B - 2 A C + D + k\endbmatrix
\\
&\hskip-10pt
   {} _{3} F _{2} \!\left [ \matrix { 1 + B - A B + 2 A C, 1 + B - A B +
    A C - D - A k, 1 + B + A C - D + A k}\\ { 2 + B - A B + 2 A C - D -
    k, 2 + 2 B - A B + 2 A C - D + k}\endmatrix ; {\displaystyle 1}\right
    ]  
\endalign$$
By rotated inversion (4.2) the identity $\sum _{k=l}
^{\infty}f_{kl}^{-1}A_k=B_l$, with $f_{kl}^{-1},A_k,B_l$ given by
(4.4)(2), (4.7), (4.5), respectively, holds. The sum
on the left-hand side splits 
into two sums, one of which is actually a double sum. In this double sum we
interchange summations and simplify. In the resulting identity we
replace $B$ by $B-2l$, $C$ by $C+(1-A)l/A$, and $D$ by $D-l$. The
effect of this last step is the same as if we would have set $l=0$.
This is the $l=0$ phenomenon which was described before. 
Thus we arrive at the summation
$$\multline 
   \sum_{k = 0}^{\infty}{{({ \textstyle B}) _{k}  
        ({ \textstyle 1 + {B\over 2}}) _{k}  
                    ({ \textstyle A B - A C}) _{(1+A)k}  
          ({ \textstyle 1 + B + A C - D}) _{A k}  
          ({ \textstyle - A C  }) _{(1-A)k}  }\over 
        {k!\,   ({ \textstyle {B\over 2}}) _{k}  
          ({ \textstyle 1 + B + A C}) _{(1+A)k}  
          ({ \textstyle 1 + A B - A C}) _{A k}  
          ({ \textstyle 1 + B - A B + A C}) _{(1-A)k}  
     }}  
\\\hskip4cm
{{          ({ \textstyle 1 + B - A B + A C - D}) _{- A k  }  
          ({ \textstyle D}) _{k} }\over
    {({ \textstyle 1 - A C}) _{- A k  }  
     ({ \textstyle 1 + B - D}) _{k} }}
\\ - 
   \Gamma \left [ \matrix 1 + B - A B + A C, 1 + B + A C, 1 + B - D, -1 - B
     + A B - 2 A C + D\\ A B - A C, - A C  , 2 + 2 B -
     A B + 2 A C - D, D\endmatrix \right ] 
\\
     \sum_{j = 0}^{\infty}
        {{({ \textstyle 1 + B - A B + 2 A C}) _{j}  
           ({ \textstyle 1 + B + A C - D}) _{j}  
           ({ \textstyle 1 + B - A B + A C - D}) _{j}  }
         \over   {j!\,
           ({ \textstyle 2 + B - A B + 2 A C - D}) _{j}  
           ({ \textstyle 2 + 2 B - A B + 2 A C - D}) _{j} }}
\\
                   \sum_{k = 0}^{\infty}
              {{({ \textstyle B}) _{k}  
                ({ \textstyle 1 + {B\over 2}}) _{k}  
                  ({ \textstyle -1 - B + A B - 2 A C + D - j}) _{k}  
                 }\over 
                {k!\,  ({ \textstyle {B\over 2}}) _{k}  
              ({ \textstyle 2 + 2 B - A B + 2 A C - D + j}) _{k} }}
\\\hskip4cm
   {{  ({ \textstyle 1 + B + A C - D + j}) _{A k}  
                  ({ \textstyle 1 + B - A B + A C - D + j}) _{- A k
                    }}\over
     {             ({ \textstyle 1 + A B - A C}) _{A k} 
                  ({ \textstyle 1 - A C}) _{- A k  }  
 }}
\\=
\Gamma \left [ \matrix 1 + B - A B + A C, 1 + B + A C, 1 + B - D, 1 + B -
   A B + 2 A C - D\\ 1 + B, 1 + B - A B + 2 A C, 1 + B - A B + A C -
   D, 1 + B + A C - D\endmatrix \right ] 
\endmultline\tag4.8$$
Unfortunately, the inner sum in the double sum cannot be evaluated in
general. (It can for $A=1$. In this case even a $q$-analogue exists,
see below.) If we terminate the first series by setting $D=-n$, $n$ a
nonnegative integer, then the term with the double sum vanishes
because of the occurence of $\Ga(D)$ in the denominator of this
expression. The resulting summation can be rewritten as
$$\multline 
  \sum_{k = 0}^{\infty}{{
        ({ \textstyle B}) _{k}  
        ({ \textstyle 1 + {B\over 2}}) _{k}  
        ({ \textstyle A B - A C}) _{(A+1) k}  
        ({ \textstyle  A C  }) _{ A k}  
        ({ \textstyle - B + A B - A C}) _{(A-1) k  } }
       \over {k!\,  ({ \textstyle {B\over 2}}) _{k}  
        ({ \textstyle 1 + B + A C}) _{(A+1)k}  
        ({ \textstyle 1 + A B - A C}) _{A k}  
        ({ \textstyle 1 + A C}) _{ (A-1) k  }  
       }}
\\
    {{        ({ \textstyle 1 + B + A C + n}) _{A k}  
        ({ \textstyle -n}) _{k}  }\over
      { ({ \textstyle - B + A B - A C - n}) _{ A k}  
        ({ \textstyle 1 + B + n}) _{k} }}
={{({ \textstyle 1 + B}) _{n}  
      ({ \textstyle 1 + B - A B + 2 A C}) _{n} }\over 
    {({ \textstyle 1 + B + A C}) _{n}  
      ({ \textstyle 1 + B - A B + A C}) _{n} }} \ .
\endmultline\tag4.9$$
If $A$ is a positive integer (4.8) and (4.9) give
summation formulas for (nonterminating, respectively terminating) 
very well-poised 
$_{4A+3}F_{4A+2}$-series. The case $A=1$ of (4.9) 
is almost equivalent to a
known $_7F_6$-summation \cite{\SlatAC, (2.4.1.5)}. In fact, both
formulas can be derived from each other by contiguous relations.
Besides, identity (4.8) for a fixed $A$ is
equivalent to (4.8) with $A$ replaced by $-A$. (Hence, the same is
true for (4.9)). In fact, replacement
of $A$ by $-A$ is the same as replacement of $C$ by $B-C$.

Letting $n$ tend to $\infty$ in (4.9) leads to the summation
$$\multline 
  \sum_{k = 0}^{\infty}{{
        ({ \textstyle B}) _{k}  
        ({ \textstyle 1 + {B\over 2}}) _{k}  
        ({ \textstyle A B - A C}) _{(A+1) k}  
        ({ \textstyle  A C  }) _{ A k}  
        ({ \textstyle - B + A B - A C}) _{(A-1) k  } }
       \over {k!\,  ({ \textstyle {B\over 2}}) _{k}  
        ({ \textstyle 1 + B + A C}) _{(A+1)k}  
        ({ \textstyle 1 + A B - A C}) _{A k}  
        ({ \textstyle 1 + A C}) _{ (A-1) k  }  
       }}
\\
=\Gamma \bmatrix { \textstyle 1 + B},
      { \textstyle 1 + B - A B + 2 A C}\\
    { \textstyle 1 + B + A C},
      { \textstyle 1 + B - A B + A C}\endbmatrix \ .
\endmultline\tag4.10$$
If $A$ is a positive integer (4.10) gives a
summation formula for nonterminating
very well-poised 
$_{3A+2}F_{3A+1}$-series.

It is only the case $A=1$, where we are able to give a $q$-analogue of
(4.8) (and hence of (4.9) and (4.10)). This is due to the absence of
a suitable bibasic analogue of the Pfaff--Saalsch\"utz formula.
(Recently a bibasic analogue of the Pfaff--Saalsch\"utz formula has
been discovered by Singer \cite{\SinDAA}, but it is not suited for our purposes.) 
To obtain the $q$-analogue of (4.8) in case $A=1$, 
start with the inverse pair
$(f_{nk})$, $(f_{kl}^{-1})$ from (4.3), with $p=q$, $a=B$, $b=B/C$. For
$B_n$ choose $B_n=q^n(D;q)_n/(D/C^2;q)_n$. Then proceed as before. 
Instead of (4.6) use the nonterminating extension of the
$q$-Pfaff--Saalsch\"utz sum \cite{\GaRaAA, (2.10.12)}.
To evaluate the inner sum of the arising double sum, use the very
well-poised $_6\phi_5$-summation \cite{\GaRaAA, (2.7.1)}. 
What is finally obtained is the summation
$$\multline 
   {} _{10} W _{9} ({\displaystyle B;  {{{\sqrt{B}}}\over {{\sqrt{C}}}},
    -{{{\sqrt{B}}}\over {{\sqrt{C}}}},  {{{\sqrt{B q}}}\over {{\sqrt{C}}}},
    -{{{\sqrt{B q}}}\over {{\sqrt{C}}}}, {{B C q}\over D}},
   C, D; q,
    {\displaystyle q})
\\
 - { {(\let \over / B q, {1\over C}, D,
       {B\over C}, {D\over {{C^2}}}, {{B C {q^2}}\over D} ;q) _\infty} \over
       {(\let \over / {{B q}\over D}, B C q, {D\over {{C^2} q}}, {D\over
       C}, q, {{B q}\over C} ;q) _\infty} } 
      {} _{3} \phi _{2} \! \left [ \matrix \let \over / {C^2} q,
       {{{C^2} q}\over D}, {{B C q}\over D}\\ \let \over /
       {{{C^2} {q^2}}\over D}, {{B C {q^2}}\over D}\endmatrix ;q,
       {\displaystyle {q\over C}} \right ]   
\\={ {(\let \over / {C^2} q, {{C q}\over D}, {{B C q}\over D}, B q ;q)
   _\infty} \over {(\let \over / C q, {{{C^2} q}\over D}, {{B q}\over D},
   B C q ;q) _\infty} } \ .
\endmultline\tag4.11$$
By comparing this identity with a special case of a transformation of
Verma and Jain \cite{\VeJaAD, (7.1); \GaRaAA, Ex.~2.25, $a=B/C$,
$b=qB/C^2$, $c=1$, $d=1$, $e=D$},
$$\multline 
  {} _{10} W _{9} ({\displaystyle B; {{{\sqrt{B}}}\over {{\sqrt{C}}}},
    -{{{\sqrt{B}}}\over {{\sqrt{C}}}}, {{{\sqrt{B q}}}\over {{\sqrt{C}}}},
    -{{{\sqrt{B q}}}\over {{\sqrt{C}}}}, C, D, {{B C q}\over D}}; q,
    {\displaystyle q})
\\
 + { {(\let \over / B q, {{C q}\over D}, {B\over C},
     D, C, {{B {q^2}}\over D}, {{B {C^3} {q^3}}\over {{D^2}}} ;q) _\infty}
     \over {(\let \over / C q, {{B q}\over D}, B C q, {{{C^2} q}\over D},
     {D\over {C q}}, {{B q}\over C}, {{B {C^2} {q^3}}\over {{D^2}}} ;q)
     _\infty} } 
\\
{} _{10} W _{9} ({\displaystyle {{B {C^2} {q^2}}\over
     {{D^2}}}; {{{\sqrt{B C}} q}\over D}, -{{{\sqrt{B C}} q}\over D},
     {{{\sqrt{B C}} {q^{{3\over 2}}}}\over D}, -{{{\sqrt{B C}} {q^{{3\over
     2}}}}\over D}, C q, {{B C q}\over D}, {{{C^2} q}\over D}}; q,
     {\displaystyle q})
\\=
{ {(\let \over / B q, {{C q}\over D}, {C^2} q, {{B C q}\over D} ;q)
   _\infty} \over {(\let \over / C q, {{B q}\over D}, B C q,
   {{{C^2} q}\over D} ;q) _\infty} } \ ,
\endmultline$$
we obtain a transformation formula between a very well-poised
$_{10}\phi_9$ and a $_3\phi_2$ series,
$$\multline 
{} _{10} W _{9} ({\displaystyle {{B {C^2} {q^2}}\over {{D^2}}}; 
    {{{\sqrt{B C}} q}\over D},
   -{{{\sqrt{B C}} q}\over D}, {{{\sqrt{B C}} {q^{{3\over 2}}}}\over D},
   -{{{\sqrt{B C}} {q^{{3\over 2}}}}\over D}},
   C q, {{{C^2} q}\over D}, {{B C q}\over D}; q, {\displaystyle q})
\\ = 
  { {(\let \over / {q\over C}, {{B C {q^2}}\over
      D}, {{{C^2} q^2}\over D}, {{B {C^2} {q^3}}\over
      {{D^2}}} ;q) _\infty} \over {(\let \over / 
       q, {{C q^2}\over D}, {{B {q^2}}\over D},
      {{B {C^3} {q^3}}\over {{D^2}}} ;q) _\infty} } 
     {} _{3} \phi _{2} \! \left [ \matrix \let \over / {C^2} q,
      {{{C^2} q}\over D}, {{B C q}\over D}\\ \let \over /
      {{{C^2} {q^2}}\over D}, {{B C {q^2}}\over D}\endmatrix ;q,
      {\displaystyle {q\over C}} \right ]  \ ,
\endmultline$$
or after having replaced $B$ by $BD^2/C^2q^2$, $C$ by $B/C$, $D$ by
$qD/C$, in that order,
$$\multline 
{} _{10} W _{9} ({\displaystyle B; 
   {\sqrt{C}}, -{\sqrt{C}},  {\sqrt{Cq}}   , -{\sqrt{Cq}}
   }, D, {{B q}\over C}, {{{B^2}}\over {C D}}
   ; q, {\displaystyle q}) 
\\= 
 { {(\let \over / {{C q}\over B}, D q, {{{B^2} q}\over {C D}}, B q
      ;q) _\infty} \over {(\let \over / q, {{B q}\over D}, {{C D q}\over B},
      {{{B^2} q}\over C} ;q) _\infty} } 
     {} _{3} \phi _{2} \! \left [ \matrix \let \over / {{{B^2} q}\over
      {{C^2}}}, {{{B^2}}\over {C D}}, D\\ \let \over / {{{B^2} q}\over {C D}},
      D q\endmatrix ;q, {\displaystyle {{C q}\over B}} \right ] 
\endmultline\tag4.12$$
Mizan Rahman (private communication) 
showed me how to derive this transformation from
standard basic hypergeometric transformation formulas by a
sequence of series manipulations.


\subhead 5. More identities\endsubhead
In this section we restrict ourselves to terminating series.
Therefore here we use the ``usual" inversion (4.1). We use special
cases of (1.4) which are different from the one, (4.3), used in the
previous section.

First, let us take $c_k=(k+c)^3$, $a_j=(j+a)^3$, $b_j=1$. The inverse
pair (1.4) in this special case reads
$$f_{nk}=\frac {(n+a)^3+(n+c)^3} {(k+a)^3+(k+c)^3}
\frac {\prod _{j=k} ^{n-1}((j+a)^3+(k+c)^3)} {\prod _{j=k+1}
^{n}((j+c)^3-(k+c)^3)}
\tag5.1)(1$$
and 
$$f_{kl}^{-1}=\frac {\prod _{j=l+1} ^{k}((j+a)^3+(k+c)^3)} {\prod _{j=l}
^{k-1}((j+c)^3-(k+c)^3)}\ .
\tag5.1)(2$$
Now choose
$$b_l=\frac {(a+c,1+a/2+c/2,d,2a-c-d+1)_l}
{(1,a/2+c/2,1+a+c-d,2c-a+d)_l}\ .\tag5.2$$
We form $a_k=\sum _{l=0} ^{k}f_{kl}^{-1}b_l$, which can be written as
$$\multline
a_k=\frac {\prod _{j=1} ^{k}((j+a)^3+(k+c)^3)} 
{\prod _{j=0}
^{k-1}((j+c)^3-(k+c)^3)}\\
{}_7V_6( a+c;1+2a-c-d,d,c-c\om-k\om,
c-c\om^2-k\om^2,-k; 1).
\endmultline$$
Here, $\om$ denotes a third root of unity.
The $_7V_6$ can be evaluated by means of Dougall's sum \cite{\SlatAC,
(2.3.4.4)}
$$\multline
{} _{7} V _{6} ({\displaystyle a; b, c, d, 1 + 2 a - b - c - d + n, -n};
   {\displaystyle 1}) \\= {{({ \textstyle 1 + a}) _{n}  
      ({ \textstyle 1 + a - b - c}) _{n}  ({ \textstyle 1 + a - b - d}) _{n}  
      ({ \textstyle 1 + a - c - d}) _{n} }\over 
    {({ \textstyle 1 + a - b}) _{n}  ({ \textstyle 1 + a - c}) _{n}  
      ({ \textstyle 1 + a - d}) _{n}  ({ \textstyle 1 + a - b - c - d}) _{n} }
    }\ ,
\endmultline\tag5.3
$$
thus obtaining
$$a_k=\frac {(1+a+c)_{2k}\,(-a+2c)_k} {(1+a+c-d)_{2k}\,(-a+2c+d)_k}
\frac {\prod _{j=1} ^{k}((j+a-d)^3+(k+c)^3)} {\prod _{j=0}
^{k-1}((j+c)^3-(k+c)^3)}
\tag5.4$$
after simplification. Now, by the inversion (4.1), the identity $\sum
_{k=0} ^{n}f_{nk}a_k=b_n$ holds, with $f_{nk}, a_k,b_n$ given in
(5.1)(1), (5.4), (5.2), respectively. This identity reads
$$\multline \sum _{k=0} ^{n}\frac {(n+a)^3+(n+c)^3} {(k+a)^3+(k+c)^3}
\frac {\prod _{j=k} ^{n-1}((j+a)^3+(k+c)^3)} {\prod _{j=k+1}
^{n}((j+c)^3-(k+c)^3)}\\
\frac {\prod _{j=1} ^{k}((j+a-d)^3+(k+c)^3)} {\prod _{j=0}
^{k-1}((j+c)^3-(k+c)^3)}
\frac {(1+a+c)_{2k}\,(-a+2c)_k} {(1+a+c-d)_{2k}\,(-a+2c+d)_k}\\
=\frac {(a+c,1+a/2+c/2,d,2a-c-d+1)_n}
{(1,a/2+c/2,1+a+c-d,2c-a+d)_n}.
\endmultline\tag5.5$$
The products which occur in the sum can be considered as cubic
analogues of the rising factorials. For example, 
$\prod _{j=k} ^{n-1}((j+a)^3+(k+c)^3)$ is a cubic analogue of
$(2k+a+c)_{n-k}=\prod _{j=k} ^{n-1}((j+a)+(k+c))$.

Next we choose $c_k=(k+c)^4$, $a_j=(j+a)^4$, $b_j=-1$. The inverse
pair (1.4) now reads
$$f_{nk}=\frac {(n+a)^4-(n+c)^4} {(k+a)^4-(k+c)^4}
\frac {\prod _{j=k} ^{n-1}((j+a)^4-(k+c)^4)} {\prod _{j=k+1}
^{n}((j+c)^4-(k+c)^4)}
\tag5.6)(1$$
and 
$$f_{kl}^{-1}=\frac {\prod _{j=l+1} ^{k}((j+a)^4-(k+c)^4)} {\prod _{j=l}
^{k-1}((j+c)^4-(k+c)^4)}\ .
\tag5.6)(2$$
For $b_l$ we take
$$b_l=\frac {(a+c,1+a/2+c/2,1-2c+2a)_l}
{(1,a/2+c/2,3c-a)_l}\ .\tag5.7$$
Then we form $a_k=\sum _{l=0} ^{k}f_{kl}^{-1}b_l$.
Again, the sum can be evaluated by means of Dougall's sum (5.3).
Thus we get
$$a_k=\frac {(1+a+c)_{2k}} {(3c-a)_{2k}}
\frac {\prod _{j=0} ^{k-1}((j+2c-a)^4-(k+c)^4)} {\prod _{j=0}
^{k-1}((j+c)^4-(k+c)^4)}\ .
\tag5.8$$
The inversion (4.1) then yields
$$\multline \sum _{k=0} ^{n}\frac {(n+a)^4-(n+c)^4} {(k+a)^4-(k+c)^4}
\frac {\prod _{j=k} ^{n-1}((j+a)^4-(k+c)^4)} {\prod _{j=k+1}
^{n}((j+c)^4-(k+c)^4)}\\
\frac {\prod _{j=0} ^{k-1}((j+2c-a)^4-(k+c)^4)} {\prod _{j=0}
^{k-1}((j+c)^4-(k+c)^4)}
\frac {(1+a+c)_{2k}} {(3c-a)_{2k}}\\
=\frac {(a+c,1+a/2+c/2,1-2c-2a)_n}
{(1,a/2+c/2,1+a+c-d,3c-a)_n}.
\endmultline\tag5.9$$
In this identity there occur quartic analogues of rising factorials.

Finally, we take $c_k=(k+c)^5$, $a_j=(j+a)^5$, $b_j=1$. The inverse
pair (1.4) then reads
$$f_{nk}=\frac {(n+a)^5+(n+c)^5} {(k+a)^5+(k+c)^5}
\frac {\prod _{j=k} ^{n-1}((j+a)^5+(k+c)^5)} {\prod _{j=k+1}
^{n}((j+c)^5-(k+c)^5)}
\tag5.10)(1$$
and 
$$f_{kl}^{-1}=\frac {\prod _{j=l+1} ^{k}((j+a)^5+(k+c)^5)} {\prod _{j=l}
^{k-1}((j+c)^5-(k+c)^5)}\ .
\tag5.10)(2$$
Now choose
$$b_l=\frac {(a+c,1+a/2+c/2)_l}
{(1,a/2+c/2)_l}\ .\tag5.11$$
Then for $c$ choose $(2a+1)/3$ and subsequently replace $a$ by $3a+1$. If we
do the same as in the previous two derivations, we finally obtain the
following identity which contains quintic analogues of rising
factorials,
$$\multline \sum _{k=0} ^{n}\frac {(n+3a+1)^5+(n+2a+1)^5} {(k+3a+1)^5+(k+2a+1)^5}
\frac {\prod _{j=k} ^{n-1}((j+3a+1)^5+(k+2a+1)^5)} {\prod _{j=k+1}
^{n}((j+2a+1)^5-(k+2a+1)^5)}\\
\frac {(5a+3)_{2k}\,(1+a+(1+2a+k)(\om+\om^2),
1+a+(1+2a+k)(\om+\om^3)} {\prod _{j=0}
^{k-1}((j+2a+1)^5-(k+2a+1)^5)}\\
\frac {1+a+(1+2a+k)(\om+\om^4))_k} {}=\frac {(5a+2,5a/2+2)_n}
{(1,5a/2+1)_n}.
\endmultline\tag5.12$$
Here, $\om$ denotes a fifth root of unity.

\remark{Acknowledgement} I thank George Gasper and Mizan Rahman for
helpful communication.
I am indebted to the referee who encouraged
me to search for new applications of the presented matrix inverse.
\endremark

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\enddocument



