\def\KrSrAA{5}
\def\KratAN{4}
\def\KratAR{3}
\def\KratBF{2}
\def\GaRaAA{1}
\input format
\topmatter 
\title A Systematic List of Two- and Three-term Contiguous Relations for 
Basic Hypergeometric Series 
\endtitle 
\author C.~Krattenthaler
\endauthor 
\affil 
Institut f\"ur Mathematik der Universit\"at Wien,\\
Strudlhofgasse 4, A-1090 Wien, Austria.\\
e-mail: KRATT\@Pap.Univie.Ac.At
\endaffil 
%\email KRATT\@Pap.Univie.Ac.At\endemail
%\dedicatory \enddedicatory
%\date \enddate
%\thanks \endthanks
%\subjclass Primary ;
% Secondary .
%\endsubjclass
%\keywords \endkeywords
%\abstract 
%\endabstract
\endtopmatter
\document
\TagsOnRight
%-------------------------------Seite 1-------------------------------
Contiguous relations base on simply two-, three-, four-term,\dots
identities of certain products. Usually these can be derived from
special cases of terminating basic hypergeometric summation or transformation
formulas. In what follows, the symbols C$\langle\text {\it number}\rangle$
refer to replacements rules of the {\sl Mathematica} package HYPQ
which is available via WWW at 
{\tt http://www.mat.univie.ac.at/\~{}kratt} (cf\.
\cite{\KratBF, \KratAR}).
\subhead 1. A two-term contiguous relation\endsubhead
This relation results from splitting the first term of a basic
hypergeometric series from the rest. Recall that
$${}_r\phi_s\!\[\matrix (A)\\(B)\endmatrix; q, z\]=\sum _{k=0}
^{\infty}\frac {(A)_k} {(q)_k(B)_k}
\((-1)^kq^{\binom{k}{2}}\)^{1+s-r}z^k,$$
where $(A)$ stands for a sequence $A_1,A_2,\dots,A_r$ and $(B)$
stands for a sequence $B_1,B_2,\mathbreak\dots,B_s$. $(A)_k$ is understood as
an abbreviation for $\prod _{i=1} ^{r}A_i$, etc. In the sequel $(A)$
and $(B)$ also may be shorter sequences, however the length will
always be clear from the context.

With these conventions we have
(C01, C02)
$${}_r\phi_s\!\[\matrix (A)\\(B)\endmatrix; q, z\]=1+\frac
{(-1)^{1+s-r}z} {(1-q)}\frac {\prodl _{i=1} ^{r}(1-A_i)} {\prodl
_{i=1} ^{s}(1-B_i)}{}_{r+1}\phi_{s+1}\!\[\matrix q,(qA)\\q^2,(qB)
\endmatrix; q, q^{1+s-r}z\] \tag1.1$$
\subhead 2. Simple three-term contiguous relations\endsubhead
Take the identity $1-q^k=(1-q^k)$. When multiplying this identity by
$z^k(A)_k/(B)_k(q)_k$ and summing over all $k$, from this identity we get
(C11--C13)
$${}_r\phi_s\!\[\matrix (A)\\(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix (A)\\(B)\endmatrix; q, qz\]=(-1)^{1+s-r}z\frac {\prodl
_{i=1} ^{r}(1-A_i)} {\prodl _{i=1} ^{s}(1-B_i)}
{}_r\phi_s\!\[\matrix (qA)\\(qB)\endmatrix; q, q^{1+s-r}z\]
\tag2.1$$
%-------------------------------Seite 2-------------------------------
Next take the identity $(1-aq^k)-(1-a)=a(1-q^k)$. This gives
(C14, C15, C16, \cite{\GaRaAA, Ex.~I.9\,(ii)})
$$\multline 
{}_r\phi_s\!\[\matrix aq, (A)\\(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix a, (A)\\(B)\endmatrix; q, z\]\\=
(-1)^{1+s-r}az \frac {\prodl _{i=1} ^{r-1}(1-A_i)} {\prodl _{i=1} ^{s}(1-B_i)}
{}_r\phi_s\!\[\matrix aq,(Aq)\\(Bq)\endmatrix; q, q^{1+s-r}z\]\\
\endmultline
\tag2.2$$
when multiplying by $z^k(aq)_{k-1}(A)_k/(q)_k(B)_k$ and summing over $k$.

When multiplying by $z^k(A)_k/(q)_k(a)_{k+1}(B)_k$ and summing over $k$
we get
(C20--C22, \cite{\GaRaAA, Ex.~I.9\,(i)})
$$\multline 
{}_r\phi_s\!\[\matrix (A)\\a,(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix (A)\\aq,(B)\endmatrix; q, z\]\\=
\frac {(-1)^{1+s-r}az} {(1-a)(1-aq)}\frac {\prodl _{i=1} ^{r}(1-A_i)}
{\prodl _{i=1} ^{s-1}(1-B_i)} {}_r\phi_s\!\[\matrix (Aq)\\aq^2,(Bq)
\endmatrix; q, q^{1+s-r}z\] \endmultline
\tag2.3$$
Now consider the similar identity $(1-aq^k)-(1-a)q^k=1-q^k$. From this
we get
(C17, C18, C19)
$$\multline 
{}_r\phi_s\!\[\matrix aq,(A)\\(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix a,(A)\\(B)\endmatrix; q, qz\]\\=
(-1)^{1+s-r}z \frac {\prodl _{i=1} ^{r-1}(1-A_i)} {\prodl _{i=1} ^{s}
(1-B_i)} {}_r\phi_s\!\[\matrix aq,(Aq)\\(Bq)\endmatrix; q, q^{1+s-r}z\]
\endmultline
\tag2.4$$
and
%-------------------------------Seite 3-------------------------------
(C23--C25)
$$\multline 
{}_r\phi_s\!\[\matrix (A)\\a,(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix (A)\\aq,(B)\endmatrix; q, qz\]\\=
\frac {(-1)^{1+s-r}z} {(1-a)(1-aq)}\frac {\prodl _{i=1} ^{r}(1-A_i)}
{\prodl _{i=1} ^{s-1}(1-B_i)} {}_r\phi_s\!\[\matrix (Aq)\\aq^2,(Bq)
\endmatrix; q, q^{1+s-r}z\] \endmultline
\tag2.5$$
Next consider $(1-aq^k)-(1-bq^k)=q^k(b-a)$. Multiplication by
$(a)_k(b)_k(A)_k/(B)_k$ gives 
(C26--C27)
$$\multline
(1-a){}_r\phi_s\!\[\matrix aq,b,(A)\\(B)\endmatrix; q, z\]-
(1-b){}_r\phi_s\!\[\matrix a,bq,(A)\\(B)\endmatrix; q, z\]\\=
(b-a){}_r\phi_s\!\[\matrix a,b,(A)\\(B)\endmatrix; q, qz\].
\endmultline
\tag2.6$$
Multiplication by $(a)_k(A)_k/(bq)_k(B)_k$ gives
(C34--C36)
$$\multline
(1-a){}_r\phi_s\!\[\matrix aq,(A)\\bq,(B)\endmatrix; q, z\]-
(1-b){}_r\phi_s\!\[\matrix a,(A)\\b,(B)\endmatrix; q, z\]\\=
(b-a){}_r\phi_s\!\[\matrix a,(A)\\bq,(B)\endmatrix; q, qz\].
\endmultline
\tag2.7$$
%-------------------------------Seite 4-------------------------------
Finally, multiplication by $(A)_k/(aq)_k(bq)_k(B)_k$ gives
(C45--C46)
$$\multline
(1-a){}_r\phi_s\!\[\matrix (A)\\a,bq,(B)\endmatrix; q, z\]-
(1-b){}_r\phi_s\!\[\matrix (A)\\aq,b,(B)\endmatrix; q, z\]\\=
(b-a){}_r\phi_s\!\[\matrix (A)\\aq,bq,(B)\endmatrix; q, z\].
\endmultline
\tag2.8$$
Now consider $(1-aq^k)\,b-(1-bq^k)\,a=b-a$. By similar considerations
we obtain
(C28--C29, \cite{\GaRaAA, Ex.~I.10\,(i)})
$$\multline
b(1-a){}_r\phi_s\!\[\matrix aq,b,(A)\\(B)\endmatrix; q, z\]-
a(1-b){}_r\phi_s\!\[\matrix a,bq,(A)\\(B)\endmatrix; q, z\]\\=
(b-a){}_r\phi_s\!\[\matrix a,b,(A)\\(B)\endmatrix; q, z\].
\endmultline
\tag2.9$$
and
(C37--C39)
$$\multline
b(1-a){}_r\phi_s\!\[\matrix aq,(A)\\bq,(B)\endmatrix; q, z\]-
a(1-b){}_r\phi_s\!\[\matrix a,(A)\\b,(B)\endmatrix; q, z\]\\=
(b-a){}_r\phi_s\!\[\matrix a,(A)\\bq,(B)\endmatrix; q, z\].
\endmultline
\tag2.10$$
and
(C47--C48)
$$\multline
b(1-a){}_r\phi_s\!\[\matrix (A)\\a,bq,(B)\endmatrix; q, z\]-
a(1-b){}_r\phi_s\!\[\matrix (A)\\aq,b,(B)\endmatrix; q, z\]\\=
(b-a){}_r\phi_s\!\[\matrix (A)\\aq,bq,(B)\endmatrix; q, z\].
\endmultline
\tag2.11$$
%-------------------------------Seite 5-------------------------------
\subhead 3. Contiguous relations resulting from the $n=1$ case of the
$q$--Saalsch\"utz sum\endsubhead
Set $n=1$ in the $q$--Saalsch\"utz \cite{\GaRaAA, (1.7.2); Appendix
(II.12)} sum. The resulting identity is equivalent
with
$$(1-C)(1-C/AB)+\frac {C} {AB}(1-A)(1-B)=(1-C/A)(1-C/B).\tag3.1$$
Now let us take $A=a, B=1/bq^k,C=1/q^k$. After some simplification
(3.1) reduces to
$$(1-aq^k)(1-b)-(1-a)(1-bq^k)=(a-b)(1-q^k).$$
From this we obtain the contiguous relations
(C30--C31, \cite{\GaRaAA, Ex.~I.9\,(iv)})
$$\multline
{}_r\phi_s\!\[\matrix aq,b,(A)\\(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix a,bq,(A)\\(B)\endmatrix; q, z\]\\=
(-1)^{1+s-r}z(a-b)\frac {\prodl _{i=1} ^{r-2}(1-A_i)}
{\prodl _{i=1} ^{s}(1-B_i)}
{}_r\phi_s\!\[\matrix aq,bq,(Aq)\\(Bq)\endmatrix; q, q^{1+s-r}z\].
\endmultline
\tag3.2$$
when multiplying by $z^k(a)_{k-1}(b)_{k-1}(A)_k/(B)_k$, and
(C40--C42, \cite{\GaRaAA, Ex.~I.9\,(iii)})
$$\multline
{}_r\phi_s\!\[\matrix aq,(A)\\bq,(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix a,(A)\\b,(B)\endmatrix; q, z\]\\=
\frac {(-1)^{1+s-r}z(a-b)} {(1-b)(1-bq)}\frac {\prodl _{i=1} ^{r-1}(1-A_i)}
{\prodl _{i=1} ^{s-1}(1-B_i)}
{}_r\phi_s\!\[\matrix aq,(Aq)\\bq^2,(Bq)\endmatrix; q, q^{1+s-r}z\].
\endmultline
\tag3.3$$
when multiplying by $z^k(a)_{k-1}(A)_k/(b)_{k+1}(B)_k$, and
%-------------------------------Seite 6-------------------------------
(C49--C50)
$$\multline
{}_r\phi_s\!\[\matrix (A)\\a,bq,(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix (A)\\aq,b,(B)\endmatrix; q, z\]\\=
\frac {(-1)^{1+s-r}z(a-b)} {(1-a)(1-aq)(1-b)(1-bq)}\frac {\prodl _{i=1}
^{r}(1-A_i)} {\prodl _{i=1} ^{s-2}(1-B_i)}
{}_r\phi_s\!\[\matrix (Aq)\\aq^2,bq^2,(Bq)\endmatrix; q, q^{1+s-r}z\].
\endmultline
\tag3.4$$
Next in (3.1) set $A=a, B=b, C=q^{-k}$. Then (3.1) reduces to
$$(1-aq^k)(1-bq^k)-q^k(1-a)(1-b)=(1-abq^k)(1-q^k).$$
In a similar manner as above, from this we get the contiguous relations
(C32--C33, C73)
$$\multline
{}_r\phi_s\!\[\matrix aq,bq,(A)\\(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix a,b,(A)\\(B)\endmatrix; q, qz\]\\=
(-1)^{1+s-r}z(1-abq)\frac {\prodl _{i=1} ^{r-2}(1-A_i)} {\prodl _{i=1} ^{s}(1-B_i)}
{}_{r+1}\phi_{s+1}\!\[\matrix aq,bq,abq^2,(Aq)\\abq,(Bq)\endmatrix; q, q^{1+s-r}z\].
\endmultline
\tag3.5$$
and
(C43--C44, C88, \cite{\KrSrAA, (6.5)})
$$\multline
{}_r\phi_s\!\[\matrix aq,(A)\\b,(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix a,(A)\\bq,(B)\endmatrix; q, qz\]\\=
\frac {(-1)^{1+s-r}z(1-abq)} {(1-b)(1-bq)}\frac {\prodl _{i=1}
^{r-1}(1-A_i)} {\prodl _{i=1} ^{s-1}(1-B_i)}
{}_{r+1}\phi_{s+1}\!\[\matrix aq,abq^2,(Aq)\\bq^2,abq,(Bq)\endmatrix; q, q^{1+s-r}z\].
\endmultline
\tag3.6$$
and
(C51--C52, C98)
$$\multline
{}_r\phi_s\!\[\matrix (A)\\a,b,(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix (A)\\aq,bq,(B)\endmatrix; q, qz\]=
\frac {(-1)^{1+s-r}z(1-abq)} {(1-a)(1-aq)(1-b)(1-bq)}\\
\frac {\prodl _{i=1}
^{r}(1-A_i)} {\prodl _{i=1} ^{s-2}(1-B_i)}
{}_{r+1}\phi_{s+1}\!\[\matrix abq^2,(Aq)\\aq^2,bq^2,abq,(Bq)\endmatrix; q, q^{1+s-r}z\].
\endmultline
\tag3.7$$
%-------------------------------Seite 7-------------------------------
Now in (3.1) we set $A=aq^k, B=c/b, C=cq^k$. After some
simplification we obtain
$$(1-cq^k)(a-b)+(1-aq^k)(b-c)+(1-bq^k)(c-a)=0.$$
From this identity we obtain the contiguous relations
(C53)
$$\multline
{}_r\phi_s\!\[\matrix a,b,cq,(A)\\(B)\endmatrix; q, z\]\frac {(a-b)}
{(1-a)(1-b)}+
{}_r\phi_s\!\[\matrix aq,b,c,(A)\\(B)\endmatrix; q, z\]
\frac {(b-c)} {(1-b)(1-c)}\\
+{}_r\phi_s\!\[\matrix a,bq,c,(A)\\(B)\endmatrix; q, z\]
\frac {(c-a)} {(1-c)(1-a)}=0
\endmultline
\tag3.8$$
and
(C54--C55)
$$\multline
{}_r\phi_s\!\[\matrix a,b,(A)\\c,(B)\endmatrix; q, z\]\frac {(a-b)}
{(1-a)(1-b)}+
{}_r\phi_s\!\[\matrix aq,b,(A)\\cq,(B)\endmatrix; q, z\]
\frac {(b-c)} {(1-b)(1-c)}\\
+{}_r\phi_s\!\[\matrix a,bq,(A)\\cq,(B)\endmatrix; q, z\]
\frac {(c-a)} {(1-c)(1-a)}=0
\endmultline
\tag3.9$$
and
(C56--C57)
$$\multline
{}_r\phi_s\!\[\matrix a,(A)\\bq,c,(B)\endmatrix; q, z\]\frac {(a-b)}
{(1-a)(1-b)}+
{}_r\phi_s\!\[\matrix aq,(A)\\bq,cq,(B)\endmatrix; q, z\]
\frac {(b-c)} {(1-b)(1-c)}\\
+{}_r\phi_s\!\[\matrix a,(A)\\b,cq,(B)\endmatrix; q, z\]
\frac {(c-a)} {(1-c)(1-a)}=0
\endmultline
\tag3.10$$
and
(C58)
$$\multline
{}_r\phi_s\!\[\matrix (A)\\aq,bq,c,(B)\endmatrix; q, z\]\frac {(a-b)}
{(1-a)(1-b)}+
{}_r\phi_s\!\[\matrix (A)\\a,bq,cq,(B)\endmatrix; q, z\]
\frac {(b-c)} {(1-b)(1-c)}\\
+{}_r\phi_s\!\[\matrix (A)\\aq,b,cq,(B)\endmatrix; q, z\]
\frac {(c-a)} {(1-c)(1-a)}=0
\endmultline
\tag3.11$$
%-------------------------------Seite 8-------------------------------
Now we set $A=b/a, B=ab/c, C=bq^k$ in (3.1) to finally get
$$(1-aq^k)(1-cq^k/a)-(1-bq^k)(1-cq^k/b)=q^k(b-a)(1-c/ab).$$
From this we derive the contiguous relations
(C59--C60)
$$\multline
{}_r\phi_s\!\[\matrix aq,cq/a,b,c/b,(A)\\(B)\endmatrix; q, z\]
(1-a)(1-c/a)\\\quad \quad -{}_r\phi_s\!\[\matrix a,c/a,bq,cq/b,(A)\\(B)
\endmatrix; q, z\](1-b)(1-c/b)\\=
{}_r\phi_s\!\[\matrix a,c/a,b,c/b,(A)\\(B)\endmatrix; q, qz\](b-a)(1-c/ab)
\endmultline
\tag3.12$$
and
(C64--C66)
$$\multline
{}_r\phi_s\!\[\matrix aq,cq/a,b,(A)\\cq/b,(B)\endmatrix; q, z\]
(1-a)(1-c/a)-{}_r\phi_s\!\[\matrix a,c/a,bq,(A)\\c/b,(B)
\endmatrix; q, z\](1-b)(1-c/b)\\=
{}_r\phi_s\!\[\matrix a,c/a,b,(A)\\cq/b,(B)\endmatrix; q, qz\](b-a)(1-c/ab)
\endmultline
\tag3.13$$
and
(C74--C75, \cite{\KrSrAA, (5.4)})
$$\multline
{}_r\phi_s\!\[\matrix aq,b,(A)\\c/a,cq/b,(B)\endmatrix; q, z\]
(1-a)(1-c/a)-{}_r\phi_s\!\[\matrix a,bq,(A)\\cq/a,c/b,(B)
\endmatrix; q, z\](1-b)(1-c/b)\\=
{}_r\phi_s\!\[\matrix a,b,(A)\\cq/a,cq/b,(B)\endmatrix; q, qz\](b-a)(1-c/ab)
\endmultline
\tag3.14$$
and
(C80--C82)
$$\multline
{}_r\phi_s\!\[\matrix aq,cq/a,(A)\\bq,cq/b,(B)\endmatrix; q, z\]
(1-a)(1-c/a)-{}_r\phi_s\!\[\matrix a,c/a,(A)\\b,c/b,(B)
\endmatrix; q, z\](1-b)(1-c/b)\\=
{}_r\phi_s\!\[\matrix a,c/a,(A)\\bq,cq/b,(B)\endmatrix; q, qz\](b-a)(1-c/ab)
\endmultline
\tag3.15$$
%-------------------------------Seite 9-------------------------------
and
(C89--C91)
$$\multline
{}_r\phi_s\!\[\matrix aq,(A)\\c/a,bq,cq/b,(B)\endmatrix; q, z\]
(1-a)(1-c/a)\\-{}_r\phi_s\!\[\matrix a,(A)\\cq/a,b,c/b,(B)
\endmatrix; q, z\](1-b)(1-c/b)\\=
{}_r\phi_s\!\[\matrix a,(A)\\cq/a,bq,cq/b,(B)\endmatrix; q, qz\](b-a)(1-c/ab)
\endmultline
\tag3.16$$
and
(C99--C100)
$$\multline
{}_r\phi_s\!\[\matrix (A)\\a,c/a,bq,cq/b,(B)\endmatrix; q, z\]
(1-a)(1-c/a)\\-{}_r\phi_s\!\[\matrix (A)\\aq,cq/a,b,c/b,(B)
\endmatrix; q, z\](1-b)(1-c/b)\\=
{}_r\phi_s\!\[\matrix (A)\\aq,cq/a,bq,cq/b,(B)\endmatrix; q, qz\](b-a)(1-c/ab)
\endmultline
\tag3.17$$
Finally, in (3.1) we set $A=aq^k, B=bq^k, C=cq^{2k}$. This leads to
$$(1-cq^k/a)(1-cq^k/b)-\frac{c}{ab}(1-aq^k)(1-bq^k)=(1-c/ab)(1-cq^{2k}).$$
From this we obtain the contiguous relations
(C61, C110)
$$\multline
{}_r\phi_s\!\[\matrix a,cq/a,b,cq/b,(A)\\(B)\endmatrix; q, z\]
(1-c/a)(1-c/b)\\-{}_r\phi_s\!\[\matrix aq,c/a,bq,c/b,(A)\\(B)
\endmatrix; q, z\]\frac {(1-a)(1-b)c} {ab}\\=
{}_{r+2}\phi_{s+2}\!\[\matrix q\sqrt c,-q\sqrt c,a,c/a,b,c/b,(A)\\
\sqrt c,-\sqrt c,(B)\endmatrix; q, z\](1-c)(1-c/ab)
\endmultline
\tag3.18$$
and
(C67--C68, C112)
$$\multline
{}_r\phi_s\!\[\matrix a,cq/a,b,(A)\\c/b,(B)\endmatrix; q, z\]
(1-c/a)(1-c/b)\\-{}_r\phi_s\!\[\matrix aq,c/a,bq,(A)\\cq/b,(B)
\endmatrix; q, z\]\frac {(1-a)(1-b)c} {ab}\\=
{}_{r+2}\phi_{s+2}\!\[\matrix q\sqrt c,-q\sqrt c,a,c/a,b,(A)\\
\sqrt c,-\sqrt c,cq/b,(B)\endmatrix; q, z\](1-c)(1-c/ab)
\endmultline
\tag3.19$$
%-------------------------------Seite 10------------------------------
and
(C76--C77, C114, \cite{\KratAN, proof of (3.10)})
$$\multline
{}_r\phi_s\!\[\matrix a,b,(A)\\c/a,c/b,(B)\endmatrix; q, z\]
(1-c/a)(1-c/b)\\-{}_r\phi_s\!\[\matrix aq,bq,(A)\\cq/a,cq/b,(B)
\endmatrix; q, z\]\frac {(1-a)(1-b)c} {ab}\\=
{}_{r+2}\phi_{s+2}\!\[\matrix q\sqrt c,-q\sqrt c,a,b,(A)\\
\sqrt c,-\sqrt c,cq/a,cq/b,(B)\endmatrix; q, z\](1-c)(1-c/ab)
\endmultline
\tag3.20$$
and
(C83, C115)
$$\multline
{}_r\phi_s\!\[\matrix a,cq/a,(A)\\bq,c/b,(B)\endmatrix; q, z\]
(1-c/a)(1-c/b)\\-{}_r\phi_s\!\[\matrix aq,c/a,(A)\\b,cq/b,(B)
\endmatrix; q, z\]\frac {(1-a)(1-b)c} {ab}\\=
{}_{r+2}\phi_{s+2}\!\[\matrix q\sqrt c,-q\sqrt c,a,c/a,(A)\\
\sqrt c,-\sqrt c,bq,cq/b,(B)\endmatrix; q, z\](1-c)(1-c/ab)
\endmultline
\tag3.21$$
and
(C92--C93, C118)
$$\multline
{}_r\phi_s\!\[\matrix a,(A)\\c/a,bq,c/b,(B)\endmatrix; q, z\]
(1-c/a)(1-c/b)\\-{}_r\phi_s\!\[\matrix aq,(A)\\cq/a,b,cq/b,(B)
\endmatrix; q, z\]\frac {(1-a)(1-b)c} {ab}\\=
{}_{r+2}\phi_{s+2}\!\[\matrix q\sqrt c,-q\sqrt c,a,(A)\\
\sqrt c,-\sqrt c,cq/a,bq,cq/b,(B)\endmatrix; q, z\](1-c)(1-c/ab)
\endmultline
\tag3.22$$
and
(C101, C120)
$$\multline
{}_r\phi_s\!\[\matrix (A)\\aq,c/a,bq,c/b,(B)\endmatrix; q, z\]
(1-c/a)(1-c/b)\\-{}_r\phi_s\!\[\matrix (A)\\a,cq/a,b,cq/b,(B)
\endmatrix; q, z\]\frac {(1-a)(1-b)c} {ab}\\=
{}_{r+2}\phi_{s+2}\!\[\matrix q\sqrt c,-q\sqrt c,(A)\\
\sqrt c,-\sqrt c,aq,cq/a,bq,cq/b,(B)\endmatrix; q, z\](1-c)(1-c/ab)
\endmultline
\tag3.23$$
%-------------------------------Seite 11------------------------------
\subhead 4. Contiguous relations resulting from the $n=1$ case of
Jackson's sum\endsubhead
Set $n=1$ in Jackson's \cite{\GaRaAA, (2.6.2); Appendix (II.22)}
$_8\phi_7$-sum. In the resulting identity replace $a$ by
$CD/q,b$ by $A,c$ by $C/A,d$ by $BD$. After some simplification this
yields
$$\multline
(1-A)(1-C/A)(1-BD)(1-CD/B)-(1-B)(1-C/B)(1-AD)(1-CD/A)\\=
(B-A)(1-C/AB)(1-D)(1-CD).\quad \quad 
\endmultline\tag4.1$$
Now we set $A=a, B=b, C=c,$ and $D=q^k$ to get
$$\multline
(1-a)(1-c/a)(1-bq^k)(1-cq^k/b)-(1-b)(1-c/b)(1-q^k)(1-cq^k/a)\\=
(b-a)(1-c/ab)(1-q^k)(1-cq^k).
\endmultline$$
From this identitiy we derive the contiguous relations
(C62, C104)
$$\multline
{}_r\phi_s\!\[\matrix a,c/a,bq,cq/b,(A)\\(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix aq,cq/a,b,c/b,(A)\\(B)\endmatrix; q, z\]\\=
(-1)^{1+s-r}z(b-a)(1-c/ab)(1-cq)\frac {\prodl _{i=1}
^{r-4}(1-A_i)} {\prodl _{i=1} ^{s}(1-B_i)}\\
{}_{r+1}\phi_{s+1}\!\[\matrix aq,cq/a,bq,cq/b,cq^2,(Aq)\\cq,(Bq)
\endmatrix; q, q^{1+s-r}z\]
\endmultline
\tag4.2$$
and
(C69--C70, C105)
$$\multline
{}_r\phi_s\!\[\matrix a,c/a,bq,(A)\\c/b,(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix aq,cq/a,b,(A)\\cq/b,(B)\endmatrix; q, z\]\\=
(-1)^{1+s-r}z\frac {(b-a)(1-c/ab)(1-cq)} {(1-c/b)(1-cq/b)}
\frac {\prodl _{i=1} ^{r-3}(1-A_i)} {\prodl _{i=1} ^{s-1}(1-B_i)}\\
{}_{r+1}\phi_{s+1}\!\[\matrix aq,cq/a,bq,cq^2,(Aq)\\cq^2/b,cq,(Bq)
\endmatrix; q, q^{1+s-r}z\]
\endmultline
\tag4.3$$
%-------------------------------Seite 12------------------------------
and
(C84--C85, C107)
$$\multline
{}_r\phi_s\!\[\matrix a,c/a,(A)\\b,c/b,(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix aq,cq/a,(A)\\bq,cq/b,(B)\endmatrix; q, z\]\\=
(-1)^{1+s-r}z\frac {(b-a)(1-c/ab)(1-cq)} {(1-b)(1-bq)(1-c/b)(1-cq/b)}
\frac {\prodl _{i=1} ^{r-2}(1-A_i)} {\prodl _{i=1} ^{s-2}(1-B_i)}\\
{}_{r+1}\phi_{s+1}\!\[\matrix aq,cq/a,cq^2,(Aq)\\bq^2,cq^2/b,cq,(Bq)
\endmatrix; q, q^{1+s-r}z\]
\endmultline
\tag4.4$$
and
(C78, C106, \cite{\KrSrAA, (5.6)})
$$\multline
{}_r\phi_s\!\[\matrix a,bq,(A)\\cq/a,c/b,(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix aq,b,(A)\\c/a,cq/b,(B)\endmatrix; q, z\]\\=
(-1)^{1+s-r}z\frac {(b-a)(1-c/ab)(1-cq)} {(1-c/a)(1-cq/a)(1-c/b)(1-cq/b)}
\frac {\prodl _{i=1} ^{r-2}(1-A_i)} {\prodl _{i=1} ^{s-2}(1-B_i)}\\
{}_{r+1}\phi_{s+1}\!\[\matrix aq,bq,cq^2,(Aq)\\cq^2/a,cq^2/b,cq,(Bq)
\endmatrix; q, q^{1+s-r}z\]
\endmultline
\tag4.5$$
and
(C94--C95, C108)
$$\multline
{}_r\phi_s\!\[\matrix a,(A)\\cq/a,b,c/b,(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix aq,(A)\\c/a,bq,cq/b,(B)\endmatrix; q, z\]\\=
(-1)^{1+s-r}z\frac {(b-a)(1-c/ab)(1-cq)} {(1-c/a)(1-cq/a)(1-b)(1-bq)(1-c/b)(1-cq/b)}
\frac {\prodl _{i=1} ^{r-1}(1-A_i)} {\prodl _{i=1} ^{s-3}(1-B_i)}\\
{}_{r+1}\phi_{s+1}\!\[\matrix aq,cq^2,(Aq)\\cq^2/a,bq^2,cq^2/b,cq,(Bq)
\endmatrix; q, q^{1+s-r}z\]
\endmultline
\tag4.6$$
and
(C102, C109)
$$\multline
{}_r\phi_s\!\[\matrix (A)\\aq,cq/a,b,c/b,(B)\endmatrix; q, z\]-
{}_r\phi_s\!\[\matrix (A)\\a,c/a,bq,cq/b,(B)\endmatrix; q, z\]\\=
(-1)^{1+s-r}z\frac {(b-a)(1-c/ab)(1-cq)} {(1-a)(1-aq)(1-c/a)(1-cq/a)
(1-b)(1-bq)(1-c/b)(1-cq/b)}\\
\frac {\prodl _{i=1} ^{r}(1-A_i)} {\prodl _{i=1} ^{s-4}(1-B_i)}
{}_{r+1}\phi_{s+1}\!\[\matrix cq^2,(Aq)\\aq^2,cq^2/a,bq^2,cq^2/b,cq,(Bq)
\endmatrix; q, q^{1+s-r}z\]
\endmultline
\tag4.7$$
%-------------------------------Seite 13------------------------------

Next in (4.1) set $A=aq^k, B=bq^k, C=q^kab/d, D=c/ab$. This results
into
$$\multline
(1-aq^k)(1-cq^k/a)(1-b/d)(1-c/bd)-(1-bq^k)(1-cq^k/b)(1-a/d)(1-c/ad)\\=
\frac {1} {d}(a-b)(1-c/ab)(1-q^kd)(1-cq^k/d).
\endmultline$$
From this identity we obtain the contiguous relations
(C63, C111)
$$\multline
{}_r\phi_s\!\[\matrix aq,cq/a,b,c/b,(A)\\(B)\endmatrix; q, z\]
\frac {(1-b/d)(1-c/bd)} {(1-b)(1-c/b)}\\-
{}_r\phi_s\!\[\matrix a,c/a,bq,cq/b,(A)\\(B)\endmatrix; q, z\]
\frac {(1-a/d)(1-c/ad)} {(1-a)(1-c/a)}\\=
{}_{r+2}\phi_{s+2}\!\[\matrix a,c/a,b,c/b,dq,cq/d,(A)\\d,c/d,(B)
\endmatrix; q, z\]\frac {(a-b)(1-c/ab)(1-d)(1-c/d)} 
{d(1-a)(1-c/a)(1-b)(1-c/b)}
\endmultline
\tag4.8$$
and
(C71--C72, C113)
$$\multline
{}_r\phi_s\!\[\matrix aq,cq/a,b,(A)\\cq/b,(B)\endmatrix; q, z\]
\frac {(1-b/d)(1-c/bd)} {(1-b)(1-c/b)}\\-
{}_r\phi_s\!\[\matrix a,c/a,bq,(A)\\c/b,(B)\endmatrix; q, z\]
\frac {(1-a/d)(1-c/ad)} {(1-a)(1-c/a)}\\=
{}_{r+2}\phi_{s+2}\!\[\matrix a,c/a,b,dq,cq/d,(A)\\cq/b,d,c/d,(B)
\endmatrix; q, z\]\frac {(a-b)(1-c/ab)(1-d)(1-c/d)}
{d(1-a)(1-c/a)(1-b)(1-c/b)}
\endmultline
\tag4.9$$
%-------------------------------Seite 14------------------------------
and
(C86--C87, C117)
$$\multline
{}_r\phi_s\!\[\matrix aq,cq/a,(A)\\bq,cq/b,(B)\endmatrix; q, z\]
\frac {(1-b/d)(1-c/bd)} {(1-b)(1-c/b)}\\-
{}_r\phi_s\!\[\matrix a,c/a,(A)\\b,c/b,(B)\endmatrix; q, z\]
\frac {(1-a/d)(1-c/ad)} {(1-a)(1-c/a)}\\=
{}_{r+2}\phi_{s+2}\!\[\matrix a,c/a,dq,cq/d,(A)\\bq,cq/b,d,c/d,(B)
\endmatrix; q, z\]\frac {(a-b)(1-c/ab)(1-d)(1-c/d)}
{d(1-a)(1-c/a)(1-b)(1-c/b)}
\endmultline
\tag4.10$$
and
(C79, C116)
$$\multline
{}_r\phi_s\!\[\matrix aq,b,(A)\\c/a,cq/b,(B)\endmatrix; q, z\]
\frac {(1-b/d)(1-c/bd)} {(1-b)(1-c/b)}\\-
{}_r\phi_s\!\[\matrix a,bq,(A)\\cq/a,c/b,(B)\endmatrix; q, z\]
\frac {(1-a/d)(1-c/ad)} {(1-a)(1-c/a)}\\=
{}_{r+2}\phi_{s+2}\!\[\matrix a,b,dq,cq/d,(A)\\cq/a,cq/b,d,c/d,(B)
\endmatrix; q, z\]\frac {(a-b)(1-c/ab)(1-d)(1-c/d)}
{d(1-a)(1-c/a)(1-b)(1-c/b)}
\endmultline
\tag4.11$$
and
(C96--C97, C119)
$$\multline
{}_r\phi_s\!\[\matrix aq,(A)\\c/a,bq,cq/b,(B)\endmatrix; q, z\]
\frac {(1-b/d)(1-c/bd)} {(1-b)(1-c/b)}\\-
{}_r\phi_s\!\[\matrix a,(A)\\cq/a,b,c/b,(B)\endmatrix; q, z\]
\frac {(1-a/d)(1-c/ad)} {(1-a)(1-c/a)}\\=
{}_{r+2}\phi_{s+2}\!\[\matrix a,dq,cq/d,(A)\\cq/a,bq,cq/b,d,c/d,(B)
\endmatrix; q, z\]\frac {(a-b)(1-c/ab)(1-d)(1-c/d)}
{d(1-a)(1-c/a)(1-b)(1-c/b)}
\endmultline
\tag4.12$$
and
(C103, C121)
$$\multline
{}_r\phi_s\!\[\matrix (A)\\a,c/a,bq,cq/b,(B)\endmatrix; q, z\]
\frac {(1-b/d)(1-c/bd)} {(1-b)(1-c/b)}\\-
{}_r\phi_s\!\[\matrix (A)\\aq,cq/a,b,c/b,(B)\endmatrix; q, z\]
\frac {(1-a/d)(1-c/ad)} {(1-a)(1-c/a)}\\=
{}_{r+2}\phi_{s+2}\!\[\matrix dq,cq/d,(A)\\aq,cq/a,bq,cq/b,d,c/d,(B)
\endmatrix; q, z\]\frac {(a-b)(1-c/ab)(1-d)(1-c/d)}
{d(1-a)(1-c/a)(1-b)(1-c/b)}
\endmultline
\tag4.13$$






\Refs

\ref\no \GaRaAA\by G.    Gasper and M. Rahman \yr 1990 
\book Basic hypergeometric series
\publ Encyclopedia of Mathematics And Its Applications~35, Cambridge University Press
\publaddr Cambridge\endref

\ref\no \KratBF\by C.    Krattenthaler \yr 1995 \paper  HYP and HYPQ
--- {\sl Mathematica\/} packages for the manipulation of binomial sums
and hypergeometric series respectively $q$-binomial sums and basic
hypergeometric series\jour J. Symbol\. Comput\.\vol 20\pages
737--744\endref 

\ref\no \KratAR\by C.    Krattenthaler \yr 2002 
\book HYPQ
\bookinfo Manual for a {\sl Mathematica} package for handling basic 
hypergeometric series\endref

\ref\no \KratAN\by C.    Krattenthaler 
\paper A new matrix inverse \jour
 Proc\. Amer\. Math\. Soc\.\vol 124\yr 1996\pages 47--59\endref

\ref\no \KrSrAA\by C.    Krattenthaler and H. M. Srivastava \yr 1993 
\paper Summations for basic hypergeometric series involving a $q$-analogue 
of the digamma function I
\jour Comput\. Math\. Appl\. \vol 32\yr 1996\pages 73--91\endref

\endRefs
\enddocument
