1 Introduction and motivation

Let \(\mathbb {N}\) be the set of natural numbers with \(\mathbb {N}_0=\{0\}\cup \mathbb {N}\). For an indeterminate x, the shifted factorial of order \(n\in \mathbb {N}_0\) with the base q is defined by \((x;q)_0=1\) and

$$\begin{aligned}(x;q)_n\,=\,(1-x)(1-qx)\cdots (1-q^{n-1}x) \quad \text {for}\quad n\in \mathbb {N}. \end{aligned}$$

Then, the Gaussian binomial coefficient reads as

$$\begin{aligned} {m\brack n}=\frac{(q;q)_m}{(q;q)_n(q;q)_{m-n}} =\frac{(q^{m-n+1};q)_n}{(q;q)_n} \quad \text {where}\quad m,\,n\in \mathbb {N}. \end{aligned}$$

In 1973, Gould and Hsu [14] discovered a useful pair of inverse series relations. The q-analogue was found immediately by Carlitz [6]. They have wide applications to classical hypergeometric series and q-series (cf. Chu [8, 11]), as well as Rogers–Ramanujan identities (cf. Chen–Chu [7]). By means of finite q-differences, Chu [12] established the duplicate form of Carlitz inversions, which can be reproduced as follows. For two complex variables \(\{x,y\}\) and four complex sequences \(\{a_k,b_k,c_k,d_k\}\), define the two polynomial sequences by

$$\begin{aligned}&\phi (x;0)=1 \quad \text {and}\quad \phi (x;n)=\prod _{i=0}^{n-1}(a_i+xb_i) \quad \text {for}\quad n\in \mathbb {N};\\&\psi (y;0)=1 \quad \text {and}\quad \psi (y;n)=\prod _{j=0}^{n-1}(c_j+yd_j) \quad \text {for}\quad n\in \mathbb {N}. \end{aligned}$$

Then, the system of equations

$$\begin{aligned} \begin{aligned} \Omega (m)=&\sum _{k\ge 0}q^{\left( {\begin{array}{c}m-2k\\ 2\end{array}}\right) }{m\brack 2k} \frac{c_k+q^{-2k}d_k}{\phi (q^{-m};k)\psi (q^{-m};k+1)}F(k)\\&-\sum _{k\ge 0}q^{\left( {\begin{array}{c}m-1-2k\\ 2\end{array}}\right) }{m\brack 2k+1} \frac{a_k+q^{-1-2k}b_k}{\phi (q^{-m};k+1)\psi (q^{-m};k+1)}G(k) \end{aligned} \end{aligned}$$
(1)

is equivalent to the system

$$\begin{aligned} F(m)=&\sum _{k=0}^{2m}(-1)^k{2m\brack k} \phi (q^{-k};m)\psi (q^{-k};m)\Omega (k), \end{aligned}$$
(2)
$$\begin{aligned} G(m)=&\sum _{k=0}^{2m+1}(-1)^k{2m+1\brack k} \phi (q^{-k};m)\psi (q^{-k};m+1)\Omega (k). \end{aligned}$$
(3)

For the inverse series relations, to prove each is to prove both. One implication is that for every identity of the form (1), there are two companion ones of the forms (2) and (3). The application along this direction has been explored by Chu [12], where a large number of classical \(_5F_4\)-series identities were derived. Instead, if there is a pair of identities fitting with (2) and (3), we would find their dual identity corresponding to (1). This will be employed in this paper to examine two particular cases of the q-Pfaff–Saalschütz theorem. What is surprising is that the dual relations result in new identities of the terminating balanced \(_4\phi _3\)-series, although it is seldom that a new q-series formula turns up.

The rest of the paper will be organized as follows. By incorporating two particular cases of the q-Pfaff–Saalschütz summation theorem, we shall first establish, in the next section, a new closed formula for the terminating balanced \(_4\phi _3\)-series and then derive some variants of it by contiguous relations. Their limiting series will be discussed in Sect. 3, where some remarkable nonterminating \(_2\phi _2\)-series identities will be deduced. Finally in Sect. 4, the polynomial argument will be utilized to obtain further balanced \(_4\phi _3\)-series identities that can be considered as “dual formulae” to those derived in Sect. 2.

Throughout the paper, the quotient of shifted factorials will be abbreviated to

$$\begin{aligned} \left[ \begin{array}{cccc}\alpha ,&{}\beta ,&{}\ldots ,&{}\gamma \\ A,&{}B,&{}\ldots ,&{}C\end{array}{\Big |q}\right] _n =\frac{\left( \alpha ;q\right) _{n}\left( \beta ;q\right) _{n}\cdots \left( \gamma ;q\right) _{n}}{\left( A;q\right) _{n}\left( B;q\right) _{n}\cdots \left( C;q\right) _{n}}. \end{aligned}$$

The q-series is defined, according to Bailey [5] and Gasper–Rahman [15], by

$$\begin{aligned} {_{1+\lambda }\phi _{\mu }} \left[ \begin{array}{cccc}a_0,&{}a_1,&{}\ldots ,&{}a_\lambda \\ &{}b_1,&{}\ldots ,&{}b_\mu \end{array}{\Big |q;z}\right] =\,\,\sum _{n=0}^{\infty } \Big \{(-1)^nq^{{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }}\Big \}^{\mu -\lambda } \left[ \begin{array}{cccc}a_0,&{}a_1,&{}\ldots ,&{}a_\lambda \\ q,&{}b_1,&{}\ldots ,&{}b_\mu \end{array}{\Big |q}\right] _n\,z^n. \end{aligned}$$

If one of its numerator parameters is of the form \(q^{-m}\) with \(m\in \mathbb {N}_0\), then we say that the \({_{1+\lambda }\phi _\mu }\)-series terminates. When \(\lambda =\mu \), the \({_{1+\lambda }\phi _\lambda }\)-series will be called k-balanced if the product of the denominator parameters exceeds that of the numerator parameters by \(q^k\), and simply balanced or Saalschützian when \(k=1\).

2 Main results and proofs

In this section, we shall prove some new identities for the terminating balanced \(_4\phi _3\)-series that may serve as the q-analogues of the classical \(_4F_3\)-series investigated by Chu and Wei [13] through the Legendre inversions.

Recall the q-Pfaff–Saalschütz summation theorem (cf. Bailey [5, §8.4]) about the terminating balanced series:

$$\begin{aligned} {_3\phi _2}\left[ \begin{array}{ccc}q^{-n},&{}a,&{}b\\ &{}c,&{}q^{1-n}ab/c\end{array}{\Big |q;q}\right] = \left[ \begin{array}{cc}c/a,c/b\\ c,c/ab\end{array}{\Big |q}\right] _n. \end{aligned}$$
(4)

We begin with its following particular case:

$$\begin{aligned} \begin{aligned} q^{\left( {\begin{array}{c}m\\ 2\end{array}}\right) +2m}\left[ \begin{array}{c}a,-c/a\\ -q^2,q^{1+m}c\end{array}{\Big |q}\right] _m&={_3\phi _2}\left[ \begin{array}{ccc}q^{-m},&{}-q^{1+m}a,&{}q^{1+m}c/a\\ &{}-q^2,&{}q^{1+m}c\end{array}{\Big |q;q}\right] \\&=\sum _{j=0}^m \left[ \begin{array}{ccc}q^{-m},&{}-q^{1+m}a,&{}q^{1+m}c/a\\ q,&{}-q^2,&{}q^{1+m}c\end{array}{\Big |q}\right] _jq^j. \end{aligned} \end{aligned}$$

Performing the replacement \(j\rightarrow k-m\), we can restate the last sum, by adding a number of initial zero terms, as the q-binomial formula

$$\begin{aligned} q^{\left( {\begin{array}{c}m\\ 2\end{array}}\right) +2m} \left[ \begin{array}{c}a,-c/a\\ -q^2,q^{1+m}c\end{array}{\Big |q}\right] _m =\sum _{k=0}^{2m} \left[ \begin{array}{ccc}q^{-m},&{}-q^{1+m}a,&{}q^{1+m}c/a\\ q,&{}-q^2,&{}q^{1+m}c\end{array}{\Big |q}\right] _{k-m}q^{k-m}. \end{aligned}$$
(5)

By making use of the two factorial expressions

$$\begin{aligned}\left[ \begin{array}{c}-q^{1+m}a,q^{1+m}c/a\\ q^{1+m}c\end{array}{\Big |q}\right] _{k-m} =\left[ \begin{array}{c}-a,c/a\\ c\end{array}{\Big |q}\right] _{k+1} \left[ \begin{array}{c}c\\ -a,c/a\end{array}{\Big |q}\right] _{m+1}\end{aligned}$$

and

$$\begin{aligned} \left[ \begin{array}{c}q^{-m}\\ q,-q^2\end{array}{\Big |q}\right] _{k-m} =&\left[ \begin{array}{c}q^{-2m}\\ q,-q^2\end{array}{\Big |q}\right] _k \left[ \begin{array}{c}q^{-m}\\ q^{1+k},-q^{2+k}\end{array}{\Big |q}\right] _{-m}\\ =&\left[ \begin{array}{c}q^{-2m}\\ q,-q^2\end{array}{\Big |q}\right] _k \left[ \begin{array}{c}q^{1+k-m},-q^{2+k-m}\\ q^{-2m}\end{array}{\Big |q}\right] _{m}\\ =&(-1)^{k}{2m\brack k} \frac{q^{\left( {\begin{array}{c}k\\ 2\end{array}}\right) +\left( {\begin{array}{c}m\\ 2\end{array}}\right) +3m}}{(-q^2;q)_k} \left[ \begin{array}{c}q^{-k},-q^{-k-1}\\ q^{m+1}\end{array}{\Big |q}\right] _{m},\end{aligned}$$

we can reformulate equality (5) as the q-binomial identity:

$$\begin{aligned}&\frac{(1+q^{m}a)(1-q^{m}c/a)}{(1-c)(1+q^{m+1})} \left[ \begin{array}{c}q,a^2,c^2/a^2\\ qc,\,q^2c\end{array}{\Big |q^2}\right] _{m}\\&\qquad =\sum _{k=0}^{2m} (-1)^{k}{2m\brack k} (q^{-k};q)_m(-q^{-k-1};q)_m \left[ \begin{array}{c}-a,c/a\\ -q,c\end{array}{\Big |q}\right] _{k+1} q^{\left( {\begin{array}{c}k+1\\ 2\end{array}}\right) }.\end{aligned}$$

The difference between this equality and its counterpart under the replacement \(a\rightarrow -a\) yields a further equation:

$$\begin{aligned} \begin{aligned}&\frac{2(a-c/a)q^m}{(1-c)(1+q^{m+1})} \left[ \begin{array}{c}q,a^2,c^2/a^2\\ qc,\,q^2c\end{array}{\Big |q^2}\right] _{m} =\sum _{k=0}^{2m} (-1)^{k}{2m\brack k} (q^{-k};q)_m(-q^{-k-1};q)_m\\&\qquad \qquad \times \bigg \{\left[ \begin{array}{c}-a,c/a\\ -q,c\end{array}{\Big |q}\right] _{k+1} -\left[ \begin{array}{c}a,-c/a\\ -q,c\end{array}{\Big |q}\right] _{k+1}\bigg \} q^{\left( {\begin{array}{c}k+1\\ 2\end{array}}\right) }.\end{aligned} \end{aligned}$$
(6)

Alternatively, by examining another particular case of (4)

$$\begin{aligned}{_3\phi _2}\left[ \begin{array}{ccc}q^{-m-1},&{}-aq^{1+m},&{}q^{1+m}c/a\\ &{}-q,&{}q^{1+m}c\end{array}{\Big |q;q}\right] =q^{\left( {\begin{array}{c}m+2\\ 2\end{array}}\right) } \left[ \begin{array}{c}a,-c/a\\ -q,q^{1+m}c\end{array}{\Big |q}\right] _{m+1}\end{aligned}$$

and carrying out the same procedure, we find the identity below

$$\begin{aligned}&\frac{1}{1-q^{2m+2}} \left[ \begin{array}{c}q,a^2,c^2/a^2\\ c,\,qc\end{array}{\Big |q^2}\right] _{m+1}\\&\qquad =\sum _{k=0}^{2m+1}(-1)^k{2m+1\brack k}(q^{-k};q)_m(-q^{-k-1};q)_{m+1} \left[ \begin{array}{c}-a,c/a\\ -q,c\end{array}{\Big |q}\right] _{k+1} q^{\left( {\begin{array}{c}k+1\\ 2\end{array}}\right) }.\end{aligned}$$

Combining this equation with its variant under the substitution \(a\rightarrow -a\), we get another equality:

$$\begin{aligned} \begin{aligned}0=&\sum _{k=0}^{2m+1}(-1)^k{2m+1\brack k}(q^{-k};q)_m(-q^{-k-1};q)_{m+1}\\&\times \bigg \{\left[ \begin{array}{c}-a,c/a\\ -q,c\end{array}{\Big |q}\right] _{k+1}-\left[ \begin{array}{c}a,-c/a\\ -q,c\end{array}{\Big |q}\right] _{k+1}\bigg \} q^{\left( {\begin{array}{c}k+1\\ 2\end{array}}\right) }.\end{aligned} \end{aligned}$$
(7)

By specifying concretely the two polynomials

$$\begin{aligned} \phi (x;m)\,\rightarrow \,(x;q)_m,\qquad \psi (y;m)\,\rightarrow \,(-yq^{-1};q)_m \end{aligned}$$

and the three sequences

$$\begin{aligned}&F(m)\rightarrow \frac{2(a-c/a)q^m}{(1-c)(1+q^{m+1})} \left[ \begin{array}{c}q,a^2,c^2/a^2\\ qc,\,q^2c\end{array}{\Big |q^2}\right] _{m},\\&G(m)\rightarrow 0,\\&\Omega (k)\,\rightarrow \bigg \{\left[ \begin{array}{c}-a,c/a\\ -q,c\end{array}{\Big |q}\right] _{k+1} -\left[ \begin{array}{c}a,-c/a\\ -q,c\end{array}{\Big |q}\right] _{k+1}\bigg \} q^{\left( {\begin{array}{c}k+1\\ 2\end{array}}\right) };\end{aligned}$$

we can see that (2) and (3) become (6) and (7), respectively. Then, the dual relation corresponding to (1) can be stated explicitly as

$$\begin{aligned}&\bigg \{\left[ \begin{array}{c}-a,c/a\\ -q,c\end{array}{\Big |q}\right] _{m+1} -\left[ \begin{array}{c}a,-c/a\\ -q,c\end{array}{\Big |q}\right] _{m+1}\bigg \} q^{\left( {\begin{array}{c}m+1\\ 2\end{array}}\right) }\\&\qquad =\sum _{k\ge 0}{m\brack 2k} \frac{q^{\left( {\begin{array}{c}m-2k\\ 2\end{array}}\right) }(1+q^{-k-1})}{(q^{-m};q)_k(-q^{-m-1};q)_{k+1}} \frac{2q^k(a-c/a)}{(1-c)(1+q^{k+1})} \left[ \begin{array}{c}q,a^2,c^2/a^2\\ qc,\,q^2c\end{array}{\Big |q^2}\right] _{k}.\end{aligned}$$

Writing the q-binomial sum in terms of \(_4\phi _3\)-series, we find the identity below.

Theorem 1

(Balanced series identity)

$$\begin{aligned}&{_4\phi _3}\left[ \begin{array}{rc}q^{-m},q^{1-m},&{}a^2,c^2/a^2\\ q^{-2m},&{}qc,q^{2}c\end{array}{\Big |q^2;q^2}\right] \\&\qquad =\frac{a-ac}{a^2-c} \bigg \{\left[ \begin{array}{c}-a,c/a\\ -1,\,c\end{array}{\Big |q}\right] _{m+1} -\left[ \begin{array}{c}a,-c/a\\ -1,\,c\end{array}{\Big |q}\right] _{m+1}\bigg \}. \end{aligned}$$

Denote by \(\Phi _m(a,c)\) the above \({_4\phi _3}\)-series. According to the linear relation

$$\begin{aligned} 1-q^{2k-2m}=q^{-m}\frac{1-q^{2m}c}{1-q^mc}(1-q^{2k-m}) -q^{-m}\frac{1-q^{m}}{1-q^mc}(1-q^{2k}c) \end{aligned}$$

we can split \(\Phi _m(a,c)\) into two \({_4\phi _3}\)-series

$$\begin{aligned} \Phi _m =&\sum _{k\ge 0}\frac{1-q^{2k-2m}}{1-q^{-2m}} \left[ \begin{array}{rc}q^{-m},q^{1-m},&{}a^2,c^2/a^2\\ q^2,q^{2-2m},&{}qc,q^{2}c\end{array}{\Big |q^2}\right] _kq^{2k}\\ =&\,\frac{(1-q^{2m}c)(1-q^{m})}{(1-q^mc)(1-q^{2m})} {_4\phi _3}\left[ \begin{array}{rc}q^{1-m},q^{2-m},&{}a^2,c^2/a^2\\ q^{2-2m},&{}qc,q^{2}c\end{array}{\Big |q^2;q^2}\right] \\&+\frac{q^{m}(1-q^{m})(1-c)}{(1-q^mc)(1-q^{2m})} {_4\phi _3}\left[ \begin{array}{rc}q^{-m},q^{1-m},&{}a^2,c^2/a^2\\ q^{2-2m},&{}c,qc\end{array}{\Big |q^2;q^2}\right] . \end{aligned}$$

Observe that the \({_4\phi _3}\)-series in the middle is just \(\Phi _{m-1}(a,c)\). Evaluating it by Theorem 1 and then simplifying the result, we obtain another balanced \(_4\phi _3\)-series identity.

Proposition 2

(Balanced series identity: \(m>0\))

$$\begin{aligned}{_4\phi _3}\left[ \begin{array}{rc}q^{-m},q^{1-m},&{}a^2,c^2/a^2\\ q^{2-2m},&{}c,\,qc\end{array}{\Big |q^2;q^2}\right] =\left[ \begin{array}{c}-a,c/a\\ -1,\,c\end{array}{\Big |q}\right] _{m} +\left[ \begin{array}{c}a,-c/a\\ -1,\,c\end{array}{\Big |q}\right] _{m}.\end{aligned}$$

By examining the linear combination

$$\begin{aligned} (1-q^{2m})(c-x)\times \text {``Theorem}~1\text {''} +(1-q^{2m}x)(1-c)\times \text {``Proposition}~2\text {''} \end{aligned}$$

and then making some simplifications, we find following balanced \(_5\phi _4\)-series identity with an extra free parameter x.

Theorem 3

(Balanced series identity: \(m>0\))

$$\begin{aligned}&{_5\phi _4}\left[ \begin{array}{ccccc}q^{-m},&{}q^{1-m},&{}q^2x,&{}a^2,&{}c^2/a^2\\ &{}q^{2-2m},&{}x,&{}qc,&{}q^{2}c\end{array}{\Big |q^2;q^2}\right] \\&\quad =\frac{(a^2-c)(1-q^mx)+(1-q^m)(ac-ax)}{(a^2-c)(1-x)} \left[ \begin{array}{c}-a,c/a\\ -1,\,qc\end{array}{\Big |q}\right] _{m}\\&\qquad +\frac{(a^2-c)(1-q^mx)-(1-q^m)(ac-ax)}{(a^2-c)(1-x)} \left[ \begin{array}{c}a,-c/a\\ -1,\,qc\end{array}{\Big |q}\right] _{m}.\end{aligned}$$

In this theorem, letting \(x\rightarrow 0\) and \(\infty \), we derive two further summation formulae.

Corollary 4

(2-Balanced series identities: \(m>0\))

$$\begin{aligned} {_4\phi _3}\left[ \begin{array}{rc}q^{-m},q^{1-m},&{}a^2,c^2/a^2\\ q^{2-2m},&{}qc,q^{2}c\end{array}{\Big |q^2;q^2}\right] =&\bigg \{1+\frac{ac(1-q^m)}{a^2-c}\bigg \} \left[ \begin{array}{c}-a,c/a\\ -1,\,qc\end{array}{\Big |q}\right] _{m}\\&+\bigg \{1-\frac{ac(1-q^m)}{a^2-c}\bigg \} \left[ \begin{array}{c}a,-c/a\\ -1,\,qc\end{array}{\Big |q}\right] _{m},\\ {_4\phi _3}\left[ \begin{array}{rc}q^{-m},q^{1-m},&{}a^2,c^2/a^2\\ q^{2-2m},&{}qc,q^{2}c\end{array}{\Big |q^2;q^4}\right] =&\bigg \{q^m+\frac{a(1-q^m)}{a^2-c}\bigg \} \left[ \begin{array}{c}-a,c/a\\ -1,\,qc\end{array}{\Big |q}\right] _{m}\\&+\bigg \{q^m-\frac{a(1-q^m)}{a^2-c}\bigg \} \left[ \begin{array}{c}a,-c/a\\ -1,\,qc\end{array}{\Big |q}\right] _{m}.\end{aligned}$$

In addition, the equalities in Theorem 1 and Proposition 2 can be reformulated as

$$\begin{aligned} \frac{a^2-c}{a-ac}{_4\phi _3}\left[ \begin{array}{rc}q^{-m},q^{1-m},&{}a^2,c^2/a^2\\ q^{-2m},&{}qc,q^{2}c\end{array}{\Big |q^2;q^2}\right] =&\left[ \begin{array}{c}-a,c/a\\ -1,\,c\end{array}{\Big |q}\right] _{m+1}\\&-\left[ \begin{array}{c}a,-c/a\\ -1,\,c\end{array}{\Big |q}\right] _{m+1},\\ {_4\phi _3}\left[ \begin{array}{rc}q^{-m},q^{-m-1},&{}a^2,c^2/a^2\\ q^{-2m},&{}c,\,qc\end{array}{\Big |q^2;q^2}\right] =&\left[ \begin{array}{c}-a,c/a\\ -1,\,c\end{array}{\Big |q}\right] _{m+1}\\&+\left[ \begin{array}{c}a,-c/a\\ -1,\,c\end{array}{\Big |q}\right] _{m+1}. \end{aligned}$$

Their sum results in the following strange summation formula.

Corollary 5

(Strange summation formula)

$$\begin{aligned} \frac{2a(1-c)}{(1+a)(a-c)} \left[ \begin{array}{cc}-a,&{}c/a\\ -1,&{}c\end{array}{\Big |q}\right] _{m+1} =&\sum _{k=0}^{\left\lceil {\frac{m}{2}}\right\rceil }q^{2k} \left[ \begin{array}{cccc}q^{-m},&{}q^{-m-1},&{}a^2,&{}c^2/a^2\\ q^2,&{}q^{-2m},&{}qc,&{}q^2c\end{array}{\Big |q^2}\right] _k\\&\times \bigg \{1- \frac{(1-q^{2k})(a^2-c-ac+q^{1+m}ac)}{(1+a)(a-c)(1-q^{1+m})}\bigg \}.\end{aligned}$$

3 Limiting series identities

Letting \(m\rightarrow \infty \) in Proposition 2, we recover the following interesting nonterminating \({_2\phi _2}\)-series identity due to Andrews and Askey [4].

Proposition 6

(Limiting series from Proposition 2)

$$\begin{aligned}{_2\phi _2}\left[ \begin{array}{c}a^2, c^2/a^2\\ c,\,qc\end{array}{\Big |q^2;q}\right] =\left[ \begin{array}{cc}-a,&{}c/a\\ -1,&{}c\end{array}{\Big |q}\right] _{\infty } +\left[ \begin{array}{cc}a,&{}-c/a\\ -1,&{}c\end{array}{\Big |q}\right] _{\infty }. \end{aligned}$$

Even the very particular case \(c=0\) of the last identity is not trivial

$$\begin{aligned}\sum _{k=0}^{\infty } \frac{(a^2;q^2)_k}{(q^2;q^2)_k}(-q)^{k^2} =\frac{(-a;q)_{\infty }}{2(-q;q)_{\infty }} +\frac{(a;q)_{\infty }}{2(-q;q)_{\infty }}.\end{aligned}$$

We offer a classical proof for Proposition 6. Recall the two transformations due to Jackson and Heine (cf. Gasper–Rahman [15, III-4 & III-2]):

$$\begin{aligned}{_2\phi _1}\left[ \begin{array}{cc}a,&{}b\\ &{}c\end{array}{\Big |q;z}\right] =&\frac{(az;q)_{\infty }}{(z;q)_{\infty }} {_2\phi _2}\left[ \begin{array}{c}a,\,c/b\\ az,\,c\end{array}{\Big |q;bz}\right] ,\\ {_2\phi _1}\left[ \begin{array}{cc}a,&{}b\\ &{}c\end{array}{\Big |q;z}\right] =&\left[ \begin{array}{c}az,c/a\\ c,\,z\end{array}{\Big |q}\right] _{\infty } {_2\phi _1}\left[ \begin{array}{cc}a,&{}abz/c\\ &{}az\end{array}{\Big |q;c/a}\right] .\end{aligned}$$

They can be utilized to reformulate the \({_2\phi _2}\)-series in Proposition 6 as follows:

$$\begin{aligned} {_2\phi _2}\left[ \begin{array}{c}a^2, c^2/a^2\\ c,\,qc\end{array}{\Big |q^2;q}\right]&=\frac{(qc/a^2;q^2)_{\infty }}{(qc;q^2)_{\infty }} {_2\phi _1}\left[ \begin{array}{cc}a^2,&{}a^2/c\\ &{}c\end{array}{\Big |q^2;qc/a^2}\right] \\&=\left[ \begin{array}{c}q,c^2/a^2\\ c,\,qc\end{array}{\Big |q^2}\right] _{\infty } {_2\phi _1}\left[ \begin{array}{cc}a^2/c,&{}qa^2/c\\ &{}q\end{array}{\Big |q^2;c^2/a^2}\right] . \end{aligned}$$

Evaluating the last \({_2\phi _1}\)-series by the q-binomial series

$$\begin{aligned}{_2\phi _1}\left[ \begin{array}{cc}a^2/c,&{}qa^2/c\\ &{}q\end{array}{\Big |q^2;c^2/a^2}\right] =&\frac{1}{2} \bigg \{{_1\phi _0}\left[ \begin{array}{c}a^2/c\\ -\end{array}{\Big |q;c/a}\right] +{_1\phi _0}\left[ \begin{array}{c}a^2/c\\ -\end{array}{\Big |q;-c/a}\right] \bigg \}\\ =&\frac{1}{2} \bigg \{\frac{(a;q)_{\infty }}{(c/a;q)_{\infty }} +\frac{(-a;q)_{\infty }}{(-c/a;q)_{\infty }}\bigg \}\end{aligned}$$

and making some simplifications, we confirm the identity in Proposition 6. \(\square \)

Two further variants of \({_2\phi _2}\)-series identities can be deduced similarly.

  • Limiting series from Theorem 1:

    $$\begin{aligned} {_2\phi _2}\left[ \begin{array}{c}a^2, c^2/a^2\\ qc,\,q^2c\end{array}{\Big |q^2;q^3}\right] =\frac{a-ac}{a^2-c} \bigg \{\left[ \begin{array}{c}-a,c/a\\ -1,\,c\end{array}{\Big |q}\right] _{\infty } -\left[ \begin{array}{c}a,-c/a\\ -1,\,c\end{array}{\Big |q}\right] _{\infty }\bigg \}.\end{aligned}$$
    (8)

  • Limiting series from Theorem 3:

    $$\begin{aligned} \begin{aligned}{_3\phi _3}\left[ \begin{array}{c}q^2x,a^2, c^2/a^2\\ x,\,qc,\,q^2c\end{array}{\Big |q^2;q}\right] =&\frac{a^2-c+ac-ax}{(a^2-c)(1-x)} \left[ \begin{array}{c}-a,c/a\\ -1,\,qc\end{array}{\Big |q}\right] _{\infty }\\&+\frac{a^2-c-ac+ax}{(a^2-c)(1-x)} \left[ \begin{array}{c}a,-c/a\\ -1,\,qc\end{array}{\Big |q}\right] _{\infty }.\end{aligned}\end{aligned}$$
    (9)

4 Polynomial argument

According to the fundamental theorem of algebra, two polynomials of degree not greater than n are identical if they agree at the \(n+1\) distinct points. This will enable us to prove some “dual formulae” derived in Sect. 2. What is remarkable is that as a result, we just exchange, in the \(_4\phi _3\)-series, the roles of a and \(q^{-m}\), where the latter is the “terminating parameter.”

We examine, for instance, the series in Theorem 1. Letting \(a=q^{-n}\) with \(n\le m\) and \(n\in \mathbb {N}_0\), we can express the resulting identity as

$$\begin{aligned} \sum _{k=0}^nq^{2k} \frac{(q^{-m};q)_{2k}(q^{-2n};q^2)_k(q^{2n}c^2;q^2)_k}{(q^2;q^2)_k(q^{-2m};q^2)_k(qc;q)_{2k}} =\frac{q^{n}(1-c)}{1-q^{2n}c} \left[ \begin{array}{c}-q^{-n},q^{n}c\\ -1,\,c\end{array}{\Big |q}\right] _{m+1}.\end{aligned}$$
(10)

According to the equalities

$$\begin{aligned} \frac{(q^{-m};q)_{2k}(-q^{-m};q)_{n}}{(q^{-2m};q^2)_k}&=(q^{k-m};q)_{k}(-q^{k-m};q)_{n-k},\\ \left[ \begin{array}{c}-q^{-n},q^{n}c\\ -1,\,c\end{array}{\Big |q}\right] _{m+1}&=\frac{(-1;q)_{1+m-n}(c;q)_{1+m+n}}{(-1;q)_{m+1}(-1;q)_{-n}(c;q)_{m+1}(c;q)_{n}}\\&=\frac{(-q^{-n};q)_{n}(q^{1+m}c;q)_{n}}{(-q^{1+m-n};q)_{n}(c;q)_{n}}\\&=\frac{(-q;q)_{n}(q^{1+m}c;q)_{n}}{(-q^{-m};q)_{n}(c;q)_{n}}q^{-mn-n};\end{aligned}$$

we have, by multiplying across (10) by \((-q^{-m};q)_{n}\) the following equality

$$\begin{aligned}&\sum _{k=0}^nq^{2k} \frac{(q^{k-m};q)_{k}(-q^{k-m};q)_{n-k}(q^{-2n};q^2)_k(q^{2n}c^2;q^2)_k}{(q^2;q^2)_k(qc;q)_{2k}}\\&\qquad =\frac{q^{-mn}(1-c)}{1-q^{2n}c} \left[ \begin{array}{c}-q,q^{1+m}c\\ c\end{array}{\Big |q}\right] _n.\end{aligned}$$

This is an equality of two polynomials of degree \(\le n\) in \(q^{-m}\), which is valid for infinitely many \(m\in \mathbb {N}_0\) with \(m\ge n\). Therefore, it is a polynomial identity in \(q^{-m}\). Making the substitution \(q^{-m}\rightarrow a\), we get the following identity.

Proposition 7

(“Dual formula” of Theorem 1)

$$\begin{aligned}{_4\phi _3}\left[ \begin{array}{cccc}q^{-2n},&{}a,&{}qa,&{}q^{2n}c^2\\ &{}qc,&{}q^{2}c,&{}a^2\end{array}{\Big |q^2;q^2}\right] =\frac{a^{n}(1-c)}{1-q^{2n}c} \left[ \begin{array}{c}-q,qc/a\\ -a,c\end{array}{\Big |q}\right] _n.\end{aligned}$$

Following the same reasoning, we can prove analogously two further identities.

  • “Dual formula” of Proposition 2:

    $$\begin{aligned}{_4\phi _3}\left[ \begin{array}{cccc}q^{-2n},&{}a,&{}qa,&{}q^{2n}c^2\\ &{}c,&{}qc,&{}q^2a^2\end{array}{\Big |q^2;q^2}\right] =a^{n}\left[ \begin{array}{c}-q,c/a\\ -qa,c\end{array}{\Big |q}\right] _n.\end{aligned}$$

  • “Dual formula” of Theorem 3:

    $$\begin{aligned} \begin{aligned}&{_5\phi _4}\left[ \begin{array}{rccc}q^{-2n},q^2y,&{}a,&{}qa,&{}q^{2n}c^2\\ y,&{}qc,&{}q^{2}c,&{}q^2a^2\end{array}{\Big |q^2;q^2}\right] \\&\quad =\bigg \{\frac{1-y/a}{1-y} +\frac{(c-y)q^{n}(1-1/a)}{(1-y)(1-q^{2n}c)}\bigg \} \frac{a^{n}(1-c)}{1-c/a} \left[ \begin{array}{c}-q,c/a\\ -qa,c\end{array}{\Big |q}\right] _n.\end{aligned}\end{aligned}$$
    (11)

For \(y=a^2\) and \(y=c\), this last identity gives rise to the two other precedent ones, found by Verma and Jain [18, Equations 5.3 & 5.4]. For different proofs, the reader can refer to Al-Salam–Verma [1], Andrews [2, 3], Chu [9, 10], Gessel–Stanton [16] and Li–Chu [17].