Another class of nonterminating $_{3}F_{2}$-series with a free argument

Abstract

By means of the linearization method, we evaluate another class of nonterminating $_{3}F_{2}$-series with a free argument x and two perturbing integer parameters m and n.

1 Introduction and motivation

There has always been a strong interest in discovering novel summation formulae for (generalized) hypergeometric series due to their broad variety of applications in mathematics, physics, and computer science (see [5–7, 13, 14, 19–21, 23]). The purpose of this paper is to evaluate, in closed forms, the following class of nonterminating $_{3}F_{2}$-series with a free variable x (with $|x|<1$ for convergence) and two perturbing integer parameters m and n:

$${\mathrm{\Omega }}_{m,n}(a,x):{=}_3{F}_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{1}{2}+m,& 3a+n\end{array}\right|x^2],$$

(1)

where, according to Bailey [2, §2.1], the classical hypergeometric series reads as

$${}_{1+p}F_p\left[\begin{array}{cc}{a}_0,& a_1,\dots ,a_p\\ & b_1,\dots ,b_p\end{array}\right|z]=\sum _{k=0}^{\mathrm{\infty }}\frac{{(a_0)}_k{(a_1)}_k\cdots {(a_p)}_k}{k!{(b_1)}_k\cdots {(b_p)}_k}{z}^k.$$

Denote by $\mathbb{Z}$ and $\mathbb{N}$, respectively, sets of integers and natural numbers with $\mathbb{N}_{0}=\{0\}\cup \mathbb{N}$. For indeterminate y and $n\in \mathbb{Z}$, the rising and falling factorials are defined by the following quotients of Euler’s Γ-function:

$$\begin{aligned} (x)_{n}=\frac{\Gamma (x+n)}{\Gamma (x)} \quad\text{and}\quad \langle {x} \rangle _{n}=\frac{\Gamma (1+x)}{\Gamma (1+x-n)}, \end{aligned}$$

where the multiparameter notation for the former one will be abbreviated to

$${\left[\begin{array}{c}A,B,\dots ,C\\ \alpha ,\beta ,\dots ,\gamma \end{array}\right]}_n=\frac{{(A)}_n{(B)}_n\cdots {(C)}_n}{{(\alpha )}_n{(\beta )}_n\cdots {(\gamma )}_n}.$$

Our work is motivated by Lambert’s binomial series (see Riordan [22, §4.5] and [1, 8–10, 15, 20]) which is well known in classical analysis. Let u and v be the two variables related through the equation $u=v/(1+v)^{\beta }$. Then

$$\begin{aligned} &\phi _{\alpha }(u):=(1+v)^{\alpha }=\sum_{k=0}^{\infty } \frac{\alpha }{\alpha +k\beta } \binom{\alpha +k\beta }{k}u^{k}, \\ &\psi _{\alpha }(u):=\frac{(1+v)^{\alpha +1}}{1+v-\beta v} =\sum_{k=0}^{ \infty } \binom{\alpha +k\beta }{k}u^{k}. \end{aligned}$$

By the bisection of series, we have further four generating functions

$$\begin{aligned} &\sum_{k=0}^{\infty } \frac{\alpha }{\alpha +2\beta k} \binom{\alpha +2\beta k}{2k}u^{2k} = \frac{\phi _{\alpha }(u)+\phi _{\alpha }(-u)}{2}, \\ &\sum_{k=0}^{\infty } \binom{\alpha +2\beta k}{2k}u^{2k} = \frac{\psi _{\alpha }(u)+\psi _{\alpha }(-u)}{2}; \\ &\sum_{k=0}^{\infty } \frac{\alpha }{\alpha +\beta (2k+1)} \binom{\alpha +\beta (2k+1)}{2k+1}u^{2k+1} = \frac{\phi _{\alpha }(u)-\phi _{\alpha }(-u)}{2}, \\ &\sum_{k=0}^{\infty } \binom{\alpha +\beta (2k+1)}{2k+1}u^{2k+1} = \frac{\psi _{\alpha }(u)-\psi _{\alpha }(-u)}{2}. \end{aligned}$$

Specifying with $\beta =\frac{3}{2}$, making the replacements $u\to \frac{2x}{3\sqrt{3}}, v\to y$, and then letting

$$\begin{aligned} \alpha \to 3a-1, \alpha \to 3a-2, \alpha \to 3a-\frac{5}{2}, \alpha \to 3a- \frac{7}{2}, \end{aligned}$$

respectively, in the above four equations, we get four hypergeometric formulae:

Here and forth, x and y are two variables related via equations

$$\begin{aligned} \pm {x}=\frac{3\sqrt{3}y_{\pm }}{2\sqrt{(1+y_{\pm })^{3}}}, \end{aligned}$$

(2)

where $y_{\pm }$ are computed from x through the fundamental algebraic relationship

$$\begin{aligned} \frac{2x}{3\sqrt{3}}=\frac{y}{(1+y)^{3/2}} \quad\text{or equivalently}\quad \biggl( \frac{2x}{y} \biggr)^{2}= \biggl( \frac{3}{1+y} \biggr)^{3}. \end{aligned}$$

Recall that the hypergeometric $_{3}F_{2}(x^{2})$-series converge (generically) only if their argument is less than 1 in magnitude. Therefore x is restricted to $(-1,1)$. There are exactly two solutions $y_{+}$ and $y_{-}$ of the above equation in the region $(-1/4,2)$ whenever x satisfies $-1< x<1$. By equating both members of the last equation to $t^{6}$, we can parameterize the algebraic “x–y curve” by rational functions:

$$\begin{aligned} x=\frac{t}{2}\bigl(3-t^{2}\bigr)\quad \text{and}\quad y= \frac{3-t^{2}}{t^{2}}. \end{aligned}$$

The portions of the curve with $t\in (-2,-1)$ and $t\in (1,2)$ lie, in the “x–y plane”, in the abovementioned region. For any x, the corresponding $y_{\pm }$ are the y-coordinates of the points $(x,y)$ that lie on these two branches that are illustrated in the Fig. 1.

Figure 1

The “$x-y$” curve

The four identities of $_{3}F_{2}$-series highlighted in the last page are not isolated examples. As we shall show, there exists a large number of closed formulae for the series $\Omega _{m,n}$. By means of the linearization method (cf. [3, 4, 11, 12, 16–18]), we shall reduce in the next section, for $m,n\in \mathbb{Z}$, the series $\Omega _{m,n}$ to $\Omega _{m',0}$ with $m'<0$. Then this last series will be evaluated in Sect. 3 via differential operators. The conclusive theorem affirms that, for all the $m,n\in \mathbb{Z}$, the nonterminating $\Omega _{m,n}$-series can be always evaluated explicitly in terms of a finite number of algebraic functions in $y_{\pm }$. Finally, by making use of Mathematica commands, 26 closed formulae are presented as exemplification.

2 Linearization method

By means of the linearization method, we shall establish, in this section, three reduction formulae that express ultimately the series $\Omega _{m,n}$ with $m,n\in \mathbb{Z}$ in terms of the series $\Omega _{m',0}$, but with $m'<0$.

2.1 $m>0$

By employing the Chu–Vandermonde formula on binomial convolutions, it is routine to prove the following linear representation lemma.

Lemma 1

(Linear representation)

For a natural number m and a variable y, the following linear relation holds:

$$\begin{aligned} \langle {y} \rangle _{m}=\sum_{i=0}^{m}(-1)^{i} \binom{m}{i} \langle {A+y} \rangle _{m-i}(A)_{i}. \end{aligned}$$

Now specifying in this lemma the parameters

$$\begin{aligned} y=k \quad\text{and}\quad A=3a-m+n-1, \end{aligned}$$

we get the equality

$$\begin{aligned} \langle {k} \rangle _{m}=\sum_{i=0}^{m}(-1)^{i} \binom{m}{i} \langle {3a-m+n-1+k} \rangle _{m-i}(3a-m+n-1)_{i}. \end{aligned}$$

By inserting this relation in the $\Omega _{m,n}$-series, we have

$$\begin{aligned} \Omega _{m,n}(a,x)={}&\sum_{k=0}^{\infty } \frac{(a)_{k}(a-\frac{1}{3})_{k}(a+\frac{1}{3})_{k}}{k!(\frac{1}{2}+m)_{k}(3a+n)_{k}}x^{2k} \\ ={}&\sum_{k=m}^{\infty } \frac{(a)_{k-m}(a-\frac{1}{3})_{k-m}(a+\frac{1}{3})_{k-m}}{(k-m)!(\frac{1}{2}+m)_{k-m}(3a+n)_{k-m}}x^{2k-2m} \\ &{}\times \sum_{i=0}^{m}(-1)^{i} \binom{m}{i} \frac{ \langle {3a-m+n-1+k} \rangle _{m-i}(3a-m+n-1)_{i}}{ \langle {k} \rangle _{m}}. \end{aligned}$$

Observing that

$$\begin{aligned} &\frac{ \langle {3a-m+n-1+k} \rangle _{m-i}}{(3a+n)_{k-m}}= \frac{(1-3a-n)_{2m}}{(3a-2m+n)_{k+i}}, \\ &\frac{(a)_{k-m}(a-\frac{1}{3})_{k-m}(a+\frac{1}{3})_{k-m}}{(k-m)! \langle {k} \rangle _{m}} =(-27)^{m} \frac{(a-m)_{k}(a-m-\frac{1}{3})_{k}(a-m+\frac{1}{3})_{k}}{k!(2-3a)_{3m}}; \end{aligned}$$

we can reformulate the double sum

$$\begin{array}{rl}{\mathrm{\Omega }}_{m,n}(a,x)=& {(-\frac{27}{x^2})}^m\frac{{(\frac{1}{2})}_m{(1-3a-n)}_{2m}}{{(2-3a)}_{3m}}\sum _{i=0}^m{(-1)}^i\left(\genfrac{}{}{0ex}{}{m}{i}\right)\frac{{(3a-m+n-1)}_i}{{(3a-2m+n)}_i}\\ & \times \sum _{k=m}^{\mathrm{\infty }}{\left[\begin{array}{ccc}a-m,& a-m-\frac{1}{3},& a-m+\frac{1}{3}\\ 1,& \frac{1}{2},& 3a-2m+n+i\end{array}\right]}_k{x}^{2k}.\end{array}$$

Expressing the last sum with respect to k in terms of $\Omega _{0,m+n+i}(a-m,x)$, we derive the first reduction formula.

Proposition 2

($m\in \mathbb{N}_{0}$ and $n\in \mathbb{Z}$)

$$\begin{aligned} \Omega _{m,n}(a,x) ={}& \biggl(-\frac{27}{x^{2}} \biggr)^{m} \frac{(\frac{1}{2})_{m}(1-3a-n)_{2m}}{(2-3a)_{3m}} \sum_{i=0}^{m}(-1)^{i} \binom{m}{i} \frac{(3a-m+n-1)_{i}}{(3a-2m+n)_{i}} \\ &{}\times \Biggl\{ \Omega _{0,m+n+i}(a-m,x) -\sum _{k=0}^{m-1} \frac{(3a-3m-1)_{3k}}{(2k)!(3a-2m+n+i)_{k}} \biggl( \frac{4x^{2}}{27} \biggr)^{k} \Biggr\} . \end{aligned}$$

2.2 $n<0$

Analogously, we can also prove, without difficulty, another linear representation lemma.

Lemma 3

(Linear representation)

For a negative integer n and a variable y, the following linear relation holds:

$$\begin{aligned} (A+y)_{-n}=\sum_{i=0}^{-n} \binom{-n}{i} \langle {B+y} \rangle _{i}(A-B+i)_{-n-i}. \end{aligned}$$

Under the parameter specification

$$\begin{aligned} y=k,\qquad A=3a+n,\qquad B=m-\frac{1}{2}, \end{aligned}$$

the equality in Lemma 3 can be restated as

$$\begin{aligned} (3a+n+k)_{-n}=\sum_{i=0}^{-n} \binom{-n}{i} \biggl\langle {m+k- \frac{1}{2}} \biggr\rangle _{i} \biggl(3a+n-m+\frac{1}{2}+i \biggr)_{-n-i}. \end{aligned}$$

By putting this relation inside the $\Omega _{m,n}$-series, we can manipulate the double sum

$$\begin{aligned} \Omega _{m,n}(a,x)={}&\sum_{k=0}^{\infty } \frac{(a)_{k}(a-\frac{1}{3})_{k}(a+\frac{1}{3})_{k}}{k!(\frac{1}{2}+m)_{k}(3a+n)_{k-n}}x^{2k} \\ &{}\times \sum_{i=0}^{-n} \binom{-n}{i} \biggl\langle {m+k-\frac{1}{2}} \biggr\rangle _{i} \biggl(3a+n-m+\frac{1}{2}+i \biggr)_{-n-i} \\ ={}&\sum_{i=0}^{-n} \binom{-n}{i} \biggl(3a+n-m+\frac{1}{2}+i \biggr)_{-n-i} \\ &{}\times \sum_{k=0}^{\infty } \biggl\langle {m+k- \frac{1}{2}} \biggr\rangle _{i} \frac{(a)_{k}(a-\frac{1}{3})_{k}(a+\frac{1}{3})_{k}}{k!(\frac{1}{2}+m)_{k}(3a+n)_{k-n}}x^{2k} \\ ={}&\sum_{i=0}^{-n}\binom{-n}{i} \frac{(\frac{1}{2}+m-i)_{i}(3a)_{n}}{(3a-m+\frac{1}{2})_{n+i}} \\ &{}\times \sum_{k=0}^{\infty } \frac{(a)_{k}(a-\frac{1}{3})_{k}(a+\frac{1}{3})_{k}}{k!(\frac{1}{2}+m-i)_{k}(3a)_{k}}x^{2k}. \end{aligned}$$

Writing the last sum by $\Omega _{m-i,0}(a,x)$, we get the second reduction formula.

Proposition 4

($m,n\in \mathbb{Z}$ with $n<0$)

$$\begin{aligned} \Omega _{m,n}(a,x)= \sum_{i=0}^{-n}(-1)^{i} \binom{-n}{i} \frac{(\frac{1}{2}-m)_{i}(3a)_{n}}{(3a-m+\frac{1}{2})_{n+i}} \Omega _{m-i,0}(a,x). \end{aligned}$$

2.3 $n>0$

The next linear relation comes from a limiting case of a known one. Dividing by $A^{m}$ equation (3.1) in [17, Lemma 3.1] and then letting $A\to \infty$, we get the following linearization lemma.

Lemma 5

(Linear representation)

For a natural number n and a variable y, the following linear relation holds:

$$\begin{aligned} 1=\sum_{i=0}^{n} \langle {B+y} \rangle _{n-i}(3C+3y)_{i} \mathrm{X}_{n}^{i}, \end{aligned}$$

(3)

where the coefficients $\mathrm{X}_{n}^{i}$ are independent of the variable y and given explicitly by the two expressions

$$\begin{aligned} \mathrm{X}_{n}^{i} &=\sum_{j=0}^{i} \frac{(-1)^{n-i+j}}{i!} \binom{i}{j} \frac{3C-3B+3n-2i}{3(C-B+\frac{j}{3})_{n-i+1}} \\ &=\sum_{j=0}^{n-i} \frac{(-1)^{n-i+j}}{(n-i)!} \binom{n-i}{j} \frac{3C-3B+3n-2i}{(3C-3B+3j)_{i+1}}. \end{aligned}$$

Specifying in Lemma 5 the parameters

$$\begin{aligned} y=k,\qquad B=3a+n-1,\qquad C=a-\frac{1}{3}, \end{aligned}$$

the equality corresponding to (3) becomes

$$\begin{aligned} 1=\sum_{i=0}^{n} \langle {3a+n+k-1} \rangle _{n-i}(3a-1+3k)_{i} \mathcal{X}_{n}^{i} \end{aligned}$$

(4)

with the coefficients $\mathcal{X}_{n}^{i}$ being determined by

$$\begin{aligned} \begin{aligned}[c] \mathcal{X}_{n}^{i} &=\sum_{j=0}^{i}\frac{(-1)^{n-i+j}}{i!} \binom{i}{j} \frac{2-6a-2i}{3(\frac{2}{3}-2a-n+\frac{j}{3})_{n-i+1}} \\ &=\sum_{j=0}^{n-i} \frac{(-1)^{n-i+j}}{(n-i)!} \binom{n-i}{j} \frac{2-6a-2i}{(2-6a-3n+3j)_{i+1}}. \end{aligned} \end{aligned}$$

(5)

By inserting this relation (5) in the $\Omega _{m,n}$-series, we get the double sum

$$\begin{array}{rl}{\mathrm{\Omega }}_{m,n}(a,x)=& \sum _{k=0}^{\mathrm{\infty }}\frac{{(a)}_k{(a-\frac{1}{3})}_k{(a+\frac{1}{3})}_k}{k!{(\frac{1}{2}+m)}_k{(3a+n)}_k}{x}^{2k}\\ & \times \sum _{i=0}^n{〈3a+n+k-1〉}_{n-i}{(3a-1+3k)}_i{\mathcal{X}}_n^i\\ =& \sum _{i=0}^n{(3a-1)}_i{(3a+i)}_{n-i}{\mathcal{X}}_n^i\\ & \times \sum _{k=0}^{\mathrm{\infty }}{\left[\begin{array}{ccc}a+\frac{i}{3}-\frac{1}{3},& a+\frac{i}{3},& a+\frac{i}{3}+\frac{1}{3}\\ 1,& \frac{1}{2}+m,& 3a+i\end{array}\right]}_k{x}^{2k}.\end{array}$$

Expressing the last sum by $\Omega _{m,0}(a+\frac{i}{3},x)$, we have the third reduction formula.

Proposition 6

Let $n\in \mathbb{N}$ and the connection coefficients $\{\mathcal{X}_{n}^{i}\}$ be given by (5). Then the following formula holds:

$$\begin{aligned} \Omega _{m,n}(a,x) =\sum_{i=0}^{n} \mathcal{X}_{n}^{i} (3a-1)_{i}(3a+i)_{n-i} \Omega _{m,0} \biggl(a+\frac{i}{3},x \biggr). \end{aligned}$$

3 Conclusive theorem and examples

For a given integer pair $\{m,n\}$, we can express the $\Omega _{m,n}$-series, by making use of Propositions 2, 4, and 6, in terms of $\Omega _{m',0}$-series with $m'\le 0$. Therefore it remains to evaluate this last series. This will be done by utilizing differential operations. Suppose that $f(x)$ is a differentiable function. Define the operator δ by

$$\begin{aligned} \delta f(x)=\frac{d}{dx} \biggl\{ \frac{f(x)}{x} \biggr\} . \end{aligned}$$

Then it is not hard to check that

$$\begin{array}{rl}& \delta {\mathrm{\Omega }}_{0,0}(a,x)=\sum _{k=0}^{\mathrm{\infty }}(2k-1){\left[\begin{array}{ccc}a,& a-\frac{1}{3},& a+\frac{1}{3}\\ 1,& \frac{1}{2},& 3a\end{array}\right]}_k{x}^{2k-2}=\frac{-1}{x^2}{\mathrm{\Omega }}_{-1,0}(a,x),\\ & {\delta }^2{\mathrm{\Omega }}_{0,0}(a,x)=\sum _{k=0}^{\mathrm{\infty }}(3-2k){\left[\begin{array}{ccc}a,& a-\frac{1}{3},& a+\frac{1}{3}\\ 1,& -\frac{1}{2},& 3a\end{array}\right]}_k{x}^{2k-4}=\frac{3}{x^4}{\mathrm{\Omega }}_{-2,0}(a,x).\end{array}$$

Proceeding by induction, we can show that

$$\begin{array}{rl}{\delta }^n{\mathrm{\Omega }}_{0,0}(a,x)& ={(-1)}^{n-1}(2n-3)!!\sum _{k=0}^{\mathrm{\infty }}{\left[\begin{array}{ccc}a,& a-\frac{1}{3},& a+\frac{1}{3}\\ 1,& \frac{3}{2}-n,& 3a\end{array}\right]}_k(2k-2n+1)x^{2k-2n}\\ & =\frac{{(-1)}^n(2n-1)!!}{x^{2n}}{\mathrm{\Omega }}_{-n,0}(a,x).\end{array}$$

Recalling that

$$\begin{aligned} \Omega _{0,0}(a,x)=\frac{1}{2} \bigl\{ (1+y_{+})^{3a-1}+(1+y_{-})^{3a-1} \bigr\} \end{aligned}$$

and then relabeling n by −m, we get the following expression.

Proposition 7

For $m<0$ and the three variables $\{x,y_{\pm }\}$ related by (2), the following formula holds:

$$\begin{aligned} \Omega _{m,0}(a,x) =\frac{(-2/x^{2})^{m}}{2(\frac{1}{2})_{-m}} \delta ^{-m} \bigl\{ (1+y_{+})^{3a-1}+(1+y_{-})^{3a-1} \bigr\} . \end{aligned}$$

As an anonymous referee pointed out, instead of Proposition 4 the case $n<0$ can be alternatively treated by repeatedly applying the operator δ to the initial function $x^{6a-1}\Omega _{0,0}(a,x)$.

Summing up, for any given pair of integers m and n, the series $\Omega _{m,n}(a,x)$ can be evaluated by carrying out the following procedure:

• Step-A: If $m>0$, write $\Omega _{m,n}(a,x)$, by means of Proposition 2, in terms of $\Omega _{0,n'}(a-m,x)$; then go to Step-B.

• Step-B: For $m\le 0$ and $n\neq0$, apply Propositions 4 and 6 to express $\Omega _{m,n}(a,x)$ as $\Omega _{m',0}(a',x)$ with $m'\le m$; then go to Step-C.

• Step-C: Finally, for $m\le 0$ and $n=0$, evaluate $\Omega _{m,0}(a,x)$, according to Proposition 7, by differentiating $\Omega _{0,0}(a,x)$.

Therefore, we have shown the following general conclusion.

Theorem 8

For all the $m,n\in \mathbb{Z}$, the nonterminating $\Omega _{m,n}$-series are always evaluable explicitly in a finite number of terms of algebraic functions in $y_{\pm }$.

Based on Propositions 2, 4, 6, and 7, we have devised appropriately Mathematica commands that are employed to evaluate $\Omega _{m,n}$ in closed forms for any specific integer pair “$m,n$”. Apart from the four formulae anticipated in the Introduction, we highlight further 26 elegant formulae as exemplification.

Example 1

($m=0$ and $n=1$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{1}{2},& 3a+1\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y)=\frac{(1+y)^{3 a-1}}{2 (6 a+1)} \{1+6 a+y-3 a y \}. \end{aligned}$$

Example 2

($m=0$ and $n=2$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{1}{2},& 3a+2\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) &=\frac{(1+y)^{3 a-1}}{4 (2 a+1) (3 a+2) (6 a+1)} \begin{Bmatrix} 4+96 a^{2}+72 a^{3}+4 y+10 a y-42 a^{2} y \\ +38 a-72 a^{3} y-a y^{2}-3 a^{2} y^{2}+18 a^{3} y^{2} \end{Bmatrix}. \end{aligned}$$

Example 3

($m=0$ and $n=-2$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{1}{2},& 3a-2\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) &=\frac{(1+y)^{3 a-1} (8-12 a-7 y+6 a y)}{(3 a-2)(y-2)^{3}}. \end{aligned}$$

Example 4

($m=0$ and $n=-3$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{1}{2},& 3a-3\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) &=\frac{2(1+y)^{3 a-1}}{(a-1) (3 a-2) (2-y)^{5}} \begin{Bmatrix} 16-40 a+24 a^{2}-29 y+58 a y \\ -24 a^{2} y+15 y^{2}-19 ay^{2}+6 a^{2} y^{2} \end{Bmatrix}. \end{aligned}$$

Example 5

($m=1$ and $n=0$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{3}{2},& 3a\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) =\frac{(1+y)^{3 a-1}(4-12 a-5 y+6 a y)}{6 (1-2a)(6 a-5) y}. \end{aligned}$$

Example 6

($m=1$ and $n=1$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{3}{2},& 3a+1\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) &=\frac{-3(1+y)^{3 a-1}}{32y(3 a-\frac{5}{2})_{4} } \begin{Bmatrix} 8 a+24 a^{2}-144 a^{3}+5 y+4 a y-132 a^{2} y \\ +144 a^{3} y+5y^{2}-31 a y^{2}+60 a^{2} y^{2}-36 a^{3} y^{2} \end{Bmatrix}. \end{aligned}$$

Example 7

($m=1$ and $n=-3$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{3}{2},& 3a-3\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) &= \frac{2 (1+y)^{3 a-1} (7-6 a-5 y+3 a y)}{3 y (a-1) (3 a-2) (y-2)^{3}}. \end{aligned}$$

Example 8

($m=-1$ and $n=0$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & -\frac{1}{2},& 3a\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) =\frac{(1+y)^{3 a-1}}{2 (2-y)} \{2+y-6 a y \}. \end{aligned}$$

Example 9

($m=-1$ and $n=1$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & -\frac{1}{2},& 3a+1\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) =\frac{1}{2} (1+y)^{3 a-1} \{1+y-3 a y \}. \end{aligned}$$

Example 10

($m=-1$ and $n=2$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & -\frac{1}{2},& 3a+2\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) &=\frac{(1+y)^{3 a-1}}{4 (3 a+2)} \bigl\{ 4+6 a+4 y-6 a y-18 a^{2} y -3 a y^{2}+9 a^{2} y^{2} \bigr\} . \end{aligned}$$

Example 11

($m=-1$ and $n=3$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & -\frac{1}{2},& 3a+3\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) &=\frac{(1+y)^{3 a-1}}{4 (a+1) (6 a+7)} \begin{Bmatrix} 14-16 a y-66 a^{2} y-36 a^{3} y-14 ay^{2}+30 a^{2} y^{2} \\ +26 a+12 a^{2}+14 y+36 a^{3} y^{2}+a y^{3}-9 a^{3} y^{3} \end{Bmatrix}. \end{aligned}$$

Example 12

($m=-1$ and $n=-1$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & -\frac{1}{2},& 3a-1\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) =\frac{(1+y)^{3 a-1}}{(2-y)^{3}} \bigl\{ 4-2 y-12 a y-3 y^{2}+6 a y^{2} \bigr\} . \end{aligned}$$

Example 13

($m=-1$ and $n=-2$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & -\frac{1}{2},& 3a-2\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) &=\frac{(1+y)^{3 a-1}}{(3 a-2) (y-2)^{5}} \begin{Bmatrix} 32-24 a y+144 a^{2} y+168 a y^{2}-144 a^{2}y^{2} \\ -48 a-48 y+35 y^{3}-72 a y^{3}+36 a^{2} y^{3} \end{Bmatrix}. \end{aligned}$$

Example 14

($m=2$ and $n=-1$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{5}{2},& 3a-1\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) &=\frac{3(1+y)^{3 a-1}}{16y^{3}(3a-\frac{11}{2})_{4} } \begin{Bmatrix} 20-192 a y+72 a^{2} y+120 a y^{2} \\ -24 a+110 y-99 y^{2}-36 a^{2} y^{2} \end{Bmatrix}. \end{aligned}$$

Example 15

($m=2$ and $n=-2$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{5}{2},& 3a-2\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) = \frac{3(1+y)^{3 a-1} (2+11 y-6 a y)}{4 y^{3}(2-3a) (3 a-\frac{11}{2})_{2}}. \end{aligned}$$

Example 16

($m=2$ and $n=-3$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{5}{2},& 3a-3\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) &= \frac{6 (1+y)^{3 a-1} (1+5 y-3 a y)}{(3 a-3)_{2} (6 a-11) (y-2) y^{3}}. \end{aligned}$$

Example 17

($m=-2$ and $n=0$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & -\frac{3}{2},& 3a\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) &=\frac{(1+y)^{3 a-1}}{6(2-y)^{3}} \begin{Bmatrix} 24-12 y-72 a y-14 y^{2}+y^{3} \\ +72 a y^{2}+72 a^{2} y^{2}-36a^{2} y^{3} \end{Bmatrix}. \end{aligned}$$

Example 18

($m=-2$ and $n=1$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & -\frac{3}{2},& 3a+1\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) &=\frac{(1+y)^{3 a-1}}{2 (2-y)} \bigl\{ 2+y-6 a y-y^{2}+a y^{2}+6 a^{2} y^{2} \bigr\} . \end{aligned}$$

Example 19

($m=-2$ and $n=-1$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & -\frac{3}{2},& 3a-1\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) &=\frac{(1+y)^{3 a-1}}{(2-y)^{5}} \begin{Bmatrix} 16-24 y-48 a y+88 a y^{2}+48 a^{2} y^{2}+20 y^{3} \\ -16 ay^{3}-48 a^{2} y^{3}+y^{4}-8 a y^{4}+12 a^{2} y^{4} \end{Bmatrix}. \end{aligned}$$

Example 20

($m=-2$ and $n=2$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & -\frac{3}{2},& 3a+2\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) =\frac{(1+y)^{3 a-1}}{4 (3a+2)} \begin{Bmatrix} 4+6 a+4 y-6 a y-18 a^{2} y \\ -5 a y^{2}+9 a^{2} y^{2}+18 a^{3}y^{2} \end{Bmatrix}. \end{aligned}$$

Example 21

($m=-2$ and $n=3$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & -\frac{3}{2},& 3a+3\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) =\frac{(1+y)^{3 a-1}}{12 (a+1)} \begin{Bmatrix} 6+6 a+6 y-12 a y-18 a^{2} y-8 a y^{2} \\ +18 a^{2} y^{2}+18 a^{3} y^{2}+a y^{3}-9 a^{3} y^{3} \end{Bmatrix}. \end{aligned}$$

Example 22

($m=-3$ and $n=1$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & -\frac{5}{2},& 3a+1\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$w(y)=\frac{{(1+y)}^{3a-1}}{10{(2-y)}^3}\left\{\begin{array}{c}40-120ay-30y^2+132a{y}^2+144a^2{y}^2+25y^3-22a{y}^3\\ -20y-180a^2{y}^3-72a^3{y}^3-5y^4-a{y}^4+36a^2{y}^4+36a^3{y}^4\end{array}\right\}.$$

Example 23

($m=-3$ and $n=2$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & -\frac{5}{2},& 3a+2\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$w(y)=\frac{{(1+y)}^{3a-1}}{20(3a+2)(2-y)}\left\{\begin{array}{c}40-90ay-180a^{2}y-20y^2-24a{y}^2+180a^2{y}^2+35a{y}^3\\ +60a+20y+216a^3{y}^2-33a^2{y}^3-180a^3{y}^3-108a^4{y}^3\end{array}\right\}.$$

Example 24

($m=-3$ and $n=3$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & -\frac{5}{2},& 3a+3\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) &=\frac{(1+y)^{3 a-1}}{20(a+1)} \begin{Bmatrix} 10-20 a y-30 a^{2} y-14 a y^{2}+30 a^{2}y^{2}+36 a^{3} y^{2} \\ +10 a+10 y+3 a y^{3}+2 a^{2} y^{3}-27 a^{3} y^{3}-18a^{4} y^{3} \end{Bmatrix}. \end{aligned}$$

Example 25

($m=3$ and $n=-3$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{7}{2},& 3a-3\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{aligned} w(y) =\frac{-45 (1+y)^{3a-1}}{8y^{5}(3a-3)_{2}(3a-\frac{17}{2})_{4}} \begin{Bmatrix} 22+187 y-168 a y+36 a^{2} y+425 y^{2} \\ -12 a-575 ay^{2}+252 a^{2} y^{2}-36 a^{3} y^{2} \end{Bmatrix}. \end{aligned}$$

Example 26

($m=3$ and $n=-2$)

$${}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{7}{2},& 3a-2\end{array}\right|x^2]=w(y_{+})+w(y_{-}),$$

where

$$\begin{array}{rl}w(y)=& \frac{45{(1+y)}^{3a-1}}{32y^5(2-3a){(3a-\frac{17}{2})}_5}\\ & \times \left\{\begin{array}{c}72-48a-624ay+144a^{2}y+1190y^2-1916a{y}^2+936a^2{y}^2\\ +612y-144a^3{y}^2-1105y^3+1342a{y}^3-540a^2{y}^3+72a^3{y}^3\end{array}\right\}.\end{array}$$

These identities are valid for all the x and y tied by (2) under the conditions $|x|<1$ and $-1/4< y<2$. When x is assigned to particular values, they may produce strange evaluation formulae. We limit ourselves to recording three groups of such formulae.

• Series with $\{x, y_{+}, y_{-} \}= \{\sqrt{ \frac{3^{5}}{7^{3}}}, \frac{3}{4}, -\frac{2}{9} \}$.

• Series with $\{x, y_{+}, y_{-} \}= \{2\sqrt{ \frac{3^{5}}{13^{3}}}, \frac{4}{9}, -\frac{3}{16} \}$.

• Series with $\{x, y_{+}, y_{-} \}= \{5\sqrt{ \frac{3^{5}}{19^{3}}}, \frac{10}{9}, -\frac{6}{25} \}$.

To our knowledge, the formulae presented in this paper for $\Omega _{m,n}(a,x)$ (when x is a free variable) have not appeared previously. Exceptions are about $\Omega _{0,0}$, $\Omega _{1,-1}$, and $\Omega _{0,1}$. Their particular cases with $\{x, y_{+}, y_{-}\}=\{1, 2, -1/4\}$ have been recorded by Milgram in his compendium [21, Equations 25, 30, 31]:

$$\begin{array}{rl}& {}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{1}{2},& 3a\end{array}\right|1]=\frac{3^{3a-1}(1+4^{1-3a})}{2},\\ & {}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{3}{2},& 3a-1\end{array}\right|1]=\frac{3^{3a-1}}{(6a-5)}\{\frac{1}{2}-4^{2-3a}\},\\ & {}_{3}F_2\left[\begin{array}{ccc}a,& a+\frac{1}{3},& a-\frac{1}{3}\\ & \frac{1}{2},& 3a+1\end{array}\right|1]=\frac{{27}^a}{6a+1}\{\frac{1}{2}+\frac{9a+1}{2^{6a+1}}\}.\end{array}$$