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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.

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 [57, 13, 14, 1921, 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:

Ω m , n (a,x): = 3 F 2 [ a , a + 1 3 , a 1 3 1 2 + m , 3 a + n | x 2 ] ,
(1)

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

F p 1 + p [ a 0 , a 1 , , a p b 1 , , b p | z ] = k = 0 ( a 0 ) k ( a 1 ) k ( a p ) k k ! ( b 1 ) k ( 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

[ A , B , , C α , β , , γ ] n = ( A ) n ( B ) n ( C ) n ( α ) n ( β ) n ( γ ) n .

Our work is motivated by Lambert’s binomial series (see Riordan [22, §4.5] and [1, 810, 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:

figurea

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 “xy 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 “xy 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
figure1

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, 1618]), 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.

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 \(m'<0\).

\(m>0\)

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

Ω m , n ( a , x ) = ( 27 x 2 ) m ( 1 2 ) m ( 1 3 a n ) 2 m ( 2 3 a ) 3 m i = 0 m ( 1 ) i ( m i ) ( 3 a m + n 1 ) i ( 3 a 2 m + n ) i × k = m [ a m , a m 1 3 , a m + 1 3 1 , 1 2 , 3 a 2 m + n + i ] k x 2 k .

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 \(n\in \mathbb{Z}\))

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

\(n<0\)

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 \(n<0\))

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

\(n>0\)

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

Ω m , n ( a , x ) = k = 0 ( a ) k ( a 1 3 ) k ( a + 1 3 ) k k ! ( 1 2 + m ) k ( 3 a + n ) k x 2 k × i = 0 n 3 a + n + k 1 n i ( 3 a 1 + 3 k ) i X n i = i = 0 n ( 3 a 1 ) i ( 3 a + i ) n i X n i × k = 0 [ a + i 3 1 3 , a + i 3 , a + i 3 + 1 3 1 , 1 2 + m , 3 a + i ] k x 2 k .

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:

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

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

δ Ω 0 , 0 ( a , x ) = k = 0 ( 2 k 1 ) [ a , a 1 3 , a + 1 3 1 , 1 2 , 3 a ] k x 2 k 2 = 1 x 2 Ω 1 , 0 ( a , x ) , δ 2 Ω 0 , 0 ( a , x ) = k = 0 ( 3 2 k ) [ a , a 1 3 , a + 1 3 1 , 1 2 , 3 a ] k x 2 k 4 = 3 x 4 Ω 2 , 0 ( a , x ) .

Proceeding by induction, we can show that

δ n Ω 0 , 0 ( a , x ) = ( 1 ) n 1 ( 2 n 3 ) ! ! k = 0 [ a , a 1 3 , a + 1 3 1 , 3 2 n , 3 a ] k ( 2 k 2 n + 1 ) x 2 k 2 n = ( 1 ) n ( 2 n 1 ) ! ! x 2 n Ω n , 0 ( a , x ) .

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 \(n=1\))

F 2 3 [ a , a + 1 3 , a 1 3 1 2 , 3 a + 1 | 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 \(n=2\))

F 2 3 [ a , a + 1 3 , a 1 3 1 2 , 3 a + 2 | 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 \(n=-2\))

F 2 3 [ a , a + 1 3 , a 1 3 1 2 , 3 a 2 | 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 \(n=-3\))

F 2 3 [ a , a + 1 3 , a 1 3 1 2 , 3 a 3 | 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 \(n=0\))

F 2 3 [ a , a + 1 3 , a 1 3 3 2 , 3 a | 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 \(n=1\))

F 2 3 [ a , a + 1 3 , a 1 3 3 2 , 3 a + 1 | 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 \(n=-3\))

F 2 3 [ a , a + 1 3 , a 1 3 3 2 , 3 a 3 | 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 \(n=0\))

F 2 3 [ a , a + 1 3 , a 1 3 1 2 , 3 a | 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 \(n=1\))

F 2 3 [ a , a + 1 3 , a 1 3 1 2 , 3 a + 1 | 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 \(n=2\))

F 2 3 [ a , a + 1 3 , a 1 3 1 2 , 3 a + 2 | 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 \(n=3\))

F 2 3 [ a , a + 1 3 , a 1 3 1 2 , 3 a + 3 | 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 \(n=-1\))

F 2 3 [ a , a + 1 3 , a 1 3 1 2 , 3 a 1 | 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 \(n=-2\))

F 2 3 [ a , a + 1 3 , a 1 3 1 2 , 3 a 2 | 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 \(n=-1\))

F 2 3 [ a , a + 1 3 , a 1 3 5 2 , 3 a 1 | 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 \(n=-2\))

F 2 3 [ a , a + 1 3 , a 1 3 5 2 , 3 a 2 | 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 \(n=-3\))

F 2 3 [ a , a + 1 3 , a 1 3 5 2 , 3 a 3 | 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 \(n=0\))

F 2 3 [ a , a + 1 3 , a 1 3 3 2 , 3 a | 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 \(n=1\))

F 2 3 [ a , a + 1 3 , a 1 3 3 2 , 3 a + 1 | 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 \(n=-1\))

F 2 3 [ a , a + 1 3 , a 1 3 3 2 , 3 a 1 | 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 \(n=2\))

F 2 3 [ a , a + 1 3 , a 1 3 3 2 , 3 a + 2 | 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 \(n=3\))

F 2 3 [ a , a + 1 3 , a 1 3 3 2 , 3 a + 3 | 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 \(n=1\))

F 2 3 [ a , a + 1 3 , a 1 3 5 2 , 3 a + 1 | x 2 ] =w( y + )+w( y ),

where

w(y)= ( 1 + y ) 3 a 1 10 ( 2 y ) 3 { 40 120 a y 30 y 2 + 132 a y 2 + 144 a 2 y 2 + 25 y 3 22 a y 3 20 y 180 a 2 y 3 72 a 3 y 3 5 y 4 a y 4 + 36 a 2 y 4 + 36 a 3 y 4 } .

Example 23

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

F 2 3 [ a , a + 1 3 , a 1 3 5 2 , 3 a + 2 | x 2 ] =w( y + )+w( y ),

where

w(y)= ( 1 + y ) 3 a 1 20 ( 3 a + 2 ) ( 2 y ) { 40 90 a y 180 a 2 y 20 y 2 24 a y 2 + 180 a 2 y 2 + 35 a y 3 + 60 a + 20 y + 216 a 3 y 2 33 a 2 y 3 180 a 3 y 3 108 a 4 y 3 } .

Example 24

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

F 2 3 [ a , a + 1 3 , a 1 3 5 2 , 3 a + 3 | 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 \(n=-3\))

F 2 3 [ a , a + 1 3 , a 1 3 7 2 , 3 a 3 | 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 \(n=-2\))

F 2 3 [ a , a + 1 3 , a 1 3 7 2 , 3 a 2 | x 2 ] =w( y + )+w( y ),

where

w ( y ) = 45 ( 1 + y ) 3 a 1 32 y 5 ( 2 3 a ) ( 3 a 17 2 ) 5 × { 72 48 a 624 a y + 144 a 2 y + 1190 y 2 1916 a y 2 + 936 a 2 y 2 + 612 y 144 a 3 y 2 1105 y 3 + 1342 a y 3 540 a 2 y 3 + 72 a 3 y 3 } .

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 \(\{x, y_{+}, y_{-} \}= \{\sqrt{ \frac{3^{5}}{7^{3}}}, \frac{3}{4}, -\frac{2}{9} \}\).

figurec

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

figured

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

figuree

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]:

F 2 3 [ a , a + 1 3 , a 1 3 1 2 , 3 a | 1 ] = 3 3 a 1 ( 1 + 4 1 3 a ) 2 , F 2 3 [ a , a + 1 3 , a 1 3 3 2 , 3 a 1 | 1 ] = 3 3 a 1 ( 6 a 5 ) { 1 2 4 2 3 a } , F 2 3 [ a , a + 1 3 , a 1 3 1 2 , 3 a + 1 | 1 ] = 27 a 6 a + 1 { 1 2 + 9 a + 1 2 6 a + 1 } .

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Acknowledgements

The authors express their sincere gratitude to the two anonymous referees for generous comments and valuable suggestions that have significantly contributed to improving the manuscript during the revision.

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Li, N.N., Chu, W. Another class of nonterminating \(_{3}F_{2}\)-series with a free argument. Adv Differ Equ 2021, 496 (2021). https://doi.org/10.1186/s13662-021-03648-7

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MSC

  • 33C20
  • 05A10

Keywords

  • Classical hypergeometric series
  • Linearization method
  • Bisection series
  • Nonterminating \(_{3}F_{2}\)-series
  • Lambert’s binomial series