Clausen’s Series 3 F 2 ( 1 ) with Integral Parameter Differences

: Ebisu and Iwassaki proved that there are three-term relations for 3 F 2 ( 1 ) with a group symmetry of order 72. In this paper, we apply some speciﬁc three-term relations for 3 F 2 ( 1 ) to partially answer the open problem raised by Miller and Paris in 2012. Given a known value 3 F 2 (( a , b , x ) , ( c , x + 1 ) ,1 ) , if f − x is an integer, then we construct an algorithm to obtain 3 F 2 (( a , b , f ) , ( c , f + n ) ,1 ) in an explicit closed form, where n is a positive integer and a , b , c and f are arbitrary complex numbers. We also extend our results to evaluate some speciﬁc forms of p + 1 F p ( 1 ) , for any positive integer p ≥ 2.


Introduction
The Clausen's hypergeometric function with unit argument [1] is defined to be the complex number 3  (a 1 ) n (a 2 ) n (a 3 ) n (b 1 ) n (b 2 ) n n! , (α) n = Γ(α + n) Γ(α) = α(α + 1) · · · (α + n − 1) denotes the Pochhammer symbol, Γ(s) is the Euler's Γ-function, and the complex numbers a i , b j are called the numerator and denominator parameters, respectively. The denominator parameters are not allowed to be zero or negative integers (b j ∈ Z ≤0 ). The classical hypergeometric function, which is often called the Gauss hypergeometric function, is defined by The most famous formula is the Gauss formula: Another interesting formula is the Ramanujan's formula ( [2] (Page 39)): 3 where n is a positive integer. When Ramanujan's collected papers were published in 1927, Equation (2) attracted great attention. There are numerous hypergeometric series identities in the mathematical literature (see [3,4]). The evaluation of the hypergeometric sum 3 F 2 (1) (the Clausenian hypergeometric function with unit argument) is of ongoing interest, since it appears ubiquitously in many physics and statistics problems [5][6][7].
There is a formula worth noting in ( [8] Equation 3.13-(37)): In 1971, Karlsson [9] has deduced the summation formula For the sake of simplicity, in this paper we assume that the parameters of the hypergeometric series 3 F 2 (1) are such that it converges and the summation formula for it makes sense. Miller and Srivastava [10] and Miller and Paris [11] have obtained the following generalized form. The value of Thus, one may ask for a similar formula for 3 F 2 (1) where at least one pair of numeratorial and denominatorial parameters differs by a negative integer. Miller and Paris [12] gave the following formula for positive integers n and p: (1 − c) k+n (1 − a) k+n (1 − b) k+n (6) Shpot and Srivastava [6] derived an elegant summation formula for Clausen's series 3 F 2 with the unit argument, with which they determined that: where B(x, y) = Γ(x)Γ(y) Γ(x+y) is the Euler beta function. Miller and Paris [12] indicated that the problem remains of deriving a summation formula for the series: where n is a positive integer and a, b, c and f are arbitrary complex numbers.
In this paper, we try to answer this question. In fact, we evaluate this Clausen's series 3 F 2 (1) under some additional conditions. That is, if we can find a known value 3 F 2 a, b, x c, x + 1 1 and f − x ∈ Z, then we can follow our algorithm to give the explicit value of 3 F 2 a, b, f c, f + n 1 , for any positive integer n. We state the Algorithm 1 in Section 4.
We will use this algorithm to obtain equivalent forms of Equations (6) and (7), and some other formulas as our examples.
Let a = (a 1 , a 2 , a 3 , a 4 , a 5 ) ∈ C 5 . Ebisu and Iwasaki [13] proved that there exist unique rational functions u(a), v(a) ∈ Q(a) such that where p = (p 1 , p 2 , p 3 , p 4 , p 5 ), q = (q 1 , q 2 , q 3 , q 4 , q 5 ) ∈ Z 5 are distinct shift vectors ([13] Theorem 1.1). They presented a systematic recipe to determine u(a) and v(a) in finite steps. The three-term relation (9) admits an S 2 (S 3 × S 3 )-symmetry of order 72. Karp and Prilepkina [14] also gave an alternative method for computing u(a), v(a) for given shifts. Some of our results could be viewed as particular cases of the results (9). However, our algorithm aims to answer the open problem raised by Miller and Paris. The method of proof in our work is also different from theirs.
Here, we present one more formula obtained by our algorithm as the last example in this introductory section. For any non-negative integer n, we know that: We will list some known 3 F 2 (1), which we could use as an initial value in our algorithm in Section 6. In the final section, we extend our results to evaluate some specific forms of p+1 F p (1), for p ≥ 2.

Preliminaries
For the sake of brevity, we sometimes denote 3 F 2 a, b, x c, x + 1 1 as F(x). We cite an explicit formula in ([8] Equation 3.13-(41)) which we need to use later.
Proof. We rewrite this hypergeometric series F(x) as: If c = 1 and c = x + 1, we regard the inner summand: as a rational function of n. Then its partial fraction decomposition is the following decomposition: if c = 1 and c = x + 1. We substitute this representation into the above formula of F(x), then we have: We know that Using Equation (11) and the Gauss formula Equation (1), we can evaluate the first two hypergeometric series in the right-hand side of the above equation. Therefore, we get Equation (12), if c = 1 and c = x + 1: .
On the other hand, if c = 1 and x = 0, we regard the inner summand (a + n)(b + n)x (n + 1) 2 (n + 1 + x) as a rational function of n. Then its partial fraction decomposition is the following decomposition We substitute this representation into the above formula of F(x), then we have We simplify the first three terms on the right-hand side of the above equation, then the resulting identity matches Equation (12) with c = 1.
We reverse Equation (12) and get another recurrence relation for our F(x).

Main Results
We will solve the recurrence relations in Lemmas 1 and 2 as explicit formulas with a specific known value F( f ).
Proof. Applying the recurrence relation in Lemma 2, a positive integer times, we get , and .
Using the mathematical induction on the integer , we know that the above formula is correct. Let x = f + m and we use the Pochhammer symbols to rewrite the function T. Then we have This result is what we want.
Similarly, if we apply the recurrence relation in Lemma 1 a positive integer times, then we get where Using the mathematical induction on the integer , we prove that the above formula is correct. Now we apply this formula to get the following theorem.

Proof.
We set x = f − m in Equation (14) and we use the Pochhammer symbols to rewrite the function T. Then we have We obtain the desired result.
For any non-negative integer n, we know that Thus, we apply the Gauss formula (i.e. Equation (1)) to get We state this result as the following lemma.

Lemma 3 (Chu-Vandermonde identity).
For any complex number x, y with (y − x) > 0, and any non-negative integer n, we have If we set y = x + 1, then Equation (15) will become n ∑ k=0 n k This formula is a special case in ( [15] Equation (14)).

Theorem 3.
Let m be a positive integer, a, b, c, f be complex numbers with f , c / ∈ Z ≤0 . Then, Proof. First, we know that Therefore, we have Applying Equation (16) with n = m − 1 and x = n + f , we have This completes the proof.

Algorithm
In this section, we will use Theorems 1-3 to give an algorithm. Under an additional condition that F( f 0 ) is known, we can use this Algorithm 1 to express 3 1 as an explicit value. Let m be a positive integer, n be a integer, and f = f 0 + n.

2:
Using Theorem 3 results in 3 F 2 a, b, f c, f + m 1 a linear combination of 3: for k ← 0 to m − 1.

6:
Applying Theorem 2 yields 3 F 2 a, b, f + k c, f + k + 1 1 as a finite sum and add a constant multiple of 3 F 2 a, b, f 0 c, f 0 + 1 1 .

7: else 8:
Using Theorem 1 results in 3 F 2 a, b, f + k c, f + k + 1 1 as a finite sum and add a constant multiple of 3 F 2 a, b, f 0 c, f 0 + 1 1 .

9: end if
10: We get 3 F 2 a, b, f 0 + n c, f 0 + n + m 1 as an explicit value.
Therefore, we have partially solved the open problem of Miller and Paris. Now we will use our algorithm to evaluate some Clausenian hypergeometric functions with unit argument. The first example is concerned with a kind of hypergeometric series, which is generated by elliptic integrals. Ramanujan stated without proof in his first letter to Hardy a particular case of R(r) with r ∈ N ( [16] (Equation (2))). Ramanujan [16] gave the explicit formulas for R(r) and R(r + 1 2 ) with r ∈ N, and also the values at r ∈ {1, 1 2 , 1 4 }.
Proof. Let us run an exact value by using our algorithm. Since − 11 4 − 1 4 = −3 is an integer, the known value used in this example is [16]: Third, we need to calculate for 0 ≤ k ≤ 4. Fourth, for 0 ≤ k ≤ 2, we have −3 ≤ n + k ≤ −1 < 0. We apply Theorem 2 to evaluate Equation (20). We obtain the following three values. 3 For the final step, if we combine all these together, we will have ( This is what we want.

More Examples
We will use our algorithm to obtain equivalent forms of Equations (6), (7) and (10) in this section.

Solution.
Since n is a positive integer, we use Equation (11) as our known value. First, we set f 0 = 1, f = n, m = p. Second, we use Theorem 3 to yield a, b, n + k c, n + k + 1 1 .
It should be noted that our formula is different from ( [12] (Equation (1.7))), but it is an equivalent form.
In 2018, Paris [18] gave two proofs to evaluate a special type of 3 F 2 (1): This type of hypergeometric series could be obtained by letting b → 1 in the above example (ref. [18] (Method 1)). Therefore, the known value we need in our algorithm in this case is the limit case of Equation (11) (see [8] (Equation 3.13-(42))): The following is the formula which is obtained by our algorithm.

Example 4. Let c, d be complex numbers with
.

(29)
Solution. We use Theorem 3 with f = 1, m = n + 1 to this hypergeometric series, For 0 ≤ k ≤ n, we use Theorem 1 with f = 1, m = k to the above identity We substitute Equation (28) into the above equation and simplify them.
We apply Equation (15) Hence, we have This formula is then obtained.
Example 5. Shpot and Srivastava [6] derived the elegant formula for Clausen's series 3 F 2 with unit argument: We still can use our algorithm to get any specific values of a, b, c, m, n, but the formula is very complicated. Therefore, we just state the process of how we apply our algorithm to get the exact value. First, the known initial value is from ([8] (Equation 3.13.(38))) with d = b and z = 1. That is, Second, the process we use is stated as below. For any non-negative integers p, q, Thus, we can write this hypergeometric series as a finite sum and add a constant multiple of the known initial value (Equation (31)).
For 0 ≤ s < 1 2 and 0 ≤ k ≤ 1, let be Ramanujan's generalized elliptic integral of the first kind of order s. The moment K n,s is defined by where n is a real number. Borwein et al. ( [19] (Theorem 2)) proved that, for 0 ≤ s < 1 2 , They also defined the generalized Catalan constant and gave the formula ( [19] (Theorem 6)) for s > 0 and 2n+1) 2 is the Catalan constant. The function R(r) (see Equation (18)) is related as follows: The odd moments of K s can be evaluated by ( [19] (Theorem 3)). If we assume that Equation (33) is evaluable, here we can use our algorithm to evaluate the even moments of K s : Borwein et al. [19] said that the results for s = 1/6 are especially interesting. Therefore, we derive the following formula, relating G 1/6 = 3 √ 3 4 log(2) as our final example.

Example 6.
For a non-negative integer n, we have:

Solution.
The initial value that we use is: First, we let a = 1 3 , b = 2 3 , c = 1, f 0 = 1 2 , and f = 1 2 − n. Second, we use Theorem 3 to yield We set n − k as the new index k in the summation and then we have Third, we apply Theorem 2 to yield Combing them all together, we get our conclusion.

Some Known Initial Values
An important part of our algorithm is to find the corresponding initial value F( f 0 ) first. Therefore, we enclose a list of some known initial values in this section. , The parameters A and B in Equations (54) and (55) are defined by: and the parameters C and D in Equations (56) and (57) are defined by: , and D = arccos Here, we list the references where all the known values mentioned above can be found: Equations (38)

Evaluations of Some Specific Forms of p+1 F p (1)
Let r be a positive integer. If all elements of the vector g = (g 1 , g 2 , . . . , g r ) are distinct, then we have the following partial fraction decomposition (ref. [24]): where This will provide us with the following lemma so that we can extend our results to evaluate some specific forms of p+1 F p (1).