FORTY “STRANGE” COMPUTER-DISCOVERED [and COMPUTER-PROVED (of course!)] HYPERGEOMETRIC SERIES EVALUATIONS

Shalosh B. EKHAD^{1}

Abstract: We state and prove forty hypergeometric (_{2}F_{1}) “strange” series evaluations that were found by a systematic search for ‘nice’ identities inspired by the simplified and annotated Zeilberger algorithm, implemented in the Maple package twoFone.

Foreword (by Human Doron Zeilberger)

There are infinitely many mathematical facts, some shallow but interesting (like 2 + 2 = 4) and some deep (like the fact that the product of two random million-digit integers is some specific twomillion-digit integer) but utterly boring (to humans, at least). A (human) theorem is an interesting (i.e. simple) fact that is not too trivial. Granted, a computer can discover, and even prove, lots of facts, but can it discover theorems?

Until now, WZ theory was able to prove ‘nice’ identities, but these had to be first conjectured (i.e. discovered) by humans. WZ theory could also discover lots of new ‘theorems’, usually much deeper than the human-discovered ones, by systematically running the Zeilberger algorithm on many different summands, finding, and simultaneously proving, recurrences satisfied by the definite sums. But the recurrences outputted are (usually) horrendous to human eyes. Humans like their facts to be simple. In [Z] I showed how to use the simplified Zeilberger algorithm [MZ] in order to look out for hypergeometric terms that yield recurrences of lower-than-expected order, in particular, of order one, which imply closed-form evaluations.

The novelty of the present compilation is not so much in its substance. It is very possible that all (or at least most) of these forty identities are either already known, or are derivable, via classical 2F_{1} transformations, and more modern quadratic ones, from known evaluations. In particular Theorems 1 and 2 have recently been discovered and proved, by a completely different (and very interesting, analytical) approach by Robert Maier[M]. Others were conjectured by Bill Gosper and first proved by Ira Gessel and Dennis Stanton[GS].

Indeed the great novelty of this article is in its form. A computer discovering ab initio, humanlynice results, worthy of the name theorems, all by itself, without any human intervention (except the initial one of programming the ansatz and the meta-algorithm). Let it serve as a paradigm and iconic example for future computer-generated interesting mathematics, as opposed to mere computer-assisted one.

1 c/o D. Zeilberger, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus,

110 Frelinghuysen Rd., Piscataway, NJ 08854-8019, USA.

E-mail: zeilberg at math dot rutgers dot edu (Subject: For Shalosh), url: http://www.math.rutgers.edu/~zeilberg/ekhad/ekhad.html . First version: Oct. 12, 2004 . Supported in part by the NSF.

Introduction

The classical hypergeometric series

F (a , b , c , x) = ∑_{k=0}^{∞} \frac{(a)_{k}(b)_{k}}{k!(c)_{k}} x^{k} ,

(where (z)_{k} := z(z + 1)(z + 2) ⋯ (z + k − 1)), that nowadays is more commonly denoted by

2F_{1} ({a, b \atop c} ; x) ,

enjoys several closed-form evaluations. The most celebrated one is the three-parameter Gauss evaluation at x = 1 (see, e.g. [AAR])

F (a , b , c , 1) = \frac{Γ(c)Γ(c − a − b)}{Γ(c − a)Γ(c − b)} .  (Gauss)

Next comes Kummer’s two-parameter exact evaluation at x = −1

F (a , b , 1 + a − b , −1) = \frac{Γ(1 + a − b)Γ(1 + a/2)}{Γ(1 + a)Γ(1 + a/2 − b)} .  (Kummer)

and two other ones at x = 1/2, also due to Gauss. In addition, Gosper conjectured, and Gessel and Stanton[GS] proved, several ‘strange’ one-parameter evaluations at other values of x. Here I will state and prove all such inequivalent strange evaluations that are of the form

F (−an , bn + b_{1} , cn + c_{1}, x)

in the range

1 ≤ a ≤ 2 , −4 ≤ b ≤ 4 , −4 ≤ c ≤ 4 ,

with a, b, c all integers. None of them is a specialization of the classical identities, and none of them can be derived from other ones in the list via the Euler and Pfaff transformations ([AAR] Theorem 2.2.5) and their iterations (up to depth 3).

They were found by running the Maple package twoFone written by my beloved mentor, Doron Zeilberger, that accompanies his article [Z]. I ran out of time and memory for such identities when b and/or c were allowed to be 5. It would be interesting to know whether there exist infinitely many such strange evaluations. Another curious surprise was that the numbers b_{1}, c_{1}, that are, in general, along with x, solutions of a system of polynomial equations, and hence are only guaranteed to be algebraic, turn out, at least in all the examples discovered so far, to be rational numbers. On the other hand, x is sometimes irrational, but so far, it was, at worst, a quadratic irrationality.

The Forty Identities

The proofs below are given in the famous, succinct, WZ (Wilf-Zeilberger) format([WZ], see also [PWZ]), that only states a rational function, the so-called certificate, of (n, k). Recall that in order to translate it into a traditional proof, one has to divide the summand by the right side, and call it F (n, k). Calling the supplied certificate R(n, k), one defines G(n, k) := R(n, k)F (n, k) and verifies the purely-routine high-school algebra identity

F (n + 1, k) − F (n, k) = G(n, k + 1) − G(n, k) .  (W Z)

(W Z) is indeed routine, in every given case, since dividing throughout by F (n, k) and simplifying, yields an identity amongst rational functions. Finally summing (W Z) over k proves that ∑_{k} F (n, k) is identically constant, as a function of n, and the proof is completed by checking that it equals 1 when n = 0.

Theorem 1 :

F (−n , −4n − 1/2 , −3n , −1) = (\frac{2^{6}}{3^{3}} )^{n} · \frac{(3/8)_{n}(5/8)_{n}}{(1/3)_{n}(2/3)_{n}} .

Proof :

\frac{−2(296n^{3} − 216n^{2}k + 771n^{2} + 651n − 372nk + 52nk^{2} − 155k + 177 − 4k^{3} + 45k^{2})(3n − k + 1)k}{(8n + 9 − 2k)(8n + 7 − 2k)(8n + 5 − 2k)(8n + 3 − 2k)(n − k + 1)}

Theorem 2 :

F (−n , −4n − 5/2 , −3n − 1 , −1) = (\frac{2^{6}}{3^{3}} )^{n} · \frac{(7/8)_{n}(9/8)_{n}}{(2/3)_{n}(4/3)_{n}} .

Proof :

(296n^{3} + 1215n^{2} − 216n^{2}k + 1660n + 52nk^{2} − 586nk + 756 − 394k − 4k^{3} + 71k^{2})·

\frac{−2(3n − k + 2)k}{(8n + 13 − 2k)(8n + 11 − 2k)(8n + 9 − 2k)(8n + 7 − 2k)(n − k + 1)}

Theorem 3 :

F (−n , −3n − 1/2 , −4n , 4) = (\frac{−3^{3}}{2^{4}} )^{n} · \frac{(1/3)_{n}(5/6)_{n}}{(1/4)_{n}(3/4)_{n}} .

Proof :

\frac{(344n^{3} − 216n^{2}k + 768n^{2} + 522n − 282nk + 34nk^{2} − 75k + 104 + 13k^{2})(4n − k + 1)k}{4(6n + 7 − 2k)(6n + 5 − 2k)(6n + 3 − 2k)(n − k + 1)(3n + 1)} .

Theorem 4 :

F (−n , −3n − 3/4 , −4n − 1 , −8) = (\frac{3^{3}}{2^{2}} )^{n} · \frac{(2/3)_{n}(7/12)_{n}}{(1/2)_{n}(3/4)_{n}} .

Proof :

\frac{−(2208n^{3} − 1488n^{2}k + 6272n^{2} + 5762n − 2620nk + 252nk^{2} − 1082k + 1707 + 167k^{2})(4n − k + 2)k}{12(12n + 15 − 4k)(12n + 11 − 4k)(12n + 7 − 4k)(n − k + 1/(3n + 2)}

Theorem 5 :

F (−n , −3n − 3/2 , −4n − 2 , −8) = (\frac{3^{3}}{2^{2}} )^{n} · \frac{(5/6)_{n}(7/6)_{n}}{(3/4)_{n}(5/4)_{n}} .

Proof :

−(1/24)\frac{(184n^{2} + 492n − 124nk + 326 + 21k^{2} − 167k)(4n − k + 3)k}{(6n + 9 − 2k)(6n + 7 − 2k)(6n + 5 − 2k)(n − k + 1)}

Theorem 6 :

F (−n , −3n − 1/4 , −4n , −8) = (\frac{3^{3}}{2^{2}} )^{n} · \frac{(1/3)_{n}(5/12)_{n}}{(1/2)_{n}(1/4)_{n}} .

Proof :

\frac{(−1/4)(736n^{3} − 496n^{2}k + 1600n^{2} + 1058n + 84nk^{2} − 624nk − 149k + 200 + 27k^{2})(4n − k + 1)k}{(12n + 13 − 4k)(12n + 9 − 4k)(12n + 5 − 4k)(n − k + 1)(3n + 1)}

Theorem 7 :

F (−n , −3n − 1/2 , −4n , −8) = (\frac{3^{3}}{2^{2}} )^{n} · \frac{(1/3)_{n}(2/3)_{n}}{(1/4)_{n}(3/4)_{n}} .

Proof : (−1/32)(2208n^{4}−1488n^{3}k+6640n^{3}+252n^{2}k^{2}−2988n^{2}k+7104n^{2}+248nk^{2}−1812nk+3156n+53k^{2}−321k+478)·

\frac{(4n − k + 1)k}{(6n + 7 − 2k)(6n + 5 − 2k)(6n + 3 − 2k)(n − k + 1)(3n + 2)(3n + 1)}

Theorem 8 :

F (−n , −3n − 5/4 , −4n − 1 , −8) = (\frac{3^{3}}{2^{2}} )^{n} · \frac{(2/3)_{n}(13/12)_{n}}{(1/2)_{n}(5/4)_{n}} .

Proof :

\frac{(−1/4)(736n^{3} − 496n^{2}k + 2336n^{2} + 2382n − 956nk + 84nk^{2} − 414k + 767 + 55k^{2})(4n − k + 2)k}{(12n + 17 − 4k)(12n + 13 − 4k)(12n + 9 − 4k)(n − k + 1)(3n + 2)}

Theorem 9 :

F (−n , −3n − 3/2 , −4n − 2 , 4) = (\frac{−3^{3}}{2^{4}} )^{n} · \frac{(4/3)_{n}(5/6)_{n}}{(3/4)_{n}(5/4)_{n}} .

Proof :

\frac{(1/12)(1032n^{3} − 648n^{2}k + 4024n^{2} + 5202n − 1690nk + 102nk^{2} − 1103k + 2228 + 135k^{2})(4n − k + 3)k}{(6n + 9 − 2k)(6n + 7 − 2k)(6n + 5 − 2k)(n − k + 1)(3n + 4)}

Theorem 10 :

F (−n , −3n − 7/4 , −4n − 2 , −8) = (\frac{3^{3}}{2^{2}} )^{n} \frac{(4/3)_{n}(11/12)_{n}}{(3/2)_{n}(3/4)_{n}} .

Proof :

\frac{(−1/12)(2208n^{3} + 9216n^{2} − 1488n^{2}k + 12806n + 252nk^{2} − 4112nk + 5924 − 2839k + 335k^{2})(4n − k + 3)k}{(12n + 19 − 4k)(12n + 15 − 4k)(12n + 11 − 4k)(n − k + 1)(3n + 4)}

Theorem 11 : Let ω be a primitive cubic root of unity (i.e. ω = (1 ± \sqrt{3}i)/2).

F (−n , −3n − 1 , −2n , ω) = (−(3/2)ω + 3/4)^{n} · \frac{(2/3)_{n}}{(1/2)_{n}} .

Proof :

(129n^{2} + 289n − 2ωn + 160 − 2ω − 88kn + 14kωn − 101k + 19kω + 13k^{2} − 5ωk^{2})·

\frac{(−1/129)(−5 + 13ω)(2n − k + 1)k}{(3n + 4 − k)(3n + 3 − k)(3n + 2 − k)(n − k + 1)}

Theorem 12 :

F (−n , −3n − 1/2 , −n + 1/2 , 4) = (−3^{3})^{n}.

Proof :

\frac{(1/9)k(2n + 1 − 2k)(84n^{2} + 172n − 60nk + 12k^{2} − 64k + 87)}{(6n + 7 − 2k)(6n + 5 − 2k)(6n + 3 − 2k)(n + 1 − k)}

Theorem 13 :

F (−n , −3n − 3/2 , −n + 1/2 , 4) = (−3^{3})^{n} · \frac{(5/6)_{n}(7/6)_{n}}{(1/2)_{n}(3/2)_{n}} .

Proof :

\frac{(1/3)(28n^{2} + 76n − 20nk + 51 − 28k + 4k^{2})(2n + 1 − 2k)k}{(6n + 9 − 2k)(6n + 7 − 2k)(6n + 5 − 2k)(n + 1 − k)}

Theorem 14 : Let α = 9 ± 4\sqrt{5}.

F (−n , −3n − 1/2 , 2n + 3/2 , α) = (\frac{2^{7}}{5^{3}} (1 + α))^{n} · \frac{(3/4)_{n}(5/4)_{n}}{(4/5)_{n}(6/5)_{n}} .

Proof : (79794 + 388827n + 706989n^{2} + 566136n^{3} + 167796n^{4} − 4518k − 4875nk − 51972k^{2} + 11429αnk+ 3754αk+2132αk^{2}+496k^{4}α+26472k^{3}−3696k^{4}−6816kn^{3}−6174kn^{2}−127434k^{2}n−72852k^{2}n^{2}+30768k^{3}n −3336n^{3}α−9466αn^{2}−8839αn−2718α+10998kαn^{2}+1060k^{2}αn^{2}−2192k^{3}αn+3336kn^{3}α+3058k^{2}αn−2584k^{3}α)·

\frac{−(1/74576)(31α − 327)k}{(6n + 7 − 2k)(6n + 5 − 2k)(6n + 3 − 2k)(n − k + 1)(4n + 3 + 2k)(α − 1)}.

Theorem 15 : Let α = 9 ± 4\sqrt{5}.

F (−n , −3n − 3/2 , 2n + 5/2 , α) = (\frac{2^{7}}{5^{3}} (1 + α))^{n} · \frac{(5/4)_{n}(7/4)_{n}}{(7/5)_{n}(8/5)_{n}} .

Proof : (384115+1282296n+1575413n^{2}+845796n^{3}+167796n^{4}−86193k−153105nk−77610k^{2}+26963αnk+

13383αk + 730αk^{2} + 496k^{4}α + 34264k^{3} − 3696k^{4} − 6816kn^{3}−

74602kn^{2} − 154458k^{2}n − 72852k^{2}n^{2} + 30768k^{3}n − 3336n^{3}α − 12126αn^{2} − 14517αn−

5745α + 17114kαn^{2} + 1060k^{2}αn^{2} − 2192k^{3}αn + 3336kn^{3}α + 1842k^{2}αn − 2984k^{3}α)·

\frac{k(−1/74576)(31α − 327)}{(6n + 9 − 2k)(6n + 7 − 2k)(6n + 5 − 2k)(n − k + 1)(4n + 5 + 2k)(α − 1)}

Theorem 16 : Let α = 2 ± \sqrt{3}.

F (−n , −2n − 1/2 , −4n − 1 , 4α) = (3/4 − (3/2)α)^{n} · \frac{(2/3)_{n}}{(1/2)_{n}} .

Proof :

(488n^{2} + 978n − 20αn + 472 − 23α + 94k^{2} − 25k^{2}α − 420kn + 52knα − 422k + 72kα)·

\frac{(1/732)(−68 + 19α)(4n + 2 − k)k}{(4n + 5 − 2k)(4n + 3 − 2k)(n − k + 1)(3n + 2)}

Theorem 17 :

F (−n , −2n − 1/3 , −3n , 9) = (−2^{2})^{n} · \frac{(1/2)_{n}}{(1/3)_{n}} .

Proof :

\frac{(1/9)k(3n − k + 1)(45n^{2} − 21nk + 72n − 11k + 25)}{(6n + 7 − 3k)(6n + 4 − 3k)(n − k + 1)(2n + 1)} .

Theorem 18 :

F (−n , −2n − 2/3 , −3n − 1 , 9) = (−2^{2})^{n} · \frac{(5/6)_{n}}{(2/3)_{n}} .

Proof :

\frac{(1/6)(15n + 19 − 7k)(3n − k + 2)k}{(6n + 8 − 3k)(6n + 5 − 3k)(n − k + 1)}

Theorem 19 :

F (−n , −n − 1/2 , 4n + 9/2 , 1/5) = (\frac{2^{14}}{5^{6}} )^{n} · \frac{(9/8)_{n}(11/8)_{n}(13/8)_{n}(15/8)_{n}}{(6/5)_{n}(9/5)_{n}(13/10)_{n}(17/10)_{n}} .

Proof :

(3036n^{4} + 1256n^{3}k + 16960n^{3} + 35275n^{2} + 4780n^{2}k − 1184n^{2}k^{2} + 6180nk− 3440nk^{2} − 544nk^{3} + 32445n + 2720k + 11154 − 2480k^{2} − 64k^{4} − 800k^{3})·

\frac{(−5/16)k}{(2n + 3 − 2k)(n − k + 1)(8n + 13 + 2k)(8n + 11 + 2k)(8n + 9 + 2k)}

Theorem 20 :

F (−n , −n + 1/2 , 4n + 3/2 , 1/5) = (\frac{2^{14}}{5^{6}} )^{n} · \frac{(3/8)_{n}(7/8)_{n}(5/8)_{n}(9/8)_{n}}{(3/5)_{n}(2/5)_{n}(9/10)_{n}(11/10)_{n}} .

Proof :

(3036n^{4} + 1256n^{3}k + 7852n^{3} + 7409n^{2} + 2844n^{2}k − 1184n^{2}k^{2} + 2252nk − 1344nk^{2} + 2993n−

544nk^{3} − 336k^{2} + 592k + 430 − 352k^{3} − 64k^{4})·

\frac{(−5/16)k}{(2n + 1 − 2k)(n − k + 1)(8n + 7 + 2k)(8n + 5 + 2k)(8n + 3 + 2k)}

Theorem 21 : Let β = (3 ± \sqrt{5})/2.

F (−n , 2n + 1 , −3n , β) = (\frac{5^{2}}{3^{3}} (1 + β))^{n} · \frac{(2/5)_{n}(3/5)_{n}}{(1/3)_{n}(2/3)_{n}} .

Proof :

(1376n + 308 + 2026n^{2} + 958n^{3} − 60k − 23k^{3}β + 38k^{2}β − 91kβ + 139βn + 87βn^{2} −201kβn − 87kβn^{2} + 38k^{2}βn + 48k^{3} + 4k^{2} − 122kn − 110kn^{2} + 4k^{2}n + 52β)·

\frac{(1/4790)k(3n − k + 1)(−78 + 17β)}{(n − k + 1)(5n + 3)(5n + 2)(n + 1)(2n + 1)}

Theorem 22 : Let β = (3 ± \sqrt{5})/2.

F (−n , 2n + 2 , −3n − 1 , β) = (\frac{5^{2}}{3^{3}} (1 + β))^{n} · \frac{(4/5)_{n}(6/5)_{n}}{(2/3)_{n}(4/3)_{n}} .

Proof : (4179n + 3463n^{2} + 958n^{3} + 24k − 317kβn − 87kβn^{2} + 48k^{3} + 54k^{2} − 109kn − 110kn^{2} + 4k^{2}n + 96β+

1674 + 183βn + 87βn^{2} + 38k^{2}βn − 23k^{3}β + 34k^{2}β − 251kβ)·

\frac{(1/4790)k(3n − k + 2)(−78 + 17β)}{(n − k + 1)(5n + 6)(5n + 4)(2n + 3)(n + 1)}

Theorem 23 : Let ϕ be the Golden ratio (1 + \sqrt{5})/2.

F (−n , 2n + 1 , 4n + 2 , ϕ^{−1}) = (\frac{2^{6}}{5^{3}} (2 − ϕ^{−1}))^{n} · \frac{(3/4)_{n}(5/4)_{n}}{(4/5)_{n}(6/5)_{n}} .

Proof :

(2815n + 3661n^{2} + 1558n^{3} + 14k^{3}ϕ^{−1} − 148ϕ^{−1}n − 84ϕ^{−1}n^{2} + 579k + 183kϕ^{−1}n+

1567kn + 1013kn^{2} + 88kϕ^{−1} + 37k^{2}ϕ^{−1}n + 34k^{2}ϕ^{−1}+
270k^{2}n + 84kϕ^{−1}n^{2} + 712 + 206k^{2} + 39k^{3} − 64ϕ^{−1})·

\frac{(−1/779)k(14ϕ^{−1} − 25)}{(n − k + 1)(4n + 4 + k)(4n + 3 + k)(4n + 2 + k)(−1 + ϕ^{−1})}

Theorem 24 : Let ϕ be the Golden ratio (1 + \sqrt{5})/2.

F (−n , 2n + 2 , 4n + 4 , ϕ^{−1}) = (\frac{2^{6}}{5^{3}} (2 − ϕ^{−1}))^{n} · \frac{(5/4)_{n}(7/4)_{n}}{(7/5)_{n}(8/5)_{n}} .

Proof :

(7628n + 5998n^{2} + 1558n^{3} + 14k^{3}ϕ^{−1} − 178ϕ^{−1}n
−84ϕ^{−1}n^{2} + 3188 + 1762k + 309kϕ^{−1}n + 2697kn + 1013kn^{2} + 253kϕ^{−1}+

37k^{2}ϕ^{−1}n + 79k^{2}ϕ^{−1} + 270k^{2}n + 84kϕ^{−1}n^{2} + 387k^{2} + 39k^{3} − 94ϕ^{−1})·

\frac{(−1/779)k(14ϕ^{−1} − 25)}{(n − k + 1)(4n + 6 + k)(4n + 5 + k)(4n + 4 + k)(−1 + ϕ^{−1})}

Theorem 25 : Let γ = 3 ± 2\sqrt{2}.

F (−2n , −4n − 1/2 , −2n + 1/2 , γ) = (−16 + 96γ)^{n} \frac{(3/8)_{n}(5/8)_{n}}{(1/4)_{n}(3/4)_{n}} .

Proof :

(25434 + 137162n + 271600n^{2} + 233440n^{3} + 73472n^{4} + 5216k^{2}γn^{2} + 10144k^{2}γn − 21148kγn−

6208kγn^{3} − 20304kγn^{2} − 1264k^{3}γn − 142224kn^{2} − 58944kn^{3} − 111324kn + 14304k^{2}n^{2}+ 22536k^{2}n − 1776k^{3}n − 1372k^{3} + 8594k^{2} + 88k^{4} + 4834k^{2}γ − 7067kγ − 1204k^{3}γ − 28145k + 3837γ+

104k^{4}γ + 13852γn + 6208γn^{3} + 16240γn^{2})·

\frac{(−1/1148)(−89 + 13γ)(4n + 1 − 2k)k}{(8n + 9 − 2k)(8n + 7 − 2k)(8n + 5 − 2k)(8n + 3 − 2k)(2n − k + 1)(2n − k + 2)(γ − 1)}

Theorem 26 : Let δ = 5 ± 3\sqrt{3}.

F (−2n , −4n − 1/3 , 2/3 , 2δ) = (3^{3}(26δ + 5))^{n} · \frac{(7/12)_{n}}{(3/4)_{n}} .

Proof :

(2059512 + 12099448n + 766292δn + 25671232n^{2} + 23410464n^{3} + 7767936n^{4} + 933000δn^{2}− 1239770kδn − 404352kn^{3}δ + 506562k^{2}δn + 268020k^{2}δn^{2} − 62100k^{3}δn − 1257180kδn^{2} −2121648k − 5589216kn^{3} − 12542136kn^{2} − 9090980kn + 1684440k^{2}n^{2} + 2508684k^{2}n

−240624k^{3}n − 56934k^{3}δ + 232953k^{2}δ − 390660kδ + 370656n^{3}δ
+205020δ + 13446k^{4} − 178812k^{3} + 904542k^{2} + 4941k^{4}δ)·

\frac{(1/364122)k(−1 + 3k)(61δ − 776)}{(12n + 13 − 3k)(12n + 10 − 3k)(12n + 7 − 3k)(12n + 4 − 3k)(2n − k + 1)(2n − k + 2)2δ − 1)}

Theorem 27 : Let δ = 5 ± 3\sqrt{3}.

F (−2n , −4n − 2/3 , 4/3 , 2δ) = (3^{3}(26δ + 5))^{n} · \frac{(5/12)_{n}}{(5/4)_{n}} .

Proof :

(3710744 + 18504312n + 607916δn + 33746032n^{2} + 26726016n^{3} + 7767936n^{4} + 661656δn^{2}−

1315958kδn − 404352kn^{3}δ + 533802k^{2}δn + 268020k^{2}δn^{2} − 62100k^{3}δn − 1280676kδn^{2}− 3169096k − 5589216kn^{3} − 14213784kn^{2} − 11770892kn + 1684440k^{2}n^{2} + 2829540k^{2}n − 240624k^{3}n

−60696k^{3}δ + 260835k^{2}δ − 439228kδ + 235872n^{3}δ + 182668δ

+13446k^{4} − 200988k^{3} + 1159494k^{2} + 4941k^{4}δ)·

\frac{(1/364122)k(1 + 3k)(61δ − 776)}{(12n + 14 − 3k)(12n + 11 − 3k)(12n + 8 − 3k)(12n + 5 − 3k)(2n − k + 1)(2n − k + 2)(2δ − 1)}

Theorem 28 : Let α = 2 ± \sqrt{3}.

F (−2n , −3n − 1/4 , −4n , 4α) = ((39/2)α − 21/4)^{n} · \frac{(5/12)_{n}}{(1/4)_{n}} .

Proof :

(171978n + 238000n^{2} + 104736n^{3} − 12496kαn^{2} − 32440kαn + 7780k^{2}αn + 17488αn^{2}+

30900αn + 8407k^{2}α − 1216k^{3}α − 18581kα + 39182 + 13262α

−81808kn^{2} + 15490k^{2} − 1792k^{3} − 43403k − 122224kn + 21112k^{2}n)·

\frac{(−1/3273)k(4n − k + 1)(397α − 1484)}{(12n + 13 − 4k)(12n + 9 − 4k)(12n + 5 − 4k)(2n − k + 1)(2n − k + 2)}

Theorem 29 :

F (−2n , −3n + 1 , −2n + 2 , −2) = (3^{3})^{n} · (−2n + 1) .

Proof :

\frac{(−1/9)k(2n − k − 1)(39n^{2} + 55n − 21kn − 14k + 3k^{2} + 18)}{(3n + 2 − k)(3n + 1 − k)(3n − k)(2n + 2 − k)}

Theorem 30 :

F (−2n , −3n , −2n + 1/2 , 1/4) = (\frac{3^{3}}{2^{6}} )^{n} · \frac{(1/3)_{n}(2/3)_{n}}{(1/4)_{n}(3/4)_{n}} .

Proof :

\frac{(−2/3)(74n^{3} + 168n^{2} − 57kn^{2} − 87kn + 124n + 12k^{2}n − 32k + 30 − k^{3} + 9k^{2})(4n + 1 − 2k)k}{(3n + 3 − k)(3n + 2 − k)(3n + 1 − k)(2n + 1 − k)(2n + 2 − k)}

Theorem 31 : Let η = −7 ± 4\sqrt{3}.

F (−2n , −3n − 1/4 , −n + 3/4 , η) = (312η + 24)^{n} · \frac{(5/12)_{n}}{(1/4)_{n}} .

Proof :

(485694n + 611008n^{2} + 248736n^{3} − 5600k^{2}ηn + 11504kηn^{2} + 23720ηnk − 11880ηn− 6512ηn^{2} + 752k^{3}η + 22744k^{2} + 11851ηk − 5432ηk^{2} − 204544nk − 79751k − 2416k^{3}

+28576k^{2}n − 126928kn^{2} − 5299η + 124358)·

\frac{(1/124368)(47η + 809)(4n + 1 − 4k)k}{(12n + 13 − 4k)(12n + 9 − 4k)(12n + 5 − 4k)(2n − k + 1)(2n − k + 2)(η − 1)}

Theorem 32 :

F (−2n , −2n + 1/3 , −n + 5/6 , −1/8 ) = (\frac{3^{3}}{2^{4}} )^{n} · \frac{(1/3)_{n}}{(1/6)_{n}} .

Proof :

\frac{(−4/27)(132n^{2} − 60nk + 202n + 74 + 9k^{2} − 45k)(6n + 1 − 6k)k}{(6n + 5 − 3k)(6n + 2 − 3k)(2n − k + 1)(2n − k + 2)}

Theorem 33 : Let ζ = −1 ± \sqrt{2}.

F (−2n , 1/2 , −4n , 2ζ) = \frac{(3/8)_{n}(5/8)_{n}}{(1/4)_{n}(3/4)_{n}} .

Proof :

\frac{(−1/2)(3 + ζ)(6ζn − 2n + 4ζ − 3k^{2} − k^{2}ζ + 8kn + 6k)(4n − k + 1)k}{(2n − k + 2)(2n − k + 1)(8n + 5)(8n + 3)}

Theorem 34 : Let ϕ be the Golden ratio (1 + \sqrt{5})/2 then

F (−2n , 2n + 1 , −3n + 1/2 , ϕ) = (\frac{5^{2}}{3^{3}} (4ϕ + 3))^{n} · \frac{(3/10)_{n}(7/10)_{n}}{(1/6)_{n}(5/6)_{n}} .

Proof :

(1678n + 2708n^{2} + 1448n^{3} + 136k^{2}ϕn − 264kϕn^{2} − 364kϕn − 118kϕ + 324ϕn

+264ϕn^{2} − 26k^{3}ϕ + 96k^{2}ϕ + 343 + 96ϕ − 267k − 19k^{2} + 24k^{3} − 42k^{2}n − 536kn^{2} − 750kn)·

\frac{(2/905)k(6n + 1 − 2k)(−43 + 16ϕ)}{(2n + 1)(2n − k + 2)(2n − k + 1)(10n + 7)(10n + 3)}

Theorem 35 : Let ϕ be the Golden ratio (1 + \sqrt{5})/2.

F (−2n , 2n + 1 , −3n , ϕ) = (\frac{5^{2}}{3^{3}} (4ϕ + 3))^{n} · \frac{(2/5)_{n}(3/5)_{n}}{(1/3)_{n}(2/3)_{n}} .

Proof :

(1334n + 1716n^{2} + 724n^{3} − 270kϕn + 68k^{2}ϕn − 132kϕn^{2} − 125kϕ + 214ϕn + 132ϕn^{2} −13k^{3}ϕ + 68k^{2}ϕ + 82ϕ − 177k − 21k^{2} + 12k^{3} − 21k^{2}n − 268kn^{2} − 433kn + 342)·

\frac{(1/905)k(3n − k + 1)(−43 + 16ϕ)}{(2n − k + 2)(2n − k + 1)(5n + 3)(5n + 2)(n + 1)}

Theorem 36 : Let ζ_{1} = 1 ± \sqrt{2}.

F (−2n , 2n + 1 , −2n + 1/2 , (1/2)ζ_{1}) = (4 + 8ζ_{1})^{n} · \frac{(3/8)_{n}(5/8)_{n}}{(1/4)_{n}(3/4)_{n}} .

Proof :

(56n^{2} + 72n + 8ζ_{1}n + 24 + 5ζ_{1} − 6k^{2} − 16kn − 12k + 4k^{2}ζ_{1} − 8ζ_{1}kn − 6ζ_{1}k)·

\frac{(2/7)k(4n + 1 − 2k)(−4 + ζ_{1})}{(2n − k + 2)(2n − k + 1)(8n + 5)(8n + 3)}

Theorem 37 : Let α = 2 ± \sqrt{3}.

F (−2n , 2n + 1 , −n + 3/4 , (1/4)α) = (3 − 6α)^{n} · \frac{(5/12)_{n}}{(1/4)_{n}} .

Proof :

\frac{(1/3)(8n + 5 + α − 2kα + 2k)(−5 + α)(4n + 1 − 4k)k}{(2n − k + 2)(2n − k + 1)(12n + 5)}

Theorem 38 :

F (−2n , 3n + 1 , −2n + 2 , 2/5) = (\frac{5^{3}}{3^{3}} )^{n} · (−2n + 1) .

Proof :

\frac{(−1/25)(147n^{2} + 33nk + 179n + 3k^{2} + 23k + 50)(2n − k − 1)k}{n(2n + 2 − k)(3n + 2)(3n + 1)}

Theorem 39 : Let ζ_{2} = 1 ± \sqrt{3}/3.

F (−2n , 4n + 2 , 4/3 , (2/3)ζ_{2}) = (−9 + 18ζ_{2})^{n} · \frac{(5/12)_{n}}{(5/4)_{n}} .

Proof :

(352n^{2} + 432n + 168ζ_{2}n + 116 + 138ζ_{2} − 2k^{2} + 9ζ_{2}k^{2}−
288ζ_{2}nk + 152nk + 126k − 237ζ_{2}k)·

\frac{(−1/88)(−16 + 9ζ_{2})(1 + 3k)k}{(2n − k + 2)(2n − k + 1)(12n + 5)(4n + 3)}

Theorem 40 : Let ζ_{2} = 1 ± \sqrt{3}/3.

F (−2n , 4n + 1 , 2/3 , (2/3)ζ_{2}) = (−9 + 18ζ_{2})^{n} · \frac{(7/12)_{n}}{(3/4)_{n}} .

Proof :

(352n^{2} + 264n + 264ζ_{2}n + 20 + 174ζ_{2} − 2k^{2} + 9ζ_{2}k^{2} − 288ζ_{2}nk + 152nk + 102k − 195ζ_{2}k)·

\frac{(−1/88)(−16 + 9ζ_{2})(−1 + 3k)k}{(2n − k + 2)(2n − k + 1)(12n + 7)(4n + 1)}

Acknowledgement: Many thanks are due to my beloved mentor, Doron Zeilberger, for writing the Maple package twoFone, and to Mohamud Mohammed for stimulating conversations with him.

REFERENCES

[AAR] G.E. Andrews, R. Askey, and R. Roy, “Special Functions”, Cambridge Univ. Press, 1999.

[GS] I. Gessel and D. Stanton, Strange evaluations of hypergeometric series, SIAM J. Math. Anal. 13(1982), 295-308.

[PWZ] M. Petkovsek, H. S. Wilf and D. Zeilberger, “A=B”, AK Peters, Wellesley, (1996). [available on-line from the authors’ websites.]

[M] R.S. Maier, A Generalization of Euler’s hypergeometric transformation, preprint. (available

from arXiv.org).

[MZ] M. Mohammed and D. Zeilberger, Sharp upper bounds for the orders outputted by the Zeilberger and q-Zeilberger algorithms, to appear in J. Symb. Comp., available from the authors’ websites.

[WZ] H.S. Wilf and D. Zeilberger, Rational functions certify combinatorial identities, J. Amer. Math. Soc. 3 (1990), 77-83.

[Z] D. Zeilberger, DECONSTRUCTING the ZEILBERGER algorithm, submitted, [available on-line from the author’s website].