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After Dark: All Night
Warning: This fic is rated NC-17 for graphic sex between two willing parties, one male, one female. If that's not your cup of tea, turn back now.
Disclaimer: Its Paramounts playground. I only wish I owned these characters. But alas! They're not mine! But this storyline IS mine, so don't copy! Please note that this story is rated NC-17 for adult content! You have been warned!
Copyright: December 1999
Dedicated to: Angelina. Did I make you proud??
:*:*:*:*:*:
Part 1 in the 'After Dark' series
By Alicia
*From now on, I promise no more affairs with strange ships* Tom's words rang through her ears as she pulled the covers down on her bed. She sat down on the edge of the bed and pulled off her turtleneck, leaving only her gray tank top. She pulled off her uniform pants and dropped them in a heap by her bed. She was exhausted and wanted to sleep, no matter what the sleepwear.
The thing that was disturbing her the most was the fact that Tom had so easily succumbed to Alice. Why couldn't he have fought just a little bit harder? Especially when she had transported herself onto the shuttle. He had actually had to debate whom to trust. *But think about it, B'Elanna* she said to herself. *He was under Alice's spell . . . *
"Computer, lights," she snapped, laying her head back on the pillow. The room was suddenly dark and she sighed. Here she was, alone in the dark, whining and complaining about an alien force that had taken over her boyfriend. It sounded so bizarre. She glanced at the chronometer by her bed and laughed. It was only 9:45 and she was in bed. "You're getting old, Torres," she muttered to herself, sitting up. "Computer, lights," she ordered again.
The bright lights flashed on and she shielded her eyes. After the initial brightness had passed, she swung her legs over the side of the bed and pulled herself into a standing position. Her tired muscles protested, but she ignored them. First order of business: get more comfortable.
She half walked, half trotted, to her dresser where she grabbed a red nightie that Tom had given her. She slipped into, memories washing over her. This might be the first night she'd keep it on until the morning . . .
Smiling at the thought, she headed to the bathroom to brush her teeth. As she swished the toothbrush around in her mouth, she looked at her reflection in the mirror, surprised at what she saw. She looked troubled. She had dark circles under her eyes (common for a Klingon, but uncommon for HER) and a sad expression lay behind them. She spit the mouthful of toothpaste into the sink and rinsed her mouth. She looked around the crowded counter and found her hairbrush. Grabbing it, she headed back into her sleeping quarters. She sat down gently on the edge of her bed, facing away from the door, toward the bathroom. She began humming softly, desperate to lift her foul mood.
She suddenly felt the bed creak and someone come up behind her. The brush was taken from her hand and set down beside her. She gasped and spun around as best she could from her position on the bed. She didn't know why, but she was surprised to find Tom sitting there, grinning like a banshee. "Thomas Eugene Paris! You scared the shit out of me!" she exclaimed, playfully swatting his arm.
"Why? Don't I always do this to you?" he asked innocently, his blue eyes sparkling. She laughed and hugged him happily.
"I didn't think you'd come tonight," she confessed, not wanting to meet his eyes.
"Why not?" Genuine curiosity shone through his voice and he cocked his head to one side, giving her a puzzled look. She shifted around in an attempt to get comfortable. She finally brought her eyes up to his.
"I wasn't sure how much time you needed to . . . .recuperate . . . .after the Alice incident," she said. Tom smiled slightly and scooted to the edge of the bed so he could sit directly beside her. He took her hand and kissed it, his warm lips brushing her cool skin, heating it on contact.
"B'Elanna, you should know by now that nothing can keep me from you for too long. I love you," he said, his voice just above a whisper. She furrowed her eyebrows and squinted her eyes, as if scrutinizing him to see if he was telling the truth. Then her face softened and she hugged him tightly.
"You know . . . that's the first time you've ever said that to me," she whispered, not trusting her voice to go any louder. Tom pulled back slightly and looked into her deep brown eyes.
"I know. And I should've said it sooner. Two years sooner . . . " he said feebly, diverting his eyes. "But, I've always felt it. Even before you said it to me the first time," he confessed. He willed his eyes back up to her face and was pleased to see the smile that lay there.
"I know. It's just nice to hear it," she said. Tom looked down at the discarded hairbrush. He picked it up and turned it over in his hands.
"Well, then I guess I'll just have to make it up to you," he murmured, moving behind her. "B'Elanna Torres, prepare yourself for a night of pampering." He gently took a handful of her chocolate colored hair and ran the brush through it, making the waves bounce and shine. Against her will, her eyelids began to lower and she became drowsy. Tom's hands worked magic through her hair and she didn't want him to stop.
She didn't know she had fallen asleep until she felt the blankets being pulled over her. She opened her eyes lazily and grabbed Tom's hand as he was departing. "Wait . . . " she said, her voice tired but firm.
"What is it?" he asked, turning back to her.
"Stay . . . " she whispered, pulling the bedcovers beside her back and sliding over to give Tom room. He smiled and slid into bed beside her. He covered himself and wrapped an arm around her. She smiled, already drifting back to sleep. He sighed as sleep claimed him too.
B'Elanna wasn't sure what had awakened her, but she was suddenly alert. Tom was laying beside her, his chest rising and falling in comfortable, easy beats. She glanced at the chronometer and sighed. It was 0123. She sat up and rubbed her eyes. "Computer, lights, half-illumination," she ordered quietly. The lights came on and she checked Tom's sleeping figure to be sure she hadn't disturbed him. He was still asleep.
Carefully untangling herself from the bed sheets, she rose, stretching as she made her way to the replicator. She replicated a small garden salad and a steaming mug of hot cocoa. She sat down on the sofa and picked up an Engineering report she hadn't finished reading. She sipped the cocoa and speared a mouthful of the salad on her fork.
"What are you doing?" a voice asked from her bed. She turned with a smile to see Tom yawning and stretching behind her, a tired yet bemused expression on his face. "I woke up alone," he explained when she was silent.
"I woke up and was hungry," she said easily, lifting her salad as if to confirm her story. Tom smiled lazily and rubbed his head, rumpling his short blond waves. B'Elanna felt a familiar stir of longing as she looked him over. He wore a drowsy look on his face, but his warp-core-blue eyes were as bright as ever. His hair was a mess and his grin was crooked. He was utterly gorgeous.
"B'Elanna?" His voice penetrated her thoughts and she looked away sheepishly, suddenly very aware of the dampness growing between her legs. Tom knew her. He knew her better than she'd like to admit, so he knew by the expression on her face what she was feeling. Right about now, she was getting . . . .wet.
"I'm okay," she half lied. Tom smirked and came up to the back of the sofa. He leaned over and brushed a kiss across her forehead, taking care to caress each ridge with his moist lips. She closed her eyes in anticipation as his lips sought out hers. She tilted her head back and he lowered his lips to hers in a lazy kiss. He came around to the other side of the sofa and she moved aside to give him room. He sat down, tucking one leg underneath him. B'Elanna put one arm around Tom and pulled his body closer, leaning in for another kiss.
She closed her eyes as the kiss deepened, and then shifted as Tom began to lower her onto her back. He rolled over on top of her and took a moment to stare down at her face, at her chocolate brown eyes and dark brown hair. At her forehead ridges and defined cheekbones. And then at her full lips, that she was moistening with her tongue. Hating for her lips to be moistened needlessly, Tom leaned in for another kiss, this time slipping his tongue into her mouth.
"Do you have any clue how terribly beautiful you are?" Tom whispered as he gazed down at B'Elanna's face. She felt a slight blush spread across her face and she quickly averted her eyes away from Tom's intense gaze. She felt his hands move to her face, his fingers caressing her lips. She looked back at him and smiled at the unspoken longing he wore on his face.
"You know what?" she asked, her voice quiet. "I never feel beautiful . . . until I'm with you," she said truthfully. Tom's smile widened as his lips replaced his fingers. She loved the feel of his kiss. So gentle and loving. Not to mention the taste of his lips. Everything about the way he loved her was perfect.
Suddenly his lips were gone from hers, as was the weight of his body. She opened her eyes and looked at him with what she knew must have been a puzzled look. The only explanation he offered was the outstretching of his hand. She narrowed her eyes but placed her hand into his. He closed his fingers around hers and pulled her gently to the bed.
She started to sit down on the edge of the bed, but Tom grabbed onto her waist and kept her on her feet. He kissed her again, all the while pulling at the hem of her nightie. She separated herself from him just long enough to raise her hands over her head. The nightie was suddenly at her feet in one silk heap. Now she was clad only in her underwear. Tom took a step back and drunk in every inch of her body. B'Elanna felt herself redden under his scrutiny.
Tom chuckled at her modesty and reached out to take her hand. He guided her to the bed and then gently lowered her onto the slightly rumpled covers. The fabric of the blanket tickled her bare back and she bit her lower lip. Tom mistook it for arousal and kissed her again, leaning over her so he was half on - half off the bed. B'Elanna pulled him fully onto the bed and rolled him onto his back. She climbed on top of him and yanked his uniform top off, taking the turtleneck with it. Her hands danced over his almost bare chest, her fingers stopping to play with the red-gold curls that poked out of his tank top. He turned his head to the side and bit on the tip of his tongue as she ripped the tank top straight down the middle with her fingernail. She pulled the remaining pieces off his chest and tossed them aside on the floor.
She took his head in her hands and turned it so he was looking at her. She leaned down to brush a kiss across his lips and her breasts brushed his chest, the friction sending electrical charges down both their spines. B'Elanna lifted herself from on top of Tom so she could have easier access to his uniform pants. Suspended in air, one arm propping her up, Tom found his window of opportunity. He reached up with one hand and caressed one of B'Elanna's bronze colored breasts, his fingers dancing over the nipple. B'Elanna gasped at the sudden warmth on her breast and almost fell on Tom's legs. She got her bearings back and growled throatily.
Her earlier task was forgotten as Tom's mouth sought out her nipples. He took one into his mouth and alternated between sucks and gentle nips. His hand payed special attention to the other one, pinching and squeezing until he got the reaction he wanted. He loved it when she got like this. She was twitching and groaning beneath his touch, meaning it wouldn't take too much more coaxing before her first orgasm came.
To make sure of that, Tom slid her underwear down to her ankles, and then over her feet. B'Elanna kicked them away, sending them onto the floor. Tom rolled her over on her back and kissed his way down her quivering stomach to the delicate area on the inside of her thighs. His mouth was getting close to her center, and she was trembling slightly beneath the warm whisper of his lips. Finally, he parted the moist folds of skin with his forefinger and slowly began to manipulate her clitoris with his other fingers. She thrashed wildly beneath him and dug her nails into the palms of her hands headily as his lips replaced his fingers. She screamed as a rush of fluid escaped her and shot onto his mouth and chin. Her body quivered for a few minutes, basking in the release. Tom lifted his head slowly and wiped his face with the back of his hand. Smiling deviously, he reached up and kissed her, making her taste herself. She was immediately aroused again, and finally resumed taking his pants off.
She rocked back on her haunches and leaned forward, yanking his pants off in one fluid motion. She grinned and slipped one hand into his boxers; of course there was plenty of room inside the 'tent' he had created with his erection. She seized it in her hand, moving her fingers up to the shaft and then sliding down to the tip. She pulled her hand back out and licked the seminal fluid that leaked from the tip. Tom groaned and felt his body jump as his boxers were slid down to his ankles, the warm air of B'Elanna's quarters hitting his exposed flesh.
B'Elanna slid down so she was eye-level with his erection, and took it in her mouth, sucking and licking the head like a lollipop. Tom's muscles tightened as her motions continued and he opened his eyes, allowing himself to watch her take his length into her mouth and pull it back out in hard, swift motions. She brought her eyes up to his, and seeing that he was watching, slipped one finger into herself and curled it like a spoon. She pulled it back out and, with the same finger, rubbed the tip of his penis where seminal fluid was still escaping. Once their juices had mixed onto her finger, she raised it to his mouth.
Looking back and forth from her face to her finger, he grabbed her wrist and slipped the finger into his mouth, the taste of them swirling in his mouth. He arched his back and released, shooting a stream of fluid into her mouth. She smiled and made a big show of swallowing and wiping the corners of her mouth with one finger. She pushed herself back up to the head of the bed and kissed Tom, slipping her tongue into his mouth and lacing her fingers around the back of his neck. She was still unfulfilled, and she ached to feel him moving around inside of her, but she reminded herself to be patient. He was, after all, only human. They did occasionally need to rest up between orgasms.
He could tell she wasn't finished with him. Just by the heat she was projecting from the damp spot between her legs and the hunger she was putting into the kiss. He felt a stirring in his groin, and obviously so did she, because before he knew it, she was sitting on top of him, their genital areas touching. Tom's virility sprung to life, grazing the wetness of B'Elanna's center. She splayed herself across him, keeping her crotch in the same place, but bending to kiss him. She pulled herself back into her original position and reached her hands around to knead his butt with her knuckles. He moaned, whispering her name, his erection throbbing painfully from its confined position.
She recognized the intensity of his whimper and raised herself from him slightly, but only long enough to grab his shaft and lead him into her well lubricated slit. He moaned as her muscles tightened around his length. She began to rock up and down, slowly at first and then with increased intensity. She took one hand away from his hips and moved it to her nipple, teasing it with her finger. She traced a line around it and then trailed it down her body, stopping where their organs were joined.
They were both rapidly approaching the inevitable release, and she couldn't help herself. While still rocking, she leaned in and bit the soft area between his shoulder blades. The coppery taste of his blood filled her mouth and she licked the blood away from his wound. He tossed his head back as he shot his seed within her, and she wasn't far behind, her muscles contracting around him. She sighed and rolled off of him, landing on her back. Her knees were still brought up, but she closed her legs, almost exhausted by their lovemaking. Tom rolled on his side to face her, propping his head up on his elbow. "You really know how to get a guy off, Torres," he teased, kissing her lazily. He could taste his blood still inside of her mouth, which served as a reminder to the pain on his shoulder. She seemed to read his mind, and gazed down at the red bite mark.
"Oh, Tom. I never realize how hard I do that," she apologized, leaning in to brush her lips across the wound. He smiled tenderly and ran his hand through her hair.
"Don't worry about it. I don't even feel it anymore."
He flipped on his other side, and she spooned up against him, crushing her breasts against his bare back. Wrapping her arms around him, she quickly fell asleep, hoping the night's events would grace her dreams . . .
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Jul 17, Wednesday
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Yesu Persaud Centre for Caribbean Studies
Ameena Gafoor lecture series
Caribbean studies conferences
Events at the Centre
Events this Year
The Walter Rodney Lecture
Gad Heuman Postgraduate Bursary
Awarded Travel Bursaries
Staff and Research Interests
Past Leverhulme Fellows
Secretarial Support
Funded PhD studentship in Caribbean Studies
Management Committee(Restricted permissions)
Estie Hugo
Estie is a PhD student in Comparative Literature. She is currently working on a project funded by the Leverhulme Trust, which investigates World Literature through the commodity frontiers of the 20th century (see more here: https://warwick.ac.uk/fac/arts/english/people/niblettmike/worldlit/). Her PhD project investigates the histories of women's work in the sugar and palm oil industries, as mediated in Nigerian, British and Caribbean fiction.
Giulia Champion
'The Empire Bites Back: Literary Cannibalism Making Kin in the 'Extractive Zone'.
Giulia is a PhD candidate in the department of English and C omparative Literary Studies.
Her research focuses on environmental decolonial studies and looks at former British, French, Portuguese and Spanish colonies in most of the Caribbean and parts of Latin America and Africa. The primary goal of her thesis is to theorise "literary cannibalism" as a set of practices through world-ecology and historical materialism, in oorder to consider past, present, and future issues related to the regions' socio-political and economic developments and environmental challenges and their representation in cultural productions.
Supervisors: Professor Neil Lazarus and Dr. Fabienne Viala Research funded by the Doctoral College and the Fund for Women Graduates.
Liz Egan (2019-), 'Constructing and Challenging Creole Whiteness in Jamaica, 1865-1938',
I am a doctoral candidate, funded by the Arts and Humanities Research Council via the Midlands4Cities consortium.
My doctoral thesis is provisionally entitled 'Constructing and Challenging Creole Whiteness in Jamaica, 1865-1938', and it explores how whiteness was represented and practiced on the island between two moments of significant public unrest.
Supervisors: Professor David Lambert at the University of Warwick and Dr Sascha Auerbach at the University of Nottingham.
PhD Scholarship students
Miriam Gordon 2021 -
Aleema Gray (2018-)
'Living in Babylon: An oral history of the Rastafarian movement in Britain 1948-2016',
My PhD represents the first ever academic contribution dedicated in its entirety to documenting the history of the Rastafari movement in Britain. An important aim of this thesis seeks to understand the history of the Rastafari through the politics of History and memory. There is a need to produce scholarship that moves beyond a focus on cultural symbolism, such as dreadlocks and roots rock reggae, to concrete expressions of Pan African social protest.
Yesu Persaud Centre for Caribbean Studies/Warwick Collaborative Postgraduate Research Scholarship student in History. Supervisors: Professor David Lambert and Dr Meleisa Ono-George.
Natasha Bondre (2016- 2021)
Reading 'Emperor Oil' in the Expanded Caribbean
Yesu Persaud Centre for Caribbean Studies (YPCCS) and the Department of English and Comparative Literary
In 2021, Natasha successfully defended her PhD thesis, entitled Reading "Emperor Oil" in the Expanded Caribbean. She is currently an Early Career Fellow at the University of Warwick (2020 - 21), working jointly in the Yesu Persaud Caribbean Centre and the Department of English and Comparative Literary Studies. Her doctoral thesis examines the ecological nature of petro-capitalism in what has been termed the 'expanded Caribbean,' through studying the literatures, in Spanish and English, of specific nation-states in the region. Her general research and teaching interests consist of (but are not limited to): post-colonial literature and theory, eco-criticism, global Anglophone and Hispanophone literatures, world-literature and world-systems theory, disaster studies, cultures of protest in the pan-Caribbean, particularly against commodity frontier complexes, and speculative fiction, particularly Afrofuturism from both sides of the Atlantic.
Yesu Persaud Centre for Caribbean Studies/Warwick Collaborative Postgraduate Research Scholarship student in English. Supervisors: Paulo de Medeiros and Dr Mike Niblett.
Laëtitia Saint-Loubert (2014-18)
Yesu Persaud Centre for Caribbean Studies/Warwick Collaborative Postgraduate Research Scholarship student in Hispanic Studies. Supervisor: Dr Fabienne Viala.
Laetitia's research investigated Caribbean literature in translation and looked at how Caribbean literature circulates within, as well as outside, the region.
Her thesis 'The Caribbean in translation: remapping thresholds of dislocation', invited a reading of translation both as a literary, linguistic practice and as a transnational expression of cross-cultural Caribbean negotiations. She conducted a research project in Puerto-Rico (funded by the Gad Heuman travel bursary) during which she was affiliated with the Instituto de Estudios del Caribe and worked for a local publisher. This field work examined the role that translation plays in the literary production of the region, (re)situating local practices within the wider, global networks of Caribbean literary circulation.
Laetitia currently teaches literature and translation at the Université de la Réunion (ATER), in addition to being a practising literary translator.
Kimberley Thomas (2013- 2019)
Yesu Persaud Centre for Caribbean Studies/Warwick Collaborative Postgraduate Research Scholarship student in History. Supervisor: Professor David Lambert.
'"Oh, the trials! the trials! they make the salt water come into my eyes": Slaves and salt in the Caribbean, 1680-1850'
Kimberley is pictured here with supervisor David Lambert and examiners Christer Petley (left) and Tim Lockley (far right).
Kandasia Kendall
Rossana Coutinho
2021 -22
My Research project, examines the (Surinamese) government, specifically their laws and how they have executed it to reinforce or prevent equality. I aim to focus on the period of colonisation to look at the history and/ or origins of systematic racism.
Supervisor, Dr John Gilmore, Department of English & Comparative Literary Studies
Christopher Maxwell
Kit Maxwell (2021-), 'Material culture of the British Caribbean in the long eighteenth century'
Supervisor, Professor David Lambert, Department of History
Page contact: Lisa Cook
Last revised: Wed 1 Dec 2021


Continue reading "God and a Woman"
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Continue reading "The Worlds Our Words Create"
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Continue reading "Here By Choice"
Love one another while you still can.
Continue reading "Love one another while you still can."
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Manhattan Sales and Contracts: Crazy Rich Asians producer's ritzy condo sells for $18.6M; March has most transactions since Fall 2019
From Future New York Manhattan Sales and Contracts: Crazy Rich Asians producer's ritzy condo sells for $18.6M; March has most transactions since Fall 2019
By CityRealty Staff Monday, March 29, 2021
There were 261 Manhattan apartment sales recorded last week amounting to $464,766,998. There were 124 condo sales with a median closing price of $1.79M ($1,448/ft2). There were 136 co-op closings with a median price of $776,250. According to CityRealty data (see below), this March has the highest number of transactions since October 2019, and the highest aggregate total since March 2020.
Nearly two and a half years since it was listed, film producer/philanthropist Sidney Kimmel (famed for such films as Crazy Rich Asians, Moneyball, and The Kite Runner) and his wife Caroline (not pictured) sold their full-floor Central Park co-op for $18.6 million. The New York Post notes that this is almost $4 million less than the $22.34 million they paid for it in 2001, but was nevertheless the top sale of the past week.
998 Fifth Avenue via Sotheby's
In the last full week of March 2021, a total of 47 luxury listings through Manhattan entered contract. This extends the streak of 30 or more luxury contracts to eight weeks, which hasn't been seen since Olshan Realty started keeping track in 2006. The aggregate total of the past week's contracts was $344,844,500.
Among the top luxury contracts, Upper East Side co-ops put up an impressive showing. A high-floor home in the Rosario Candela-designed 998 Fifth Avenue entered contract with an asking price of $24 million, and a massive home at 30 East 71st Street also put up an impressive showing. The siren song of Billionaires' Row could not be ignored, though; a half-floor home at One57 had the second-highest contract of the past week at $20.9 million, and a tower condo at 111 West 57th Street wasn't far behind with an asking price of $13.5 million. A combination unit at 150 East 78th Street, an amenity-rich new condo in the heart of the Upper East Side, rounded out the top five contracts.
1. 50 Central Park South at The Ritz-Carlton, 50 Central Park South, #27 – $18,600,000 ($2,724 / ft2)
Midtown West | Condo | 2 beds, 6 baths | 6,829 ft2
50 Central Park South at The Ritz-Carlton via CityRealty
Interiors via Douglas Elliman
From the listing:
As seen on the pages of Architectural Digest, this exquisite full floor home has panoramic views showcasing Central Park and the New York Skyline. Rarely does a home of this quality and caliber become available.
The living room is centered on the antique marble fireplace and is connected to a wood paneled library through a bar, with Lalique panels, making the combined spaces perfect for entertaining. These rooms face North with a wall of French doors that lead to the grand terrace overlooking Central Park and the iconic skylines of the Upper East and Upper West Sides. The awe-inspiring 53-foot terrace is one of the features unique to this home, and the front and center view of Central Park is gorgeous in every season, day or night.
1 Available Listing for $49M
2. The White House, 262 Central Park West, #15A – $16,875,000
Central Park West | Co-op
The White House via CityRealty
3. 111 Murray Street, #54W – $9,990,000 ($2,974 / ft2)
Tribeca | Condo | 4 beds, 5.5 baths | 3,359 ft2
All images of 111 Murray Street via Douglas Elliman
At 3,330 square feet, this exceptional 4-bedroom, 5.5-bathroom residence with ceilings up to approximately 10'7" comprises an entire half floor, offering expansive north, south, and west-facing views that include cinematic panoramas of the Hudson River and beyond. From a private elevator vestibule, a grand set of double doors reveals a formal entry foyer that leads to a breathtaking corner living and dining room. Floor-to-ceiling glass wraps the gentle curve of this sprawling 28'4" × 23'4" space for exceptional natural light and spectacular north- and west-facing vistas. The adjacent eat-in kitchen, with an entire wall of floor-to-ceiling windows, center island breakfast bar, and dual points of entry, allows for seamless entertaining and intimate daily living alike.
5 Available Listings from $3.95M
4. 960 Fifth Avenue, #9B – $8,600,000
Upper East Side | Co-op | 2 beds, 2.5 baths
960 Fifth Avenue via Warburg
1 Available Listing for $10.25M
5. 40 East End Avenue, #7/8B – $8,835,000 ($2,375 / ft2)
Yorkville | Condo | 5 beds, 5.5 baths | 3,530 ft2
All images of 40 East End Avenue via Corcoran
Duplex 7/8B is a gracious five-bedroom, 5.5-bath residence designed by renowned architect Deborah Berke Partners. Spanning approximately 3,530 square feet with 375 square feet of outdoor space and expansive city views, this generously scaled residence features 10-foot ceilings, oversized casement windows, custom moldings, 8-foot doorways and white oak flooring.
On the lower level, a formal entry gallery leads into a large great room with access to a 375 square foot outdoor terrace looking out to the Midtown skyline. The adjacent windowed chef's kitchen boasts high-gloss lacquer painted Italian cabinetry complemented by hand-selected Italian Arabescato Cervaiole marble slabs and a gracious kitchen island for entertaining. Upper cabinet doors have a unique woven metal mesh panel and beveled trim, adding a refined pattern to the kitchen composition. The kitchen is complemented by sophisticated appliances by Gaggenau including a double oven, two dishwashers and a full height wine refrigerator. A connected butler's pantry allows for even more storage space. Also present on this floor is an en-suite guest bedroom with two closets.
6 Available Listings from $4.7M
Top Contracts
1. 998 Fifth Avenue, #11W – $24,000,000
Carnegie Hill | Co-op | 4 beds, 5.5 baths
998 Fifth Avenue via CityRealty
Interiors via Sotheby's
This magnificent high floor apartment features stunning Central Park views, sun flooded interiors, tremendously high ceilings, lovely and grandly proportioned rooms, and exceptional architectural details. A massive windowed gallery with original paneling opens to the fabulous entertaining rooms, laid out in enfilade format, with a fireplace on both the northern and southern flank. These extraordinary public rooms are distinguished by broad sweeping views of the Metropolitan Museum, Central Park, and across to the West Side skyline.
There is a baronial living room which spans more than 38 feet, as well as a wonderfully scaled dining room. Beyond, the vast back of house is comprised of a breakfast area, massive chef's kitchen, pantry and laundry. There are also two big bedrooms plus a staff room.
2. One57, 157 West 57th Street, #66B – $20,900,000
Midtown West | Condo | 3 beds, 3.5 baths | 4,193 ft2
All images of One57 via Compass
Half-floor residence on the 66th floor with unobstructed Central Park views, this 3-bedroom residence features floor to ceiling windows creating an expansive living space of extraordinary scale and light. Enter through the inviting gallery leading into the Grand Salon, the centerpiece of this exceptional floor plan, with soaring ceilings and a grand salon measuring 43 by 30 feet. The ideal setting for formal entertaining and displaying fine art, this residence boasts over 40 feet of direct Central Park frontage to the north and panoramic city views to the south. Adjacent to the Grand Salon is the eat-in chef's kitchen adorned in Miele appliances, dual built-in refrigerators/freezers, dual stainless steel wall ovens, dual dishwashers, warming drawer, premium coffee system, Subzero wine cooler, and custom designed integrated kitchen island stools and cabinetry handcrafted by Smallbone of Devizes.
17 Available Listings from $3M
3. 150 East 78th Street, #8A/8B – $14,300,000
Lenox Hill | Condo | 7 beds, 7.5 baths | 5,071 ft2
All renderings of 150 East 78th Street via Hayes Davidson/M18
The sponsor sold 2 units --8A and 8B to one buyer to be combined by the sponsor. 8A originally was configured as 2,881 square feet with 4 bedrooms and 4.5 bathrooms. 8B was 2,190 square feet and originally configured as 3 bedrooms and 3.5 bathrooms. The unit will now encompass the entire 8th floor and will have 5,071 square feet. This is in a new condo designed by Robert A.M. Stern with interiors by Robert Couturier. Amenities are on multiple levels and include a concierge, fitness center, children's activity room, a regulation squash court and a golf simulator, a catering kitchen, and a rooftop terrace with an outdoor kitchen and glorious views.
10 Available Listings from $5.5M
4. 111 West 57th Street, #14S – $13,500,000
All images of 111 West 57th Street via Douglas Elliman
Steinway Tower condo with 3 bedroom and 4.5 bathroom and a study. The building has a private porte-cochere on 58th Street, and a lobby featuring opulent accents in limestone and marble. Amenities include include an 82-foot 2-lane swimming pool with cabanas, separate sauna and treatment rooms, a double-height fitness center with a mezzanine terrace, private dining room and chef's catering kitchen. There is 24-hour doormen and concierge.
5. 30 East 71st Street, #11A – $12,500,000
All images of 30 East 71st Street via Sotheby's
This magnificent and high floor renovated 10-room apartment is located in one of New York's finest prewar cooperatives, in the East 70's and Madison, just steps from Central Park. This apartment underwent an 18 month+ renovation and includes large new windows, 6 zoned central air conditioning, hardwood floors, new bathrooms, soaring beamed ceilings and moldings. Designed by Amanda Nisbet, this four-bedroom apartment is in move-in condition and combines original architectural details with a color palette that takes center stage. A semi private elevator landing leads to an elegant entry gallery which opens to a grand living room with high ceilings and a wood burning fireplace overlooking the majestic Spire of St. James Church. Adjacent, there is a comfortable library with custom built-ins and a bar with wine refrigerator.
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Please join us for a Fair Housing for Housing Providers Training, featuring our partner, Legal Aid Society of San Diego, on April 30, 2019. Please see below and attached for the details of the training.
As space is limited, please make sure to register to reserve your place by calling Donald Estrada at 619-471-2644, or you can email Mr. Estrada at doonalde@lassd.org.


text:
HomeDirty TackleFifpro member wants Jose Mourinho jailed for bullying Bastian Schweinsteiger
Fifpro member wants Jose Mourinho jailed for bullying Bastian Schweinsteiger
https://www.gettyimages.com/
Jose Mourinho has deemed Bastian Schweinsteiger surplus to requirements at Man United, but with two years left on his three-year contract, the 32-year-old German midfielder reportedly doesn't want to move, so Mourinho has cast him out into the wilderness of training with the reserves. According to footballer turned lawyer Dejan Stefanovic, who is a member of world players' union Fifpro, this constitutes bullying and is worthy of a prison sentence for Mourinho.
From BBC Sport:
"It's clear bullying," Slovenia FifPro member Dejan Stefanovic told the BBC.
"In Slovenia, we would have indicted Mourinho and asked for the highest penalty — three years in prison." […]
Speaking to BBC World Football, lawyer Stefanovic, who is also president of the Slovenian Footballers and Sportsmen Union, added: "This is clearly meant to discourage Bastian.
"Mourinho is setting a bad example. I would take it to the ordinary court. In our case, the penal code is quite clear. If there is bullying and harassment, there's a prison sanction, a maximum of three years in prison.
"Schweinsteiger should turn to the Professional Footballers' Association and he should file a complaint and I would also seek penalty for Mourinho."
Meanwhile, Juan Mata is just happy that Mourinho is too preoccupied with Schweinsteiger to notice him.
Zidane: "The Champions League made me dump other competitions so we can stay together"
Alexis Sanchez and Henrikh Mkhitaryan accidentally realize their dreams at the same time
Man United's biggest January signing: This tractor


text:
Bentley starts building brand new 1929 Blowers
By Loz Blain
The 1929 Bentley Blower: that supercharger on the front represents one very stubborn man's vision for racing Bentleys of the late 1920s
Bentley's own Blower, from which these reissues will be copied, is in beautiful condition
The 1929 Team Blower: a famous race car that never took a win
The old girl gets a run at Goodwood
Bentley occasionaly enters the 1929 Team Blower in the Mille Miglia
12 new Blower Bentleys will be built, by the company's bespoke Mulliner division
This car, originally converted to use a supercharger by Tim Birkin, is now part of Bentley's heritage fleet
This is not a fancy or luxury interior, it's a race car
Drunk on nostalgia after completing its recent recreation of the "lost" 1939 Corniche, Bentley has commissioned its Mulliner division to build 12 reproductions of one of the brand's most famous pre-WW2 race cars.
The "Blower" was built in 1929, originally by superstar "Bentley Boys" racer Sir Henry Ralph Stanley "Tim" Birkin, who had just won Le Mans behind the wheel of a Speed Six, but who knew the Bentleys were in for a hiding if they didn't come up with something to challenge the supercharged Mercedes they were to compete against the next year.
The top brass at Bentley hated the idea of forced induction, preferring to make bigger motors instead, but Birkin felt that only a supercharger could take on the Mercedes. Eventually, he decided to make a prototype of his own that boosted horsepower from 130 up to around 240. It was hideously unreliable, and the whole jaunt nearly bankrupted his wealthy family, but it was unarguably flat-out faster than anything else Bentley was making, and Birkin finally convinced Bentley to manufacture 55 "Blower" cars with 4½-liter engines – 55 being enough to homologate the car for production endurance racing.
At Le Mans in 1930, Birkin put the Blower to work against Rudolf Caracciola's Mercedes SSK, pushing the German car to its limits and forcing Caracciola to overuse his own supercharger to the point where it failed and he was forced to retire. The Bentley Blower's failed as well, mind you, but it was reported as a ding-dong race up until that point, and Bentley's un-blown Speed Sixes swooped in to take the win for England in one of the most exciting races of the era.
So despite the fact that "Blower" was a four letter word to W.O. Bentley back when they were built, and the fact that they never won a single one of the 12 races they competed in, the supercharged race cars have a special place in Bentley's history. Indeed, the company even owns one. Chassis HB 3403 was the second of the four "team Blowers" Birkin built himself after convincing a wealthy heiress to throw her cash after his dream (nearly bankrupting herself in the process), and Bentley picked it up for the brand's heritage fleet.
And now, the company has decided it needs more Blowers. So it's planning to disassemble its heritage car down to the last bolt, then 3D scan the lot. Then, using original molds, jigs and hand tools that have survived from the 1920s (with a bit of help from 21st century technology), the Mulliner team will built 12 sets of parts, and then assemble them into 12 cars. The process will take two years.
Bentley says the 12 new Blowers will be "identical wherever possible to the original – mechanically, aesthetically and spiritually – with only minimal hidden changes dictated by modern safety concerns." Price is available on application, and will be a lot. Doubtless Bentley remembers well that its original run of 55 Blowers nearly bankrupted the company, along with Birkin's family and his heiress friend.
Birkin would surely be proud to see his sacrilegious creations become a part of Bentley's heritage celebrations. That is, if he hadn't died penniless in 1933 at the ripe old age of 36, a personification of the old adage that if you want to make a million dollars out of racing, the best way is to start out with a billion. Godspeed, you moustachioed madman, you gave it a gumboot full.
Source: Bentley
AutomotiveBentleyClassic Motor CarsSupercharged
Loz Blain
Loz has been one of our most versatile contributors since 2007, and has since proven himself as a photographer, videographer, presenter, producer and podcast engineer, as well as a senior features writer. Joining the team as a motorcycle specialist, he's covered just about everything for New Atlas, concentrating lately on eVTOLs, hydrogen, energy, aviation, audiovisual, weird stuff and things that go fast.
guzmanchinky September 9, 2019 06:29 PM
How fun would it be to assemble these machines? So cool.
buzzclick September 10, 2019 01:32 PM
Saw one in the flesh at Limerock racetrack in the 80's. What a beast. To my eye just about the best looking and imposing car Britain produced in the 20's - owlbeyou
Gannet September 10, 2019 06:33 PM
If you miss out on one of those, this company has been making tasty replicas for a while. They did some by rebodying Bentleys from the 1950s, so it had a genuine motor, chassis with numbers etc. Not sure if they make some from scratch
http://www.vintageracinggreen.com/racing_green/racing-green-cars/b-special-speed-8


text:
\section{\bf Introduction and statement of results}\label{secone}

In \cite{greene}, Greene introduced the notion of hypergeometric functions over finite fields or
\emph{Gaussian hypergeometric series} which are analogous to the classical hypergeometric series.
Gaussian hypergeometric series possess many interesting properties analogous to the classical hypergeometric series.
Finding the number of solutions of a polynomial equation over a finite field has been of interest to
mathematicians for many years. Many mathematicians have studied this problem and found interesting relations to Gaussian
hypergeometric series. For example see \cite{BK, BK2, Fuselier, koike, Lennon, mccarthy, ono, rouse, vega}.
\par
We will now restate some definitions from \cite{greene}. Let $q=p^e$ be a power of an odd prime and $\mathbb{F}_q$
the finite field of $q$ elements. Throughout this paper, $A, B, C, S, \chi, \phi, \varepsilon$ will denote complex multiplicative characters on $\mathbb{F}_q^{\times}$.
The notation $\varepsilon, \phi$ will always be reserved
for the trivial and quadratic characters, respectively. Extend each character $\chi \in \widehat{\mathbb{F}_q^{\times}}$ to all of $\mathbb{F}_q$ by setting $\chi(0):=0$. The binomial coefficient
${A \choose B}$ is defined by
\begin{align}\label{eq0}
{A \choose B}:=\frac{B(-1)}{q}J(A,\overline{B})=\frac{B(-1)}{q}\sum_{x \in \mathbb{F}_q}A(x)\overline{B}(1-x),
\end{align}
where $J(A, B)$ denotes the usual Jacobi sum and $\overline{B}$ is the inverse of $B$.
The following special case is known from \cite{greene}
\begin{align}\label{eq01}
{A \choose \varepsilon}={A \choose A}=-\frac{1}{q}+\frac{q-1}{q}\delta(A),
\end{align}
where $\delta(A)=0$ if $A\neq\varepsilon$ and $\delta(A)=1$ if $A=\varepsilon$. With this notation, for characters $A_0, A_1,\ldots, A_n$ and $B_1, B_2,\ldots, B_n$ of $\mathbb{F}_q$,
the Gaussian hypergeometric series ${_{n+1}}F_n\left(\begin{array}{cccc}
                A_0, & A_1, & \cdots, & A_n\\
                 & B_1, & \cdots, & B_n
              \end{array}\mid x \right)$ over $\mathbb{F}_q$ is defined as
\begin{align}\label{eq00}
{_{n+1}}F_n\left(\begin{array}{cccc}
                A_0, & A_1, & \cdots, & A_n\\
                 & B_1, & \cdots, & B_n
              \end{array}\mid x \right):
              =\frac{q}{q-1}\sum_{\chi}{A_0\chi \choose \chi}{A_1\chi \choose B_1\chi}
              \cdots {A_n\chi \choose B_n\chi}\chi(x),
\end{align}
where the sum is over all characters $\chi$ of $\mathbb{F}_q$.
\par
Let $\lambda \in \mathbb{Q}\setminus \{0, -1\}$ and $l \geq 2$. Denote by
$C_{l,\lambda}$ the nonsingular projective algebraic curve over $\mathbb{Q}$ with affine equation given by
\begin{align}\label{eq1}
y^l=(x-1)(x^2+\lambda).
\end{align}
\begin{definition}
Suppose $p$ is a prime of good reduction for $C_{l, \lambda}$. Let $q=p^e$.
Define the integer $a_q(C_{l, \lambda})$ by
\begin{align}\label{eq2}
a_q(C_{l, \lambda}):=1+q-\#C_{l, \lambda}(\mathbb{F}_q),
\end{align}
where $\#C_{l, \lambda}(\mathbb{F}_q)$ denotes the number of points that the curve $C_{l, \lambda}$ has over
$\mathbb{F}_q$.
\end{definition}
It is clear that a prime $p$ not dividing $l$ is of good reduction for $C_{l, \lambda}$ if and only if
$\text{ord}_p(\lambda(\lambda+1))=0.$
\par
We now state a remark about the number of $\mathbb{F}_q$-points on $C_{l, \lambda}$.
For details, see \cite[Remark 1.1]{BK}.
\begin{remark}
Let $l \neq 3$. Then
\begin{align}\label{eq11}
\#C_{l, \lambda}(\mathbb{F}_q)=1+\#\{(x, y)\in \mathbb{F}_q^2: y^l=(x-1)(x^2+\lambda)\}.
\end{align}
Again, let $l=3$ and $p\equiv 1$ $($\emph{mod} $3)$. Then
\begin{align}\label{eq12}
\#C_{l, \lambda}(\mathbb{F}_q)=3+\#\{(x, y)\in \mathbb{F}_q^2: y^l=(x-1)(x^2+\lambda)\}.
\end{align}
\end{remark}
\begin{remark}\label{remark1}
If $l=3$, $C_{l, \lambda}$ is an elliptic curve. The change of variables $$X-Z\rightarrow X,~Y\rightarrow Y~~\text{and}~~ X\rightarrow X$$ transforms the projective curve $$C_{3, \lambda}: Y^3=(X-Z)(X^2+\lambda Z^2)$$ to
\begin{align}\label{projective-1}
Y^3&=X(X^2+2XZ+(1+\lambda)Z^2).
\end{align}
Now dehomogenizing \eqref{projective-1} by putting $X=1$ and then making the substitution $$Y\rightarrow (1+\lambda)x, Z\rightarrow (1+\lambda)y-\frac{1}{1+\lambda},$$ we find that $C_{3, \lambda}$ is isomorphic over $\mathbb{Q}$ to the elliptic curve
\begin{align}\label{200}
y^2=x^3-\frac{\lambda}{(1+\lambda)^4}.
\end{align}
\end{remark}
Ono \cite[Thm. 5]{ono}, proved that if $\lambda \in \mathbb{Q}\setminus \{0, -1\}$ and $p$ is an odd prime for which
$\text{ord}_p(\lambda(\lambda+1))=0$ then
\begin{align}\label{eq122}
{_{3}}F_2\left(\begin{array}{cccc}
                \phi, & \phi, & \phi\\
                   & \varepsilon, & \varepsilon
              \end{array}\mid \frac{1+\lambda}{\lambda} \right)
              =\frac{\phi(-\lambda)(a_p(C_{2,\lambda})^2-p)}{p^2}.
\end{align}
Note that a change of variables in Theorem 5 of \cite{ono} is required to arrive at \eqref{eq122}. In this paper, we give a proof of the following result which generalizes \eqref{eq122} to the algebraic curve $C_{l, \lambda}$ over $\mathbb{F}_q$.
\begin{theorem}\label{thm4}
Let $p$ be a prime such that \emph{ord}$_p(\lambda(\lambda+1))=0$ and $q=p^e \equiv 1$ $($\emph{mod} $l)$.
If $l\geq 2$ is such that $3 \nmid l$ or $4\nmid l$, then
\begin{align}
a_q(C_{l, \lambda})^2 &=q^2\cdot \sum_{i=1}^{l-1}\frac{J(S^{3i}, S^{-i})}{S^i(-4\lambda^3)J(S^i, S^i)}\cdot {_{3}}F_2
\left(\begin{array}{cccc}
                S^{3i}, & S^i, & S^{2i}\phi\\
                   & S^{4i}, & S^{2i}
              \end{array}\mid \frac{1+\lambda}{\lambda} \right)\notag\\
              &\hspace{.5cm}+q\cdot \sum_{i=1}^{l-1}
              \frac{\phi(-\lambda)J(S^{3i}, S^{-i})}{S^i(-4\lambda(1+\lambda)^2)J(S^i, S^i)}+Q,\notag
\end{align}
where
\begin{align}
Q=\left\{
    \begin{array}{ll}
      (l-1)(q-1)-(l-3)a_q(C_{l,\lambda}), & \hbox{if $l$ is odd;} \\
      (l-2)(q-1)-(l-2)a_q(C_{l,\lambda})\notag\\-2q\cdot \displaystyle\sum_{i=1}^{\frac{l}{2}-1}\frac{J(\phi,S^{-2i})}{J(S^{i}\phi,S^{-3i})}\cdot{_{2}}F_1\left(\begin{array}{cc}
                S^{3i}, & S^{3i}\phi\\
                   & S^{4i}
              \end{array}\mid 1+\lambda \right), & \hbox{if $l$ is even}
    \end{array}
  \right.
\end{align}
and $S$ is a character on $\mathbb{F}_q$ of order $l$.
\end{theorem}
In addition, we will also prove the following results about the number of $\mathbb{F}_q$-points on the curve $C_{l, \lambda}$.
\begin{theorem}\label{thm22}
Suppose that $q=p^e \equiv 1$ $($\emph{mod} $l)$ and \emph{ord}$_p(\lambda(\lambda+1))=0$.
If $3\nmid l$ and $\frac{q-1}{l}$ is even, then
\begin{align}\label{eq114}
-a_q(C_{l, \lambda})=q\cdot \displaystyle\sum_{i=1}^{l-1}\frac{J(\phi, S^{-i})}{J(\sqrt{S^i}, \sqrt{S^{-3i}}\phi)}
\cdot {_{2}}F_1\left(\begin{array}{cccc}
                \sqrt{S^{3i}}\phi, & \sqrt{S^{3i}}\\
                 & S^{2i}
              \end{array}\mid 1+\lambda \right)
\end{align}
and for $l=3$,
\begin{align}\label{eq115}
-a_q(C_{l, \lambda})=2+q\cdot\displaystyle\sum_{i=1}^{2}{_{2}}F_1\left(\begin{array}{cccc}
                \phi, & \varepsilon\\
                 & S^i
              \end{array}\mid 1+\lambda \right),
\end{align}
where $S$ is a character of order $l$ on $\mathbb{F}_q$.
\end{theorem}
\begin{theorem}\label{thm6}
If $q\equiv 1$ $($\emph{mod} $l)$, then for $\lambda=\frac{1}{3}$, we have
\begin{align}
-a_q(C_{l, \lambda})
=\left\{
\begin{array}{lll}
0, & \hbox{if $l\neq 3$ and $q\equiv 2$ $($\emph{mod} $3)$;}\\
q\cdot \displaystyle\sum_{i=1}^{l-1}S^i(\frac{27}{8})
\left[{\chi_3 \choose S^i}+{\chi_3^2 \choose S^i}\right], & \hbox{if $l\neq 3$ and $q\equiv 1$ $($\emph{mod} $3)$;} \\
2+q\cdot \displaystyle\sum_{i=1}^{2}
\left[{\chi_3 \choose \chi_3^i}+{\chi_3^2 \choose \chi_3^i}\right], & \hbox{if $l=3$,}
\end{array}
\right.\notag
\end{align}
where $S$ and $\chi_3$ are characters on $\mathbb{F}_q$ of order $l$ and $3$ respectively.
\end{theorem}
We also give an alternate proof of the following result of D. McCarthy.
\begin{theorem}\cite[Thm. 2.3]{mccarthy}\label{thm44}
If $q\equiv 1~($\emph{mod} $3)$, then
\begin{align}
-\frac{\phi(-2)}{q}\cdot a_q(C_{2,\frac{1}{3}})=2\emph{Re}{\chi_3 \choose \phi}\notag
\end{align}
and
\begin{align}
-\phi(-2)\cdot a_q(C_{2,\frac{1}{3}})=2\emph{Re}\left[\frac{G(\chi_3)G(\phi)}{G(\chi_3\phi)}\right],\notag
\end{align}
where $\chi_3$ is a character of order $3$ of $\mathbb{F}_q$ and $G(\chi)$ is a Gauss sum.
\end{theorem}
In section 4, we will prove the following results on special values of ${_{3}}F_2$ and ${_{2}}F_1$ hypergeometric series.
In \cite[Thm. 1.3]{evans}, R. Evans and J. Greene gave an expression for ${_{3}}F_2(\frac{1}{4})$ which was an extension of a result of K. Ono\cite{ono}.
The following result gives the value of ${_{3}}F_2(4)$ which also extends another result of K. Ono\cite{ono}.
\begin{theorem}\label{thm5}
If $S$ is a character on $\mathbb{F}_q$ with order not equal to $1$, $3$, or $4$, then
\begin{align}
{_{3}}F_2\left(\begin{array}{cccc}
                S^{-3}, & S^{-1}, & S^{-2}\phi\\
                   & S^{-4}, & S^{-2}
              \end{array}\mid 4 \right)
              =\left\{
              \begin{array}{ll}
\displaystyle-\frac{\phi(-3)S(16)}{q}, &\hspace{-4cm} \hbox{if $q\equiv 2$ $($\emph{mod} $3)$;}\\
\displaystyle\frac{S(-\frac{16}{27})J(S^{-1}, S^{-1})}{J(S^{-3}, S)}\left[{S \choose \chi_3}
+{S \choose \chi_3^2}\right]^2-\frac{\phi(-3)S(16)}{q},\notag\\
&\hspace{-4cm}\hbox{if $q\equiv 1$ $($\emph{mod} $3)$,}
\end{array}
\right.\notag
\end{align}
where $\chi_3$ is a character of order $3$ of $\mathbb{F}_q$.
\end{theorem}
The result of K. Ono can be obtained by putting $S=\phi$, thus solving a problem posed by M. Koike \cite[p. 465]{koike}. We remark that in view of \cite[Thm. 4.2]{greene}, there is a result similar to Theorem \ref{thm5} in which the argument $4$ is replaced by $\frac{1}{4}$. However, our result about ${_{3}}F_2(\frac{1}{4})$ will be different from the result obtained by R. Evans and J. Greene.
\begin{theorem}\label{thm8}
Let $S$ be a character on $\mathbb{F}_q$ whose order is not equal to $3$. If $S$ is square of some character on $\mathbb{F}_q$, then
\begin{align}
&(i)~{_{2}}F_1\left(\begin{array}{cccc}
                \sqrt{S^{-3}}\phi, & \sqrt{S^{-3}}\\
                 & S^{-2}
              \end{array}\mid \frac{4}{3} \right)
              =\left\{
              \begin{array}{ll}
              0, &\hspace{-2cm} \hbox{if $q\equiv 2$ $($\emph{mod} $3)$;} \\
              \displaystyle \frac{S(\frac{8}{27})J(\sqrt{S^{-1}},\sqrt{S^3}\phi)}{J(\phi,S)}
              \left[\displaystyle {S \choose \chi_3}+{S \choose \chi_3^2}\right],\\
              &\hspace{-2cm} \hbox{if $q\equiv 1$ $($\emph{mod} $3)$.}
              \end{array}
              \right.\notag\\
&(ii)~{_{2}}F_1\left(\begin{array}{cccc}
                \sqrt{S^{-3}}\phi, & \sqrt{S^{-3}}\\
                 & S^{-1}\phi
              \end{array}\mid -\frac{1}{3} \right)
              =\left\{
              \begin{array}{ll}
              0, & \hspace{-2.5cm}\hbox{if $q\equiv 2$ $($\emph{mod} $3)$;} \\
              \displaystyle\frac{S(\frac{8}{27})J(\sqrt{S^{-1}},\sqrt{S^3}\phi)}{\sqrt{S}\phi(-1)J(\phi,S)}
              \displaystyle\left[{S \choose \chi_3}+{S \choose \chi_3^2}\right],\\
              & \hspace{-2.5cm}\hbox{if $q\equiv 1$ $($\emph{mod} $3)$.}
              \end{array}
              \right.\notag\\
&(iii)~{_{2}}F_1\left(\begin{array}{cccc}
                \sqrt{S^{-3}}\phi, & \sqrt{S^{-1}}\\
                 & S^{-2}
              \end{array}\mid 4 \right)=\left\{
              \begin{array}{ll}
              0, &\hspace{-2.9cm} \hbox{if $q\equiv 2$ $($\emph{mod} $3)$;} \\
              \displaystyle\frac{\sqrt{S}(-\frac{64}{27})J(\sqrt{S^{-1}},\sqrt{S^3}\phi)}{\phi(-3)J(\phi,S)}
              \displaystyle\left[{S \choose \chi_3}+{S \choose \chi_3^2}\right],\\
              & \hspace{-2.9cm}\hbox{if $q\equiv 1$ $($\emph{mod} $3)$.}
              \end{array}
              \right.\notag\\
&(iv)~{_{2}}F_1\left(\begin{array}{cccc}
                \sqrt{S^{-3}}\phi, & \sqrt{S}\phi\\
                 & S^{-1}\phi
              \end{array}\mid \frac{1}{4} \right)=\left\{
              \begin{array}{ll}
              0, & \hspace{-2.9cm}\hbox{if $q\equiv 2$ $($\emph{mod} $3)$;} \\
              \displaystyle\frac{\sqrt{S}(-\frac{1}{27})J(\sqrt{S^{-1}},\sqrt{S^3}\phi)}{\phi(3)J(\phi,S)}
              \displaystyle\left[{S \choose \chi_3}+{S \choose \chi_3^2}\right],\\
              &\hspace{-2.9cm} \hbox{if $q\equiv 1$ $($\emph{mod} $3)$.}
              \end{array}
              \right.\notag
\end{align}
\end{theorem}

\section{\bf Preliminaries}
We start with a result which enables us to count the number of points on a curve using multiplicative characters
on $\mathbb{F}_q$ where $q=p^e$ (see \cite{vega}).
\begin{lemma}\label{lemma1}
Let $a\in\mathbb{F}_q^{\times}$. If $n|(q-1)$, then $$\#\{x\in\mathbb{F}_q: x^n=a\}=\sum \chi(a),$$
where the sum runs over all characters $\chi$ on $\mathbb{F}_q$ of order dividing $n$.
\end{lemma}
The \emph{orthogonality relations} for multiplicative characters are (see \cite[Chapter 8]{ireland}):
\begin{align}\label{lemma11}
\sum_{x\in\mathbb{F}_q}\chi(x)=\left\{
                                  \begin{array}{ll}
                                    q-1 & \hbox{if~ $\chi=\varepsilon$;} \\
                                    0 & \hbox{if ~~$\chi\neq\varepsilon$.}
                                  \end{array}
                                \right.
\end{align}
We now restate some results of R. Evans and J. Greene \cite{evans,evans1}.
The function $F(A, B; x)$ is defined by \cite{evans1}
\begin{align}\label{eq3}
F(A, B; x):=\frac{q}{q-1}
\sum_{\chi}
{ A\chi^2 \choose \chi }{A\chi \choose B\chi }\chi\left(\frac{x}{4}\right),
\end{align}
and its normalization as
\begin{align}\label{eq4}
F^*(A, B; x):=F(A, B; x)+AB(-1)\frac{\overline{A}(\frac{x}{4})}{q}.
\end{align}
Another character sum from \cite{evans} that we will need is
\begin{align}\label{eq5}
g(A, B; x):=\sum_{t\in \mathbb{F}_q}A(1-t)B(1-xt^2), ~~~~~~~x\in \mathbb{F}_q.
\end{align}
\begin{theorem}\cite[Thm. 2.2]{evans}\label{thm1}
If $A \neq C$ and $x \notin \{0, 1\}$, then
$$F^*\left(A, C; \frac{x}{x-1}\right)=\frac{A(2)A\overline{C}(1-x)}{q}\cdot g(A\overline{C}^2, \overline{A}C; 1-x).$$
\end{theorem}
\begin{theorem}\cite[Thm. 2.5]{evans}\label{thm2}
Let $C\neq \phi$, $A\notin \{\varepsilon, C, C^2\}$, $x\neq 1$. Then
\begin{align}
{_{3}}F_2\left(\begin{array}{ccc}
                A, & \overline{A}C^2, & C\phi\\
                   & C^2, & C
              \end{array}\mid x \right) &=-\frac{\overline{C}(x)\phi(1-x)}{q} \nonumber\\
              &+C(-1)A\overline{C}(4)A\overline{C}^2(1-x)
              \cdot\frac{J(\overline{A}C^2, A\overline{C})}{q^2 J(A, \overline{A}C)}
              \cdot g(A\overline{C}^2, \overline{A}C; 1-x)^2.\nonumber
              \end{align}
\end{theorem}
\begin{theorem}\cite[Thm. 1.2]{evans1}\label{thm3}
Let $R^2 \notin \{\varepsilon, C, C^2\}$. Then
$$F^*(R^2, C; x)=R(4) \frac{J(\phi, C\overline{R}^2)}{J(\overline{R}C, \overline{R}\phi)}\cdot
{_{2}}F_1\left(\begin{array}{cccc}
                R\phi, & R\\
                 & C
              \end{array}\mid x \right).$$
\end{theorem}
We now prove a result similar to the above theorem.
\begin{proposition}\label{propo1}
We have
\begin{align}
F^*(\varepsilon, C; x)=\left\{
                 \begin{array}{ll}
                   {_{2}}F_1\left(\begin{array}{cccc}
                \phi, & \varepsilon\\
                 & C
              \end{array}\mid x \right), & \hbox{if $C\neq\varepsilon$;} \\
                   -(q-2)\cdot {_{2}}F_1\left(\begin{array}{cccc}
                \phi, & \varepsilon\\
                 & C
              \end{array}\mid x \right), & \hbox{if $C=\varepsilon$.}
                 \end{array}
               \right.
\end{align}
\end{proposition}
\begin{proof} We prove the result following the technique used in \cite{evans1}. Putting $B=\varepsilon$ in the relation \cite[(4.21)]{greene}, we have
\begin{align}\label{eq119}
{\chi^2 \choose \chi}={\phi \chi \choose \chi}{\chi \choose \chi}{\phi \choose \phi }^{-1}\chi(4).
\end{align}
From \eqref{eq119} and \eqref{eq3}, we obtain
\begin{align}\label{eq111}
F(\varepsilon,C;x)&=\frac{q}{q-1}\sum_{\chi}{\chi^2 \choose \chi}{\chi \choose C\chi}\chi\left(\frac{x}{4}\right)\notag\\
&=\frac{q}{q-1}\sum_{\chi}{\chi \choose C\chi}{\phi\chi \choose \chi}{\chi \choose \chi}
{\phi \choose \phi}^{-1}\chi(4)\chi\left(\frac{x}{4}\right)\notag\\
&={\phi \choose \phi}^{-1}{_{3}}F_2\left(\begin{array}{ccc}
                \phi, & \varepsilon, & \varepsilon\\
                   & C, & \varepsilon
              \end{array}\mid x \right).
\end{align}
By \cite[Thm. 3.15 (v)]{greene}, \eqref{eq111} reduces to
\begin{align}\label{eq112}
{\phi \choose \phi}F(\varepsilon,C;x)={C \choose C}{_{2}}F_1\left(\begin{array}{cccc}
                \phi, & \varepsilon\\
                 & C
              \end{array}\mid x \right)-\frac{C(-1)}{q}{\phi \choose \varepsilon}.
\end{align}
From \eqref{eq4}, we have
\begin{align}\label{eq113}
{\phi \choose \phi}F(\varepsilon,C;x)&={\phi \choose \phi}F^*(\varepsilon,C;x)-\frac{C(-1)}{q}{\phi \choose \phi}.
\end{align}
Comparing equations \eqref{eq112} and \eqref{eq113}, we obtain
\begin{align}
F^*(\varepsilon,C;x)={C \choose C}{\phi \choose \phi}^{-1}{_{2}}F_1\left(\begin{array}{cccc}
                \phi, & \varepsilon\\
                 & C
              \end{array}\mid x \right).
\end{align}
Using \eqref{eq01}, we complete the proof of the result.
\end{proof}
The following result is due to J. Greene.
\begin{theorem}\cite[Thm. 4.4 (i) \& (ii)]{greene}\label{thm7}
For $x\in \mathbb{F}_q$,
\begin{align}
\hspace{-3cm}(i)\hspace{1cm} {_{2}}F_1\left(\begin{array}{cccc}
                A, & B\\
                 & C
              \end{array}\mid x \right)&=A(-1){_{2}}F_1\left(\begin{array}{cccc}
                A, & B\\
                 & AB\overline{C}
              \end{array}\mid 1-x \right)\notag\\
              &\hspace{.5cm}+A(-1)\displaystyle{B \choose \overline{A}C}
              \delta(1-x)-\displaystyle{B \choose C}\delta(x),\notag\\
\hspace{-3cm}(ii)\hspace{1cm} {_{2}}F_1\left(\begin{array}{cccc}
                A, & B\\
                 & C
              \end{array}\mid x \right)&=C(-1)\overline{A}(1-x){_{2}}F_1\left(\begin{array}{cccc}
                A, & C\overline{B}\\
                 & C
              \end{array}\mid \frac{x}{x-1} \right)\notag\\
              &\hspace{.5cm}+A(-1)\displaystyle{B \choose \overline{A}C}\delta(1-x),\notag
\end{align}
where $\delta$ is the function $\delta(0)=1$ and $\delta(x)=0$ for $x\neq 0$.
\end{theorem}

\section{\bf Proof of results}
\begin{pf}{\bf \ref{thm4}.}
Putting $A=S^i, B=S^i$ and $x= -\frac{1}{\lambda}$ in \eqref{eq5}, we obtain
\begin{align}
g\left(S^i, S^i; -\frac{1}{\lambda}\right)&=\sum_{t\in \mathbb{F}_q}S^{-i}(-\lambda)S^i((t-1)(t^2+\lambda))\notag
\end{align}
which gives
\begin{align}\label{eq6}
\sum_{t\in \mathbb{F}_q}S^i((t-1)(t^2+\lambda))&=S^{i}(-\lambda)g\left(S^i, S^i; -\frac{1}{\lambda}\right).
\end{align}
Moreover,
\begin{align}
&\#\{(x, y)\in \mathbb{F}_q^2: y^l=(x-1)(x^2+\lambda)\}\notag\\
&=\sum_{t\in\mathbb{F}_q}\#\{y\in\mathbb{F}_q: y^l=(t-1)(t^2+\lambda)\}\nonumber\\
&=\sum_{t\in\mathbb{F}_q, (t-1)(t^2+\lambda)\neq0}
\#\{y\in\mathbb{F}_q: y^l=(t-1)(t^2+\lambda)\}+\#\{t\in\mathbb{F}_q: (t-1)(t^2+\lambda)=0\}.\nonumber
\end{align}
Applying Lemma \ref{lemma1}, we obtain
\begin{align}
&\#\{(x, y)\in \mathbb{F}_q^2: y^l=(x-1)(x^2+\lambda)\}\nonumber\\
&=\sum_{t\in\mathbb{F}_q}\sum_{i=0}^{l-1}S^i((t-1)(t^2+\lambda))+
\#\{t\in\mathbb{F}_q: (t-1)(t^2+\lambda)=0\}\nonumber\\
&=q+\sum_{t\in\mathbb{F}_q}\sum_{i=1}^{l-1}S^i((t-1)(t^2+\lambda)).\nonumber
\end{align}
Since ord$_p(\lambda(\lambda+1))=0$, \eqref{eq11} yields
\begin{align}\label{eq13}
-a_q(C_{l, \lambda})=\sum_{i=1}^{l-1}\sum_{t\in\mathbb{F}_q}S^i((t-1)(t^2+\lambda)).
\end{align}
Squaring both sides of \eqref{eq13}, we obtain
\begin{align}
a_q(C_{l, \lambda})^2=\sum_{i=1}^{l-1}\left[\sum_{t\in\mathbb{F}_q}S^i((t-1)(t^2+\lambda))\right]^2
+\sum_{i,j=1,i\neq j}^{l-1}\sum_{t\in\mathbb{F}_q}S^{i+j}((t-1)(t^2+\lambda)).\notag
\end{align}
Again using \eqref{eq6} and \eqref{lemma11}, we deduce that
\begin{align}\label{eq7}
a_q(C_{l, \lambda})^2=\sum_{i=1}^{l-1}S^i(\lambda^2)g
\left(S^i, S^i; -\frac{1}{\lambda}\right)^2&+2(q-1)\cdot\lfloor\frac{l-1}{2}\rfloor\notag\\&+\sum_{i,j=1,i\neq j, i+j\neq l}^{l-1}\sum_{t\in\mathbb{F}_q}S^{i+j}((t-1)(t^2+\lambda)).
\end{align}
Since $3\nmid l$ or $4\nmid l$, taking $A=S^{-3i}$, $C=S^{-2i}$ and
$x=\frac{1+\lambda}{\lambda}$ in Theorem \ref{thm2}, we obtain
\begin{align}\label{eq001}
g\left(S^i, S^i; -\frac{1}{\lambda}\right)^2&=q^2\cdot\frac{S^i(-4\lambda)J(S^{-3i}, S^{i})}{J(S^{-i}, S^{-i})}
\cdot {_{3}}F_2\left(\begin{array}{cccc}
                S^{-3i}, & S^{-i}, & S^{-2i}\phi\\
                   & S^{-4i}, & S^{-2i}
              \end{array}\mid \frac{1+\lambda}{\lambda} \right)\notag\\
              &\hspace{.5cm}+q\cdot
              \frac{\phi(-\lambda)S^i(-\frac{4(1+\lambda)^2}{\lambda})J(S^{-3i}, S^{i})}{J(S^{-i}, S^{-i})}.
\end{align}
Now we find the value of $$\sum_{i,j=1,i\neq j,i+j\neq l}^{l-1}\sum_{t\in\mathbb{F}_q}S^{i+j}((t-1)(t^2+\lambda)).$$
Let $P(i_k)$ be the set of all possible values of $i$ such that $i+j\equiv k~($mod $l)$, $1\leq i,j \leq l-1$ and $i\neq j$. Then for odd values of $l$ $$\#P(i_k)=l-3$$ and for even values of $l$ $$\#P(i_k)=\left\{
                                                                         \begin{array}{ll}
                                                                           l-2, & \hbox{if $k$ is odd;} \\
                                                                           l-4, & \hbox{if $k$ is even.}
                                                                         \end{array}
                                                                       \right.$$
Therefore,
\begin{align}
&\sum_{i,j=1,i\neq j,i+j\neq l}^{l-1}\sum_{t\in\mathbb{F}_q}S^{i+j}((t-1)(t^2+\lambda))\notag\\&=\left\{
                                                                               \begin{array}{ll}
                                                  (l-3)\displaystyle\sum_{i=1}^{l-1}\sum_{t\in\mathbb{F}_q}S^i((t-1)(t^2+\lambda)), & \hbox{if $l$ is odd;} \\
                                                  (l-2)\displaystyle\sum_{i=1}^{l-1}\sum_{t\in\mathbb{F}_q}S^i((t-1)(t^2+\lambda)) - 2\sum_{i=1}^{\frac{l}{2}-1}\sum_{t\in \mathbb{F}_q}S^{2i}((t-1)(t^2+\lambda)), & \hbox{if $l$ is even.}
                                                                               \end{array}
                                                                             \right.
\end{align}
From \eqref{eq13}, \eqref{eq5} and Theorem \ref{thm1}, we deduce that
\begin{align}\label{eq002}
&\sum_{i,j=1,i\neq j,i+j\neq l}^{l-1}\sum_{t\in\mathbb{F}_q}S^{i+j}((t-1)(t^2+\lambda))\notag\\&=\left\{
                                                                                \begin{array}{ll}
                                                                                -(l-3)a_q(C_{l,\lambda}), & \hbox{if $l$ is odd;}\\
                                                                                -(l-2)a_q(C_{l,\lambda})-2q\cdot \displaystyle\sum_{i=1}^{\frac{l}{2}-1}\frac{J(\phi,S^{2i})}{J(S^{-i},S^{3i}\phi)}\cdot{_{2}}F_1\left(\begin{array}{cc}
                S^{-3i}\phi, & S^{-3i}\\
                   & S^{-4i}
              \end{array}\mid 1+\lambda \right), & \hbox{if $l$ is even.}
                                                                                \end{array}
                                                                                \right.
\end{align}
Using \eqref{eq001} and \eqref{eq002} in \eqref{eq7}, we complete the proof.
\end{pf}
\begin{remark}
Putting $l=2$ in Theorem \ref{thm4}, we obtain
\begin{align}\label{eq33}
a_q(C_{2,\lambda})^2=q^2\phi(-\lambda)\cdot{_{3}}F_2\left(\begin{array}{cccc}
                \phi, & \phi, & \phi\\
                   & \varepsilon, & \varepsilon
              \end{array}\mid \frac{1+\lambda}{\lambda} \right)+q,
\end{align}
which yields \eqref{eq122} over $\mathbb{F}_q$.
\end{remark}
\begin{pf}{\bf \ref{thm22}.}
Following the proof of Theorem \ref{thm4}, we obtain
\begin{align}\label{eq8}
\sum_{t\in \mathbb{F}_q}S^i((t-1)(t^2+\lambda))&=S^{i}(-\lambda)g\left(S^i, S^i; -\frac{1}{\lambda}\right)
\end{align}
and
\begin{align}\label{eq9}
\#\{(x, y)\in \mathbb{F}_q^2: y^l=(x-1)(x^2+\lambda)\}&=q+\sum_{t\in\mathbb{F}_q}\sum_{i=1}^{l-1}S^i((t-1)(t^2+\lambda)).
\end{align}
Since $S^i\neq \varepsilon$, we have $S^{-2i}\neq S^{-3i}$. Putting $A=S^{-3i},~C=S^{-2i}$
and $x=\frac{1+\lambda}{\lambda}$ in Theorem \ref{thm1}, we deduce that
\begin{align}\label{eq116}
g(S^i, S^i; -\frac{1}{\lambda})=q S^i(-\frac{8}{\lambda})F^*(S^{-3i}, S^{-2i}; 1+\lambda).
\end{align}
As $\frac{q-1}{l}$ is even, $S^i$ is a square. Also, $3\nmid l$ implies that $S^i\neq\varepsilon$.
So applying Theorem \ref{thm3}, we obtain
\begin{align}\label{eq10}
g(S^i, S^i; -\frac{1}{\lambda})=\frac{q S^i(-\frac{1}{\lambda})
J(\phi, S^{i})}{J(\sqrt{S^{-i}}, \sqrt{S^{3i}}\phi)}\cdot{_{2}}F_1\left(\begin{array}{cccc}
                \sqrt{S^{-3i}}\phi, & \sqrt{S^{-3i}}\\
                 & S^{-2i}
              \end{array}\mid 1+\lambda \right).
\end{align}
From \eqref{eq8}, \eqref{eq9}, and \eqref{eq10}, we have
\begin{align}
\#\{(x, y)\in \mathbb{F}_q^2&: y^l=(x-1)(x^2+\lambda)\}\notag\\&=q+q\cdot\sum_{i=1}^{l-1}
\frac{J(\phi, S^{i})}{J(\sqrt{S^{-i}}, \sqrt{S^{3i}}\phi)}\cdot{_{2}}F_1\left(\begin{array}{cccc}
                \sqrt{S^{-3i}}\phi, & \sqrt{S^{-3i}}\\
                 & S^{-2i}
              \end{array}\mid 1+\lambda \right).
\end{align}
Since ord$_p(\lambda(\lambda+1))=0$, \eqref{eq11} completes the proof of \eqref{eq114}.

Again for $l=3$, using Proposition \ref{propo1} in \eqref{eq116} and then combining with \eqref{eq8} and \eqref{eq9}, we obtain
\begin{align}
\#\{(x, y)\in \mathbb{F}_q^2&: y^l=(x-1)(x^2+\lambda)\}=q+
                           q\cdot\displaystyle\sum_{i=1}^{2}{_{2}}F_1\left(\begin{array}{cccc}
                \phi, & \varepsilon\\
                 & S^i
              \end{array}\mid 1+\lambda \right)\notag
\end{align}
which yields the result because of \eqref{eq12}.
\end{pf}
\begin{corollary}
Let $p$ is an odd prime for which \emph{ord}$_p(\lambda(\lambda+1))=0$. If $p\equiv 1~($\emph{mod} $3)$ and $x^2+3y^2=p$, then
\begin{align}
a_p(C_{3,-\frac{1}{2}})=\phi(2)(-1)^{x+y-1}\left(\frac{x}{3}\right)\cdot 2x\notag
\end{align}
and
\begin{align}
p\cdot\displaystyle\sum_{i=1}^{2}{_{2}}F_1\left(\begin{array}{cccc}
                \phi, & \varepsilon\\
                 & \chi_3^i
              \end{array}\mid \frac{1}{2} \right)=\phi(2)(-1)^{x+y}\left(\frac{x}{3}\right)\cdot 2x-2,\notag
\end{align}
where $\chi_3$ is a character on $\mathbb{F}_p$ of order $3$.
\end{corollary}
\begin{proof}
As mentioned in Remark \ref{remark1}, $C_{3,-\frac{1}{2}}$ is isomorphic over $\mathbb{Q}$ to the elliptic curve $y^2=x^3+2^3$, which is $2$-quadratic twist of $y^2=x^3+1$. It is known that if $E(d)$ is the $d$-quadratic twist of the elliptic curve $E$ and gcd$(p,d)=1$, then $$a_p(E)=\phi(d)a_p(E(d)).$$ Since gcd$(p,2)=1$, hence by \cite[Proposition 2]{ono}, we have $$a_p(C_{3,-\frac{1}{2}})=\phi(2)(-1)^{x+y-1}\left(\frac{x}{3}\right)\cdot 2x.$$
\par Again combining this result with the equation \eqref{eq115}, we complete the second part of the corollary.
\end{proof}
\begin{corollary}
Let $p$ be an odd prime for which \emph{ord}$_p(\lambda(\lambda+1))=0$. If $q = p^e \equiv 1~($\emph{mod} $4)$, then
\begin{align}
{_{3}}F_2\left(\begin{array}{cccc}
                \phi, & \phi, & \phi\\
                   & \varepsilon, & \varepsilon
              \end{array}\mid \frac{1+\lambda}{\lambda} \right)=\phi(\lambda){_{2}}F_1\left(\begin{array}{cccc}
                \overline{\chi_4}, & \chi_4\\
                 & \varepsilon
              \end{array}\mid 1+\lambda \right)^2-\frac{\phi(\lambda)}{q},\notag
\end{align}
where $\chi_4$ is a character of order $4$ on $\mathbb{F}_q$.
\end{corollary}
\begin{proof}
Putting $l=2$ in Theorem \ref{thm22} and then squaring both sides, we have
\begin{align}
a_q(C_{2,\lambda})^2=q^2\cdot\frac{J(\phi,\phi)^2}{J(\chi_4,\overline{\chi_4})^2}\cdot{_{2}}F_1\left(\begin{array}{cccc}
                \overline{\chi_4}, & \chi_4\\
                 & \varepsilon
              \end{array}\mid 1+\lambda \right)^2.\notag
\end{align}
Using \eqref{eq01} and then comparing with \eqref{eq33}, we complete the proof.
\end{proof}
\begin{pf}{\bf \ref{thm6}.}
Putting $\lambda=\frac{1}{3}$ in \eqref{eq1} and making the change of variables
$(x, y)\rightarrow (\frac{x}{9}+\frac{1}{3}, y)$, and then replacing $-\frac{x}{6}$ by $x$ we obtain the equivalent equation as
\begin{eqnarray*}
y^l=-\frac{8}{27}(1+x^3).
\end{eqnarray*}
Therefore,
\begin{align}
\#\{(x, y)\in \mathbb{F}_q^2: y^l=(x-1)(x^2+\frac{1}{3})\}
&=\#\{(x, y)\in \mathbb{F}_q^2: y^l=-\frac{8}{27}(1+x^3)\}\notag\\
&=q+\sum_{i=1}^{l-1}\sum_{x\in\mathbb{F}_q}S^i(-\frac{8}{27})S^i(1+x^3).\notag
\end{align}
Now recall that the binomial theorem (see \cite{greene}) for a character $A$ on $\mathbb{F}_q$
is given by
\begin{align}\label{eq121}
A(1+x)=\delta(x)+\frac{q}{q-1}\sum_\chi{A \choose \chi}\chi(x),
\end{align}
where $\delta(x)=1~($resp. $0)$
if $x=0~($resp. $x\neq 0).$ Hence
\begin{align}\label{eq120}
\#\{(x, y)\in \mathbb{F}_q^2: y^l=(x&-1)(x^2+\frac{1}{3})\}\notag\\
&=q+\sum_{i=1}^{l-1}\sum_{x\in\mathbb{F}_q}S^i(-\frac{8}{27})
\left[\delta(x^3)+\frac{q}{q-1}\sum_\chi{S^i \choose \chi}\chi(x^3)\right]\notag\\
&=q+\sum_{i=1}^{l-1}S^i(-\frac{8}{27})+\frac{q}{q-1}\sum_{i=0}^{l-1}S^i(-\frac{8}{27})
\sum_\chi{S^i \choose \chi}\sum_{x\in\mathbb{F}_q}\chi^3(x).
\end{align}
By \eqref{lemma11}, $\sum_{x\in\mathbb{F}_q}\chi^3(x)$ is nonzero if $\chi^3=\varepsilon$,
which is possible only for $\varepsilon$, $\chi_3$ and $\chi_3^2$. Therefore, \eqref{eq120} reduces to
\begin{align}
\#\{(x, y)\in \mathbb{F}_q^2: y^l=(x&-1)(x^2+\frac{1}{3})\}\notag\\
&=\left\{
\begin{array}{ll}
q, & \hbox{if $q\equiv 2$ $($mod $3)$;} \\
q+q\cdot\displaystyle\sum_{i=1}^{l-1}S^i(\frac{27}{8})
\left[{\chi_3 \choose S^i}+{\chi_3^2 \choose S^i}\right], & \hbox{if $q\equiv 1$ $($mod $3)$,}
\end{array}
\right.\notag
\end{align}
which completes the proof of the result because of \eqref{eq11} and \eqref{eq12}.
\end{pf}
We now give an alternate proof of a result of McCarthy.
\begin{pf}{\bf \ref{thm44}.}
Since $q\equiv 1~($mod $3)$, putting $l=2$ in Theorem \ref{thm6} we find that
\begin{align}
-a_q(C_{2,\frac{1}{3}})=q\phi(6)\left[{\chi_3 \choose \phi}+{\chi_3^2 \choose \phi}\right].\notag
\end{align}
Since $\phi(-3)=1$ if and only if $q\equiv 1~($mod $3)$ and
$\overline{{\chi_3 \choose \phi}}={\chi_3^2 \choose \phi}$, the first part follows.

Again the second part follows from the fact that if $\chi\psi$ is nontrivial, then
\begin{align}
J(\chi, \psi)=\frac{G(\chi)G(\psi)}{G(\chi\psi)},\notag
\end{align}
where $J(\chi, \psi)$ and $G(\chi)$ are Jacobi and Gauss sums respectively.
\end{pf}
Simplifying the expression for $a_q(C_{l,\lambda})$ given in Theorem \ref{thm6},
we obtain the following result which generalizes the case $l=2$ treated in Theorem \ref{thm44}.
\begin{corollary}
Let $d=$ \emph{lcm}$(3, l)$. If $q\equiv 1$ $($\emph{mod} $d)$, then
\begin{align}\label{coro1}
-a_q(C_{l, \frac{1}{3}})=
\left\{
\begin{array}{lll}
2+2q\cdot\emph{Re}\displaystyle\left[{\chi_3 \choose \chi_3}+{\chi_3^2 \choose \chi_3}\right],
\hspace{3.27cm} \hbox{if $l=3$;}\\
2q\cdot \displaystyle\sum_{i=1}^{\frac{l-1}{2}}\emph{Re}
\left[S^i\left(\frac{27}{8}\right)\left\{{\chi_3 \choose S^i}+{\chi_3^2 \choose S^i}\right\}\right],
\hspace{1cm}\hbox{if $l$ is odd, $l>3$;} \\
2q\cdot \displaystyle\left[\phi(-2)\emph{Re}{\chi_3 \choose \phi}
+\displaystyle\sum_{i=1}^{\frac{l-2}{2}}\emph{Re}
\left\{S^i\left(\frac{27}{8}\right)
\left({\chi_3 \choose S^i}+{\chi_3^2 \choose S^i}\right)\right\}\right],\\ \hspace{8.5cm}  \hbox{if $l$ is even;}
                           \end{array}
                         \right.\notag
\end{align}
where $S$ and $\chi_3$ are characters on $\mathbb{F}_q$ of order $l$ and $3$ respectively.
\end{corollary}

\section{Values of ${_{3}}F_2$ and ${_{2}}F_1$ Gaussian hypergeometric series}
In this section, we will give the proof of Theorem \ref{thm5} and Theorem \ref{thm8}. We now prove the following lemmas from which Theorem \ref{thm5} and Theorem \ref{thm8} follow directly.
\begin{lemma}\label{lemma2}
If $S$ is a character on $\mathbb{F}_q$ whose order is not equal to $1$, $3$ or $4$, then
\begin{align}
{_{3}}F_2\left(\begin{array}{cccc}
                S^{-3}, & S^{-1}, & S^{-2}\phi\\
                   & S^{-4}, & S^{-2}
              \end{array}\mid \frac{1+\lambda}{\lambda} \right)&
              =\frac{J(S^{-1}, S^{-1})}{q^2 S(-4\lambda^3) J(S^{-3}, S)}
              \left[\sum_{x\in\mathbb{F}_q}S((x-1)(x^2+\lambda))\right]^2\notag\\
              &-\frac{S^2(\frac{1+\lambda}{\lambda})}{q}\phi(-\lambda).\notag
\end{align}
\end{lemma}
\begin{proof}
Since $S$ is a character on $\mathbb{F}_q$ whose order is not equal to $1$, $3$ or $4$, so applying
Theorem \ref{thm2} directly for $A=S^{-3}$, $C=S^{-2}$ and $x=\frac{1+\lambda}{\lambda}$, we get
\begin{align}
{_{3}}F_2\left(\begin{array}{cccc}
                S^{-3}, & S^{-1}, & S^{-2}\phi\\
                   & S^{-4}, & S^{-2}
              \end{array}\mid \frac{1+\lambda}{\lambda} \right)
              &=\frac{J(S^{-1}, S^{-1})}{q^2 S(-4\lambda) J(S^{-3}, S)}g(S, S;-\frac{1}{\lambda})^2\notag\\
              &-\frac{S^2(\frac{1+\lambda}{\lambda})}{q}\phi(-\lambda),\notag
\end{align}
which yields the result because of
\begin{align}
g(S, S; -\frac{1}{\lambda})=\sum_{x\in \mathbb{F}_q}S^{-1}(-\lambda)S((x-1)(x^2+\lambda)).\notag
\end{align}
\end{proof}
\begin{lemma}\label{lemma3}
Let $S$ be any character on $\mathbb{F}_q$. For $\lambda=\frac{1}{3}$, we have
\begin{align}
\sum_{x\in\mathbb{F}_q}S((x-1)(x^2+\lambda))
=\left\{
\begin{array}{ll}
0, & \hbox{if $q\equiv 2$ $($\emph{mod} $3)$;} \\
qS(-\frac{8}{27})\displaystyle\left[{S \choose \chi_3}+
{S \choose \chi_3^2}\right], & \hbox{if $q\equiv 1$ $($\emph{mod} $3)$,}
\end{array}
\right.\notag
\end{align}
where $\chi_3$ is a character of order $3$ on $\mathbb{F}_q$.
\end{lemma}
\begin{proof}
As shown in the proof of Theorem \ref{thm6}, $$y^l=(x-1)(x^2+\frac{1}{3})$$
is equivalent to $$y^l=-\frac{8}{27}(1+x^3).$$
Hence using \eqref{eq121}, we have
\begin{align}
\sum_{x\in\mathbb{F}_q}S((x-1)(x^2+\frac{1}{3}))&=\sum_{x\in\mathbb{F}_q}S(-\frac{8}{27})S(1+x^3)\notag\\
&=S(-\frac{8}{27}) + \frac{q}{q-1}S(-\frac{8}{27})\sum_{x\in\mathbb{F}_q}\sum_{\chi}{S \choose \chi}\chi^3(x).\notag
\end{align}
Following the proof of Theorem \ref{thm6}, the result follows easily.
\end{proof}
\begin{lemma}\label{lemma4}
If $S$ is square of some character on $\mathbb{F}_q$ and $S$ is not of order $3$, then
\begin{align}
\sum_{x\in\mathbb{F}_q}S((x-1)(x^2+\lambda))
=\frac{qJ(\phi, S)}{J(\sqrt{S^{-1}}, \sqrt{S^3}\phi)}\cdot {_{2}}F_1\left(\begin{array}{cccc}
                \sqrt{S^{-3}}\phi, & \sqrt{S^{-3}}\\
                 & S^{-2}
              \end{array}\mid 1+\lambda \right)\notag
\end{align}
\end{lemma}
\begin{proof}
We have
\begin{align}
\sum_{x\in \mathbb{F}_q}S((x-1)(x^2+\lambda))=S(-\lambda)g(S, S; -\frac{1}{\lambda}).\notag
\end{align}
Since $S$ is a square of some character of $\mathbb{F}_q$, applying Theorem \ref{thm1}, we obtain
\begin{equation}\label{eq118}
\sum_{x\in \mathbb{F}_q}S((x-1)(x^2+\lambda))=qS^3(2)F^*(S^{-3}, S^{-2}; 1+\lambda).
\end{equation}
Also $S$ is not of order $3$. Using Theorem \ref{thm3} we complete the proof.
\end{proof}
Following the proof of Lemma \ref{lemma4} and applying Proposition \ref{propo1} in \eqref{eq118},
we have the following result.
\begin{lemma}
If $S$ is a character of order $3$ on $\mathbb{F}_q$, then
\begin{align}
\sum_{x\in\mathbb{F}_q}S((x-1)(x^2+\lambda))=q\cdot{_{2}}F_1\left(\begin{array}{cccc}
                \phi, & \varepsilon\\
                 & S
              \end{array}\mid x \right).
\end{align}
\end{lemma}
\begin{pf}{\bf \ref{thm5}.}
Putting $\lambda=\frac{1}{3}$ in Lemma \ref{lemma2} and then combining it with Lemma \ref{lemma3} we complete the proof.
\end{pf}
\begin{pf}{\bf \ref{thm8}.}
(i) Putting $\lambda=\frac{1}{3}$ in Lemma \ref{lemma4} and then using Lemma \ref{lemma3}, we complete the proof.\\

(ii) Taking $x=\frac{4}{3}$ in Theorem \ref{thm7} (i), we obtain
\begin{align}
{_{2}}F_1\left(\begin{array}{cccc}
                \sqrt{S^{-3}}\phi, & \sqrt{S^{-3}}\\
                 & S^{-1}\phi
              \end{array}\mid -\frac{1}{3} \right)=\sqrt{S}\phi(-1){_{2}}F_1\left(\begin{array}{cccc}
                \sqrt{S^{-3}}\phi, & \sqrt{S^{-3}}\\
                 & S^{-2}
              \end{array}\mid \frac{4}{3} \right).\notag
\end{align}
Now using (i), we complete the proof.\\

(iii) Applying Theorem \ref{thm7} (ii) for $x=\frac{4}{3}$, we have
\begin{align}
{_{2}}F_1\left(\begin{array}{cccc}
                \sqrt{S^{-3}}\phi, & \sqrt{S^{-1}}\\
                 & S^{-2}
              \end{array}\mid 4 \right)=\sqrt{S^3}\phi(-3){_{2}}F_1\left(\begin{array}{cccc}
                \sqrt{S^{-3}}\phi, & \sqrt{S^{-3}}\\
                 & S^{-2}
              \end{array}\mid \frac{4}{3} \right)\notag
\end{align}
and the result follows from (i).\\

(iv) Using Theorem \ref{thm7} (ii) for $x=-\frac{1}{3}$, we find that
\begin{align}
{_{2}}F_1\left(\begin{array}{cccc}
                \sqrt{S^{-3}}\phi, & \sqrt{S}\phi\\
                 & S^{-1}\phi
              \end{array}\mid \frac{1}{4} \right)=\phi(-1)\sqrt{S^3}\phi(\frac{3}{4}){_{2}}F_1\left(\begin{array}{cccc}
                \sqrt{S^{-3}}\phi, & \sqrt{S^{-3}}\\
                 & S^{-1}\phi
              \end{array}\mid -\frac{1}{3} \right)\notag
\end{align}
and then proof follows from the proof of (ii).
\end{pf}

\section*{Acknowledgment}
We thank Ken Ono for many helpful suggestions during the preparation of the article.
The second author acknowledges the financial support of Department of Science and Technology,
Government of India for supporting a part of this work under INSPIRE Fellowship.

redpajama_set_name: RedPajamaArXiv


text:
30 years of Unreliable Evidence
Featured by Legal Reporter
Washington, D.C. (JusticeNewsFlash.com – News Report) – If the FBI presented evidence in a courtroom, would you be inclined to believe what they present as truth? Many would reply a strong 'yes' to this question, it's a reliable institution, they know what they're doing – don't they? According to a new study revealed by 60 minutes and the Washington Post, the FBI crime laboratory in Quantico, Virginia has been using completely faulty analysis.
Bullet lead analysis has been used by the FBI for longer than 40 years in thousands of cases, many of which are now being re-addressed to examine whether or not those convicted were unjustly imprisoned. According to a CBS article, 'the FBI believed that lead in bullets had unique chemical signatures, and that by breaking them down and analyzing them, it was possible to match bullets, not only to a single batch of ammunition coming out of a factory, but to a single box of bullets. And that is what the FBI did in the case of Lee Wayne Hunt, tying a bullet fragment found where the murders took place to a box of bullets the prosecutors linked to Hunt.' Lee Wayne Hunt has been in jail for 22 years and 6 months for a double murder crime he claims he did not commit. But isn't that what everyone says in jail?
But they could be right according to a study conducted by William Tobin, a former chief metallurgist for the FBI. He reported that they bullet lead analysis has 'never actually been scientifically tested.' How can a tool that is one of the major deciding factors on whether or not a person is convicted of a crime not have been tested? It seems too absurd to be true, a case of too much blind faith. Tobin's study examined different bullets and discovered that bullets from the same group weren't always the exactly the same. What is more wrong in this story is that the FBI had this knowledge, yet refrained from ever reporting it to the convicted, lawyers or judges involved.
So what is being done? One organization called the 'Innocence Project" has helped free more than 200 wrongly convicted defendants. Since this technique has been used in the early 1980's, there were around 2,500 cases that the FBI analyzed. This is an alarming amount of individuals who could be serving time in the midst of their innocence, back to Hunt, for example. One man knows the truth. Staples Hughes, the appellate defender for the state of North Carolina revealed his 20-year secret. Jerry Cashwell, a client of Hughes, told him how he alone had committed the double murder. He admitted that Lee Wayne Hunt wasn't even there. Hughes had to hold this knowledge because of the attorney-client privilege, but when he committed suicide, Hughes testified that Hunt was innocent. But the North Carolina Supreme Court will not hear his appeal. The injustice of it all lies in the hands of the FBI to retract their bullet lead testimonies and make right what is wrong. Freeing Hunt would be a good start.
Written by: Jana Simard – Legal News Reporter
Cepac – Lawyer Marketing launches Justice News Flash – Press Release Distribution Service for the legal community.
Featured Mesothelioma Asbestos
Mesothelioma research revealed- Men are at greater risk, Boston Massachusetts


text:
Q: "HTTP 403 forbidden" with Jersey 2.2 in GlassFish 4.1.2 I have an issue with GlassFish using IntellJ IDEA. Using in that environment Jersey library and restful architecture, when I try to verify CRUD's operation, I don't find issues with GET and POST. Instead with PUT and DELETE, server answer with error 403. What can I do? Is there a way to solve issue?

This is my REST class:
//restituisce la lista di utenti
@GET
@Path("/getlistanodi")
@Produces({"application/json", "application/xml"})
public Response getListaNodi() {

    return Response.ok(NodiEdge.getInstance()).build();
}

@Path("/add")
@POST
@Consumes({"application/json", "application/xml"})
public Response addNodo(Nodo u) {

    NodiEdge.getInstance().addNodo(u);
    return Response.ok().build();
}

@Path("/remove/{idnodo}")
@DELETE
@Produces({"application/json", "application/xml"})
public Response removeNodo(@PathParam("idnodo") String id) {

    NodiEdge.getInstance().removeNodo(id);
    return Response.ok().build();
}

This is file web.xml:
<filter>
    <filter-name>UploadFilter</filter-name>
    <filter-class>com.sun.webui.jsf.util.UploadFilter</filter-class>
    <init-param>
        <param-name>maxSize</param-name>
        <param-value>-1</param-value>
    </init-param>
</filter>
<filter-mapping>
    <filter-name>UploadFilter</filter-name>
    <servlet-name>FacesServlet</servlet-name>
</filter-mapping>
<servlet>
    <servlet-name>FacesServlet</servlet-name>
    <servlet-class>javax.faces.webapp.FacesServlet</servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>
<servlet>
    <servlet-name>ThemeServlet</servlet-name>
    <servlet-class>com.sun.webui.theme.ThemeServlet</servlet-class>
    <load-on-startup>2</load-on-startup>
</servlet>
<servlet>
<servlet-name>DownloadServlet</servlet-name>
<servlet-class>org.glassfish.admingui.common.servlet.DownloadServlet</servlet-class>
<init-param>
    <param-name>ContentSources</param-name>
    <param-value>
            org.glassfish.admingui.common.servlet.LBConfigContentSource,
            org.glassfish.admingui.common.servlet.ClientStubsContentSource,
            org.glassfish.admingui.common.servlet.LogFilesContentSource
            org.glassfish.admingui.common.servlet.LogViewerContentSource
    </param-value>
</init-param>
<init-param>
    <param-name>contentSourceId</param-name>
    <param-value>LBConfig</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>DownloadServlet</servlet-name>
<url-pattern>/download/*</url-pattern>
</servlet-mapping>
<servlet-mapping>
    <servlet-name>FacesServlet</servlet-name>
    <url-pattern>/resource/*</url-pattern>
</servlet-mapping>
<servlet-mapping>
    <servlet-name>FacesServlet</servlet-name>
    <url-pattern>/html/*</url-pattern>
</servlet-mapping>
<servlet-mapping>
    <servlet-name>FacesServlet</servlet-name>
    <url-pattern>/faces/*</url-pattern>
</servlet-mapping>
<servlet-mapping>
    <servlet-name>FacesServlet</servlet-name>
    <url-pattern>*.jsf</url-pattern>
</servlet-mapping>
<servlet-mapping>
    <servlet-name>ThemeServlet</servlet-name>
    <url-pattern>/theme/*</url-pattern>
</servlet-mapping>
<session-config>
    <cookie-config>
        <http-only>true</http-only>
    </cookie-config>
</session-config>
<!-- following mapping added to avoid JSF warning. refer to GLASSFISH-19403 -->
<mime-mapping>
    <extension>jsp</extension>
    <mime-type>text/html</mime-type>
</mime-mapping>

<welcome-file-list>
    <welcome-file>/index.jsf</welcome-file>
</welcome-file-list>
<error-page>
    <exception-type>javax.faces.application.ViewExpiredException</exception-type>
    <location>/</location>
</error-page>

<login-config>
    <auth-method>FORM</auth-method>
    <realm-name>admin-realm</realm-name>
    <form-login-config>
        <form-login-page>/login.jsf</form-login-page>
        <form-error-page>/loginError.jsf</form-error-page>
    </form-login-config>
</login-config>
<security-role>
    <role-name>admin</role-name>
</security-role>

And I add Methods'class:
@XmlElement(name="nodi")
private List<Nodo> listaNodi;
private static NodiEdge instance;

private int distNodi;

public NodiEdge() {

    listaNodi = new ArrayList<Nodo>();
}

//singleton
public synchronized static NodiEdge getInstance(){

    if(instance==null)
        instance = new NodiEdge();
    return instance;
}

public synchronized List<Nodo> getListaNodi() {

    return new ArrayList<>(listaNodi);
}

public synchronized List<Nodo> addNodo(Nodo u)
{
    List<Nodo> copiaLista = getListaNodi();

    if (listaNodi.size()==0)
        listaNodi.add(u);
    else {
        for (Nodo n : copiaLista) {
            if (n.getID().toLowerCase().equals(u.getID().toLowerCase())) {
                System.err.println("Inserimento negato! Identificatore già esistente.");
                break;
            }
            else {
                distNodi = Math.abs((u.getX() - n.getX()) + (u.getY() - n.getY()));
                if (distNodi < 20) {
                    System.err.println("Inserimento negato! Posizione troppo vicina ad altri nodi.");
                    break;
                } else {
                    listaNodi.add(u);
                    System.out.println("ID nodo: " + n.getID() + ", indirizzo IP: " + n.getIPaddress() +
                            ", n° porta: " + n.getPNd());
                }
            }

        }

    }

    return copiaLista;

}

public synchronized void removeNodo(String id)
{
    List<Nodo> copiaLista = getListaNodi();

    if (listaNodi.size()==0)
        System.err.println("Operazione impossibile! Nessun nodo nella griglia");
    else {
        for (Nodo n : copiaLista) {
            if (!n.getID().toLowerCase().equals(id.toLowerCase()))
                System.err.println("Nodo non presente nella griglia.");
            else {
                listaNodi.remove(n);
                System.out.println("Nodo eliminato.");
                break;
            }

        }

    }

redpajama_set_name: RedPajamaStackExchange


An awareness pack for delivery to staff and learners on appropriate health and social care courses (teaching and training delivery). It is also suitable for training delivery in care settings (including domiciliary). You can adapt it if you wish to focus on generic risk enablement (simply delete references to dementia).
Download only (not available on CD).
Risk assessment tools, Essential training, Mental Capacity Act, 5 core principles (MCA), Deprivation of liberty safeguards, Understanding dementia, What is risk?


Have you ever been in a situation where you thought things cannot get worse, and then they do? We see this happening in the story of Jesus' crucifixion. He had already been hung on the cross. Dying, he said he was thirsty; a sponge soaked in vinegar was held to his mouth. "It is finished," he said and died. Could part of what he was saying have been, "Enough!"?
Yet enough was not enough. Already he had endured the trial before Pilate, the demonstrations and insults of the crowds, the way to the cross, the gruesome hanging with criminals on either side. Already dead on the cross, the soldiers stripped Jesus even of his clothes and divided them among themselves. Was that enough? They were about to break the bones of his legs to make sure he would die, only to discover that he was already dead. Was that enough? No! One of them then pierced his side with a spear to spill his blood.
Those who were witnessing the gruesome tragedy knew Jesus had died. It is finished, they thought. For them, that was the end of the story. Sad and broken, they accepted reality and made preparations for his burial.
So it is with any life situation when there seems to be absolutely no hope. But John 20:1-18, teaches us this essential Christian lesson: The status quo is NOT the end of the story. The Disciples "had not yet known" the promise of the Scriptures (v. 9). Even Mary, looking at the Risen Jesus, "did not know that he was Jesus" (v. 14.).
Believe the good news of the gospel: CHRIST IS RISEN! We too shall rise with him! Alleluia! Let us, therefore, invest our "alleluias" in living and serving as the people of the Resurrection.
Sara Makari serves with the Evangelical Lutheran Church in America with the Diyar Consortium of the Evangelical Lutheran Church in Jordan and the Holy Land. Her appointment is made possible by your gifts to Disciples Mission Fund, Our Church's Wider Mission, and your special gifts.