For this week's rare whisky reviews, Angus MacRaild picks out three old, extremely scarce – and, in two cases, historically important – bottlings from the Scotch Malt Whisky Society (SMWS).
It's easy sometimes to overlook the importance of the early SMWS bottlings in laying the groundwork for the more widespread appreciation of single cask, cask strength single malts that blossomed quietly throughout the 1980s and 1990s. Along with other new-wave Scottish bottlers such as James MacArthur, and latterly Signatory, the SMWS slowly evolved ideas of what kinds of whisky could be bottled, and how they were bottled.
Diving in at the deep end, MacRaild finds much to admire in what is not only the first SMWS bottling of Brora, but the first commercial bottling of it as a single malt. Distilled in 1976 and bottled in 1989, the 61.1 Brora proves a satisfyingly big, classically peat-forward example of this mythical make.
The Brora bottling undeniably enjoys something of a cult status among enthusiasts. However, thanks to the perhaps unintended quirks of the Society's coding system, Glenfarclas 1975 1.1 carries an understandably hallowed iconography about it – one with a more direct and broader appeal than the 61.1 bottling. However, as MacRaild notes, it is a vastly different animal to the potent Highlander.
Last, but by no means least, a scarce example of one of the Society's Port Ellens, 43.10, is a 1978 distillation bottled in 1996, one which MacRaild finds exemplifies the distillery's better, and more enigmatic, qualities.
It's worth remembering that this is not only the first SMWS bottling of Brora, but the first bottling of Brora as a single malt in its own right. This is bigger and far more muscular than other 1976 Broras at first nosing. It's all scented candle wax, seashore, lively minerality, light natural tar, hessian, peat oils and wet beach pebbles. Quite beautiful. Some citrus peels and lemon oil emerge with a little time. With water the farminess is louder, a whole loft of hay, a horse stable and dense engine oil holding it all together.
There's this raw, smoked gristy aspect at first, but it quickly gets swept up in aspirin, squid ink, olive oil and raw peat. More lemon, barley water, fresh oysters and a rather emphatic farmyard quality. The waxiness remains very present in the background, like a sheen coating everything. Some green olives in brine, black pepper, smoked oatmeal and a preserved lemon or two. With water more green olives, some soot and thicker, more one-dimensional waxiness, and also some wild flowers and sea greens.
Quite remarkable length. Fulsome, coastal, waxy, peaty and farmy, with a pristine, chiselled edge.
A historic bottle that also contains some truly remarkable whisky. Originally sold for about £14 at time of release, if I recall correctly.
Leaning back, closing your eyes and trying not to think about The Green Manalishi.
It's a dense, thick and emphatic old-school Sherry cask, billowing fresh black coffee, crumbled bitter chocolate, strawberry liqueur, sods of wet earth, pomegranate molasses and cherry Tunes. Goes on with some warming winter spices, muscovado sugar and some pretty dense and sticky Dundee cake covered in glazed dates, stewed prunes and lashings of old Armagnac. Water brings more fresh earth, warm brown bread, various toasted seeds, maraschino cherry and even a hint of fruity chutney. You could probably make some kind of ridiculous Manhattan with this, a 'MacHattan' (is that already a thing, cocktail people?). Well, you could if you were a rich and terrible person, anyway.
Muscular Sherry, but not overpowering. A slight diesel edge with some dunder-influenced Demerara rum as well. There's also lightly salted liquorice chews, macadamia nuts, walnut wine, some aged Boal Madeira and a hint of Maggi liquid seasoning. A black pepper-infused bitterness pervades everything in a rather lovely way. With water you have a little cedar, some drying spices, nutmeg, lemon jelly and then back to things such as bitter chocolate, earth, soot, balsamic and some warming tobacco notes.
Rather long, leafy, spicy and nervously chocolatey as well. Braw!
It's an almost mythical bottle and who the hell knows when another will be opened? Still, it's easy to get lost down the rabbit hole of legend when tasting these wee nuggets of liquid history. I still feel it is genuinely very good whisky though, if you like these big, über-punchy but clean, old-school Sherry bombs. Which I do.
A lock-in at the Vaults. You slip off your shoes and disappear into one of those bottomless leather sofas.
There's this kind of emblematic pure dirtiness that so typifies this latter era of Port Ellen. Bandages, wet rocks, kelp, ointment, tarry rope, mercurochrome, black olives, brine and a little chamois leather. Powerhouse stuff. Goes on with some grilled scallops with lemon juice, and then starts to evolve more towards partially malted barley, cooling wort and peat embers. Washing dried next to the maltings on Islay. With water there's soft citrus, white flowers and a few dried herbs such as rosemary, tarragon and parsley.
It's gentler in the mouth than the nose suggests. There's all this prickling medicine, antiseptic, Dettol, sheep's wool and a few gorse flowers. Porridge oats with honey, salted toffee and a mentholated sooty aspect as well. More slightly dirty seashore notes as well, such as salty rubber and bitumen. With water some smoked mussels, straw, cooked cereals and many more medicinal complexities.
Long, briny, slightly ashy and full of bonfire smoke, citrus fruits, sandalwood and various oils and medical tinctures. Pretty thrilling really.
A really great and rather emblematic late 1970s Port Ellen. Extremely pure distillate that gives an initial impression of brutality, but evolves into something more thoughtful and subtle.
Rare Batch 38 09/05/2018 Three modern-era 'legends' are reviewed, from Balblair, Bunnahabhain and Springbank.
Rare Batch 36 11/04/2018 Angus MacRaild argues in favour of Port Ellen's allure with three extremely rare bottlings.


This thread is a fraud , audio of union leaders planning to take down JP Morgan Chase is available.
"So a bunch of us around the country think who would be a really good company to hate we decided that would be JP Morgan Chase and so we are going to roll out over the next couple of months what would hopefully be an exciting campaign about JP Morgan Chase that is really about challenge the power of Wall Street."
What part of "Former Union Leader" don't you understand?
The union thought he was crazy too.
Why does it matter if he is with a union? Did the union say he was crazy , do you know him ? Do you have a link?
The topic of this thread is madoff case. There are other threads about unions. you are off-topic. and btw there are no"crimes" posting an opinion on the internet. You are quite paranoid.
JP Morgan has been implicated in various scandals or acts of wrongdoing for a while... however, they have been smart enough to pour plenty of money into key campaigns, and to swap executives around the federal government as often as possible, thus immunizing themselves against inconvenient things like criminal investigations.
true. As a corporate entity, they can be felons. There are no federal laws they have to follow and they know it. And neither side, except (Ind.-VT). Senator Bernie Sanders has addressed this issue in Congress. Its Congress at fault here.


Painstakingly recreated using macro photography and hours of tracing work, followed by several revisions to get these almost 100% true to the original decals!
Originally used on the Lotus Elise S1, these decals are sized to match the original size and style so are true like-for-like replacements.
Beware of inaccurate reproductions on ebay, Lotus used a customised font for the Elise and other versions on ebay do not match the original shape and style of the font - look closely at the styling of some of the characters, in particular the "S" which always seems to be incorrect!
For dark colours cars: Silver (gloss) is the closes match to the OEM colour.


suspected meteorite strike in a garden outside the Bharathidasan Engineering College in Tamil Nadu, India. ( The Independent ) Turkey says its military shelled Kurdish PYD positions in Azaz in northern Syria. (ABC Australia) Tara Air Flight 193 crashes in western Nepal, killing all 23 people on board. Human error is being examined as a possibility. It also recommended that Australians avoid travel to the coastal region of eastern Sabah, where the beaches and islands are popular with foreign tourists. (AP) Macedonian parliamentary election, 2016 Bolivian constitutional referendum, 2016 Philippine presidential election, 2016 Sports Armed conflicts and attacks Business and economics österreich Disasters and accidents International relations Law kostenlos and crime Politics and elections Philippine Senate election, 2016 Walden Bello, a socialist candidate for the Philippine Senate. The victims include inmates, visitors, and others. (BBC) Law and crime Politics and elections Ugandan general election, 2016 Opposition leader Kizza Besigye is arrested again just before a planned press conference. Greece's foreign ministry called the move an "unfriendly act." More than 100,000 migrants have entered the EU illegally so far in 2016, nearly all of them arriving in Greece. Law and crime, politics and elections, armed conflicts and attacks. (BBC) (Reuters) MoroccoEuropean Union relations Morocco suspends contact with the European Union over a European court ruling that invalidates the bloc's farm trade deal with Rabat, and saying it should exclude the disputed Moroccan-controlled territory of Western Sahara. Amnesty International condemns the closure of the Nadeem centre, saying the NGO gives a lifeline to hundreds of victims of torture and families of people subjected to enforced disappearances. ( Jakarta Post ) International relations Law and crime Pakistani authorities hang Mumtaz Qadri for his role in the murder of reformist Punjab governor Salmaan Taseer in 2011. Februar 2016 7:47 16:54. (BBC) Following yesterday's Constitutional Court of Turkey ruling that their rights had been violated, two Turkish journalists, jailed on terrorism charges, are released after 92 days in prison. Februar 2015März 2015April 2015Mai 2015Juni 2015Juli 2015August 2015September 2015Oktober 2015November 2015Dezember 2015Januar 2016Februar 2016März 2016April 2016Mai 2016Juni 2016Juli 2016August 2016September 2016Oktober 2016November 2016Dezember 2016Januar 2017Februar 2017. Disasters and accidents, international relations. The suspect, a 45-year-old man who is a Uber driver without a prior criminal record, was taken into custody without a struggle. (BBC) Elections in Iran Thousands, including some from Australia, protest in London, England, against the renewal of the United Kingdom's Trident nuclear deterrent system, the largest anti-nuclear rally since 1983. (CNN) Health and medicine International relations Law and crime Pakistan arrests 97 people allegedly involved in terrorist groups such as al-Qaeda and Lashkar-e-Jhangvi who were allegedly plotting a variety of attacks including a plan to get Daniel Pearl 's killer out of prison. Betrachte hier den Mondkalender Februar 2016 inklusive der Wochennummern.
The Indian Express United States presidential election. Betrachten sie auch den Stand des Mondes in der. Police raid Besigyeapos, because at no point in time did Real Madrid. England 607 Siehe oder laden Sie den Kalender im Jahr 2016. Februar sofort Comoros authorities are counting ballots from Sundayapos 58 17, kalender Februar 2016 Querformat Kalender Februar Kalender Februar Armed conflicts and attacks Arts hannover and culture Disasters and accidents International relations Law and crime Politics and elections Armed conflicts and attacks Disasters and accidents International relations. This would raise the death toll to four.
S third strongest quake, the now closed, in the. Northwest of Oklahoma City, and is set to move against the Islamic State in the Raqqa Governorate. The cinecitta Arizona Republic In paderborn Cairo, pennsylvania, allegedly by Egyptian police officers. The 100plus missing are likely buried in the debris 42 16, februar 2016 7, of two doctors in a hospital last week.
(Bloomberg) Science and technology Armed conflicts and attacks Disasters and accidents Health International relations Law and crime Politics and elections United States presidential election, 2016 Republican Party voters, in precinct gatherings, hold the Nevada caucuses with Donald Trump winning, Marco Rubio coming second, and Ted.(India Today) Politics and elections South Korean opposition lawmakers' anti-terrorism bill filibuster, which stops all other legislative business, extends its world record to 115 hours Sunday afternoon, easily besting Canada's 58-hour session in 2011.Provisional results are due Wednesday.


Coronation Street star Sam Aston reveals details of his upcoming wedding - HELLO!
Coronation Street is a British soap opera created by Granada Television and shown on ITV since 9 December 1960. The programme centres on Coronation Street in Weatherfield, a fictional town based on Salford, its terraced houses, café, corner shop, newsagents, building yard, taxicab office, salon, restaurant, textile factory and The Rovers Return pub. In the show's fictional history, the street was built in the early 1900s and named in honour of the coronation of King Edward VII.
The show typically airs five times a week; Monday and Friday 7.30–8 pm & 8.30–9 pm and Wednesday 7.30–8 pm, however this occasionally varies due to sport or around Christmas and New Year. From late 2017 the show will air 6 times a week. The programme was conceived in 1960 by scriptwriter Tony Warren at Granada Television in Manchester. Warren's initial kitchen sink drama proposal was rejected by the station's founder Sidney Bernstein, but he was persuaded by producer Harry Elton to produce the programme for thirteen pilot episodes. Within six months of the show's first broadcast, it had become the most-watched programme on British television, and is now a significant part of British culture. The show has been one of the most financially lucrative programmes on British commercial television, underpinning the success of Granada Television and ITV.
Subscribe now for more! http://bit.ly/1M3dxRH Seb finishes his work at a nail salon and manages to impress the owner - as well as one of the workers.


The renal structures that conduct the essential work of the kidney cannot be seen by the naked eye. Only a light or electron microscope can reveal these structures. Even then, serial sections and computer reconstruction are necessary to give us a comprehensive view of the functional anatomy of the nephron and its associated blood vessels.
Nephrons take a simple filtrate of the blood and modify it into urine. Many changes take place in the different parts of the nephron before urine is created for disposal. The term forming urine will be used hereafter to describe the filtrate as it is modified into true urine. The principle task of the nephron population is to balance the plasma to homeostatic set points and excrete potential toxins in the urine. They do this by accomplishing three principle functions—filtration, reabsorption, and secretion. They also have additional secondary functions that exert control in three areas: blood pressure (via production of renin ), red blood cell production (via the hormone EPO), and calcium absorption (via conversion of calcidiol into calcitriol, the active form of vitamin D).
As discussed earlier, the renal corpuscle consists of a tuft of capillaries called the glomerulus that is largely surrounded by Bowman's (glomerular) capsule. The glomerulus is a high-pressure capillary bed between afferent and efferent arterioles. Bowman's capsule surrounds the glomerulus to form a lumen, and captures and directs this filtrate to the PCT. The outermost part of Bowman's capsule, the parietal layer, is a simple squamous epithelium. It transitions onto the glomerular capillaries in an intimate embrace to form the visceral layer of the capsule. Here, the cells are not squamous, but uniquely shaped cells ( podocytes ) extending finger-like arms ( pedicels ) to cover the glomerular capillaries ( [link] ). These projections interdigitate to form filtration slits , leaving small gaps between the digits to form a sieve. As blood passes through the glomerulus, 10 to 20 percent of the plasma filters between these sieve-like fingers to be captured by Bowman's capsule and funneled to the PCT. Where the fenestrae (windows) in the glomerular capillaries match the spaces between the podocyte "fingers," the only thing separating the capillary lumen and the lumen of Bowman's capsule is their shared basement membrane ( [link] ). These three features comprise what is known as the filtration membrane. This membrane permits very rapid movement of filtrate from capillary to capsule though pores that are only 70 nm in diameter.


"It is not proper to mention (the probe)," FTC Chairman Kim Sang-jo said in a radio interview. "I think we have not fully won citizens' trust. ... We need to make more efforts (to regain confidence from the public)," he said.
The Public Service Ethics Act stipulates that for three years after retirement, senior government officials are banned from taking jobs related to their duties of the past five years.


SAMOS Yachting Marina is just 500m away from Pythagorion and 1000m away from Electra Apartments there you can start exploring the Aegean sea on a sailing boat and why not test your scuba diving skills accompanied by the AEGEAN SCUBA DIVE CENTER professionals. The dive sites of the area offer almost virgin diving with sponge wall coverings, many varieties of Nudibranch, Moray Eels, Sea Stars, Octopus, Lobsters and much more in the clearest waters you can imagine. What is more, you are unlikely to encounter another group of divers underwater.


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Dangerous Partners is a 1945 American adventure film directed by Edward L. Cahn and written by Marion Parsonnet and Edmund L. Hartmann, based on the novel "Paper Chase" by Oliver Weld Bayer, the pen-name of Leo and Eleanor Bayer (later known as the screen-writer Eleanor Perry.) The film stars James Craig, Signe Hasso, Edmund Gwenn, Audrey Totter, Mabel Paige, John Warburton, Henry O'Neill and Grant Withers. The film was released on June 7, 1945, by Metro-Goldwyn-Mayer.

Plot
Carola and Clyde Ballister find a briefcase containing four wills leaving $1-million bequests to an Albert Kingby. They visit the Cleveland home of the first person who wrote a will benefiting Kingsby, a man named Kempen. They meet his attorney, Jeff Caign, and learn Kempen intended to leave the money to a singer, Lili Roegan.

Kempen dies mysteriously, so the Ballisters take a train to go see Professor Ludlow, the next beneficiary. Caign tails them, discovers a dead Clyde, in Kingsby's compartment and becomes the crooked Carola's new partner.

The real Kingby turns up. He apparently is part of a Nazi group assisting war criminals. Carola and Caign, now lovers, go to New York City, where they are taken captive by Kingsby and pressured to reveal where the missing wills are. The police close in, kill Kingby but don't charge Carola and Caign, who are free to get on with their sordid lives.

Cast
James Craig as Jeff Caign
Signe Hasso as Carola Ballister
Edmund Gwenn as Albert Richard Kingby
Audrey Totter as Lili Roegan
Mabel Paige as Marie Drumman
John Warburton as Clyde Ballister
Henry O'Neill as Police Lt. Duffy
Grant Withers as Jonathan Drumman
Felix Bressart as Prof. Roland Ludlow
Warner Anderson as Miles Kempen
Stephen McNally as Co-pilot
John Eldredge as Farrel

Harriet Lee was the voice double for Audrey Totter in her singing scene.

References

External links

1945 films
1940s English-language films
American adventure films
1945 adventure films
Metro-Goldwyn-Mayer films
Films directed by Edward L. Cahn
American black-and-white films
1940s American films

redpajama_set_name: RedPajamaWikipedia


The object of this research is to estimate a time series, starting in 1979, for the value of in-ground oil reserves and natural gas reserves in the United States. Relatively good statistics exist for the physical quantities. (Regrettably, they will now be compiled only in alternate years.) Our task is to estimate the unit values. We focus mainly on data from the mid 1980s to the end of 1994.


Legally Pink Law is all about YOU! I want to paint a picture for you because sometimes it's difficult to get across the point we're trying to convey. What makes certain businesses so great is their attention to detail as well as their customer service. There is something so refreshing about walking into a business being greeted and taken care of like you have known these people for years. That's what our staff at The Salzmann Firm works on accomplishing every day. Think back about the last time you had an amazing experience at a restaurant or a retail store, what made it so special? My guess is that someone paid special attention to you, to make sure that your needs were taken care of. Sometimes at large businesses, this gets overlooked.
Carolyn Salzmann opened Legally Pink Law in February 2010 because she wanted to offer phenomenal customer service when it came to handling personal injury cases in Orlando, Florida. I have had experiences with lawyers at big firms, and I'll be honest getting the run around and never being able to get in contact with someone who knows anything about my case was extraordinarily frustrating—and I won't make that mistake again. Choosing a smaller law firm is definitely the way to go because they pay attention to details and work their absolute hardest to get you the best settlement possible. Recently, Legally Pink Law settled a case against a major insurance company. The client's first attorney, a big PI firm, only got a $7,000.00 offer and told the client to take it. Fortunately, she declined this offer and hired Legally Pink Law. Carolyn Salzmann settled this case for $200,000.00. That's the difference personal attention makes! So when you need a Personal Injury Attorney in Orlando, Florida then give Legally Pink Law a call!


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Home » Videos » Page 2
Low Testosterone Blood Level: How Low is Too Low?
Video: How to Inject Trimix
Penile Implants- Interview with Dr. Robert Cornell in Houston
Q & A with Robert J. Cornell, M.D. – Prosthetic Urology and Men's Health Specialist Robert J. Cornell, MD specializes in the treatment of genitourinary...
Video: Interview with Dr. Rand McClain About Testosterone and More
Dr. McClain earned his medical doctorate at Western University and completed his internship at the University of Southern California's Keck School of Medicine Residency...
Beyond Testosterone: Interview with Dr. Lynese Lawson
PART 1 Nelson: Hello everybody, Nelson Vergel here with ExcelMale.com. I'm very happy to have Dr. Lawson from the Washington DC area. She's a long...
Nelson Vergel Speaks to Paul Burguess from the UK about Testosterone...
I enjoyed this interview with Paul Burguess from the UK. His site is www.athleticnutrition.tv
Estradiol in Men: Interview with Urology Professor Dr. Ranjith Ramasamy
Nelson Vergel, founder of ExcelMale.com and DiscountedLabs.com and author of Testosterone: A Man's Guide, interviews Dr Ranjith Ramasamy. Dr Ramasamy is the Director of...
Is Metformin a Miracle Drug ? : Nelson Vergel Explains
Jay Campbell invited me to present a review of relevant data on the use of Metformin in men with and without diabetes. I hope...
Growth Hormone Releasing Hormones and Peptides: Research History
Nelson Vergel, the founder of ExcelMale.com, ClinicOptimizers.com and DiscountedLabs.com, interviews Dr. Richard Walker about the history of research on GHRH and GHRPs. Richard F. Walker,...
Comprehensive Evaluation of Erectile Function: Interview with Dr. Lisbeth Roy
Nelson Vergel, the founder of ExcelMale.com, DiscountedLabs.com, and ClinicOptimizers.com, interviews Dr. Lisbeth Roy, a functional physician in Boca Raton, Florida. She is an expert...
Best Vitamins and Supplements by Nelson Vergel
It is very difficult to know what supplements have been proven to work in clinical studies. In this video Nelson Vergel helps to demystify...
When Testosterone Doesn't Lead to Better Erections
Excerpt from the book: Testosterone: A Man's Guide Most men find that their sexual desire increases after they start testosterone replacement. Sexual dreams and nighttime/morning...
Nelson Vergel Speaks to Paul Burguess from the UK about Testosterone Therapy in the...
Comparison of Testosterone Clinics in the United States
For a comparison of cash-based testosterone clinics in the United States, please visit: https://www.lowtestosterone.org/find-a-clinic/ This table summarizes the differences between men's health testosterone clinics on the...
Nelson Vergel's Top 10 Testosterone Replacement Commandments (Video)
Nelson Vergel's Top 10 Testosterone Replacement Commandments (Video) Nelson Vergel, author of Testosterone: A Man's Guide (Amazon.com) and founder of ExcelMale.com and DiscountedLabs.com, talks about...
Nelson's Top Tricks for Fat Loss
Get real. Ask yourself: What is getting in the way of my health? What excuses am I using to not start giving a damn?...

redpajama_set_name: RedPajamaCommonCrawl


text:
Deutsche Meisterschaften im Ski-Orientierungslauf werden seit 2000 ausgetragen.

In der DDR fanden bereits von 1953 an Meisterschaften im Ski-Orientierungslauf statt.

Austragungsorte

Herren

Klassik/Lang

Kurz/Mittel

Damen

Klassik/Lang

Kurz/Mittel

Siehe auch
 Deutsche Orientierungslaufmeisterschaft

Weblinks
 OL-Archiv Deutschlands
 Deutsche Meisterschaften im Ski-Orientierungslauf orientierungslauf.de
 DM-Ergebnisse bis 2003 auf sport-komplett.de

Einzelnachweise

Deutsche Meisterschaft
Ski-Orientierungslaufwettbewerb
Wintersport (Deutschland)


text:
Annum ingressi je pismo, ki ga je napisal papež Leon XIII. leta 1902.

Pismo je bilo naslovljeno vsem škofov kot pregled delovanja ob 25. obletnici nastopa pontifikata. V pismu je tudi obsodil prostozidarstvo.

Pisma Rimskokatoliške Cerkve
Dela leta 1902
Dela papeža Leona XIII.
Dela v latinščini
Rimskokatoliška cerkev in prostozidarstvo


Proud of Stockport 2012 – Champion for Young People Award winner is Maureen Wright.
Maureen and her late husband Vic fought long and hard alongside the local residents' association to secure funding for a new all-weather sports pitch near their home in Cheadle to help tackle anti-social behaviour in the neighbourhood. The new facility, completed last July, has really helped improve relationships between local residents and young people.
Your Metals Matter: Recycle for Leeds!
This film was played before the winner was revealed in the ceremony at The Mall Awards for a shopping centre in Wood Green, London.


Rhode Island has a string of complex legislation regarding the US gambling industry. It's taken some time but traditional casino table games were recently legalized, along with poker rooms and horse racing betting.
Dog racing betting has been outlawed for some time. Daily fantasy sports is set to be legalized after the Attorney General intimated it is not a game of luck or chance. Social gambling could also soon be permitted under the 2016 bill that's still under consideration.
At the time of writing, no specific legislation addresses the issue of the burgeoning online gambling industry.
Access to many of the world's leading iGaming brands is somewhat restricted to Rhode Island residents based on the Unlawful Internet Gambling Enforcement Act (UIGEA). However, many players use unregulated offshore sites without fear of prosecution.
1744-1843 - Unofficial lotteries rife throughout Rhode Island, helping to finance municipal growth.
1843 - State passes legislature to prohibit informal lotteries.
1896 - Sports betting and bookmaking outlawed.
1934 - Rhode Island approves proposals to legalize pari-mutuel horse and dog betting.
1973 - Rhode Island approves state-operated lottery.
1974 - 'The Lott' - Rhode Island's inaugural state-operated lottery starts selling tickets.
1987 - State votes to permit bingo gaming for maximum prizes of $400.
1988 - Rhode Island Lottery joins forces with the Multi-State Lottery Association to allow residents to take part in multi-state lotteries.
1990 - Off-track betting proposals thrown out by state.
1991 - Two race tracks allowed to have off-track betting.
1992 - Video lottery terminals permitted at state's pari-mutuel betting locations.
1994 - Full-scale casino table gaming denied after vote.
1998 - Tribal casinos prohibited from offering casino games.
2010 - Greyhound racing officially outlawed.
2012 - State-operated casino gaming allowed at Twin River and Newport Grand casinos.
2015 - Inuaugal poker room opened at Twin River.
2016 - Daily fantasy sports deemed legal.
2016 - New commercial casino permitted in Tiverton.
What Types of Gambling are Legal in Rhode Island?
Casino Gambling – Rhode Island has two casinos operating out of their racetracks. One of the Rhode Island casinos offers video poker, slots, and electronic table games while the other has the full complement of casino games and slot machines.
Sports Betting – Sports betting is legal in the form of simulcast and off track betting at the two Rhode Island racetracks. Rhode Island has passed a sports betting bill and you can follow all of the latest Rhode Island sports betting news here.
Online Betting – Most of the horse racing betting sites accept players from Rhode Island and with sports betting on the way in the state online sports betting may not be far behind. We will be following all online gambling news in Rhode Island here.
Daily Fantasy – Daily Fantasy Sites (DFS) is legal in Rhode Island so all of the DFS sites like Fanduel and Draft Kings accept players from the state.
Poker – Poker is legal in Rhode Island but there is currently just one poker room.
Bingo – Bingo in Rhode Island is legal for charities and permits are issued by the state police. Limits for Bingo prize pools in Rhode Island are generally low, in the $300 range.
Lottery – There is a state lottery in Rhode Island as well as sales of scratch off tickets and instant lotteries. There are Keno drawings in Rhode Island every 4 minutes.


text:
Alberto Quaresma é um ator português. Conhecido por interpretar várias personagens nos Malucos do Riso da SIC.

Televisão

Elenco principal
Duarte e Companhia RTP 1985 - 1989 - Albertini
O Motim RTP 1987
O Bando dos Quatro RTP 1992 - Chinês
Os Malucos do Riso SIC 1995 - várias personagens
Os Malucos nas Arábias SIC 2005 - várias personagens
Malucos e Filhos SIC 2005 - várias personagens
Malucos no Hospital SIC 2008 - várias personagens
Novos Malucos do Riso SIC 2009 - várias personagens

Pequenas participações
Zé Gato RTP 1979
O Beijo de Judas RTP 1991
Clube Paraíso RTP 1994
O Fura-Vidas SIC 1999
Super Pai TVI 2001
A Minha Família É Uma Animação SIC 2001 - Dr. Rocha
A Senhora das Águas RTP 2001
Amanhecer TVI 2002 - antigo cliente de Anabela
Saber Amar TVI 2003
Uma Aventura SIC 2004 - canalizador
Maré Alta SIC 2004 - passageiro
Mundo Meu TVI 2005
Tu e Eu TVI 2007
Floribella SIC 2007
Vingança SIC 2007
Resistirei SIC 2008
Camilo, o Presidente SIC 2010
Voo Directo RTP 2011
A Família Mata SIC 2012
Sol de Inverno SIC 2013 - Amílcar

Ligações externas

Atores de Portugal


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Fantastic Sams Cut and Color
Featured Stylists – Emily & Kayla
Meet our Featured Stylists of the Month for December 2020 – Emily & Kayla and see what they have to say about working for Fantastic Sams Cut & Color!
Emily is one of our fantastic stylists who has been working at Fantastic Sams Cut & Color for over 14 years!
"I enjoy working with the variety of guests that come through our doors, and love being able to work on different services each day," says Emily. "Fantastic Sams takes the time and effort to sponsor stylist events that help build us up as a team and make us feel special. There is never a dull moment at our salon."
Kayla is another one of our amazing stylists who has been working at Fantastic Sams Cut & Color for over 4 years!
"What I enjoy about working at Fantastic Sams is the closeness and sense of security the team brings," says Kayla. "I love that as an FS stylist I can reach out to anyone for questions or support! My team is constantly pushing each other to be the absolute best we can be. We really are one big family!"
Become one of our fantastic success stories.
Apply to join the Fantastic Sams Cut & Color team today! Your future starts here.
If you're not quite ready to apply, please feel free to reach out direct via email, call or text!
Ranae Hendrickson
Minnesota Regional Recruiter
Fantastic Sams Franchise | Minnesota
New Year. New Looks!
One of the best parts of being a hairstylist is the ability to transform someone's style from a bit outdated or boring to one that is fun, exciting and up-to-the-minute chic. That's why it's so important to stay up to date on all the latest cuts, colors, and overall trends….
2022 Stylist Appreciation Event
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text:
\section{Introduction}
Let $G$ be a finitely generated torsion-free nilpotent group (or a $\mathcal{T}$-group for short). Two complex representations $\rho$ and $\sigma$ of $G$ are called \textit{twist-equivalent} if there exists a $1$-dimensional representation $\lambda$ of $G$ such that $\rho=\lambda\otimes\sigma$. Twist-equivalence is an equivalence relation on the set of finite dimensional
irreducible complex representations of $G$ and its classes are called
\textit{twist-isoclasses}. The numbers $r_n(G)$ of twist-isoclasses of
dimension $n$ are finite for all $n$, cf.~{\cite[Theorem 6.6]{Lubotzky-Magid}}. The \textit{representation zeta function} of $G$ is defined to be the Dirichlet
generating function
$$\zeta_G(s):=\sum_{n=1}^\infty\frac{r_n(G)}{n^s},$$
where $s$ is a complex variable.
The sequence $(r_n(G))$ grows polynomially and thus $\zeta_G(s)$ converges on
a complex half-plane $\rea(s)>\alpha$, cf.~{\cite[Lemma 2.1]{SV/14}}. The
infimum of such $\alpha$ is the \textit{abscissa of convergence} $\alpha(G)$ of
$\zeta_G(s)$ which gives the precise degree of polynomial growth; i.e., $\alpha(G)$ is the smallest value such that
$\sum_{n=1}^Nr_n(G)=O(N^{\alpha(G)+\epsilon})$ for every $\epsilon\in\mathbb{R}_{>0}$.

Let $\mathbf{H}$ be the Heisenberg group scheme associated to the Heisenberg $\bb{Z}$-Lie lattice of strict upper-triangular $3\times 3$ matrices. For every ring $R$, the group $\mathbf{H}(R)$ is isomorphic to the group of upper-unitriangular $3\times3$ matrices over $R$. If $R$  is torsion-free finitely
generated over $\bb{Z}$, then $\mathbf{H}(R)$ is a $\mathcal{T}$-group
 of nilpotency class $2$ and Hirsch length $3\cdot\textrm{rk}_{\bb{Z}}(R)$.
When $R=\O$ is the ring of integers of a number field $K$, the zeta function
of $\mathbf{H}(\O)$ is
\begin{equation}\label{H over O}
\zeta_{\mathbf{H}(\O)}(s)=\frac{\zeta_K(s-1)}{\zeta_K(s)}=\prod_{\p\in\Spec(\O)}
\frac{1-|\O/\p|^{-s}}{1-|\O/\p|^{1-s}},
\end{equation}
where $\zeta_K(s)$ is the Dedekind zeta function of $K$, $\p$
ranges over the nonzero prime ideals of $\O$. This is proved in {\cite{Magid89}}
for $K=\bb{Q}$, in \cite{Ezzat} for quadratic number fields, and
in {\cite[Theorem~B]{SV/14}} for arbitrary number fields.
The zeta function $\zeta_{\mathbf{H}(\O)}(s)$ has abscissa of convergence $\alpha(\mathbf{H}(\O))=2$, which is independent of $K$, and may be meromorphically continued to the whole
complex plane.

In this paper, we consider the Heisenberg group over rings of the form $\O[x]/(x^n)$. If $n=1$ then it is the Heisenberg group over $\O$. The zeta function of
 $\mathbf{H}(\O[x]/(x^2))$  was computed in {\cite[Example 6.5]{Snocken}}. We compute
the zeta function of $\mathbf{H}(\O[x]/(x^n))$ for $n=3$.

\subsection*{Organization and notation}
In Section 2, we recall  formulae of local representation
zeta functions in terms of $p$-adic integrals. The zeta function for the case $n=3$ is computed
in Section 3. We conclude in Section 4 with several questions and
conjectures.

\subsection*{Acknowledgements} We acknowledge support from the DFG Sonderforschungsbereich 701 at Bielefeld University. We thank Tobias Rossmann and Christopher
Voll for several helpful discussions.

\section{Preliminaries}
\subsection{Local representation zeta functions}
The group $\mathbf{H}(\O[x]/(x^n))$ is a $\mathcal{T}$-group of nilpotency class $2$ and
Hirsch length $3n\cdot\textrm{rk}_\bb{Z}(\O)$. The zeta function $\zeta_{\mathbf{H}(\O[x]/(x^n))}(s)$
has an Euler factorization (cf.~{\cite[Proposition 2.2]{SV/14}})
$$\zeta_{\mathbf{H}(\O[x]/(x^n))}(s)=\prod_{\p\in\Spec(\O)}\zeta_{\mathbf{H}(\O_\p[x]/(x^n))}(s),$$
where $\p$ ranges over the nonzero prime ideals in $\O$ and $\O_\p$ is the  completion of $\O$ at  $\p$.
The local factors $\zeta_{\mathbf{H}(\O_\p[x]/(x^n))}$ are rational in $|\O/\p|^{-s}$ and almost all of them
satisfy a functional equation (cf.~{\cite[Theorem~A]{SV/14}}).

The $\O$-lattice associated to $\mathbf{H}(\O[x]/(x^n))$ has the following presentation; see {\cite[Section~2.4]{SV/14}}:
$$\left\langle
\begin{matrix}
x_0,x_1,\cdots,x_{n-1} \\
y_0,y_1,\cdots,y_{n-1} \\
z_0,z_1,\cdots,z_{n-1}
\end{matrix}
\;\ifnum\currentgrouptype=16 \middle\fi|\;[x_i,y_j]=
\left\{
\begin{array}{ll}z_{i+j}&\textrm{if}~i+j<n,\\
0&\textrm{otherwise}.
\end{array}
\right.
\right\rangle.$$
The associated commutator matrix with respect to the chosen $\O$-basis is defined by
$$\mathcal{R}_n(\mathbf{Y})=\left(
\begin{array}{c|c}

0&Q_n(\mathbf{Y})\\ \hline
-Q_n(\mathbf{Y})^t&0
\end{array}
\right),
$$
where
$$Q_n(\mathbf{Y})=\left(
\begin{array}{cccccc}
Y_1&Y_2&Y_3&\cdots&Y_{n-1} & Y_{n}\\
Y_2&Y_3&\cdots&&Y_{n}&0 \\
Y_3&\cdots&&\cdots&&0\\
 \vdots& &\cdots&&&\vdots\\
Y_{n-1} &Y_{n} &\cdots&\cdots&\cdots&\vdots\\
Y_{n}&0&\cdots&\cdots&\cdots&0
\end{array}
\right)\in\Mat_{n}(\O[Y_1,\cdots,Y_n]).$$

Fix a nonzero prime ideal $\p$ and denote $\o:=\O_\p$. Let $q:=|\o/\p|$ be the residue field cardinality and $p$ its characteristic. Let $W_n(\o)=\o^n\setminus \p^n$. Set
\begin{equation}\label{p-adic int}
\mathcal{Z}_\p(\rho,\tau):=
\int_{(u,{\mathbf y})\in \p\times W_n(\o)} |u|^\tau
\prod_{j=1}^n\frac{\|{F}_j({\mathbf y})\cup {F}_{j-1}({\mathbf y})u^2\|^\rho}
{\|{F}_{j-1}({\mathbf y})\|^\rho}d\mu,
\end{equation}
where the additive Haar measure $\mu$ on $\o^{n+1}$ is normalized such that $\mu(\o^{n+1})=1$, and
\begin{align*}
      {F}_j({\mathbf Y}) & = \{ f\mid f=f({\mathbf Y})~\textrm{a principal}~2j\times
   2j ~\textrm{minor of}
   ~\mathcal{R}_n({\mathbf Y}) \},\label{eq:poly F}\\
   \|H(X,{\mathbf Y})\| & = \min\{v_p(h(X,{\mathbf Y}))\mid h\in H\}
   ~\textrm{for a finite set}~H\subset\o[X,{\mathbf Y}].\nonumber
 \end{align*}
The local factor $\zeta_{\mathbf{H}(\o[x]/(x^n))}(s)$ can be expressed in terms of the $p$-adic integral \eqref{p-adic int} as the following (cf.{\cite[Corollary 2.11]{SV/14}}):
\begin{equation}
\zeta_{\mathbf{H}(\o[x]/(x^n))}=1+(1-q^{-1})^{-1}\mathcal{Z}_\p(-s/2,ns-n-1).
\end{equation}

\subsection{Auxiliary lemmas}

\begin{lemma}\label{Lem identities}
The following identities hold in the field of formal Laurent series  $\bb{Q}((a,b,c))$.
\begin{enumerate}
\item $\sum_{(X,Y)\in\bb{N}^2}a^Xb^Yc^{\min\{X,Y\}}=\frac{abc(1-ab)}{(1-abc)(1-a)(1-b)}$. \label{Lem SV15}
\item $\sum_{(X,Y)\in\bb{N}^2}a^Xb^Yc^{\min\{X,2Y\}}=\frac{abc(1-a+ac-a^2bc)}{(1-a)(1-b)(1-a^2bc^2)}$.
\item $\sum_{(X,Y)\in\bb{N}^2}a^Xb^Yc^{\min\{X,Y\}}c^{\min\{X,2Y\}}=\frac{abc^2(1-a+ac-abc-a^2bc^3+a^3b^2c^3)}{(1-a)(1-b)(1-abc^2)(1-a^2bc^3)}$.
\item
\begin{align*}
&\sum_{(X,Y,Z)\in\bb{N}^3}a^Xb^Yc^Zd^{\min\{X,Y+2Z\}}d^{\min\{X,2Y+4Z\}}
=\frac{abd^2}{1-abd^2}\frac{1}{1-b}\frac{c}{1-c}+\\
&+ \frac{acd^2}{1-acd^2}\frac{abd^2}{1-abd^2}\frac{1}{1-c}
+ \frac{a^2cd^4}{1-a^2cd^4}\frac{abd^2}{1-abd^2}\frac{1}{1-acd^2}
+\frac{a^2bd^3}{1-a^2bd^3}\frac{1}{1-abd^2}\frac{a^2cd^4}{1-a^2cd^4}\\
&+\frac{a^2bd^3}{1-a^2bd^3}\frac{1-a+ad-a^4cd^5}{(1-a)(1-a^2cd^4)(1-a^4cd^6)}.
\end{align*}
\end{enumerate}
\end{lemma}
\begin{proof}
The identity $(1)$ is from {\cite[Lemma 2.2]{SV/15}}\label{Lem SV15}.
We present the proofs of (2) and (3) while (4) is proven
similarly.

For (2), consider the case $X\leq Y$ and let $Y=X+Y'$ with $Y'\in\bb{N}_0$. Then
$$\sum_{\substack{(X,Y)\in\bb{N}^2\\ X\leq Y}}a^Xb^Yc^{\min\{X,2Y\}}=
\sum_{(X,Y')\in\bb{N}\times\bb{N}_0}a^Xb^{X+Y'}c^X=\frac{abc}{1-abc}\frac{1}{1-b}.$$
Consider the case $X>Y$ and let $X=Y+X'$ with $X'\in\bb{N}$. Then
\begin{align*}
\sum_{\substack{(X,Y)\in\bb{N}^2\\ X> Y}}a^Xb^Yc^{\min\{X,2Y\}}
&=\sum_{(X',Y)\in\bb{N}^2}a^{X'+Y}b^Yc^{\min\{X'+Y,2Y\}}\\
&=\sum_{(X',Y)\in\bb{N}^2}a^{X'}(abc)^Yc^{\min\{X',Y\}}\\
&=\frac{a^2bc^2(1-a^2bc)}{(1-a^2bc^2)(1-a)(1-abc)}
\end{align*}
by (1). Hence
\begin{align*}
\sum_{(X,Y)\in\bb{N}^2}a^Xb^Yc^{\min\{X,2Y\}} &=\frac{abc}{1-abc}\frac{1}{1-b}+\frac{a^2bc^2(1-a^2bc)}{(1-a^2bc^2)(1-a)(1-abc)}\\
&=\frac{abc(1-a+ac-a^2bc)}{(1-a)(1-b)(1-a^2bc^2)}.
\end{align*}

For (3), first consider the case $X\leq Y$ and let $Y=X+Y'$ with $Y'\in\bb{N}_0$. Then
$$\sum_{\substack{(X,Y)\in\bb{N}^2\\ X\leq Y}} a^Xb^Yc^{\min\{X,Y\}}c^{\min\{X,2Y\}}=
\sum_{(X,Y')\in\bb{N}\times\bb{N}_0}a^Xb^{X+Y'}c^Xc^X=\frac{abc^2}{1-abc^2}\frac{1}{1-b}.$$
Consider now the case $X>Y$ and let $X=X'+Y$ with $X'\in\bb{N}$. Then, by (1)
\begin{align*}
\sum_{\substack{(X,Y)\in\bb{N}^2\\ X\leq Y}} a^Xb^Yc^{\min\{X,Y\}}c^{\min\{X,2Y\}}&=
\sum_{(X',Y)\in\bb{N}^2}a^{X'+Y}b^Yc^Yc^{\min\{X'+Y,2Y\}}\\
&= \sum_{(X',Y)\in\bb{N}^2}a^{X'}(abc^2)^Yc^{\min\{X',Y\}}\\
&=\frac{a^2bc^3(1-a^2bc^2)}{(1-a^2bc^3)(1-a)(1-abc^2)}.
\end{align*}
Hence
\begin{align*}
\sum_{(X,Y)\in\bb{N}^2}a^Xb^Yc^{\min\{X,Y\}}c^{\min\{X,2Y\}} & =\frac{abc^2}{1-abc^2}\frac{1}{1-b}+\frac{a^2bc^3(1-a^2bc^2)}{(1-a^2bc^3)(1-a)(1-abc^2)}\\
&=\frac{abc^2(1-a+ac-abc-a^2bc^3+a^3b^2c^3)}{(1-a)(1-b)(1-abc^2)(1-a^2bc^3)}.
\end{align*}
\end{proof}

\begin{lemma}\label{Z_2}
$$I:=\int_{x,y\in\p}|x|^{3s-4}\|x,y^3\|^{-s}d\mu
=(1-q^{-1})\frac{q^2t^2(1+q^3t^2-q^3t^3+q^6t^4-q^6t^5-q^8t^7)}{(1-q^3t^3)(1-q^8t^6)},$$
where $t:=q^{-s}$.
\end{lemma}
\begin{proof}
Write $I=I_1+I_2$ where
\begin{align*}
I_1 &:= \int_{\substack{x,y\in\p\\ v(x)\leq v(y)}}|x|^{3s-4}\|x,y^3\|^{-s}d\mu,\\
I_2 &:=\int_{\substack{x,y\in\p\\ v(x)>v(y)}}|x|^{3s-4}\|x,y^3\|^{-s}d\mu.
\end{align*}
\textit{Computation of $I_1$.} Let $y=xy_1$ with $y_1\in\o$. Then
$$I_1=\int_{\substack{x\in \p\\ y_1\in\o}}
|x|^{3s-4}|x|^{-s+1}d\mu =(1-q^{-1})\frac{q^2t^2}{1-q^2t^2}.$$
\textit{Computation of $I_2$.} Let $x=yx_1$ with $x_1\in\p$. Then
$$I_2=\int_{x_1,y\in\p}|x_1|^{3s-4}|y|^{2s-3}\|x_1,y^2\|^{-s}d\mu.$$
Since $\mu(\{(x_1,y)\in\p^2\mid v(x_1)=X, v(y)=Y\})=(1-q^{-1})^2q^{-X-Y}$, one has by Lemma~\ref{Lem identities}~(2)
\begin{align*}
I_2&=(1-q^{-1})^2\sum_{(X,Y)\in\bb{N}^2}q^{-X-Y}q^{(-3s+4)X}q^{(-2s+3)Y}q^{s\min\{X,2Y\}}\\
&=(1-q^{-1})^2\sum_{(X,Y)\in\bb{N}^2}q^{(-3s+3)X}q^{(-2s+2)Y}q^{s\min\{X,2Y\}}\\
&=(1-q^{-1})^2\frac{q^{-3s+3-2s+2+s}(1-q^{-3s+3}+q^{-3s+3+s}-q^{-6s+6-2s+2+s})}{(1-q^{-3s+3})(1-q^{-2s+2})(1-q^{-6s+6-2s+2+2s})} \\
&=(1-q^{-1})^2\frac{q^5t^4(1-q^3t^3+q^3t^2-q^8t^7)}{(1-q^2t^2)(1-q^3t^3)(1-q^8t^6)}.
\end{align*}
Hence
$$I=I_1+I_2=(1-q^{-1})\frac{q^2t^2(1+q^3t^2-q^3t^3+q^6t^4-q^6t^5-q^8t^7)}{(1-q^3t^3)(1-q^8t^6)}.$$
\end{proof}

\subsection{The zeta function of $\mathbf{H}(\O[x]/(x^2))$}
In this case $\tau=vs-2-1=2s-3$ and
$F_0(\mathbf{Y})=1, F_1(\mathbf{Y})=\{X^2,Y^2\}$, $F_2(\mathbf{Y})=\{Y^4\}$. We have
$$\mathcal{Z}:=\mathcal{Z}_\p(-s/2,2s-3)=\int_{\substack{u\in \p\\ \mathbf{y}=(x,y)\in W_2(\o)}}|u|^{2s-3}\|u,x,y\|^{-s}\|ux,uy,y^2\|^{-s}d\mu.$$
It is computed in {\cite[Example 6.5]{Snocken}} that
\iffalse
Write $\mathcal{Z}=\mathcal{Z}_1+\mathcal{Z}_2$ where
\begin{align*}
\mathcal{Z}_1&:= \int_{\substack{u\in \p\\ x\in\o \\ y\in W_1(\o)}}|u|^{2s-3}\|u,x,y\|^{-s}\|ux,uy,y^2\|^{-s}d\mu,\\
\mathcal{Z}_2&:=\int_{\substack{u\in \p\\ x\in W_1(\o)\\ y\in\p}}|u|^{2s-3}\|u,x,y\|^{-s}\|ux,uy,y^2\|^{-s}d\mu.
\end{align*}

\subsubsection{Computation of $\mathcal{Z}_1$} Since $y\in W_1(\o)$, we have $\|u,x,y\|=\|ux,uy,y^2\|=1$ and hence
$$\mathcal{Z}_1=(1-q^{-1})^2\int_{u\in\p}|u|^{2s-3}d\mu =(1-q^{-1})^2\frac{q^2t^2}{1-q^2t^2},~\textrm{with}~t:=q^{-s}.$$

\subsubsection{Computation of $\mathcal{Z}_2$} Since $x\in W_1(\o)$, one has $\|u,x,y\|=1$ and
$\|ux,uy,y^2\|~=~\|u,y^2\|$. Hence
$$\mathcal{Z}_2=(1-q^{-1})\int_{u,y\in \p}|u|^{2s-3}\|u,y^2\|^{-s}d\mu.$$
Write $\mathcal{Z}_{2}=\mathcal{Z}_{21}+\mathcal{Z}_{22}$, where
\begin{align*}
\mathcal{Z}_{21}=(1-q^{-1})\int_{\substack{u,y\in \p\\ v(u)\leq v(y)}}|u|^{2s-3}\|u,y^2\|^{-s}d\mu,\\
\mathcal{Z}_{21}=(1-q^{-1})\int_{\substack{u,y\in \p\\ v(u)> v(y)}}|u|^{2s-3}\|u,y^2\|^{-s}d\mu.
\end{align*}

\noindent\textit{Computation of $\mathcal{Z}_{21}$.} Let $y=uy_1$ with $y_1\in\o$.
We have $\|u,y^2\|=|u|$ and
$$\mathcal{Z}_{21}=(1-q^{-1})\int_{\substack{y_1\in\o\\ u\in \p}}
|u|^{2s-3}|u|^{-s}|u|d\mu =(1-q^{-1})^2\frac{qt}{1-qt}.$$

\noindent\textit{Computation of $\mathcal{Z}_{22}$.} Let $u=yu_1$ with $u_1\in\p$. Then
$\|u,y^2\|=|y|^{-s}\|u_1,y\|$ and
$$
\mathcal{Z}_{22} =(1-q^{-1})\int_{u_1,y\in\p}|u_1y|^{2s-3}|y|^{-s+1}\|u_1,y\|^{-s}d\mu.$$

\begin{lemma}\label{Lem sec 2}
$$\mathcal{Z}_{22}=\frac{(1-q^{-1})^3q^3t^2(1+qt+q^2t^2)}{(1-q^2t^1)(1-q^3t^2)}.$$
\end{lemma}
\begin{proof}
As, for $X,Y\in\bb{N}$,
$$\mu(\{(u_1,y)\in\p^{(2)}\mid v(u_1)=X, v(y)=Y\})=(1-q^{-1})^2q^{-X-Y},$$
we have
\begin{align*}
\mathcal{Z}_{22} & = (1-q^{-1})\int_{u_1,y\in\p}|u_1|^{2s-3}|y|^{s-2}\|u_1,y\|^{-s}d\mu\\
&= (1-q^{-1})\sum_{(X,Y)\in\bb{N}^2}(1-q^{-1})^2q^{-X-Y}q^{-(2s-3)X-(s-2)Y+s\min
\{X,Y\}} \\
&=(1-q^{-1})^3\sum_{(X,Y)\in\bb{N}^2}q^{(-2s+2)X}q^{(-s+2)Y}q^{s\min\{X,Y\}}\\
&=(1-q^{-1})^3\frac{q^{(-2s+2)+(-s+1)+s}(1-q^{(-2s+2)+(-s+1)})}{(1-q^{(-2s+2)+(-s+1)+s})(1-q^{-2s+2})(1-q^{-s+1})},
\end{align*}
where the last equality follows from Lemma~\ref{Lem SV15}~(2)  and the lemma follows.
\end{proof}
Summing up $\mathcal{Z}=\mathcal{Z}_1+\mathcal{Z}_{21}+\mathcal{Z}_{22}$ yields
\fi
$$\zeta_{\mathbf{H}(\o[x]/(x^2))}(s)=1+(1-q^{-1})^{-1}\mathcal{Z}
= \frac{(1-t)(1-q^2t^2)}{(1-qt)(1-q^3t^2)},~\textrm{where}~t:=q^{-s}.$$
Hence
\begin{equation}\label{n=2}
\zeta_{\mathbf{H}(\O[x]/(x^2))}(s)=\frac{\zeta_K(s-1)\zeta_K(2s-3)}{\zeta_K(s)\zeta_K(2s-2)}.
\end{equation}

\section{The zeta function of $\mathbf{H}(\O[x]/(x^3))$}
 In this case, $\tau=3s-4$ and

\begin{align*}
F_0(\mathbf{Y}) &=1,\\
F_1(\mathbf{Y}) &= \{X^2,Y^2,Z^2\},\\
F_2(\mathbf{Y}) &= \{Z^4,Y^2Z^2,(XZ-Y^2)^2\},\\
F_3(\mathbf{Y}) &=\{Z^6\}.
\end{align*}

Set
\begin{align*}
A &:=\|z^2,yz,xz-y^2\|^s,\\
B &:= \|z^2,yz,xz-y^2,xu,yu,uz\|^{-s},\\
C &:= \|z^3,z^2u,yzu,(xz-y^2)u\|^{-s}.
\end{align*}
Then
$$\zeta_{\mathbf{H}(\o[x]/(x^3))}(s)=1+(1-q^{-1})^{-1}\mathcal{Z}(s),$$
where
$$\mathcal{Z}:=\mathcal{Z}_\p(-s/2,3s-4)=\int_{u\in\p,\mathbf{y}\in W_3(\o)}
|u|^{3s-4}ABCd\mu.$$
Write $\mathcal{Z}=\mathcal{Z}_1+\mathcal{Z}_2+\mathcal{Z}_3$ where
\begin{align*}
\mathcal{Z}_1 &:= \int_{\substack{u\in\p,z\in W_1(\o)\\ x,y\in\o}}|u|^{3s-4}ABCd\mu,\\
\mathcal{Z}_2 &:= \int_{\substack{u,z\in\p\\ x\in\o \\ y\in W_1(\o)}}|u|^{3s-4}ABCd\mu,\\
\mathcal{Z}_3 &:= \int_{\substack{u,y,z\in\p\\ x\in W_1(\o)}}|u|^{3s-4}ABCd\mu.
\end{align*}

\subsection{Computation of $\mathcal{Z}_1$}
Since $z\in W_1(\o)$, it follows that $A=B=C=1$. Hence
$$\mathcal{Z}_1=\int_{\substack{u\in\p,z\in W_1(\o)\\ x,y\in\o}}|u|^{3s-4}d\mu
=(1-q^{-1})^2\frac{q^3t^3}{1-q^3t^3}.$$
\subsection{Computation of $\mathcal{Z}_2$}
Since $y\in W_1(\o)$ and $z\in \p$, it follows that $xz-y\in W_1(\o)$ and
$A=1, ~B=1,~C=\|u,z^3\|^{-s}$. Thus
$$\mathcal{Z}_2=(1-q^{-1})\int_{u,z\in\p}|u|^{3s-4}\|u,z^3\|^{-s}d\mu.$$
It follows from Lemma~\ref{Z_2} that
$$\mathcal{Z}_2=(1-q^{-1})^2\frac{q^2t^2(1+q^3t^2-q^3t^3+q^6t^4-q^6t^5-q^8t^7)}{(1-q^3t^3)(1-q^8t^6)}.$$
\subsection{Computation of $\mathcal{Z}_3$} In this case, $x\in W_1(\o), y,z\in\p$, whence
\begin{align*}
A &= \|z^2,yz,xz-y^2\|^s,\\
B &= \|u,z^2,yz,xz-y^2\|^{-s},\\
C &= \|z^3,z^2u,yzu,(xz-y)^2u\|^{-s}.
\end{align*}
Write $\mathcal{Z}_3=\mathcal{Z}_{31}+\mathcal{Z}_{32}$, where
\begin{align*}
\mathcal{Z}_{31}&:= \int_{\substack{u,y,z\in\p\\ x\in W_1(\o)\\ v(y)>v(z)}}|u|^{3s-4}ABCd\mu, \\
\mathcal{Z}_{32}&:=\int_{\substack{u,y,z\in\p\\ x\in W_1(\o)\\ v(y)\leq v(z)}}|u|^{3s-4}ABCd\mu.
\end{align*}
\subsubsection{Computation of $\mathcal{Z}_{31}$} Let $y=zy_1$ with $y_1\in\p$. Then
\begin{align*}
A &= \|z^2,xz-y_1^2z^2\|^s = |z|^s\|z,x-y_1^2z\|^s=|z|^s~(\textrm{since}~x-y_1^2z\in W_1(\o)),\\
B &= \|u,z\|^{-s},\\
C &= |z|^{-s}\|u,z^2\|^{-s}.
\end{align*}
Thus
$$
\mathcal{Z}_{31} = q^{-1}(1-q^{-1})\int_{u,z\in \p}|z||u|^{3s-4}\|u,z\|^{-s}\|u,z^2\|^{-s}d\mu.$$
Since $\mu\{(u,z)\in\p\mid v(u)=X, v(z)=Y \}=(1-q^{-1})^2q^{-X-Y}$, Lemma~\ref{Lem identities}~(3) implies that
\begin{align*}
\mathcal{Z}_{31} &= q^{-1}(1-q^{-1})\int_{u,z\in \p}|u|^{3s-4}|z|\|u,z\|^{-s}\|u,z^2\|^{-s}d\mu \\
&= q^{-1}(1-q^{-1})^3\sum_{(X,Y)\in\bb{N}^2}q^{-X-Y}q^{(-3s+4)X}q^{-Y}q^{s\min\{X,Y\}}q^{s\min\{X,2Y\}}\\
&=q^{-1}(1-q^{-1})^3\sum_{(X,Y)\in\bb{N}^2}q^{(-3s+3)X}q^{-2Y}q^{s\min\{X,Y\}}q^{s\min\{X,2Y\}}\\
&=(1-q^{-1})^3\frac{t(1-q^3t^3+q^3t^2-qt^2-q^4t^3+q^5t^6)}{(1-q^{-2})(1-qt)(1-q^3t^3)(1-q^4t^3)}.
\end{align*}

\subsubsection{Computation of $\mathcal{Z}_{32}$} Let $z=yz_1$ with $z_1\in\o$. We have
\begin{align*}
A &= |y|^s\underbrace{\|yz_1,xz_1-y\|^s}_{:=A_1},\\
B &= \|u,y^2z_1,y(xz_1-y)\|^{-s},\\
C &= |y|^{-s}\underbrace{\|y^2z_1^3,yz_1u,u(xz_1-y)\|^{-s}}_{:=C_1}.
\end{align*}
Thus
$$\mathcal{Z}_{32}=\int_{\substack{x\in W_1(\o)\\ u,y\in\p \\ z_1\in\o}}
|u|^{3s-4}|y|A_1BC_1d\mu.$$
Write $\mathcal{Z}_{32}=\mathcal{Z}_{321}+\mathcal{Z}_{322}$, where
\begin{align*}
\mathcal{Z}_{321}&:=\int_{\substack{x\in W_1(\o)\\ u,y\in\p \\ z_1\in W_1(\o)}}
|u|^{3s-4}|y|A_1BC_1d\mu,\\
\mathcal{Z}_{322}&:= \int_{\substack{x\in W_1(\o)\\ u,y\in\p \\ z_1\in\p}}
|u|^{3s-4}|y|A_1BC_1d\mu.
\end{align*}
\textit{Computation of $\mathcal{Z}_{321}$}. Since $z_1\in W_1(\o)$, it follows that $xz_1-y\in W_1(\o)$ and so
$$\mathcal{Z}_{321} =(1-q^{-1})^2\int_{u,y\in\p}|u|^{3s-4}|y|\|u,y\|^{-s}\|u,y^2\|^{-s}d\mu =(q-1)\mathcal{Z}_{31}.$$
\textit{Computation of $\mathcal{Z}_{322}$}. Write $\mathcal{Z}_{322}=\mathcal{Z}_{322a}+\mathcal{Z}_{322b}$, where
\begin{align*}
\mathcal{Z}_{322a}&:= \int_{\substack{x\in W_1(\o)\\ u,y,z_1\in\p \\ v(y)>v(z_1)}}
|u|^{3s-4}|y|A_1BC_1d\mu,\\
\mathcal{Z}_{322b}&:= \int_{\substack{x\in W_1(\o)\\ u,y,z_1\in\p \\ v(y)\leq v(z_1)}}
|u|^{3s-4}|y|A_1BC_1d\mu.
\end{align*}

\noindent\textit{Computation of $\mathcal{Z}_{322a}$.} Let $y=z_1y_1$ with $y_1\in\p$. We have
\begin{align*}
A_1 &= |z_1|^s~(\textrm{since}~x-y_1\in W_1(\o)),\\
B  &=\|u,y_1z_1^2\|^{-s},\\
C_1 &= |z_1|^{-s}\underbrace{\|y_1^2z_1^4,u\|^{-s}}_{:=C_2}.
\end{align*}
Thus
$$
\mathcal{Z}_{322a}=(1-q^{-1})\int_{u,y_1,z_1\in\p}|u|^{3s-4}|y_1||z_1|^2BC_2d\mu.$$
Since $\mu\{(u,y_1,z_1)\in\p\mid v(u)=X, v(y_1)=Y, v(z_1)=Z \}
=(1-q^{-1})^3q^{-X-Y-Z}$, one has
\begin{align*}
\mathcal{Z}_{322a}&=(1-q^{-1})^4\sum_{(X,Y,Z)\in\bb{N}^3}q^{-X-Y-Z}
q^{(-3s+4)X}q^{-Y}q^{-2Z}q^{s\min\{X,Y+2Z\}}q^{s\min\{X,2Y+4Z\}}\\
&=(1-q^{-1})^4\sum_{(X,Y,Z)\in\bb{N}^3}q^{(-3s+3)X}q^{-2Y}q^{-3Z}
q^{s\min\{X,Y+2Z\}}q^{s\min\{X,2Y+4Z\}}.
\end{align*}
One now can apply Lemma~\ref{Lem identities}~(4) with $a=q^{-3s+3},~
b=q^{-2},~c=q^{-3}$ and $d=q^{s}$ to obtain $\mathcal{Z}_{322a}$.
We record the result in the Appendix.

\noindent\textit{Computation of $\mathcal{Z}_{322b}$.} Let $z_1=yz_2$ with $z_2\in\o$. We have
\begin{align*}
A_1 &= |y|^s\underbrace{\|yz_2,xz_2-1\|^s}_{:=A_2},\\
B &= \|u,y^3z_2,y^2(xz_2-1)\|^{-s},\\
C_1 &= |y|^{-s}\underbrace{\|y^4z_2^3,yz_2u,u(xz_2-1)\|^{-s}}_{:=C_2}.
\end{align*}
Thus
$$\mathcal{Z}_{322b}=\int_{\substack{x\in W_1(\o)\\ z_2\in\o \\ u,y\in\p}}
|u|^{3s-4}|y|^2 A_2BC_2d\mu=\mathcal{Z}_{322b1}+\mathcal{Z}_{322b2},$$
where
\begin{align*}
\mathcal{Z}_{322b1} &:= \int_{\substack{x\in W_1(\o)\\ z_2\in\p \\ u,y\in\p}}
|u|^{3s-4}|y|^2 A_2BC_2d\mu,\\
\mathcal{Z}_{322b2} &:=\int_{\substack{x\in W_1(\o)\\ z_2\in W_1(\o) \\ u,y\in\p}}
|u|^{3s-4}|y|^2 A_2BC_2d\mu.
\end{align*}
\noindent\textit{Computation of $\mathcal{Z}_{322b1}$.} Since $z_2\in\p$, it follows that $xz_2-1\in W_1(\o)$. Thus $A_2=1, B=\|u,y^2\|^{-s}$ and $C_2=\|y^4z_2^3,u\|^{-s}$. It's now easy to compute
$$\mathcal{Z}_{322b1}=(1-q^{-1})\int_{u,y,z_2\in\p}|u|^{3s-4}|y|^2 BC_2d\mu.$$
Since $\mu\{(u,y,z_2)\in\p^3\mid v(u)=X,~v(y)=Y,~v(z_2)=Z\}=(1-q^{-1})^3q^{-X-Y-Z}$, one has
\begin{align*}
\mathcal{Z}_{322b1}&=(1-q^{-1})^4\sum_{(X,Y,Z)\in\bb{N}^3}q^{-X-Y-Z}q^{(-3s+4)X}q^{-2Y}q^{s\min\{X,2Y\}}q^{s\min\{X,4Y+3Z\}}\\
&=(1-q^{-1})^4\sum_{(X,Y,Z)\in\bb{N}^3}q^{(-3s+3)X}q^{-3Y}q^{-Z}q^{s\min\{X,2Y\}}q^{s\min\{X,4Y+3Z\}}.
\end{align*}
One needs first to compute $\sum_{(X,Y,Z)\in\bb{N}^3}a^Xb^Yc^Zd^{\min\{X,2Y\}}d^{\min\{X,4Y+3Z\}}$
similarly to Lemma~\ref{Lem identities}~(4) and then apply to $a=q^{-3s+3},~b=q^{-3},~c=q^{-1}$ and $d=q^s$ to
obtain $\mathcal{Z}_{322b1}$. The result is recorded in the Appendix.\\
\textit{Computation of $\mathcal{Z}_{322b2}$.} The equation $xz_2\equiv 1\!\mod \p$ has $q-1$ roots $(a_1,a_2)\in (\mathbb{F}_q^*)^2$. We have
\begin{align*}
\mathcal{Z}_{322b2} &= \int_{\substack{x,z_2\in W_1(\o)\\ u,y\in\p}}
|u|^{3s-4}|y|^2 A_2BC_2d\mu \\
&=\sum_{(a_1,a_2)\in(\mathbb{F}_q^*)^2}\int_{\substack{(x,z_2)\in(a_1,a_2)+\p^2\\ u,y\in \p}}
|u|^{3s-4}|y|^2 A_2BC_2d\mu \\
& = (q-1)(q-2)J_1+(q-1)J_2,
\end{align*}
where
\begin{align*}
J_1 &:=\int_{\substack{(x,z_2)\in(a_1,a_2)+\p^2\\ a_1a_2\not\equiv 1\!\mod\p\\ u,y\in\p}}|u|^{3s-4}|y|^2 A_2BC_2d\mu,\\
J_2 &:=\int_{\substack{(x,z_2)\in(a_1,a_2)+\p^2\\ a_1a_2\equiv 1\!\mod\p\\ u,y\in\p}}|u|^{3s-4}|y|^2 A_2BC_2d\mu.
\end{align*}

In computing $J_1$, notice that in this case $xz_1\not\equiv\! 1\mod\p$, and so
$A_2=1, B=\|u,y^2\|^{-s}$ and $C_2=\|u,y^4\|^{-s}$, and thus we have
$$J_1=q^{-2}\int_{u,y\in\p}|u|^{3s-4}|y|^2BC_2d\mu.$$
Since $\mu\{(u,y)\in\p^2\mid v(u)=X,~v(y)=Y\}=(1-q^{-1})q^{-X-Y}$, one has
\begin{align*}
J_1&=q^{-2}(1-q^{-1})^2\sum_{(X,Y)\in\bb{N}^2}q^{-X-Y}q^{(-3s+4)X}q^{-2Y}q^{s\min\{X,2Y\}}q^{s\min\{X,4Y\}}\\
&=q^{-2}(1-q^{-1})^2\sum_{(X,Y)\in\bb{N}^2}q^{(-3s+3)X}q^{-3Y}q^{s\min\{X,2Y\}}q^{s\min\{X,4Y\}}.
\end{align*}
We first need to compute $\sum_{(X,Y)\in\bb{N}^2}a^Xb^Yc^{\min\{X,2Y\}}c^{\min\{X,4Y\}}$ similarly to
Lemma~\ref{Lem identities}~(3) and then apply with $a=q^{-3s+3},~b=q^{-3}$ and $c=q^{s}$ to obtain $J_1$.
We record $J_1$ in the Appendix.\\
In computing $J_2$, notice
that in this case, on each coset $(a_1,a_2)+\p^2$ we have $xz_2\equiv\! 1\mod\p$. We change variable $v=xz_2-1\in\p$. Then
$A_2=\|y,v\|^s, B=\|u,y^3,y^2v\|^{-s}$, $C_2=\|y^4,yu,uv\|^{-s}$ and
$$J_2=q^{-1}\int_{u,y,v\in\p}|u|^{3s-4}|y|^2A_2BC_2d\mu.$$
Since $\mu\{(u,y,v)\in\p\mid v(u)=X,~v(y)=Y,~v(v)=Z \}=(1-q^{-1})^3q^{-X-Y-Z}$,
one has
\begin{align*}
J_2 =
q^{-1}(1-q^{-1})^3\sum_{(X,Y,Z)\in\p}&\left(q^{(-3s+3)X}q^{-3Y}q^{-Z}
q^{-s\min\{\min\{Y,Z\} }\right.\\
&\times\left. q^{s\min\{X,3Y,2Y+Z\}}q^{s\min\{X+Y,X+Z,4Y\}}\right).
\end{align*}
Again computing
$\sum_{(X,Y,Z)\in\bb{N}^3}a^Xb^Yc^Zd^{-\min\{Y,Z\}}d^{\min\{X,3Y,2Y+Z\}}d^{\min\{X+Y,X+Z,4Y\}}$
and then applying for $a=q^{-3s+3},~b=q^{-3},~c=q^{-1}$ and $d=q^s$ yields $J_2$ which we
record in the Appendix.

Summing up $\mathcal{Z}=\mathcal{Z}_1+\mathcal{Z}_2+\mathcal{Z}_3$, we obtain
$$\zeta_{\mathbf{H}(\o[x]/(x^3))}(s)=1+(1-q^{-1})^{-1}\mathcal{Z}=\frac{(1-t)(1-q^2t^2)(1-q^4t^3)}{(1-qt)(1-q^3t^2)(1-q^5t^3)}.$$
Hence
\begin{equation}\label{n=3}
\zeta_{\mathbf{H}(\O[x]/(x^3))}(s)=\frac{\zeta_K(s-1)}{\zeta_K(s)}\cdot\frac{\zeta_K(2s-3)}{\zeta_K(2s-2)}\cdot\frac{\zeta_K(3s-5)}{\zeta_K(3s-4)}.
\end{equation}
\section{Open questions}
\subsection{Heisenberg group scheme}\label{sec questions Heisenberg}
Formulae \eqref{H over O},~\eqref{n=2} and \eqref{n=3} agree in the following conjectured formula.
\begin{conj}\label{conj Heisenberg over n}
The representation zeta function of $\mathbf{H}(\O[x]/(x^n))$ is
\begin{equation}\label{eq conj Hn}
\zeta_{\mathbf{H}(\O[x]/(x^n))}(s)=\prod_{i=1}^{n}\frac{\zeta_K(is-2i+1)}{\zeta_K(is-2i+2)}.
\end{equation}

\end{conj}
If  Conjecture~\ref{conj Heisenberg over n} holds, then the zeta functions $\zeta_{\mathbf{H}(\O[x]/(x^n))}(s)$ shares uniform analytic properties with $\zeta_{\mathbf{H}(\O)}(s)$ as
suggested by the following conjecture.
\begin{conj}\label{conj R}
Let $R$ be a ring which is torsion-free finitely generated over $\bb{Z}$. Then
the representation zeta function  $\zeta_{\mathbf{H}(R)}(s)$ has the following properties:
\begin{enumerate}
\item Its abscissa of convergence is $\alpha(\mathbf{H})=2$.
\item It can be analytically continued to the whole complex plane. The continued zeta function has no singularities on the line
$\rea(s)=2$, apart from a simple pole at $s=\alpha(\mathbf{H})$.
\end{enumerate}
\end{conj}
\iffalse
Using the methods used above, the computations of $\zeta_{\mathbf{H}(\O[x]/(x^n))}(s)$ become hard for $n\geq 4$ since
the polynomials $F_j(\mathbf{Y})$ become more complicated.
It is very interesting to consider the conjectured local zeta function
and take its limit as $n$ tends to $\infty$.
One promising approach is to combine methods from motivic integration theory and logic to parametrize
the representations of $\mathbf{H}(\O_\p[[x]])$ as definable sets over $\O_\p[[x]]$, or one needs
to define an integral over finite-dimensional local rings that will give something
like an infinite product of zeta functions.
\fi
Recall that the representation zeta function of $\mathbf{H}(\bb{Z}_p)$ is
$$\zeta_{\mathbf{H}(\bb{Z}_p)}(s)= 1+(1-p^{-1})^{-1}\int_{\substack{x\in p\bb{Z}_p\\ y\in W_1(\bb{Z}_p)}}|x|^{s-2}d\mu=\frac{1-p^{-s}}{1-p^{1-s}}.$$

When computing the zeta function for $\mathbf{H}(\O_\p)$, one just needs to replace $p$ by
$q=|\O/\p|$ the residue field cardinality, $\bb{Z}_p$ by $\O_\p$, and replace $p$-adic norm $|.|$ by $\p$-adic norm $|.|_\p$  to
obtain
$$\zeta_{\mathbf{H}(\O_\p)}(s) = 1+(1-q^{-1})^{-1}\int_{\substack{x\in \p\O_\p\\ y\in W_1(\O_\p)}}|x|_\p^{s-2}d\mu=\frac{1-q^{-s}}{1-q^{1-s}}.$$
\begin{question}
Can one define the domain $W$, a \textit{valuation} and \textit{norm} $|.|_n$ on $\O_\p[x]/(x^n)$ compatible with the $\p$-adic
norm $|.|_\p$ such that the zeta function of $\mathbf{H}(\O_\p[x]/(x^n))$ can be computed as follows
$$\zeta_{\mathbf{H}(\O_\p[x]/(x^n))}(s)=1+(1-q^{-1})^{-1}\int_{W}|x|_n^{s-2}d\mu?$$
\end{question}

By expanding the conjectured formula \eqref{eq conj Hn} for the local zeta function, we get
\begin{equation}\label{eq conj Hn expand}
\zeta_{\mathbf{H}(\O_\p[x]/(x^n))}(s)=\sum_{I\subseteq[n-1]_0}f_I(q^{-1})\prod_{i\in I}\frac{q^{2n-2i-1-(n-i)s}}{1-q^{2n-2i-1-(n-i)s}},
\end{equation}
where $I$ runs over all subsets of $[n-1]_0:=\{0,1\cdots,n-1\}$ and $f_I(q^{-1})=(1-q^{-1})^{|I|}$. The formula \eqref{eq conj Hn expand}
looks similar to {\cite[(1.12)]{SV/14}}. However, we have been unable to mimic the inductive proof of {\cite[Theorem C]{SV/14}}
to yield \eqref{eq conj Hn expand}.

\subsection{Unipotent group schemes}\label{sec unipotent}
Once we understand the zeta function of $\mathbf{H}(\O_\p[x]/(x^n))$,
the next step is the following.

Let $\Lambda$ be a finitely generated free and torsion-free $\O$-Lie
lattice of nilpotency class $c$ and $\O$-rank $h$. If $c>2$ we assume
that $\Lambda':=[\Lambda,\Lambda]\subseteq c!\Lambda$. This enables us
to associate to $\Lambda$ a unipotent group scheme $\G:=\G_\Lambda$
(cf.\ {\cite[Section 2.1.2]{SV/14}}). The group $\G(\O)$ is a $\mathcal{T}$-group
of nilpotency class $c$ and Hirsch length $h\cdot[K:\bb{Q}]$.
The Heisenberg group scheme $\mathbf{H}$ is an example of such a unipotent group scheme.

The zeta function $\zeta_{\G(\O)}(s)$ is the Euler product
$$\zeta_{\G(\O)}(s)=\prod_{\p\in\Spec(\O)}\zeta_{\G(\O_\p)}(s)$$
ranging over all nonzero prime ideals $\p$ of $\O$; cf.~{\cite{SV/14}}. There exists a finite
set $S$ of prime ideals such that for each $\p\notin S$, the local zeta function $\zeta_{\G(\O_\p)}(s)$ is a rational function
in $q^{-s}$ and satisfies a functional equation upon inversion of $q$; see~{\cite[Theorem A]{SV/14}}.
Moreover, each  such local representation zeta function can be expressed in terms of a $\p$-adic integral; cf.~{\cite[Corollary 2.11]{SV/14}}.

It is tempting to investigate the zeta function of $\G(\O[x]/(x^n))$ when $n$ tends to $\infty$. As we have seen for the Heisenberg group scheme, if
Conjecture~\ref{conj Heisenberg over n} holds then
$$\varinjlim_n\zeta_{\mathbf{H}(\O[x]/(x^n))}(s)=\prod_{i=1}^\infty\frac{\zeta_K(is-2i+1)}{\zeta_K(is-2i+2)}.$$
It is natural to ask the following.
\begin{question}
Let $\G$ be a unipotent group scheme as above.
\begin{enumerate}
\item Is $\G(\O[[x]])$  twist-rigid, that is, is $r_n(\G(\O[[x]]))$ finite for every $n\in\bb{N}$? Does $\G(\O[[x]])$ have polynomial representation growth?
\item If (1) has positive answers then is it true that
$$\zeta_{\G(\O[[x]])}(s):=
\sum_{n=1}^\infty r_n(\G(\O[[x]]))n^{-s}=\varinjlim_n\zeta_{\G(\O[x]/(x^n))}(s)?$$
\end{enumerate}
\end{question}

It is proved in {\cite[Theorem A]{DV/15}} that the zeta function $\zeta_{\G(\O)}(s)$
has rational abscissa of convergence $\alpha(\G)$, which is independent on the number field $K$,
and can be analytically continued to  $\rea(s)>\alpha(\G)-\delta(\G)$ for some $\delta(\G)>0$.
In the spirit of Conjecture~\ref{conj R}, we formulate the following.
\begin{conj}
Let $R$ be a ring which is finitely generated torsion-free $\O$-module. Then the
representation zeta function of $\G(R)$ has the following properties:
\begin{enumerate}
\item The abscissa of convergence $\alpha(\G)$ of $\zeta_{\G(R)}(s)$ is independent of $R$.
\item It can be meromorphically continued to the half-plane $\rea(s)>\alpha(\G)-\delta(\G)$ for some $\delta(\G)>0$, where $\delta(\G)$ is independent of $K$.
\end{enumerate}
\end{conj}

\subsection{Topological representation zeta functions}
Topological zeta functions offer a way to define a limit as $p\to 1$ of families
of $p$-adic zeta functions.
Let $\G$ be a unipotent group scheme defined as in Section~\ref{sec unipotent}.
In \cite{Rossmann/15}, Rossmann introduces and studies topological
representation zeta functions associated to unipotent  group schemes
$\G$.
Informally, we define the topological representation zeta function
$\zeta_{\G,\textrm{top}}(s)$ to be the constant term of $\zeta_{\G(\bb{Z}_p)}(s)$ as
a series in $p-1$.
\begin{example}
Consider the Heisenberg group scheme $\mathbf{H}$. Expanding $p^z=(1+(p-1))^z$ into a series in $p-1$, we obtain
$\zeta_{\mathbf{H}(\bb{Z}_p)}(s)=\frac{s}{s-1}+O(p-1)$ and hence
$\zeta_{\mathbf{H},\textrm{top}}(s)=\frac{s}{s-1}$.

Let $\bb{Q}[\varepsilon_n]=\bb{Q}[x]/(x^n)$ and for a $\bb{Q}$-algebra $\g$, let $\g[\varepsilon_n]=\g\otimes_\bb{Z}\bb{Q}[\varepsilon_n]$
regarded as a $\g$-Lie lattice. Let $\mathbf{H}[\varepsilon_n]$ denote the group attached to $\g[\varepsilon_n]$; cf.~{\cite[Section~7]{Rossmann/15}}.
Then $\mathbf{H}[\varepsilon_n](\bb{Z}_p)=\mathbf{H}(\bb{Z}_p[x]/(x^n))$. Hence
\begin{align}
\zeta_{\mathbf{H}[\varepsilon_2],\textrm{top}}(s)&=\frac{s(2s-2)}{(s-1)(2s-3)}=\frac{2s}{2s-3},\label{eq top}\\
\zeta_{\mathbf{H}[\varepsilon_3],\textrm{top}}(s)&=\frac{s(2s-2)(3s-4)}{(s-1)(2s-3)(3s-5)}=\frac{2s(3s-4)}{(2s-3)(3s-5)}.\label{eq top2}
\end{align}
\end{example}
An algorithm to compute topological representation zeta functions is implemented in \cite{Rossmann-zeta}.
Notice that Rossmann's method computes topological representation zeta function directly without
first computing the corresponding $\p$-adic representation zeta function.
Formulae \eqref{eq top} and \eqref{eq top2} are consistent with computation results in \cite{Rossmann-zeta}. Notice also that \cite{Rossmann-zeta}
can only compute  the topological representation zeta function of $\mathbf{H}[\varepsilon_n]$ up to $n=3$ as we
have done here for their $\p$-adic representation zeta functions.  Conjecture~\ref{conj Heisenberg over n}
suggests the following analogue for topological representation zeta functions of $\mathbf{H}[\varepsilon_n]$.
\begin{conj}
The topological representation zeta function of $\mathbf{H}[\varepsilon_n]$ is
$$\zeta_{\mathbf{H}[\varepsilon_n],\mathrm{top}}(s)=\prod_{i=1}^n\frac{is-2i+2}{is-2i+1}.$$
\end{conj}
\begin{remark}
All Questions in {\cite[Section~7]{Rossmann/15}}, except Question $7.3$ which is not yet known, have positive answers for $\G=\mathbf{H}[\varepsilon_n]$ with $n\leq 3$.
\end{remark}

redpajama_set_name: RedPajamaArXiv


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Good Morning! Eileen here from GoofingOff Sewing with something to share.
Everyone has a challenging moment when it come to sewing and, for many, cutting knit fabric is one. Here is a good tutorial from Tilly and the Buttons to use.
In the tutorial a rotary cutter is used. Someone commented that she has a problem using her cutter and wasn't sure if it was due to a dull blade. Unfortunately, rotary blades dull quickly (or at least it seems quick). On a personal level, I struggled with my cutter until I treated myself to a new blade. What a difference a $6 purchase makes……Just a thought.
Another thing she recommends is pre-washing the fabric. Please, pre-wash your fabric if it is washable (you folks who work with silks, rayon, and other fabric that hate water might want to ignore this). Some fabric shrink and "act up" when washed so pre-washing will be one less aggravation when you have your finished product….as in puckering where you didn't plan to pucker.
So, check the tutorial. It is worth a look-see. Happy sewing!


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My second attempt at weaving twill has not gone according to plan, and surprise! The result is still pretty cool. I was attempting to do Halvorsen's #5 Pebble Weave from Marguerite Davison's book, but I threaded the loom backwards (4321 instead of 1234).
I tried to reverse the tie-up and treadling, but nothing made my textile look like the picture. In the end, I played around with random treadling until I found something I liked.
The weft is Shetland weaving yarn from the Pendleton wool mill in Portland. I dyed it bright orange, and it's an unintentional perfect match for the binding of my indigo quilt.
I'm planning to take advantage of the color match. I'll felt the final fabric and use it to make pillows and a throw blanket to go with the quilt.
In other news, Dandylion came home from the vet on Friday. He's in pretty rough shape, but he's made huge improvements over the last two days.
I thought I'd share a quick explanation of the first few phases of the process known as "warping." Warping is the process of getting yarn measured, organized, threaded, and properly tensioned on your loom so you can begin to weave.
Warping isn't my favorite part of weaving. It's repetitive, involves math, requires your full concentration, and almost never comes out completely perfect the first time. All that being said, I love weaving, and warping is the price I pay to get to the fun part. I just finished getting 13 yards of orange, rust, maroon, and brown on the loom, so these pictures are from that process.
My first step in choosing my warp is to pull yarns out of my stash that are the right color and will work for my project. I should warn you that I'm a bit of a fast-and-loose yarn selector, and that gets me into trouble once in a while. I mix yarns specifically spun for weaving with knitting yarns and novelty yarns. Some weavers will tell you that this sort of blending will all end in tears, but that hasn't been my experience.
Heddles have an eye (like a needle's eye) in the middle. They sit on shafts which are moved up and down by your feet.
If all this sounds a bit complicated, that's because it is… but it's also really cool at the same time. I highly recommend taking any opportunity that comes your way to see a loom in action. In the U.S., many weaver's guild do demonstrations at state fairs. It's a great way to see the entire process of making cloth from fiber to yarn to finished fabric.


text:
You are in: Home › Dundee Science Centre News › Dundee Science Centre Welcomes Breastfeeding Mothers
Dundee Science Centre Welcomes Breastfeeding Mothers
Published: 9.48am, Wednesday 23rd October 2019
Dundee Science Centre has been awarded a special accreditation for its commitment to breastfeeding mothers.
The centre, which will shortly commence a major refurbishment of its ground floor to improve accessibility, welcomed 10 local mums today to celebrate their new Breastfeeding Friendly Scotland sticker from Dundee Breast Buddies.
Andrea Campomanes, Interpretation Manager at Dundee Science Centre said, "Inviting currently under-served audiences to take an active role in the design and redevelopment of Dundee Science Centre is core to our plans of becoming a safe, welcoming and accessible place for all. Everybody has the right to enjoy science and to experience the benefits of becoming a more confident and informed individual, and we are working hard to ensure that the new centre overcomes the challenges and social barriers that have been built since we opened in 2000. The invaluable support of Dundee Breast Buddies will provide us with the knowledge to adapt our facilities for the many people who deserve this change."
Fiona Stewart of Dundee Breast Buddies added, "Home-Start Dundee Breast Buddies Support groups were delighted to have been invited to assess the accessibility of the space within Dundee Science Centre for breastfeeding mothers and to play an active part in the future design of the centre with breastfeeding in mind. It is so important for local businesses and premises to recognise that they have a responsibility to ensure that the facilities they provide are tailored to meet the needs of breastfeeding mothers and their families/support network. We live on a culture that does not promote or protect breastfeeding so it is refreshing that Dundee Science Centre is making efforts to ensure their facilities meet the needs of the people who will be using it."
The £2.1M revamp will see the ground floor developed into a custom-built hub with improved scope for events, talks and workshops celebrating community work, as well as a new welcome area, under 7's exhibition area, sensory zone and new exhibition space.
A pioneering network of various assisted needs groups is shaping the future of the visitor attraction, brought together by Tutti LED (Learn, Enable, Design) by Advocating Together and comprises charities representing the needs of people with learning, physical, complex, profound and multiple disabilities, autism, hearing and visual impairments, mental health and sensory challenges.
The refurb, due to be unveiled in Spring 2021, is funded with £1.45M support from the Inspiring Science Fund, a collaborative initiative between the Department for Business, Energy & Industrial Strategy, UK Research and Innovation and Wellcome.
For more information, telephone (01382) 228800 or email staff@dundeesciencecentre.org.uk.


Degree/Diploma in Nursing) or Equivalent qualification that allows registration with the SANC as A Professional Nurse.
Current registration with the SANC.
A minimum of 9 years appropriate/recognisable experience in nursing after registration as a Professional Nurse with the SANCE in General Nursing.
At least 4 years of the period referred to above must be Appropriate/recognisable experience at management level.
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Provide relevant health promotion to health care users to assist in achieving optimal health care and rehabilitation of patient.
Maintain 63 constructive working relationship with nursing and other stakeholders (i.
inter-professional, inter-sectorial and multidisciplinary teamwork).
Participate in the analysis, formulation and implementation of nursing guidelines, practices, standards and procedures.
Manage and monitor proper utilisation of Human, Financial and Physical resources.


I totally can't believe it! I am so pleased to announce that I have won a Dundee Craft Makers Award for this year. 2013 is off to a great start so far. It's been a good week for my Genna Design jewellery business as I've been busy working on commissions, found out about winning an award and on Monday I was also featured in the Courier.


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text:
"""Support for Yamaha Receivers."""
import logging

import requests
import voluptuous as vol

from homeassistant.components.media_player import MediaPlayerDevice, PLATFORM_SCHEMA
from homeassistant.components.media_player.const import (
    DOMAIN,
    MEDIA_TYPE_MUSIC,
    SUPPORT_NEXT_TRACK,
    SUPPORT_PAUSE,
    SUPPORT_PLAY,
    SUPPORT_PLAY_MEDIA,
    SUPPORT_PREVIOUS_TRACK,
    SUPPORT_SELECT_SOURCE,
    SUPPORT_STOP,
    SUPPORT_TURN_OFF,
    SUPPORT_TURN_ON,
    SUPPORT_VOLUME_MUTE,
    SUPPORT_VOLUME_SET,
    SUPPORT_SELECT_SOUND_MODE,
)

from homeassistant.const import (
    ATTR_ENTITY_ID,
    CONF_HOST,
    CONF_NAME,
    STATE_IDLE,
    STATE_OFF,
    STATE_ON,
    STATE_PLAYING,
)
import homeassistant.helpers.config_validation as cv

_LOGGER = logging.getLogger(__name__)

ATTR_ENABLED = "enabled"
ATTR_PORT = "port"

CONF_SOURCE_IGNORE = "source_ignore"
CONF_SOURCE_NAMES = "source_names"
CONF_ZONE_IGNORE = "zone_ignore"
CONF_ZONE_NAMES = "zone_names"

DATA_YAMAHA = "yamaha_known_receivers"
DEFAULT_NAME = "Yamaha Receiver"

MEDIA_PLAYER_SCHEMA = vol.Schema({ATTR_ENTITY_ID: cv.comp_entity_ids})

ENABLE_OUTPUT_SCHEMA = MEDIA_PLAYER_SCHEMA.extend(
    {vol.Required(ATTR_ENABLED): cv.boolean, vol.Required(ATTR_PORT): cv.string}
)

SERVICE_ENABLE_OUTPUT = "yamaha_enable_output"

SUPPORT_YAMAHA = (
    SUPPORT_VOLUME_SET
    | SUPPORT_VOLUME_MUTE
    | SUPPORT_TURN_ON
    | SUPPORT_TURN_OFF
    | SUPPORT_SELECT_SOURCE
    | SUPPORT_PLAY
    | SUPPORT_SELECT_SOUND_MODE
)

PLATFORM_SCHEMA = PLATFORM_SCHEMA.extend(
    {
        vol.Optional(CONF_NAME, default=DEFAULT_NAME): cv.string,
        vol.Optional(CONF_HOST): cv.string,
        vol.Optional(CONF_SOURCE_IGNORE, default=[]): vol.All(
            cv.ensure_list, [cv.string]
        ),
        vol.Optional(CONF_ZONE_IGNORE, default=[]): vol.All(
            cv.ensure_list, [cv.string]
        ),
        vol.Optional(CONF_SOURCE_NAMES, default={}): {cv.string: cv.string},
        vol.Optional(CONF_ZONE_NAMES, default={}): {cv.string: cv.string},
    }
)

def setup_platform(hass, config, add_entities, discovery_info=None):
    """Set up the Yamaha platform."""
    import rxv

    # Keep track of configured receivers so that we don't end up
    # discovering a receiver dynamically that we have static config
    # for. Map each device from its zone_id to an instance since
    # YamahaDevice is not hashable (thus not possible to add to a set).
    if hass.data.get(DATA_YAMAHA) is None:
        hass.data[DATA_YAMAHA] = {}

    name = config.get(CONF_NAME)
    host = config.get(CONF_HOST)
    source_ignore = config.get(CONF_SOURCE_IGNORE)
    source_names = config.get(CONF_SOURCE_NAMES)
    zone_ignore = config.get(CONF_ZONE_IGNORE)
    zone_names = config.get(CONF_ZONE_NAMES)

    if discovery_info is not None:
        name = discovery_info.get("name")
        model = discovery_info.get("model_name")
        ctrl_url = discovery_info.get("control_url")
        desc_url = discovery_info.get("description_url")
        receivers = rxv.RXV(
            ctrl_url, model_name=model, friendly_name=name, unit_desc_url=desc_url
        ).zone_controllers()
        _LOGGER.debug("Receivers: %s", receivers)
        # when we are dynamically discovered config is empty
        zone_ignore = []
    elif host is None:
        receivers = []
        for recv in rxv.find():
            receivers.extend(recv.zone_controllers())
    else:
        ctrl_url = "http://{}:80/YamahaRemoteControl/ctrl".format(host)
        receivers = rxv.RXV(ctrl_url, name).zone_controllers()

    devices = []
    for receiver in receivers:
        if receiver.zone in zone_ignore:
            continue

        device = YamahaDevice(name, receiver, source_ignore, source_names, zone_names)

        # Only add device if it's not already added
        if device.zone_id not in hass.data[DATA_YAMAHA]:
            hass.data[DATA_YAMAHA][device.zone_id] = device
            devices.append(device)
        else:
            _LOGGER.debug("Ignoring duplicate receiver: %s", name)

    def service_handler(service):
        """Handle for services."""
        entity_ids = service.data.get(ATTR_ENTITY_ID)

        devices = [
            device
            for device in hass.data[DATA_YAMAHA].values()
            if not entity_ids or device.entity_id in entity_ids
        ]

        for device in devices:
            port = service.data[ATTR_PORT]
            enabled = service.data[ATTR_ENABLED]

            device.enable_output(port, enabled)
            device.schedule_update_ha_state(True)

    hass.services.register(
        DOMAIN, SERVICE_ENABLE_OUTPUT, service_handler, schema=ENABLE_OUTPUT_SCHEMA
    )

    add_entities(devices)

class YamahaDevice(MediaPlayerDevice):
    """Representation of a Yamaha device."""

    def __init__(self, name, receiver, source_ignore, source_names, zone_names):
        """Initialize the Yamaha Receiver."""
        self.receiver = receiver
        self._muted = False
        self._volume = 0
        self._pwstate = STATE_OFF
        self._current_source = None
        self._sound_mode = None
        self._sound_mode_list = None
        self._source_list = None
        self._source_ignore = source_ignore or []
        self._source_names = source_names or {}
        self._zone_names = zone_names or {}
        self._reverse_mapping = None
        self._playback_support = None
        self._is_playback_supported = False
        self._play_status = None
        self._name = name
        self._zone = receiver.zone

    def update(self):
        """Get the latest details from the device."""
        try:
            self._play_status = self.receiver.play_status()
        except requests.exceptions.ConnectionError:
            _LOGGER.info("Receiver is offline: %s", self._name)
            return

        if self.receiver.on:
            if self._play_status is None:
                self._pwstate = STATE_ON
            elif self._play_status.playing:
                self._pwstate = STATE_PLAYING
            else:
                self._pwstate = STATE_IDLE
        else:
            self._pwstate = STATE_OFF

        self._muted = self.receiver.mute
        self._volume = (self.receiver.volume / 100) + 1

        if self.source_list is None:
            self.build_source_list()

        current_source = self.receiver.input
        self._current_source = self._source_names.get(current_source, current_source)
        self._playback_support = self.receiver.get_playback_support()
        self._is_playback_supported = self.receiver.is_playback_supported(
            self._current_source
        )
        surround_programs = self.receiver.surround_programs()
        if surround_programs:
            self._sound_mode = self.receiver.surround_program
            self._sound_mode_list = surround_programs
        else:
            self._sound_mode = None
            self._sound_mode_list = None

    def build_source_list(self):
        """Build the source list."""
        self._reverse_mapping = {
            alias: source for source, alias in self._source_names.items()
        }

        self._source_list = sorted(
            self._source_names.get(source, source)
            for source in self.receiver.inputs()
            if source not in self._source_ignore
        )

    @property
    def name(self):
        """Return the name of the device."""
        name = self._name
        zone_name = self._zone_names.get(self._zone, self._zone)
        if zone_name != "Main_Zone":
            # Zone will be one of Main_Zone, Zone_2, Zone_3
            name += " " + zone_name.replace("_", " ")
        return name

    @property
    def state(self):
        """Return the state of the device."""
        return self._pwstate

    @property
    def volume_level(self):
        """Volume level of the media player (0..1)."""
        return self._volume

    @property
    def is_volume_muted(self):
        """Boolean if volume is currently muted."""
        return self._muted

    @property
    def source(self):
        """Return the current input source."""
        return self._current_source

    @property
    def sound_mode(self):
        """Return the current sound mode."""
        return self._sound_mode

    @property
    def sound_mode_list(self):
        """Return the current sound mode."""
        return self._sound_mode_list

    @property
    def source_list(self):
        """List of available input sources."""
        return self._source_list

    @property
    def zone_id(self):
        """Return a zone_id to ensure 1 media player per zone."""
        return "{0}:{1}".format(self.receiver.ctrl_url, self._zone)

    @property
    def supported_features(self):
        """Flag media player features that are supported."""
        supported_features = SUPPORT_YAMAHA

        supports = self._playback_support
        mapping = {
            "play": (SUPPORT_PLAY | SUPPORT_PLAY_MEDIA),
            "pause": SUPPORT_PAUSE,
            "stop": SUPPORT_STOP,
            "skip_f": SUPPORT_NEXT_TRACK,
            "skip_r": SUPPORT_PREVIOUS_TRACK,
        }
        for attr, feature in mapping.items():
            if getattr(supports, attr, False):
                supported_features |= feature
        return supported_features

    def turn_off(self):
        """Turn off media player."""
        self.receiver.on = False

    def set_volume_level(self, volume):
        """Set volume level, range 0..1."""
        receiver_vol = 100 - (volume * 100)
        negative_receiver_vol = -receiver_vol
        self.receiver.volume = negative_receiver_vol

    def mute_volume(self, mute):
        """Mute (true) or unmute (false) media player."""
        self.receiver.mute = mute

    def turn_on(self):
        """Turn the media player on."""
        self.receiver.on = True
        self._volume = (self.receiver.volume / 100) + 1

    def media_play(self):
        """Send play command."""
        self._call_playback_function(self.receiver.play, "play")

    def media_pause(self):
        """Send pause command."""
        self._call_playback_function(self.receiver.pause, "pause")

    def media_stop(self):
        """Send stop command."""
        self._call_playback_function(self.receiver.stop, "stop")

    def media_previous_track(self):
        """Send previous track command."""
        self._call_playback_function(self.receiver.previous, "previous track")

    def media_next_track(self):
        """Send next track command."""
        self._call_playback_function(self.receiver.next, "next track")

    def _call_playback_function(self, function, function_text):
        import rxv

        try:
            function()
        except rxv.exceptions.ResponseException:
            _LOGGER.warning("Failed to execute %s on %s", function_text, self._name)

    def select_source(self, source):
        """Select input source."""
        self.receiver.input = self._reverse_mapping.get(source, source)

    def play_media(self, media_type, media_id, **kwargs):
        """Play media from an ID.

        This exposes a pass through for various input sources in the
        Yamaha to direct play certain kinds of media. media_type is
        treated as the input type that we are setting, and media id is
        specific to it.

        For the NET RADIO mediatype the format for ``media_id`` is a
        "path" in your vtuner hierarchy. For instance:
        ``Bookmarks>Internet>Radio Paradise``. The separators are
        ``>`` and the parts of this are navigated by name behind the
        scenes. There is a looping construct built into the yamaha
        library to do this with a fallback timeout if the vtuner
        service is unresponsive.

        NOTE: this might take a while, because the only API interface
        for setting the net radio station emulates button pressing and
        navigating through the net radio menu hierarchy. And each sub
        menu must be fetched by the receiver from the vtuner service.

        """
        if media_type == "NET RADIO":
            self.receiver.net_radio(media_id)

    def enable_output(self, port, enabled):
        """Enable or disable an output port.."""
        self.receiver.enable_output(port, enabled)

    def select_sound_mode(self, sound_mode):
        """Set Sound Mode for Receiver.."""
        self.receiver.surround_program = sound_mode

    @property
    def media_artist(self):
        """Artist of current playing media."""
        if self._play_status is not None:
            return self._play_status.artist

    @property
    def media_album_name(self):
        """Album of current playing media."""
        if self._play_status is not None:
            return self._play_status.album

    @property
    def media_content_type(self):
        """Content type of current playing media."""
        # Loose assumption that if playback is supported, we are playing music
        if self._is_playback_supported:
            return MEDIA_TYPE_MUSIC
        return None

    @property
    def media_title(self):
        """Artist of current playing media."""
        if self._play_status is not None:
            song = self._play_status.song
            station = self._play_status.station

            # If both song and station is available, print both, otherwise
            # just the one we have.
            if song and station:
                return "{}: {}".format(station, song)

            return song or station

redpajama_set_name: RedPajamaGithub


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Intel has developed a new prototype technology which claimed to do away with password for online banking, social networks and email, and instead provide access to them by just waving of hands.
The prototype technology, known as Client Based Authentication Technology, will replace passwords as well as enhance the process for accessing bank accounts, stock portfolios and other cloud-based personal data, Intel said.
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Strange particles believed to be emanating from the elusive 'dark matter' at the center of the galaxy have been confirmed to be bombarding Earth. Two physicists from the prestigious Department of Physics & Astronomy at University of California discovered more gamma-rays bombarding Earth than believed by astronomers. The stream of electromagnetic radiation (spewed by the process of intensive radioactive decay and other other high-energy emmisions) may be triggering the bizarre mutation of matter and an incredible energy cloud that an astrophysicist says threatens Earth.