text:
G.M. Says New Wave of Trump Tariffs Could Force U.S. Job Cuts
A GMC Sierra truck on display in Detroit in March. General Motors said tariffs under consideration by the Trump administration could force individual vehicle prices up thousands of dollars.
General Motors warned Friday that if President Trump pushed ahead with another wave of tariffs, the move could backfire, leading to "less investment, fewer jobs and lower wages" for its employees.
The automaker said that the president's threat to impose tariffs on imports of cars and car parts — along with an earlier spate of penalties — could drive vehicle prices up by thousands of dollars. The "hardest hit" cars, General Motors said in comments submitted to the Commerce Department, are likely to be the ones bought by consumers who can least afford an increase. Demand would suffer and production would slow, all of which "could lead to a smaller G.M."
The president has promoted tariffs as a way to protect American businesses and workers, aiming at dozens of nations with metal tariffs, as well as bringing broader levies against Chinese goods. But companies, which rely on other markets for sales, production and materials, have been increasingly vocal about the potential damage from his policies.
The warning by G.M., echoed in comments by trade groups and other automakers, could test the president's aggressive approach to trade and his commitment to business. In the past, Mr. Trump has lauded General Motors for its job creation and vowed to defend the auto industry.
A G.M. spokeswoman, Dayna Hart, said that the company had no contingency plans calling for job cuts, but that such a move was "something that could happen."
"We are still assessing the impact," she added.
The White House did not respond to a request for comment.
G.M. and other industry leaders are caught in the middle of an escalating trade war that has prompted retaliation from the European Union, Mexico, Canada and China.
Last month, Mr. Trump ordered an investigation into whether imported cars and automotive components pose a national security risk, calling for penalties expected to be as high as 25 percent. The administration has already put levies on imported steel and aluminum, and is about place tariffs on $34 billion worth of Chinese goods.
G.M. and other automakers rely heavily on parts and materials from overseas to build their cars. The president's threat to pull out of the North American Free Trade Agreement could hurt the industry's supply chain, which integrates operations in the United States, Canada and Mexico.
"If there's a full-blown trade war, it will be pretty tough for the auto industry and consumers," said Michelle Krebs, an analyst at AutoTrader.com.
"Consumers are already facing headwinds in credit and average prices going up," she said. "If you add a tariff, my guess is a lot of people just won't buy new cars."
The increasingly global nature of automotive supply chains has left manufacturers especially exposed.
This week, Harley-Davidson said it would move some of its production outside of the United States to avoid retaliatory measures by the European Union. The company said it was the only "sustainable option" to "maintain a viable business in Europe."
The decision invoked the ire of the president, who quickly threatened punitive taxes. He accused the Wisconsin-based motorcycle maker, which he had cited as a poster child of American manufacturing, of having "surrendered."
Fiat Chrysler Automobiles, which has several plants in the United States, is also considering other manufacturing options, according to a Bloomberg interview with Bob Lee, an executive overseeing powertrains for the company.
Is Harley-Davidson Still an 'American Icon' to Trump?
President Trump has often praised the Wisconsin-based motorcycle manufacturer, but its recent plans to shift some of its production overseas has angered Mr. Trump.
"Harley-Davidson is a true American icon — one of the greats. So, thank you Harley-Davidson for building things in America." "They told me without even complaining that it's very hard to do business with other countries because they tax our goods at such a high rate." "President Trump is attacking one of his favorite American manufacturers: Harley-Davidson." "President Trump has a new target in the growing trade war and it is not a foreign adversary." "He is going to war with Harley-Davidson." "So, Harley-Davidson is using that as an excuse. And I don't like that because I've been very good to Harley-Davidson and they used that as an excuse. And I think the people that ride Harleys are not happy with Harley-Davidson and I wouldn't be either."
President Trump has often praised the Wisconsin-based motorcycle manufacturer, but its recent plans to shift some of its production overseas has angered Mr. Trump.CreditCredit...Carlos Barria/Reuters
"It's contingency planning on a massive scale — supply-based planning, logistics planning, vehicle-build location planning," Mr. Lee told Bloomberg on Thursday. "This is not trivial, and it's been going on for awhile."
G.M. and its chief executive, Mary T. Barra, have been at odds before with the Trump administration.
Last year, after Mr. Trump blamed both white supremacists and the groups protesting them for violence in Charlottesville, Va., Ms. Barra and other executives weighed disbanding an elite council formed to advise the president on economic issues. Mr. Trump shut down the group before they could dissolve it.
In January, Ms. Barra said Nafta should be modernized, but not scrapped entirely. Mr. Trump has suggested that the United States withdraw from the trade pact or renegotiate it.
In criticizing prospective auto tariffs, G.M. played heavily on its position as one of the country's largest employers. The company said it had 47 manufacturing locations, 25 service-part facilities and 110,000 employees in the United States, where it conducts most of its research and development, design, engineering and other work.
G.M. suggested that additional tariffs would put American companies at a disadvantage in the midst of a "fast-paced transportation revolution led by cutting-edge technologies." Its investments in jobs and operations at home, the carmaker said, are critical to this effort. "The economic fortitude of companies like ours directly supports the economic strength of the nation, which, in turn, contributes to the security posture of the United States," the company said.
The Commerce Department will hold public hearings in a few weeks on the auto tariffs, before it releases the results of its investigation. The government said it had received 2,500 comments already and expected more by the deadline on Friday.
"The purpose of the comment period," the commerce secretary, Wilbur Ross, said in a statement, was "to make sure that all stakeholders' views are heard, both pro and con."
"That will enable us to make our best informed recommendation to the president," he added.
In a highly intertwined, global car industry, a trade war can play out in unexpected and costly ways. American automakers, which export an estimated two million vehicles, increasingly rely on global sales as a buffer in tough times. Retaliatory tariffs from Europe or China could weaken their overseas business.
They also depend heavily on parts from foreign suppliers. G.M. has said that its supply chain sprawls across 20,000 businesses worldwide and is an operation of "great breadth, scope and complexity."
European and Japanese automakers are major employers in the United States, supplying hundreds of thousands of jobs. Daimler, the maker of Mercedes-Benz cars, recently issued a profit warning, in part blaming tit-for-tat measures for weakness in its sales of S.U.V.s, which are built in Alabama.
The German automaker BMW, which has a large factory in South Carolina, said in a comment about tariffs that the protectionist approach would make American companies less incentivized to improve their productivity.
"The deeper the economic ties from trade and integrated value chains, the more costly conflict would be, and therefore the more unlikely it is to occur," BMW said.
Toyota wrote in its submission to the Commerce Department that the cost of its popular Camry sedan would rise $1,800 if subject to new tariffs. The car is built in Kentucky while sourcing 30 percent of its materials from abroad.
The Japanese automaker said that the tariff, if approved, would be terribly timed. The auto industry is at a vulnerable point in its sales cycle, exiting a long stretch of record growth and heading into a period of weaker demand.
The Alliance of Automobile Manufacturers, a Washington-based trade group that represents foreign and American companies, was more specific.
With a 25 percent tariff, the average price of an imported vehicle would rise by $5,800, the group said in a statement. Annual sales would slide by one to two million vehicles. And production would slip 1.5 percent, causing 195,000 workers in the United States to lose their jobs.
No, Tariffs Aren't Going to Pay Down the National Debt
Listening to Bad Pitches and Asking One Question Too Many
As Fuel Prices Rise, Airlines Warn of Higher Fares
Instagram Account That Sought Harassment Tales May Be Unmasked
The Atlantic Cuts Ties With Conservative Writer Kevin Williamson
Martin Sorrell Is Gone From WPP, but That Doesn't Mean He's Gone Quiet


text:
cantilever table
Started as a cross collaboration between artist and designers Megan and Nicholas Tamas, the cantilever table was a project that had a very simple goal; to make a nice work table for their new studio space. Though both artists had experience in metal fabrication, they chose to divide the labor of design and fabrication between themselves. Nicholas Tamas worked as the principle designer of the project, layout out sketches and measurements for the table as well as cut lists and assembly orders that could then be handed off to Megan. Megan Tamas worked as the fabricator and principle craftsman of the project, using her years of experience and the blueprints for the design created by her brother, she set about crafting the table from reclaimed materials. Together they finished the table and successfully installed it within their new studio space, Neon Void Studio & Gallery.
megan tamas
Megan Tamas began her career as an artist by first attending the Tyler School of Art where she received her BA in Studio Sculpture. She followed this by then attending the prestigious Rhode Island School of Design where she received her MFA in Sculpture in 2015. Following this impressive academic trek, she then attended multiple studio residencies before landing her current job at the Massachusetts Museum of Contemporary Art as senior art fabricator. Among her many responsibilities, Megan has overseen the production and fabrication of many shows for many notable artists including: Nick Cave, Alex Dacorte, Elizabeth King, and many more. While continuing her career at Mass MoCA, Megan also found time to start her own studio gallery with her brother Nicholas, Neon Void Studio, in neighboring Adams, Massachusetts.
Although she has a variably impressive portfolio, Megan's' current work combines organic and inorganic forms to create bio-organic specimens that are reminiscent of creatures and forms found in Cult Sci-Fi and Mid-Century cinema. She presents her work by transforming the space to one more accommodating to her specimens and to give them the illusion of being factual scientific eperiments. It is her hope, that through her work she can inspire scientific and technological change as well as foster creativity in a growing scientific community focused on the synthetic growing and creation of new forms of life.
nicholas tamas
Nicholas Tamas began his professional artistic career at an early age when he attended the Tennessee Governor School for the Arts while still in high school. Following his graduation, he attended the University of Tennessee School of Art in Knoxville Tennessee where he majored in sculpture and printmaking. During his time at the University, he maximized his time by learning about as many materials and processes as he could. He also spent his summers furthering his academic career with residencies and internships in Italy and New York, and was instrumental in the planning of the Southern Graphics Council Conference that took place at his University his senior year. He finished his undergraduate degree in the winter of 2015 where he received Magna Cum Laude honors. Following his graduation, He began working at the Massachusetts Museum of Contemporary Art as an art fabricator and assistant lighting designer where he has built, designed, and lit shows for the Museums resident artists. The artists featured at the museum form an illustrious list and include artists like Nick Cave, Aled DaCorte, James Turrell and many others.
Nicholas Tamas' sculptural work is heavily inspired by cybernetic and bionic pop culture from the 80's and 90's as well as transhumanist philosophy. This is represented in his work as faux-cybernetics and bionics that are displayed in a manner that gives them the illusion of being functional and purchasable prosthesis/cybernetics. His goal with his work is to inspire the creation and advancement of real cybernetics and bionics as well as to further transhumanist ideals; to this end, he is currently pursuing a career and education as an industrial designer at the Rhode Island School of Design in Providence, Rhode Island.

redpajama_set_name: RedPajamaCommonCrawl


well… vacation was lovely. My family rocks… I got back tuesday and the trip was delightfully uneventful.
While I was gone, my co-worker friends were busy. Jordan went and made himself a blog (I knew he couldn't resist long) and he and Shroud are now my virtual neighbors at 1 and 1 hosting Oh and sedotn moved to a real host.


text:
Massive NFT collection crashes after creator's 'racist' cartoons unearthed
It may be the first time a high-profile NFT collection has been damaged by accusations of racism.
The Jungle Freaks NFTs have crashed in price.
Jungle Freaks
NFTs are steadily becoming less niche and more mainstream -- and with more exposure comes more responsibility. In what was the first example of a major nonfungible token brand contending with allegations of racism, a popular NFT collection saw a huge price crash on Nov. 1 after the designer behind it was accused of drawing racist cartoons in the 1970s.
The project in question is called Jungle Freaks, and is what's known as a "profile pic collection." It's a set of 10,000 hand-drawn NFTs, each with different properties that, like Pokemon cards, make some more valuable than others. Launching on Oct. 16, it had a "floor price" of 1.45 Ether as of Oct. 31. That means the cheapest, most common Jungle Freaks were selling for $6,260. Rarer NFTs were trading for much higher, like this Jungle Freak that sold for $95,000.
Much of the project's success was tied to George Trosley, the designer of the NFTs. Trosley is a famous cartoonist, whose works have appeared in Hustler magazine since its inception in 1974. On Nov. 1, some of those '70s Hustler works were circulated on Twitter and on several Discord servers. Due to their confronting racist imagery, which includes but is not limited to depictions of the Ku Klux Klan, I won't embed the cartoons in this story. You can find them here, but be warned. They are discomforting.
Their online circulation caused the price of Jungle Freaks to drop precipitously. The floor price dropped about 80%, to between 0.2 and 0.3, in the day following the cartoons' circulation. One owner to sell was actor Elijah Wood, who was known to hold several Jungle Freaks. He tweeted out a statement calling the racist cartoons "disturbing," adding that he's sold his Jungle Freaks NFTs and donated the funds to Black Lives Matter. At the time of this update, on Dec. 6, the floor is .299 ether ($1,209).
In a YouTube video published Nov. 2, Trosley and his son George III, who manages the Jungle Freaks collection, said the comics were intended to shine a light on racism in America during the '70s. "The illustrations that I was contacted to drawn, some over 40 years ago, have been taken out of context," Trosley wrote in a statement on Twitter. "Today's generation may not have an understanding of what was taking place in journalism and the world during my time at Hustler magazine."
In a followup Medium post, the senior Trosley wrote: "I am not a racist, nor was I ever a rogue racist cartoon artist during my time at Hustler Magazine; which unfortunately is the current narrative being portrayed within the NFT community." He added that Hustler publisher Larry Flynt was shot in 1978 by a white supremacist who was upset that the publication featured a spread including an interracial couple.
On Friday, Dec. 3., the Trosley's announced they had donated $100,000 to Florida Memorial University, an institution focused on scholarships and education for African Americans.
https://t.co/ODuYVSbK27 pic.twitter.com/DxYQRgeYTn
— TROSLEY // KING OF THE JUNGLE (@TrosleyNFT) November 2, 2021
pic.twitter.com/NAmoOuhsNw
— Elijah Wood (@elijahwood) November 1, 2021
JF x @FLMemorialUniv
❤🚀 pic.twitter.com/t22kc2ByvA
— JUNGLE FREAKS BY TROSLEY (@JungleFreaksNFT) December 3, 2021
While cryptocurrencies live or die based on hype or market sentiment, community involvement is a strong element of successful NFT collections. This is because a collection's value is to a large extent determined by the lowest price one owner is willing to sell it for. The price fluctuations of Jungle Freaks in the coming weeks and months will test how a high-profile, profit-seeking NFT community responds to accusations of racism.
Hustler didn't respond to a request for comment.
Update, Dec. 6: Adds Medium post and news of donation to Florida Memorial University.


I'm this month's talent on Light Knocks!
Light Knocks is an awesome website for creative people, run by Jerome and Tiffany, who describe their site as "a collective community of creative minds that are pursuing a common goal, living a purposeful life, collaborating our talents, showcasing our passions, and bringing awareness to our generation." At Light Knocks, you will find all kinds of awesome talent across the board of the creative industries. There's music, film, writing, and more.
In addition to artist spotlights, you will find inspirational quotes, advice, and a lot of thoughtful articles to help keep your creative juices flowing. There's something at Light Knocks for everyone.
I'm absolutely thrilled to be featured on Light Knocks this month! A big thanks to Jerome and Tiffany for having me!
OK, so head on over to Light Knocks and check it out!


View Poll Results: Do you find this ROM better than others considering GPS, batery life and feel?
This can be installed over 2.2.
Reverted gps drivers to JPY - please report on performance.
Disclaimer: If this shall brick your device, make you break up with your girlfriend, make the cat fly, the TV explode and wipe the mankind from the surface of the Earth, I am not responsible!
The ROM was tested on my Captivate. I installed several times to make sure it behaves.
To exclude one cause of battery drain issues always flash this ROM with FULL BATTERY!
This will wipe your /system, /data, /dbdata, /cache.
Internal/External SDcard will remain untouched.
1. Pull out your SIM and external sdcard!
2. Return to stock (those without 3 button combo, use this) <- If you are coming from another i9000 ROM and if you dont have lagfix enabled, this step is not required.
4. Flash from CWM Recovery.
Basic ROM comes with K12H speedmod kernel and I9000UGJL2 modem.
This is JPY deodexed and rooted. Nothing else.
First, if you install a different kernel, don't blame the ROM if you have issues.
If you changed system apps (apks) or framework other than what is posted in this thread, don't blame the ROM or the addon pack for making your phone an useless piece of plastic.
If you changed system binaries (drivers, applications -non apks-), again, don't blame the ROM for temporary killing your device.
You may put up a post with your problem and with the changes you made to the OS by yourself. I will do my best to get you out of the situation but don't first rate the ROM as worse and then leave. Everything can be fixed so give me a chance!
I cannot uninstall system apps with TitaniumBackup. Why? Because TiBu does not mount the /system as RW. Will discuss this with Joel. Workaround: go to SGS Toolbox and tap on mount /system as rw (read/write) then uninstall the applications you want from TiBu. Restart the phone and you're done.
My bluetooth doesn't work: Yes it does! I'm using it myself.
I cannot send MMS. Make sure you have your phone configured with the correct Access Point Name and the correct outgoing server number.
Stock email doesn't work/doesn't support exchange. There are several reports that the email app works fine and this ROM was the only one they could run Exchange on so make sure your provider doesn't play with the towers configuration.
The phone is telling me that I was 50% of the time without signal. Why? This could be related to the modem. I am now testing JK4. If you have this bug download and flash UGJK4 modem . As well, make sure that you have WCDMA 850MHz selected by typing *#2263# - COMBI BANDS - WCDMA 850 (FDD V) [*]. If the star is not there, tap on that line and that should be all.
The ROM feels really smooth on RFS, I am surprised how much improvement Samsung made with their file system in 2.2.1.
Thanks and donations go to R64, ericshieh, bedwa, DocRambone, Fr44g0r, creator2456, s0niqu3, Fabolous, camalot, khiyatesh, glitterballs, neldar, Guardian_Bob, MikeyMike01, designgears and others. If you notice someone not mentioned, dont hesitate to let me know.
Special thanks to MaxRabbit for offering a hosting space for all files!!
Cezar's Sensation - Home Button haptic feedback fix, optimized extended power menu (no slider, all options are visible) and 4 lockscreens mod.
Looks interesting. Do you plan to add ad blocking?
Sent from my liberated Captivate via XDA App.
Is already added in the Customized version. If you want it in the Stock one you will have to install AdFree Android from the Market.
I just change the kernel for a smoking one!
Which kernel are you using?.


text:
Speech recognition in nonstationary noise based on parallel HMMs and spectral subtraction
Ryuji Mine*, Tetsunori Kobayashi, Katsuhiko Shirai
School of Fundamental Science and Engineering
This paper proposes a method of speech recognition in a nonstationary noisy environment, combining the parallel HMMs and the spectral subtraction. In the proposed method, a set of hypothesis is generated with respect to the combination of the speech and the noise that can produce the observed data by a series of subtraction processes. Using HMMs prepared separately for the speech and the noise, the probabilities of occurrence are calculated. The 100-word recognition in the noisy environment in an ordinary car running in an urban area, is defined as the task in the experiment. Comparative experiments, are made for the proposed method, the ordinary spectral subtraction method and other parallel HMM methods. Then, the effectiveness of the proposed method is verified.
Systems and Computers in Japan
https://doi.org/10.1002/scj.4690271405
Noise robustness
Nonstationary noise
Parallel HMM
Spectral subtraction
10.1002/scj.4690271405
Dive into the research topics of 'Speech recognition in nonstationary noise based on parallel HMMs and spectral subtraction'. Together they form a unique fingerprint.
Speech Recognition Mathematics 100%
Subtraction Mathematics 79%
Speech recognition Engineering & Materials Science 61%
Spectrality Mathematics 49%
Speech Mathematics 34%
Urban Areas Mathematics 21%
Mine, R., Kobayashi, T., & Shirai, K. (1996). Speech recognition in nonstationary noise based on parallel HMMs and spectral subtraction. Systems and Computers in Japan, 27(14), 37-44. https://doi.org/10.1002/scj.4690271405
Speech recognition in nonstationary noise based on parallel HMMs and spectral subtraction. / Mine, Ryuji; Kobayashi, Tetsunori; Shirai, Katsuhiko.
In: Systems and Computers in Japan, Vol. 27, No. 14, 12.1996, p. 37-44.
Mine, R, Kobayashi, T & Shirai, K 1996, 'Speech recognition in nonstationary noise based on parallel HMMs and spectral subtraction', Systems and Computers in Japan, vol. 27, no. 14, pp. 37-44. https://doi.org/10.1002/scj.4690271405
Mine R, Kobayashi T, Shirai K. Speech recognition in nonstationary noise based on parallel HMMs and spectral subtraction. Systems and Computers in Japan. 1996 Dec;27(14):37-44. doi: 10.1002/scj.4690271405
Mine, Ryuji ; Kobayashi, Tetsunori ; Shirai, Katsuhiko. / Speech recognition in nonstationary noise based on parallel HMMs and spectral subtraction. In: Systems and Computers in Japan. 1996 ; Vol. 27, No. 14. pp. 37-44.
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title = "Speech recognition in nonstationary noise based on parallel HMMs and spectral subtraction",
abstract = "This paper proposes a method of speech recognition in a nonstationary noisy environment, combining the parallel HMMs and the spectral subtraction. In the proposed method, a set of hypothesis is generated with respect to the combination of the speech and the noise that can produce the observed data by a series of subtraction processes. Using HMMs prepared separately for the speech and the noise, the probabilities of occurrence are calculated. The 100-word recognition in the noisy environment in an ordinary car running in an urban area, is defined as the task in the experiment. Comparative experiments, are made for the proposed method, the ordinary spectral subtraction method and other parallel HMM methods. Then, the effectiveness of the proposed method is verified.",
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AU - Kobayashi, Tetsunori
AU - Shirai, Katsuhiko
N2 - This paper proposes a method of speech recognition in a nonstationary noisy environment, combining the parallel HMMs and the spectral subtraction. In the proposed method, a set of hypothesis is generated with respect to the combination of the speech and the noise that can produce the observed data by a series of subtraction processes. Using HMMs prepared separately for the speech and the noise, the probabilities of occurrence are calculated. The 100-word recognition in the noisy environment in an ordinary car running in an urban area, is defined as the task in the experiment. Comparative experiments, are made for the proposed method, the ordinary spectral subtraction method and other parallel HMM methods. Then, the effectiveness of the proposed method is verified.
AB - This paper proposes a method of speech recognition in a nonstationary noisy environment, combining the parallel HMMs and the spectral subtraction. In the proposed method, a set of hypothesis is generated with respect to the combination of the speech and the noise that can produce the observed data by a series of subtraction processes. Using HMMs prepared separately for the speech and the noise, the probabilities of occurrence are calculated. The 100-word recognition in the noisy environment in an ordinary car running in an urban area, is defined as the task in the experiment. Comparative experiments, are made for the proposed method, the ordinary spectral subtraction method and other parallel HMM methods. Then, the effectiveness of the proposed method is verified.
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it relaxes, helps you fall asleep.
The product is intended for adults. Drink once a day one cup of the infusion prepared from 1 tea bag per 200 ml of water, preferably after a heavy meal or between meals. A daily dose (1 tea bag) contains 1,5 g of St. John's wort herb.
Do not exceed the recommended dose for consumption during the day. Do not use if you are allergic to the St. John's wort herb and during pregnancy and breastfeeding. During use, avoid direct exposure to sunlight and UV rays (e.g. in the solarium). This product may cause allergic skin reactions.
It not recommended for those taking antidepressants.
Put the teabag in a cup, pour hot water (200 ml), cover the cup.
To obtain profound taste and aroma, the tea should be brewed for 8-10 minutes.
This product cannot be used as a substitute (replacement) of a varied diet.
Keeping a healthy lifestyle and having a balanced and varied diet are essential for maintaining good health.


text:
Anna Nicole to get Bahamas burial. Who'll get Britney's tell-all interview? Plus: Olbermann calls Cooper "a marketing experiment."
By Scott Lamb
Published February 23, 2007 2:30PM (EST)
First Word
Bouncing Britney back in rehab: We know it's hard to follow, but Britney Spears is now -- at least for the moment -- back in rehab in Malibu, Calif. She returned to the Promises center there Thursday after Kevin Federline canceled an emergency court meeting to discuss custody of their two children, fueling rumors that she went back into treatment for the sake of her kids. Before heading in, though, Spears had one last freakout outside the house where Federline is reportedly staying, using an umbrella to attack an SUV while paparazzi looked on. (Associated Press, New York Daily News)
No Borat, no Oscar presenter: The producers behind Sunday's Oscars asked Sacha Baron Cohen to present an award as part of Sunday night's show, which might have added a refreshing bit of unpredictability to what is usually a staid evening -- his acceptance speech at the Golden Globes was one of the evening's highlights. But when Cohen said he wanted to appear as his character Borat Sagdiyev, reports the Los Angeles Times, the producers said no, and Cohen declined. (L.A. Times via Starpulse)
Smith gets Bahamas burial: The long, weird court battle to decide the final resting place of Anna Nicole Smith ended Thursday with a surprising outcome -- Florida Circuit Court Judge Larry Seidlin gave Richard Milstein, Smith's daughter Dannielynn's court-appointed guardian, custody of Smith's remains, and tearfully asked that she be buried with her son, Daniel, in the Bahamas: "I want her buried with her son in the Bahamas," Seidlin said. "I want them to be together." Milstein, after conferring with Smith's mother, Virgie Arthur, ex-boyfriend Larry Birkhead and companion Howard K. Stern, said the burial will indeed take place in the Bahamas, though no date has been set. (People)
White noise ... Brad Pitt and Angelina Jolie are rumored to be in the process of adopting another child, this time a boy from Vietnam -- a source tells Us Weekly they've already filed papers with the U.S. Bureau of Citizenship and Immigration Services. (Daily Dish) ... There's been talk recently that Terrence Howard and Naomi Campbell are an item, and a source tells Gatecrasher they'll become official at the Oscars: "They are making their debut as a couple this weekend and will attend all the Oscar parties together." (Gatecrasher) ... Producers in London are reportedly planning to bring "Gone With the Wind" to the stage as a musical; Hugh Jackman (right) is being eyed for the role of Rhett Butler. (Fox 411) ... Amanda Peet gave birth to a baby girl on Wednesday (or possibly Thursday); it's her first child with screenwriter husband David Benioff. (Celebrity Baby Blog)
Sidestepping swag: You didn't really think that just because the taxman has started paying attention to those lavish awards season gift bags that the swag orgy that surrounds the Oscars would disappear, did you? Of course not. In addition to all the "gifting lounges" that have sprung up this week in Los Angeles, where stars get complimentary designer jeans and face creams, the latest trend in marketers giving away stuff to rich actors is something called the "branded retreat," which the New York Times describes today as "an invitation-only, luxury destination where free products are just a corollary to the more subtle (and possibly tax-free) pleasures of food, drink, entertainment and spa treatments, on the house." As the man behind SoHo House, the swanky private club that's ground zero of branded retreats this Oscar season, tells the paper, "I think goodie bags are a bit last year, aren't they?" ("Let Them Eat Foie Gras (Gift Bags Are So Last Year)," N.Y. Times)
Buzz Index
Most viewed on Yahoo News: "Eight Places Every Woman Should Go," Rolf Potts
Top story on Reddit: "Hunting Chimps May Change View of Human Evolution," Reuters
Top search term on Sphere: Washington's birthday
Most popular on Google News: "Body Going to Bahamas as Paternity Fight Looms," Palm Beach Post
"Temperature of Coffee Expected to Rise Nine Degrees by End of 21st Century" (The Onion)
Looking back at Gordon Matta-Clark: The Whitney Museum has just opened a retrospective of the work of Gordon Matta-Clark, the experimental '70s SoHo artist perhaps best known for his "building cuts" -- chopping up abandoned buildings -- and for launching Food, an artist-run restaurant. The New York Times, which calls the show "heavenly," sees it as a representation of the golden old days of art in New York, "wherein a cheeky young man, if so inclined, might dangle from the outside of a tall building for a while and not attract undue attention. If there's a metaphor for what the downtown art world was, but no longer is, that's it." The review notes that the timing -- the show's opening coincides with the start of the Armory Show and other art fairs in New York -- also carries significance: "Matta-Clark represents a whole different notion of the art world from the youth-besotted, money-sick one we've got today. How sweet that this retrospective opens at the same time the art fairs do." ("Cross Sections of Yesterday," N.Y. Times)
"He is the only person who has not been informed that he is a marketing experiment."
-- Keith Olbermann talking about Anderson Cooper. (Eat the Press)
Associated Press television writer David Bauder pins down the odds on who is likely to score the inevitable TV tell-all interview with Britney Spears:
Matt Lauer: 3 to 1
Barbara Walters: 5 to 2
Oprah Winfrey: 4 to 1
Anderson Cooper: 5 to 1
Diane Sawyer: 10 to 1
Katie Couric: 10 to 1
On Friday night, "Nanny 911" (Fox, 8 p.m. EST) and "Trading Spouses" (Fox, 9 p.m. EST) both air their last episodes before going on hiatus, and Ayaan Hirsi Ali, Steven Weber and David Mamet appear on "Real Time With Bill Maher" (HBO, 11 p.m. EST). On Sunday, it's the annual pre-Oscar chatfest "The Barbara Walters Special" (ABC, 7 p.m. EST), followed by the "79th Annual Academy Awards" (ABC, 8:30 p.m. EST).
Regis and Kelly (ABC, 9 a.m. EST) Billy Bob Thornton, Jesse L. Martin, toy fair roundup with Chris Byrne, guest co-host Geraldo Rivera
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Ellen (Syndicated, check local listings) Jennifer Hudson, Cuba Gooding Jr., William H. Macy
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David Letterman (CBS, 11:30 p.m. EST) Samuel L. Jackson, Jim Gaffigan
Jay Leno (NBC, 11:35 p.m. EST) Rep. Nancy Pelosi, William H. Macy, John Mellencamp
Jimmy Kimmel (ABC, 12:05 a.m. EST) Kathie Lee Gifford, Ken Davitian, Mickey Avalon
Conan O'Brien (NBC, 12:35 a.m. EST) Billy Bob Thornton, J. Alexander, G. Love
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Creation Station is the name of the ministry area designed for our youngest kids. Children will discover that God created everyone and everything, and He wants us to create too!
We believe that the parent is the primary faith trainer of the child and that it is never too early or too late to start training your child. God's intention is for the church and family to work together as a team. Sienna Ranch partners with parents in nurturing children with the love of Jesus Christ. We teach biblical truths in a loving, safe, and fun environment. Caring teachers engage the kids with songs, Bible stories, games and other hands-on activities.
All of our staff and volunteers are background-screened and trained. We have a pager system for signing your child in and out, and to alert you anywhere in the church if needed. Safety and security are our utmost concern.
Train Station is the name of the ministry area designed for our elementary school-aged kids.
Children will learn more about God, Jesus, and the Holy Spirit as they discover that the Bible is one big story and is as relevant today as it was when it was written! There are so many practical examples of how God wants us to act in every-day situations.
9:15am to 10:15am - Your children will study the same Bible lesson the adults study each week, providing opportunities for great discussions at home.
*Note – We encourage children to attend both classes on Sunday mornings. They are never bored because the activities, games, skits and videos are different! The double session each week reinforces God's timeless truth being taught so they will remember for life.
We call our Wednesday night group "Kids Apps" because the kids learn life applications: from gardening to astronomy, and everything in between, children learn about the world they live in through hands-on activities and with a biblical perspective. Children will enjoy the various topics through the school year with their friends and expert teachers from various fields.
All of our staff and volunteers are background-screened and trained. Safety and security are our utmost concern.
As students are immersed into worldly culture, our student ministry exists to engage and equip students to immerse themselves in Jesus. Our desire is that every student leaves knowing the truth, living the truth, and with the ability to share truth.
Your youth will study the same biblical lesson that the adults study each week, tailored to their level. This gives families a common foundation for lively discussions throughout the week.
Students meet to hang out, play impactful games, and study the Bible through small group discussion.
Midweek nights are designed to open up a very practical application of God's word. Once a month we also put our faith into practice by serving somewhere in the community.
Immersion events are designed to take students outside the church and into real world situations to experience their faith and prepare them to articulate their faith. These are one-day events held throughout the year.
Convergence XP events are shared vocational experiences with students and parents. We believe that every Christian is called, and that our work is part of fulfilling that calling. Throughout the year business professionals, restaurant owners, artists, and other professionals meet with students and parents to share how their faith impacts and shapes their work.
We believe spiritual growth happens best when people are connected and in community with each other. Our desire is to help everyone get connected into a Group where they can be encouraged and equipped to know truth, pursue goodness, and create beauty. Our groups are usually comprised of about 8-10 adults who meet consistently for Bible study and fellowship.
You are welcome to come to any group at any of the times they are meeting, or you can fill out a visitor's card and you will be contacted by someone from our staff to help you get connected to a group.


text:
First Day of Trial for UBS Executive
Blows Dealt to Department of Justice in Offshore Bank Account Enforcement
Stolen Identity Tax Refund Fraud Stealing Headlines
The IRS's tax return-related identity theft problem has officially crossed from item of interest to law enforcement and tax professionals to a disturbing problem worthy of the national primetime news. On September 21, 2014, tax return-related identity theft (also known as "stolen identity refund fraud") was the focus of a 15-minute segment on CBS's 60 Minutes, Biggest IRS Scam Around: Identity Tax Refund Fraud (www.cbsnews.com/news/irs-scam-identity-tax-refund-fraud-60-minutes). Among those interviewed in the segment were IRS Commissioner John Koskinen, Deputy IRS Commissioner Steve Miller, Southern Florida United States Attorney Wilfredo Ferrer, and Miami Field Office FBI Special Agent In Charge George Piro.
The seriousness of the matter certainly came through to anyone watching, with United States Attorney Ferrer using a geographically-appropriate water analogy to describe the situation as "a tsunami of fraud." Florida has been the hardest hit by these fraud schemes, and Ferrer was also refreshingly blunt on several key points: that the scheme is surprisingly simple and requires no more than a computer, a social security number, and a date of birth; that the IRS was not ready to deal with the large number of fraudulent returns; and that clearing up a victim's identity theft refund claim with the IRS can be "a nightmarish process."


text:
\section{Introduction}\label{se-1}
An interconnection network is always represented by a graph $\Gamma=(V,E)$, where $V$ denotes the node set and $E$ denotes the edge set. Now various interconnections are proposed. The hypercubes network obtained considerable attention in virtue of its perfect properties, such as symmetry, regular structure, strong connectivity, and small diameter \cite{Bossard, Fink}. An $n$-dimensional hypercube denoted by $Q_{n}$ has $2^{n}$ vertices, and the vertex set is $V(Q_{n})=\{x_{1}x_{2}\cdots x_{n} \mid x_{i}=0~~ or~~ 1, i=1,2,\ldots, n\}$. Two vertices $X=x_{1}x_{2}\cdots x_{n}$ and $Y=y_{1}y_{2}\cdots y_{n}$ are adjacent if and only if there exists  $1\le i \le n$, such that $x_{i}=\overline{y_{i}}$, where $\overline{y_{i}}$ denoted the complement of binary digit $y_{i}$, and $x_{j}=y_{j}$ for all $j\ne i$. As the importance of the hypercubes networks, many variants of it were presented, among which, for instance, are enhanced hypercube, augmented hypercube, folded hypercube \cite{Tzeng, El-Amawy, Choudum}. The $n$-dimensional enhanced hypercube is one of the important variants of hypercube introduced by Tzeng in \cite{Tzeng} which is defined as follows.
\begin{defi}\label{defi-1}
For $n\ge 2$ and $1\le k\le n-1$, the enhanced hypercube $Q_{n, k}=(V,E)$ is an undirected simple graph with vertex set  $V=\{x_{1}x_{2}\cdots x_{n}\mid x_{i}=0~~ \emph {or}~~ 1, i=1,2,\ldots,n\}$. Two vertices $X=x_{1}x_{2}\cdots x_{n}$ and $Y=y_{1}y_{2}\cdots y_{n}$ are adjacent if $Y$ satisfies one of the following two conditions:
\begin{enumerate}
\item[(1)] $Y=x_{1}x_{2}\cdots x_{i-1}\overline x_{i}x_{i+1}\cdots x_{n}$ for some $1\le i \le n$;
\item[(2)] $Y=x_{1}x_{2}\cdots x_{k-1}\overline x_{k} \overline x_{k+1}\cdots \overline x_{n}$.
\end{enumerate}
\end{defi}
According to the above definition, we can see that $Q_{n, k}$ contains $Q_{n}$ as its subgraph. In fact,  $Q_{n, k}$ is a $(n+1)$-regular graph with $2^{n}$ vertices and $(n+1)2^{n-1}$ edges. Its special case of $k=1$ is the well-known folded hypercube denoted by $FQ_{n}$. As a variant of the hypercube, the n-dimensional folded hypercube $FQ_{n}$, proposed first by El-Amawy and Latifi \cite{El-Amawy}, is a graph obtained from the hypercube $Q_{n}$ by adding some edges, called a complementary edges, between  vertices $X=x_{1}x_{2}\cdots x_{n}$ and $\overline X=\overline x_{1}\overline x_{2}\cdots \overline x_{n}$. The enhanced hypercube is superior to the hypercube in many aspects. For example, the diameter of the enhanced hypercube is almost half of the hypercube. The hypercube is $n$-regular and $n$-connected, whereas the enhanced hypercube is $(n+1)$-regular and $(n+1)$-connected \cite{Liu-1}.

Let $G$ be a finite group, and let $S$ be a symmetric subset of $G$ such that $\mathbf{1}\not\in S$ and $S^{-1}=\{s^{-1}\mid s\in S\}=S$. The \emph{Cayley graph} on $G$ with respect to a symmetric subset $S$ of $G$, denoted by $Cay(G,S)$, is the undirected graph with vertex set $G$ and with an edge $\{g,h\}$ connecting $g$ and $h$ if $hg^{-1}\in S$, or equivalently $gh^{-1}\in S$. It is well known that $Cay(G,S)$ is connected if and only if $S$ generates $G$. Particularly, the group $Z_{2}^{n}=Z_{2}\times Z_{2}\times \cdots\times Z_{2}$ can be viewed as a vector space of dimension $n$ over the Galois filed $F_{2}$. Suppose that $\{e_{1},\ldots,e_{n}\}$ is the standard  basis of $Z_2^n$ and $\epsilon_{k}$ is the vector with first $k-1$ entries equal to $0$ and other entries equal to $1$ for some  $1\le k\le n-1$. Let $S=\{e_{1},\ldots, e_{n},\epsilon_{k}\}$ be the subset of $Z_2^n$. It is clear that  the so called enhanced hypercube $Q_{n, k}$ is just the Cayley graph $Cay(Z_2^n,S)$.

The concept of resistance distance of a graph $\Gamma=(V, E)$ was first introduced by Klein and Randi\'{c} \cite{Klein-1}. Let $\Gamma$ be a connected graph. The resistance distance between vertices $v_{i}$ and $v_{j}$ of $\Gamma$, denoted by $r_{ij}$, is defined  to be the effective resistance between the nodes $v_{i}$ and $v_{j}$ as computed with ohm's law when all the edges of $\Gamma$ are considered to be unit resistors. The traditional distance between vertices $v_{i}$ and $v_{j}$, denoted by $d_{ij}$, is the length of a shortest path connecting them. The Wiener index $W(\Gamma)$ was given by $W(\Gamma)=\sum\limits_{i<j}d_{ij}$ in \cite{Wiener}. As an analogue to the Wiener index, the sum $Kf(\Gamma)=\sum\limits_{i<j}r_{ij}$ was proposed in \cite{Klein-1}, later called the Kirchhoff index of $\Gamma$ in \cite{Bonchev}. Klein and Randi\'{c} \cite{Klein-1} proved that $r_{ij}\le d_{ij}$ and thus $Kf(\Gamma)\le W(\Gamma)$ with equality if and only if $\Gamma$ is a tree.

The Kirchhoff index has wide applications in physical interpretations, electric circuit, graph theory, and chemistry. For example, Gutman and Mohar \cite{Gutman} and Zhu, Klein et al. \cite{H.Y} proved that the Kirchhoff index of a connected graph $\Gamma$ with $n(n\ge2)$ vertices is the sum of reciprocal nonzero Laplacian eigenvalues of the graph multiplied by the number of the vertices. Like the Wiener index, the Kirchhoff index is a structure descriptor \cite{W.J}. The Kirchhoff index has been computed for cycles \cite{Klein-2, Lukovits}, complete graphs \cite{Lukovits}, geodetic graphs \cite{Palacios}, distance transitive graphs \cite{Palacios}, and so on. The Kirchhoff index of certain composite operations between two graphs was studied, such as product, lexicographic product \cite{H.Xu} and join, corona, cluster \cite{H.Zhang}.

In \cite{Liu-2}, we can see that the exact formula of the Kirchhoff index of the hypercubes networks $Q_{n}$ and related complex networks had been provided. Motivated by previous results, in this paper we present the formulae for the Kirchhoff index of the enhanced hypercube $Q_{n, k}$ and such formulae are brief for $Q_{n, 1}$ and $Q_{n, n-1}$. Moreover, we  prove that $Kf(Q_{n, k})$ is increased as $k$ increases for a given $n$. Additionally, the bounds of $Kf(Q_{n, k})$ and its Limit function is obtained, that is, $\lim\limits_{n\to\infty}\frac{Kf(Q_{n, k})}{\frac{2^{2n}}{n+1}}=1$.

\section{Preliminaries}\label{se-2}

Let $\Gamma$ be a simple graph with vertex set $V$ and edge set $E$. The \emph{adjacency matrix} $A(\Gamma)$ of $\Gamma$ is the $n\times n$ matrix with the $(i, j)$-entry equal to $1$ if vertices $v_{i}$ and $v_{j}$ are adjacent and $0$ otherwise. For $v_i\in V$, let $N(v_i)$ denote the set of neighbours of $v_i$, that is, $N(v_i)=\{v_j\in V\mid v_j\sim v_i\}$. The size of $N(v_i)$ is called the \emph{degree} of $v_i$, denoted by $d_i$. Let $D(\Gamma)$ be the diagonal matrix with $i$-th diagonal entry equal to $d_i$. The \emph{Laplacian matrix} of $\Gamma$ is defined by $L(\Gamma)=D(\Gamma)-A(\Gamma)$. The eigenvalues of $A(\Gamma)$ and $L(\Gamma)$ are called the \emph{adjacency eigenvalues} and \emph{Laplacian eigenvalues} of $\Gamma$, respectively. The multiset of adjacency (resp. Laplacian) eigenvalues together with their multiplicities is called the \emph{adjacency (resp. Laplacian) spectrum} of $\Gamma$, denoted by $\mathrm{Spec}_A(\Gamma)$ (resp. $\mathrm{Spec}_L(\Gamma)$).

In \cite{Babai}, Babai derived an expression for the spectrum of  the Cayley graph $Cay(G,S)$ in terms of irreducible characters of $G$. Here, we need only to know the eigenvalues of the Cayley graph of an abelian group.

\begin{lem}[\cite{Steinberg}]\label{lem-1}
Let $G=\{g_{1},g_{2},\ldots,g_{n}\}$ be an abelian group and $S\subseteq G$ is a symmetric set. Let $\chi_{1},\ldots,\chi_{n}$ be the irreducible characters of $G$ and Let $A$ be the adjacency matrix of the Cayley graph of $G$ with respect to $S$. Then the eigenvalues of the adjacency matrix $A$ are the real numbers
\[\lambda_{i}=\sum\limits_{s\in S}\chi_{i}(s)\]
where $1\le i\le n$.
\end{lem}

\begin{lem}[\cite{Steinberg}]\label{lem-2}
Let $G_{1},G_{2}$ be abelian groups and suppose that $\chi_{1},\ldots,\chi_{m}$ and $\varphi_{1},\ldots,\varphi_{n}$ are the irreducible representations of $G_{1}, G_{2}$, respectively. In particular, $m=|G_{1}|$ and $n=|G_{2}|$. Then the functions $\alpha_{ij}:G_{1}\times G_{2}\rightarrow \mathbb{C^{*}}$ with $1\le i\le m, 1\le j\le n$ given by
\[\alpha_{ij}(g_{1}, g_{2})=\chi_{i}(g_{1})\varphi_{j}(g_{2})\]
form a complete set of irreducible representations of $G_{1}\times G_{2}$.
\end{lem}

It is well known that $Z_2$ has two irreducible characters $\chi_0(a)=1(\forall a\in Z_2)$ and $\chi_1(a)=(-1)^{a}(\forall a\in Z_2)$. As a direct consequence of Lemma \ref{lem-2}, the irreducible characters $\chi_{i_{1},i_{2},\ldots,i_{n}}$ of $Z_{2}^{n}$ are given by
\begin{equation}\label{ch-eq-1}\chi_{i_{1},\ldots,i_{n}}((a_{1},\ldots,a_{n}))=(-1)^{i_{1}a_{1}+\cdots+i_{n}a_{n}}\end{equation}
where $(a_{1},\ldots,a_{n})\in Z_{2}^{n}$ and $i_{j}\in\{0,1\}$ for $1\le j\le n$. From (\ref{ch-eq-1}), Lemma \ref{lem-1} and Lemma \ref{lem-2}, we get the eigenvalues of $Cay(Z_2^n,S)$.

\begin{thm}\label{thm-1}
Let $S$ be a subset of  $Z_{2}^{n}$. Then the adjacency eigenvalues of the Cayley graph $Cay(Z_{2}^{n}, S)$ are given by
\[\lambda_{i_{1},i_{2},\ldots,i_{n}}=\sum\limits_{(s_{1},s_{2},\ldots,s_{n})\in S}(-1)^{i_{1}s_{1}+\cdots+i_{n}s_{n}}\]
where $(i_{1},\ldots,i_{n})\in Z_{2}^{n}$.
\end{thm}

Gutman and Mohar \cite{Gutman} and Zhu, Klein et al. \cite{H.Y} obtained the Kirchhoff index of a graph in terms of Laplacian eigenvalues as follows:

\begin{lem}[\cite{Gutman, H.Y}]\label{lem-3}
Let $\Gamma$ be a connected graph with $n\ge 2$ vertices. Then
\[Kf(\Gamma)=n\sum \limits_{i=1}^{n-1}\frac{1}{\mu_{i}},\]
where $0=\mu_{0}<\mu_{1}\le\cdots\le\mu_{n-1}$ are the Laplacian eigenvalues of $\Gamma$.
\end{lem}

\begin{lem}[\cite{Godsil}]\label{lem-4}
Let $A$ be the adjacency matrix of a graph $\Gamma$, and let $\rho$ be its spectral radius. Then the following are equivalent:\\
(1) $\Gamma$ is bipartite.\\
(2) The spectrum of $A$ is symmetric about the origin, i.e., for any $\lambda$, the multiplicities of $\lambda$ and $-\lambda$ as eigenvalues of $A$ are the same.\\
(3) $-\rho$ is an eigenvalue of $A$.
\end{lem}

\section{The eigenvalues of enhanced hypercubes}\label{se-3}

In this section, we focus on the eigenvalues of enhanced hypercubes. Let $\{e_{1},\ldots,e_{n}\}$ be the standard basis of $Z_{2}^{n}$, and let $\epsilon_{k}=e_k+e_{k+1}+\cdots+e_n$ for some $1\le k\le n-1$. The $n$-dimensional enhanced hypercube $Q_{n, k}(1\le k\le n-1)$ is just the Cayley graph $Cay(Z_{2}^{n}, S)$, where $Z_{2}^{n}=Z_{2}\times Z_{2}\times\cdots\times Z_{2}$ and $S=\{e_{1},\ldots, e_{n},\epsilon_{k}\}$.

\begin{lem}\label{lem-5}
For  $1\le k\le n-1$, the eigenvalues of $Q_{n, k}$ are given by $\zeta_{t}=n-2t-1$ with multiplicity $\alpha_{t}=\sum\limits_{j=1}^{n}{n-k+1\choose 2j-1}{k-1\choose t-2j+1}$ for  $t=1,2,...,n$, and $\xi_{t}=n-2t+1$ with multiplicity $\beta_{t}=\sum\limits_{j=0}^{n}{n-k+1\choose 2j}{k-1\choose t-2j}$ for $t=0,1,2,...,n$.
\end{lem}
\begin{proof}
Since $Q_{n, k}\cong Cay(Z_2^n, S)$, we need only to calculate the eigenvalues of $Cay(Z_{2}^{n}, S)$. For $v=(v_{1},v_{2},\ldots,v_{n})^T\in Z_2^n$, according to (\ref{ch-eq-1}), the value of the irreducible character  $\chi_{v}$ at $a=(a_1,a_2,...,a_n)^T\in Z_2^n$ is $\chi_{v}(a)=(-1)^{v_{1}a_{1}+v_{2}a_{2}+\cdots+v_{n}a_{n}}$. Thus
\begin{equation}\label{ch-eq-2}
\left\{\begin{array}{ll}\chi_{v}(e_i)=(-1)^{v_{i}}\\
\chi_{v}(\epsilon_k)=(-1)^{v_{k}+v_{k+1}+\cdots+v_n}
\end{array}\right.
\end{equation}
From (\ref{ch-eq-2}) and Theorem \ref{thm-1}, the eigenvalue corresponding to $v=(v_{1},v_{2},\ldots,v_{n})^T\in Z_2^n$ is
\begin{equation}\label{ch-eq-3}\lambda_v=\sum_{s\in S}\chi_v(s)=(-1)^{v_{1}}+(-1)^{v_{2}}+\cdots+(-1)^{v_{n}}+(-1)^{v_{k}+v_{k+1}+\cdots+v_n}
\end{equation}
Given $0\le r\le t\le n$, let
$\Lambda(t, r)=\{v\in Z_2^n\mid v_{1}+v_{2}+\cdots+v_{n}=t, v_{k}+v_{k+1}+\cdots+v_n=r\}$. It is easy to see  that $|\Lambda(t, r)|={n-k+1\choose r}{k-1\choose t-r}$. From (\ref{ch-eq-3}), for any $v\in\Lambda(t, r)$, we have
\begin{equation}\label{ch-eq-4}\lambda_v=\sum_{s\in S}\chi_v(s)=n-2t+(-1)^r\end{equation}
Clearly, $Z_2^n=\cup_{0\le r\le t\le n}\Lambda(t, r)$ is a partition of $Z_2^n$. By taking $r=2j-1$ for $1\le j\le \left\lfloor \frac{t+1}{2}\right\rfloor$, from (\ref{ch-eq-4}) we get the eigenvalue $\zeta_{t}=n-2t-1$, where $t=1,2,...,n$,  with multiplicity $|\Lambda(t, r)|=\sum\limits_{j=1}^{\left\lfloor \frac{t+1}{2}\right\rfloor}{n-k+1\choose 2j-1}{k-1\choose t-2j+1}$ which equals $\alpha_{t}=\sum\limits_{j=1}^{n}{n-k+1\choose 2j-1}{k-1\choose t-2j+1}$;  by taking $r=2j$ for $0\le j\le \left\lfloor \frac{t}{2} \right\rfloor$, we get the eigenvalue $\xi_{t}=n-2t+1$, where $t=0,1,2,...,n$, with multiplicity $|\Lambda(t, r)|=\sum\limits_{j=0}^{\left\lfloor \frac{t}{2} \right\rfloor}{n-k+1\choose 2j}{k-1\choose t-2j}$ which equals $\beta_{t}=\sum\limits_{j=0}^{n}{n-k+1\choose 2j}{k-1\choose t-2j}$.

This completes the proof.
\end{proof}

Note that some of the two kind of eigenvalues described in Lemma \ref{lem-5} may be equal, in fact, $\zeta_t=\xi_{t+1}$ for $t=1,2,...,n-1$. Thus  we can give the spectrum of $Q_{n, k}$ from Lemma \ref{lem-5}.

\begin{thm}\label{thm-2}
Let $1\le k\le n-1$, the spectrum of $Q_{n, k}$ is given by \[\mathrm{Spec}_{A}(Q_{n, k})=\{n+1, [n-1]^{k-1}, [n-2t-1]^{\gamma_t},[-n-1]^{\gamma_n} \mid t=1,2,...,n-1\},\] where $\gamma_t=\sum\limits_{j=0}^{n}{n-k+2\choose 2j}{k-1\choose t+1-2j}$ for $1\le t\le n-1$ and $\gamma_n=\sum\limits_{j=1}^{n}{n-k+1\choose 2j-1}{k-1\choose n-2j+1}$.
\end{thm}
\begin{proof}
For convenience, denote by $\lambda_0>\lambda_1>\lambda_{2}>\cdots$ the distinct eigenvalues of $Q_{n, k}$. By Lemma \ref{lem-5}, we know that $\lambda_0=\xi_{0}=n+1$ with multiplicity $\beta_{0}=\sum\limits_{j=0}^{n}{n-k+1\choose 2j}{k-1\choose 0-2j}=1$. $\lambda_1=\xi_{1}=n-1$ with multiplicity $\beta_{1}=\sum\limits_{j=0}^{n}{n-k+1\choose 2j}{k-1\choose 1-2j}={k-1\choose 1}=k-1$. In addition, we see that $\lambda_{1+t}=\zeta_t=\xi_{t+1}=n-2t-1$ for $t=1,2,...,n-1$, so the multiplicity of the eigenvalue $n-2t-1$ is $\alpha_{t}+\beta_{t+1}$, where $\gamma_t=\alpha_{t}+\beta_{t+1}$ can be simplified as
\[\begin{array}{lll}
\gamma_{t}&=&\alpha_{t}+\beta_{t+1}=\sum\limits_{j=1}^{n}{n-k+1\choose 2j-1}{k-1\choose t-2j+1}+\sum\limits_{j=0}^{n}{n-k+1\choose 2j}{k-1\choose t-2j+1}\\
&=&\sum\limits_{j=1}^{n}\left({n-k+1\choose 2j-1}+{n-k+1\choose 2j}\right){k-1\choose t-2j+1}+{k-1\choose t+1}\\
&=&\sum\limits_{j=0}^{n}{n-k+2\choose 2j}{k-1\choose t+1-2j}.\\
\end{array}\]
At last, $\lambda_{n+1}=\zeta_n=-n-1$ with multiplicity $\gamma_n=\alpha_{n}=\sum\limits_{j=1}^{n}{n-k+1\choose 2j-1}{k-1\choose n-2j+1}$.

This completes the proof.
\end{proof}
Since the enhanced hypercube $Q_{n, k}$ is regular, the Laplacian spectrum of $Q_{n, k}$ can be reduced by its $A$-spectrum. From Theorem \ref{thm-2}, we get the following result.

\begin{thm}\label{thm-3}
Let $1\le k\le n-1$, the Laplacian spectrum of $Q_{n, k}$ is given by
\[\mathrm{Spec}_{L}(Q_{n, k})=\{0,[2]^{k-1}, [2t+2]^{\gamma_{t}},[2n+2]^{\gamma_n} \mid t=1,2,...,n-1\},\]
where $\gamma_t=\sum\limits_{j=0}^{n}{n-k+2\choose 2j}{k-1\choose t+1-2j}$ for $1\le t\le n-1$ and $\gamma_n=\sum\limits_{j=1}^{n}{n-k+1\choose 2j-1}{k-1\choose n-2j+1}$.

\end{thm}

\begin{proof}
Since $Q_{n, k}$ is $n+1$ regular, we have $L(Q_{n, k})=(n+1)I-A(Q_{n, k})$. Thus every eigenvector of $A(Q_{n, k})$ with eigenvalue $\lambda$ is an eigenvector of $L(Q_{n, k})$  with eigenvalue $(n+1)-\lambda$. By Theorem \ref{thm-2}, the result yields immediately.
\end{proof}

In general, the summations of $\alpha_{t}$ and $\beta_{t}$ can not be simply calculated. However, we can find the simple expression of the spectrum for some special $k$. By Theorem \ref{thm-2} and Theorem \ref{thm-3}, we give the $A$-spectrum and Laplacian spectrum of $Q_{n, k}$ for $k=1$.

\begin{cor}\label{cor-1}
For $k=1$ and $n\ge2$, the $A$-spectrum and Laplacian spectrum of $Q_{n, 1}$ is given by
\begin{itemize}
\item[(1)] $ \mathrm{Spec}_{A}(Q_{n,1})=\left\{\begin{array}{ll}
\{n+1, [n-3]^{\binom{n+1}{2}},\ldots,[-n+3]^{\binom{n+1}{n-1}},-n-1\} &\mbox{if $n$ is odd}\\
\{n+1, [n-3]^{\binom{n+1}{2}},\ldots,[-n+5]^{\binom{n+1}{n-2}},[-n+1]^{n+1}\} &\mbox{if $n$ is even}.
\end{array}
\right. $
\item[(2)] $ \mathrm{Spec}_{L}(Q_{n,1})=\left\{\begin{array}{ll}
\{0, [4]^{\binom{n+1}{2}},[8]^{\binom{n+1}{4}},\ldots,[2n-2]^{\binom{n+1}{n-1}},2n+2\} &\mbox{if $n$ is odd}\\
\{0, [4]^{\binom{n+1}{2}},[8]^{\binom{n+1}{4}},\ldots,[2n-4]^{\binom{n+1}{n-2}},[2n]^{n+1}\} &\mbox{if $n$ is even}.
\end{array}
\right.$
\end{itemize}
\end{cor}
\begin{remark}
The folded hypercube $FQ_{n}$ introduced in \cite{Chen} is just the enhanced hypercube $Q_{n, k}$ for $k=1$, i.e., $FQ_{n}=Q_{n, 1}$. Thus Corollary \ref{cor-1} gives the $A$-spectrum and Laplacian spectrum of the folded hypercube $FQ_{n}$.
\end{remark}

We give the following two Corollaries at the last of this section, which can be regarded as an application of Theorem \ref{thm-2} .
Denote  the adjacency eigenvalues of $\Gamma$ by $\lambda_{1}\ge \lambda_{2}\ge\cdots\ge\lambda_{n}$, and $\lambda_{1}-\lambda_{2}$ is the so called \emph{spectral gap} of the adjacency matrix, which is also an important spectral parameter related to the expander property \cite{Shlomo}.
\begin{cor}\label{cor-2}
For $2\le k\le n-1$, the second largest eigenvalue of $Q_{n,k}$ is $n-1$, and so the adjacency spectral gap of $Q_{n, k}$ is 2. Especially, the second largest eigenvalue of $Q_{n,1}$ is $n-3$, the corresponding adjacency spectral gap of $Q_{n, 1}$ is 4.
\end{cor}
\begin{proof}
From Theorem \ref{thm-2}, we known that $n+1$ is the first largest eigenvalue of $Q_{n, k}$. Moreover, for $k\ge2$, the second largest eigenvalue is $n-1$, for $k=1$, the second largest eigenvalue of $Q_{n, 1}$ is $n-3$.
\end{proof}
H. M. Liu in \cite{Liu-3} characterized the bipartite graphs among enhanced hypercubes $Q_{n, k}$. Here we give it another simple  proof by the spectrum of $Q_{n, k}$.
\begin{cor}\label{cor-3}
$Q_{n, k}$ is a bipartite graph if and only if $n$ and $k$ have the same parity for $1\le k\le n-1$.
\end{cor}
\begin{proof}
By Lemma \ref{lem-3}, $Q_{n, k}$ is a bipartite graph if and only if $-n-1$ is an eigenvalue with multiplicity one, by theorem \ref{thm-1}, if and only if $\gamma_{n}=\sum\limits_{j=1}^{n}{n-k+1\choose 2j-1}{k-1\choose n-2j+1}=1$.  Note that the valid term in the summation of $\gamma_n$ must satisfy $n-k+1\ge2j-1$ and $k-1\ge n-2j+1$. Hence $n-k+1=2j-1$, that is, $j=\frac{n-k+2}{2}$. So we have $\gamma_{n}=\sum\limits_{j=1}^{n}{n-k+1\choose 2j-1}{k-1\choose n-2j+1}={n-k+1\choose n-k+1}{k-1\choose k-1}=1$ if  $n$ and $k$ have the same parity and $\gamma_{n}=0$ otherwise. It immediately follows our result.
\end{proof}

\section{The Kirchhoff index of enhanced hypercubes}\label{se-4}

In this section, we focus on our main results. First we give the formula of the Kirchhoff index of the enhanced hypercube $Q_{n, k}$ where $1\le k\le n-1$. Next we will show the monotonicity of $Kf(Q_{n, k})$. At last, we get a  limiting function for $Kf(Q_{n, k})$.

\begin{thm}\label{thm-4}
Let $1\le k\le n-1$. The Kirchhoff index of the enhanced hypercube $Q_{n, k}$ is given by
\[Kf(Q_{n, k})=\left\{\begin{array}{ll}
2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+2\choose 2j}{k-1\choose t+1-2j} &\mbox{if $n\equiv k~(\mathrm{mod}~2)$}\\
2^{n-1}\sum\limits_{t=0}^{n-1}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+2\choose 2j}{k-1\choose t+1-2j} &\mbox{if $n\not\equiv k~(\mathrm{mod}~2)$}.
\end{array}\right.\]
\end{thm}
\begin{proof}
We denote the Laplacian eigenvalues of $Q_{n, k}$ by $\mu_{i}$ for $i=0,1,...,2^n-1$. By Lemma \ref{lem-3} and Theorem \ref{thm-3} we have
\[\begin{array}{lll}
Kf(Q_{n, k})&=&\displaystyle 2^{n}\sum\limits_{i=1}^{2^{n}-1}\frac{1}{\mu_{i}}=2^{n}\sum\limits_{\mu\in \mathrm{Spec}_{L}(Q_{n, k})\backslash 0 }\frac{1}{\mu}\\
&=&2^{n}\left(\frac{k-1}{2}+\frac{\gamma_{n}}{2n+2}+\sum\limits_{t=1}^{n-1}\frac{\gamma_{t}}{2t+2}\right)\\
&=&2^{n-1}\left(k-1+\frac{\gamma_n}{n+1}+\sum\limits_{t=1}^{n-1}\frac{\gamma_{t}}{t+1}\right).
\end{array}\]
By Corollary \ref{cor-3}, we known that  $\gamma_{n}=1$ if  $n$ and $k$ have the same parity, and $\gamma_{n}=0$ otherwise. By Theorem \ref{thm-2}, we have $\gamma_{t}=\sum\limits_{j=0}^{n}{n-k+2\choose 2j}{k-1\choose t+1-2j}$ for $t=1,2,...,n-1$. Therefore, we have
\begin{equation}\label{Q-eq-1}Kf(Q_{n, k})=\left\{\begin{array}{ll}
2^{n-1}\left(k-1+\frac{1}{n+1}+\sum\limits_{t=1}^{n-1}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+2\choose 2j}{k-1\choose t+1-2j}\right)&\mbox{if $n\equiv k~(\mathrm{mod}~2)$}\\
2^{n-1}\left(k-1+\sum\limits_{t=1}^{n-1}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+2\choose 2j}{k-1\choose t+1-2j}\right) &\mbox{if $n\not\equiv k~(\mathrm{mod}~2)$}.
\end{array}\right.\end{equation}
It is easy to verify that  $\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+2\choose 2j}{k-1\choose t+1-2j}=k-1$ if $t=0$. On the other hand, if $t=n$, we have $\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+2\choose 2j}{k-1\choose t+1-2j}=\frac{1}{n+1}\sum\limits_{j=0}^{n}{n-k+2\choose 2j}{k-1\choose n+1-2j}=\frac{1}{n+1}$ for $n\equiv k~(\mathrm{mod}~2)$. Thus (\ref{Q-eq-1}) becomes
\[Kf(Q_{n, k})=\left\{\begin{array}{ll}
2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+2\choose 2j}{k-1\choose t+1-2j} &\mbox{if $n\equiv k~(\mathrm{mod}~2)$}\\
2^{n-1}\sum\limits_{t=0}^{n-1}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+2\choose 2j}{k-1\choose t+1-2j} &\mbox{if $n\not\equiv k~(\mathrm{mod}~2)$}.
\end{array}\right.\]

This completes the proof.
\end{proof}

Theorem \ref{thm-4} gives an exact formula of the Kirchhoff index of $Q_{n, k}$. However, its representation has some complicated. Now we give a lemma that will be used to simplify the Kirchhoff index of $Q_{n, k}$ for some $k$.

\begin{lem}\label{lem-6}
For a positive integer $n$, we have $\sum\limits_{i=1}^{n}\frac{1}{2i}{n+1\choose 2i}=\sum\limits_{s=1}^{n}\frac{2^{s}-1}{s+1}$.
\end{lem}
\begin{proof}
By the fundamental identity of combination ${n-1\choose 2i-1}+{n-1\choose 2i}={n\choose 2i}$ and $\frac{1}{2i}{n-1\choose 2i-1}=\frac{1}{n}{n\choose 2i}$, we get
\[\begin{array}{lll}
\sum\limits_{i=1}^{n}\frac{1}{2i}{n+1\choose 2i}&=&\sum\limits_{i=1}^{n}\left(\frac{1}{2i}{n\choose2i-1}+\frac{1}{2i}{n\choose 2i}\right)=\sum\limits_{i=1}^{n}\left(\frac{1}{n+1}{n+1\choose2i}+\frac{1}{2i}{n\choose 2i}\right)\\
&=&\sum\limits_{i=1}^{n}\left(\frac{1}{n+1}{n+1\choose2i}+\frac{1}{2i}{n-1\choose 2i-1}+\frac{1}{2i}{n-1\choose2i}\right)\\
&=&\sum\limits_{i=1}^{n}\left(\frac{1}{n+1}{n+1\choose2i}+\frac{1}{n}{n\choose 2i}+\frac{1}{2i}{n-2\choose2i-1}+\frac{1}{2i}{n-2\choose2i}\right)\\
&=&\sum\limits_{i=1}^{n}\left[\frac{1}{n+1}{n+1\choose2i}+\frac{1}{n}{n\choose 2i}+\cdots+\frac{1}{n-(n-2i)}{n-(n-2i)\choose2i}+\frac{1}{2i}{n-(n-2i)-1\choose2i}\right]\\
&=&\sum\limits_{i=1}^{n}\sum\limits_{r=0}^{n}\frac{1}{n+1-r}{n+1-r\choose2i}=\sum\limits_{r=0}^{n}\frac{1}{n+1-r}\sum\limits_{i=1}^{n}{n+1-r\choose2i}\\
&=&\sum\limits_{r=0}^{n}\frac{1}{n+1-r}(2^{n-r}-1)=\sum\limits_{s=1}^{n}\frac{2^{s}-1}{s+1}\\
\end{array}\]
This completes the proof.
\end{proof}

By Lemma \ref{lem-6}, we arrive at the following result, which is a special case of Theorem \ref{thm-4}. In fact, the formula of Kirchhoff index of the folded hypercube $FQ_n$ have been presented in \cite{Liu-4}. Corollary \ref{cor-4} simplifies the formula.

\begin{cor}\label{cor-4}
The Kirchhoff index of the enhanced hypercube $Q_{n, 1}$ is given by
\[\begin{array}{ll}Kf(Q_{n, 1})=2^{n-1}\sum\limits_{t=1}^{n}\frac{2^{t}-1}{t+1}\end{array}.\]
\end{cor}

\begin{proof}
By Theorem \ref{thm-4}, the Kirchhoff index of $Q_{n, 1}$ can be presented by
\[Kf(Q_{n, 1})=\left\{\begin{array}{ll}
2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n+1\choose 2j}{0\choose t+1-2j} &\mbox{if $n$ is odd}\\
2^{n-1}\sum\limits_{t=0}^{n-1}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n+1\choose 2j}{0\choose t+1-2j}&\mbox{if $n$ is even}.
\end{array}\right.\]
It is easy to check that
\[\begin{array}{lll}
\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n+1\choose 2j}{0\choose t-2j+1}=\sum\limits_{j=1}^{n}\frac{1}{2j}{n+1\choose 2j}{0\choose 0}=\sum\limits_{j=1}^{n}\frac{1}{2j}{n+1\choose 2j}
\end{array}\]
and
\[\begin{array}{lll}
\sum\limits_{t=0}^{n-1}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n+1\choose 2j}{0\choose t-2j+1}=\sum\limits_{j=1}^{n}\frac{1}{2j}{n+1\choose 2j}{0\choose 0}=\sum\limits_{j=1}^{n}\frac{1}{2j}{n+1\choose 2j}
\end{array}.\]
Combined with the above two situations, we get
\[\begin{array}{lll}
Kf(Q_{n, 1})&=&2^{n-1}\sum\limits_{j=1}^{n}\frac{1}{2j}{n+1\choose2j}
\end{array}.\]
By lemma \ref{lem-6}, we have
\[\begin{array}{ll}
Kf(Q_{n, 1})=2^{n-1}\sum\limits_{j=1}^{n}\frac{1}{2j}{n+1\choose2j}=2^{n-1}\sum\limits_{t=1}^{n}\frac{2^{t}-1}{t+1}.\\
\end{array}\]

This completes the proof.
\end{proof}
\begin{cor}\label{cor-5}
The Kirchhoff index of the enhanced hypercube $Q_{n, n-1}$ is given by
\[\begin{array}{ll}
Kf(Q_{n, n-1})=2^{n-1}\left(\sum\limits_{t=1}^{n-2}\frac{2^{t}-1}{t}+3\frac{(n-2)2^{n-1}+1}{n(n-1)}\right)
\end{array}.\]
\end{cor}
\begin{proof}
Let $k=n-1$, then $n$ and $k$ have the different parity. By Theorem \ref{thm-4}, we have
\[\begin{array}{lll}
Kf(Q_{n, n-1})&=&2^{n-1}\sum\limits_{t=0}^{n-1}\sum\limits_{j=0}^{n}\frac{1}{t+1}{3\choose2j}{n-2\choose t+1-2j}\\
&=&2^{n-1}\sum\limits_{t=0}^{n-1}\frac{1}{t+1}\left({3\choose 0}{n-2\choose t+1}+{3\choose 2}{n-2\choose t-1}\right)\\
&=&2^{n-1}\left(\sum\limits_{t=0}^{n-1}\frac{1}{t+1}{n-2\choose t+1}+3\sum\limits_{t=0}^{n-1}\frac{1}{t+1}{n-2\choose t-1}\right)\\
\end{array}\]
By Lemma \ref{lem-6}, it follows that
\[\begin{array}{ll}
 \sum\limits_{t=0}^{n-1}\frac{1}{t+1}{n-2\choose t+1}=\sum\limits_{r=1}^{n}\frac{1}{r}{n-2\choose r}=\sum\limits_{s=1}^{n-2}\frac{2^{s}-1}{s}
 \end{array}.\]
Besides,
\[\begin{array}{lll}
\sum\limits_{t=0}^{n-1}\frac{1}{t+1}{n-2\choose t-1}&=&\sum\limits_{t=0}^{n-1}\frac{1}{t+1}\left({n-1\choose t}-{n-2\choose t}\right)=\sum\limits_{t=0}^{n-1}\frac{1}{t+1}{n-1\choose t}-\sum\limits_{t=0}^{n-1}\frac{1}{t+1}{n-2\choose t}\\
&=&\frac{1}{n}\sum\limits_{t=0}^{n-1}{n\choose t+1}-\frac{1}{n-1}\sum\limits_{t=0}^{n-1}{n-1\choose t+1}=\frac{1}{n(n-1)}[(n-2)2^{n-1}+1]
\end{array}\]
Hence we obtain that
\[\begin{array}{ll}
Kf(Q_{n,n-1})=2^{n-1}\left(\sum\limits_{t=1}^{n-2}\frac{2^{t}-1}{t}+3\frac{(n-2)2^{n-1}+1}{n(n-1)}\right)
\end{array}.\]

This completes the proof.
\end{proof}

\begin{table}[h]{\footnotesize
\caption{\label{tab-1}{Kirchhoff index of $Q_{n, k}$ ($1\le k\le n-1$).}}
\resizebox{\textwidth}{26mm}{
\begin{tabular}{c|ccccccccccc}
\hline
\diagbox{n}{$Kf(Q_{n, k})$}{k}&$1$&$2$&$3$&$4$&$5$&$6$&$7$&$8$&$9$\\
\hline
$2$&3&--&--&--&--&--&--&--&--\\
\hline
$3$&13&14&--&--&--&--&--&--&--\\
\hline
$4$&50&51.6&54&--&--&--&--&--&--\\
\hline
$5$&182.7&185.3&189.1&194.9&--&--&--&--&--\\
\hline
$6$&653.3&657.9&664&672.8&687.5&--&--&--&--\\
\hline
$7$&2322.7&2330.7&2341.0&2355.0&2376.3&2413.6&--&--&--\\
\hline
$8$&8272&8286.2&8304&8327.4&8360.4&8412.4&8509.4&--&--\\
\hline
$9$&29626&29651&29682&29722&29776&29854&29984&30242&--\\
\hline
$10$&106870&106910&106970&107040&107130&107250&107440&107770&108480\\
\hline
\end{tabular}}}
\end{table}

We calculate the Kirchhoff index of $Q_{n, k}$ for some $n$ and $k$ in Tabla \ref{tab-1}, from which we see that the Kirchhoff index of $Q(n, k)$ is increased with the increase of $k$. Fortunately, it is true in general. To prove this property, we lead-in the following lemma.
\begin{lem}\label{lem-7}
 Let $1\le k\le n-1$. We have
$\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n-k\choose 2j}{k-1\choose t-2j-1}-{n-k\choose 2j-1}{k-1\choose t-2j}\right]\ge0$.
\end{lem}
\begin{proof}
Since $1\le k\le n-1$, we have $1\le n-k\le n-1$ and $0\le k-1\le n-2$. It is clear that
$F(n,k)=\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n-k\choose 2j}{k-1\choose t-2j-1}-{n-k\choose 2j-1}{k-1\choose t-2j}\right]=\sum\limits_{t=0}^{n}\frac{1}{t+1}\sum\limits_{r=0}^{n}(-1)^{r}{n-k\choose r}{k-1\choose (t-1)-r}$, from which
 we define a function $F_{n, k}(x)=\sum\limits_{t=0}^{n}\frac{1}{t+1}\sum\limits_{r=0}^{n}(-1)^{r}{n-k\choose r}{k-1\choose (t-1)-r}x^{t}$. Obviously, $F_{n, k}(1)=F(n,k)$.
Now we introduce another function,
\[\begin{array}{lll}
G(x)&=&x(1-x)^{n-k}(1+x)^{k-1}=x\sum\limits_{i=0}^{n-k}(-1)^{i}{n-k\choose i}x^{i}\sum\limits_{j=0}^{k-1}{k-1\choose j}x^{j}\\
&=&x\sum\limits_{i=0}^{n-k}\sum\limits_{j=0}^{k-1}(-1)^{i}{n-k\choose i}{k-1\choose j}x^{i+j}=x\sum\limits_{t=1}^{n}\sum\limits_{r=0}^{n}(-1)^{r}{n-k\choose r}{k-1\choose(t-1)-r}x^{t-1}\\
&=&\sum\limits_{t=1}^{n}\sum\limits_{r=0}^{n}(-1)^{r}{n-k\choose r}{k-1\choose(t-1)-r}x^{t}=\sum\limits_{t=0}^{n}\sum\limits_{r=0}^{n}(-1)^{r}{n-k\choose r}{k-1\choose(t-1)-r}x^{t}.
\end{array}\]
Then we have
\[\begin{array}{ll}
H(x)=\int_{0}^{x}G(u)\,du&=\int_{0}^{x}u(1-u)^{n-k}(1+u)^{k-1}\,du\\
&=\int_{0}^{x}\sum\limits_{t=0}^{n}\sum\limits_{r=0}^{n}(-1)^{r}{n-k\choose r}{k-1\choose(t-1)-r}u^{t}\,du\\
&=\sum\limits_{t=0}^{n}\sum\limits_{r=0}^{n}(-1)^{r}{n-k\choose r}{k-1\choose(t-1)-r}\int_{0}^{x}u^{t}\,du\\
&=\sum\limits_{t=0}^{n}\frac{1}{t+1}\sum\limits_{r=0}^{n}(-1)^{r}{n-k\choose r}{k-1\choose (t-1)-r}x^{t+1}.
\end{array}\]
Hence, $H(1)=F_{n, k}(1)=F(n,k)$. On the other hand,
\[\begin{array}{lll}
H(1)=\int_{0}^{1}G(u)\,du=\int_{0}^{1}u(1-u)^{n-k}(1+u)^{k-1}\,du.
\end{array}\]
Since $G(u)=u(1-u)^{n-k}(1+u)^{k-1}\ge0$ while $0\le u\le1$, we have
\[\begin{array}{lll}
F(n, k)=\int_{0}^{1}G(u)\,du=\int_{0}^{1}u(1-u)^{n-k}(1+u)^{k-1}\,du\ge0.
\end{array}\]
This completes the proof.
\end{proof}
\begin{thm}\label{thm-5}
Let $1\le k\le n-1$. For a given $n$, the Kirchhoff index of the enhanced hypercube $Q_{n, k}$ is increased with the increase of $k$ .
\end{thm}
\begin{proof}
 Denote by $\triangledown_{k}=Kf(Q_{n, k+1})-Kf(Q_{n, k})$ for $1\le k<n-1$. We will show that $\triangledown_{k}>0$ by induction on $k$.
By Theorem \ref{thm-4}, if $n\not\equiv k~(\mathrm{mod}~2)$, we have
\begin{equation}\label{delta-1}\begin{array}{ll}
\triangledown_{k}&=2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+1\choose 2j}{k\choose t+1-2j}-2^{n-1}\sum\limits_{t=0}^{n-1}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+2\choose 2j}{k-1\choose t+1-2j}\\
&=2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+1\choose 2j}{k\choose t+1-2j}-2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+2\choose 2j}{k-1\choose t+1-2j}
\end{array}
\end{equation}
The second equality holds because $\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+2\choose 2j}{k-1\choose t+1-2j}=0$ if $t=n$( in fact, by $n-k+2\ge 2j$ and $k-1\ge  n+1-2j$, we obtain that $n-k=2j-2$, which contradict to $n\not\equiv k~(\mathrm{mod}~2)$). Thus (\ref{delta-1}) can be simplified as
\begin{equation}\label{delta-2}\begin{array}{ll}
\triangledown_{k}&=2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n-k+1\choose 2j}{k\choose t-2j+1}-{n-k+2\choose 2j}{k-1\choose t-2j+1}\right]\\
&=2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n-k+1\choose 2j}\left({k-1\choose t+1-2j}+{k-1\choose t-2j}\right)-\left({n-k+1\choose 2j-1}+{n-k+1\choose 2j}\right){k-1\choose t+1-2j}\right]\\
&=2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n-k+1\choose 2j}{k-1\choose t-2j}-{n-k+1\choose 2j-1}{k-1\choose t-2j+1}\right].
\end{array}
\end{equation}
If $n\equiv k~(\mathrm{mod}~2)$, we have
\begin{equation}\label{delta-3}\begin{array}{ll}
\triangledown_{k}&=2^{n-1}\sum\limits_{t=0}^{n-1}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+1\choose 2j}{k\choose t+1-2j}-2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+2\choose 2j}{k-1\choose t+1-2j}\\
&=2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n-k+1\choose 2j}{k\choose t-2j+1}-{n-k+2\choose 2j}{k-1\choose t-2j+1}\right]\\
&=2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n-k+1\choose 2j}\left({k-1\choose t+1-2j}+{k-1\choose t-2j}\right)-\left({n-k+1\choose 2j-1}+{n-k+1\choose 2j}\right){k-1\choose t+1-2j}\right]\\
&=2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n-k+1\choose 2j}{k-1\choose t-2j}-{n-k+1\choose 2j-1}{k-1\choose t-2j+1}\right].
\end{array}
\end{equation}
The second equality holds because $\sum\limits_{j=0}^{n}\frac{1}{t+1}{n-k+1\choose 2j}{k\choose t-2j+1}=0$ if $t=n$ ( in fact, by $n-k+1\ge 2j$ and $k\ge n-2j+1$, we can imply that
$n-k=2j-1$, which contradict to $n\equiv k~(\mathrm{mod}~2)$).

The last representations of $\triangledown_{k}$ in  (\ref{delta-2}) and (\ref{delta-3}) are the same. Hence we need not to distinguish the above  two cases in what follows.

For $k=1$, from (\ref{delta-2}) we have
\begin{equation}\label{qq-1}\begin{array}{ll}
\triangledown_{1}&=2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n\choose 2j}{0\choose t-2j}-{n\choose 2j-1}{0\choose t-2j+1}\right]\\
&=2^{n-1}\left(\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n\choose 2j}{0\choose t-2j}-\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}{n\choose 2j-1}{0\choose t-2j+1} \right)\\
&=2^{n-1}\left(\sum\limits_{j=0}^{n}\frac{1}{2j+1}{n\choose 2j}-\sum\limits_{j=1}^{n}\frac{1}{2j}{n\choose 2j-1}\right)\\
&=2^{n-1}\left(\frac{1}{n+1}\sum\limits_{j=0}^{n}{n+1\choose 2j+1}-\frac{1}{n+1}\sum\limits_{j=1}^{n}{n+1\choose 2j}\right)=\frac{2^{n-1}}{n+1}>0.
\end{array}
\end{equation}
Now we assume that $\triangledown_{k}=2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n-k+1\choose 2j}{k-1\choose t-2j}-{n-k+1\choose 2j-1}{k-1\choose t-2j+1}\right]>0$ holds for  $1\le k < n-2$. Next, we will prove that $\triangledown_{k+1}>0$. By regarding $k$ as $k+1$, from (\ref{delta-2}) we get

\begin{equation}\label{delta-5}\begin{array}{ll}
\triangledown_{k+1}=2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n-k\choose 2j}{k\choose t-2j}-{n-k\choose 2j-1}{k\choose t-2j+1}\right].
\end{array}\end{equation}
Notice that the general term of (\ref{delta-5}) can be simplified as
\[\begin{array}{lll}
&&{n-k\choose 2j}{k\choose t-2j}-{n-k\choose 2j-1}{k\choose t-2j+1}\\
&=&\left[{n-k+1\choose 2j}-{n-k\choose 2j-1}\right]\left[{k-1\choose t-2j}+{k-1\choose t-2j-1}\right]
-\left[{n-k+1\choose 2j-1}  -  {n-k\choose 2j-2}\right]\left[{k-1\choose t-2j}  +  {k-1\choose t-2j+1}\right]\\
&  =  &  \left({n-k+1\choose 2j}{k-1\choose t-2j} -  {n-k+1\choose 2j-1}{k-1\choose t-2j+1}\right)+\left({n-k\choose 2j-2}{k-1\choose t-2j+1}-{n-k\choose2j-1}{k-1\choose t-2j}\right)\\
&&+\left({n-k+1\choose 2j}{k-1\choose t-2j-1} - {n-k\choose2j-1}{k-1\choose t-2j-1}\right)+\left({n-k\choose 2j-2}{k-1\choose t-2j}  -  {n-k+1\choose 2j-1}{k-1\choose t-2j}\right)\\
&  =  &  \left({n-k+1\choose 2j}{k-1\choose t-2j}-{n-k+1\choose 2j-1}{k-1\choose t-2j+1}\right)+\left({n-k\choose 2j-2}{k-1\choose t-2j+1}-{n-k\choose2j-1}{k-1\choose t-2j}\right)+{n-k\choose2j}{k-1\choose t-2j-1}-{n-k\choose 2j-1}{k-1\choose t-2j}\\
&  =  &  \left({n-k+1\choose 2j}{k-1\choose t-2j}-{n-k+1\choose 2j-1}{k-1\choose t-2j+1}\right)+{n-k\choose 2j-2}{k-1\choose t-2j+1}+{n-k\choose2j}{k-1\choose t-2j-1}-2{n-k\choose 2j-1}{k-1\choose t-2j}.
\end{array}\]
We can rewrite $\triangledown_{k+1}$ as follows.

\[\begin{array}{lll}
\triangledown_{k+1}&=&2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n-k\choose 2j}{k\choose t-2j}-{n-k\choose 2j-1}{k\choose t-2j+1}\right]\\
&=&2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n-k+1\choose 2j}{k-1\choose t-2j}-{n-k+1\choose 2j-1}{k-1\choose t-2j+1}\right]\\
&&+2^{n-1}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n-k\choose 2j}{k-1\choose t-2j-1}+{n-k\choose 2j-2}{k-1\choose t-2j+1} - 2{n-k\choose 2j-1}{k-1\choose t-2j}\right]\\
&=&\triangledown_{k}+2^{n-1}\sum\limits_{t=0}^{n}\frac{1}{t+1}\sum\limits_{j=0}^{n}\left[{n-k\choose 2j}{k-1\choose t-2j-1}+{n-k\choose 2j-2}{k-1\choose t-2j+1}-2{n-k\choose 2j-1}{k-1\choose t-2j}\right]\\
&=&\triangledown_{k}+2^{n-1}\sum\limits_{t=0}^{n}\frac{1}{t+1}\sum\limits_{j=0}^{n}\left[2{n-k\choose 2j}{k-1\choose t-2j-1}-2{n-k\choose 2j-1}{k-1\choose t-2j}\right]\\
&=&\triangledown_{k}+2^{n}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n-k\choose 2j}{k-1\choose t-2j-1}-{n-k\choose 2j-1}{k-1\choose t-2j}\right]\\
&=&\triangledown_{k}+2^{n}\sum\limits_{t=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{t+1}\left[{n-k\choose 2j}{k-1\choose t-2j-1}-{n-k\choose 2j-1}{k-1\choose t-2j}\right].\\
\end{array}\]
By the induction hypothesis, we have $\triangledown_{k}>0$. From Lemma \ref{lem-7}, we know that the above last term are no less than zero. It implies  that $\triangledown_{k+1}=Kf(Q_{n, k+2})-Kf(Q_{n, k+1})> 0$, i.e., $Kf(Q_{n, k+1})> Kf(Q_{n, k})$ for any $1\le k< n-1$.

This completes the proof.
\end{proof}

\begin{remark}
From Theorem \ref{thm-5} we have an interesting observation that $Q_{n,1}, Q_{n,2}$,..., $Q_{n, n-1}$ have different Kirchhoff indexes. It implies that they have different spectra and so they are not isomorphic from each other.
\end{remark}

By Theorem \ref{thm-5}, we know that  the Kirchhoff index of $Q_{n, k}$ is increased as $k$ increases. It follows that $Kf(Q_{n,1})\le Kf(Q_{n,k})\le Kf(Q_{n,n-1})$ for $1\le k\le n-1$. Thus we obtain the lower  and upper bounds of $Kf(Q_{n, k})$ below.
\begin{cor}\label{cor-6}
$\sum\limits_{t=1}^{n}\frac{2^{t}-1}{t+1}\le \frac{Kf(Q_{n,k})}{2^{n-1}}\le \sum\limits_{t=1}^{n-2}\frac{2^{t}-1}{t}+3\frac{(n-2)2^{n-1}+1}{n(n-1)}$.
\end{cor}

From the above Corollary we have the following limit function for $Kf(Q_{n, k})$.

\begin{thm}\label{thm-7}
$\lim\limits_{n\to\infty}\frac{Kf(Q_{n, k})}{\frac{2^{2n}}{n+1}}=1$.
\end{thm}

\begin{proof}
By Corollary \ref{cor-6}, we have $\sum\limits_{t=1}^{n}\frac{2^{t}-1}{t+1}\le \frac{Kf(Q_{n,k})}{2^{n-1}}\le \sum\limits_{t=1}^{n-2}\frac{2^{t}-1}{t}+3\frac{(n-2)2^{n-1}+1}{n(n-1)}$.\\
Hence
\[\begin{array}{lll}
\frac{\sum\limits_{t=1}^{n}\frac{2^{t}-1}{t+1}}{\frac{2^{n+1}}{n+1}}\le \frac{Kf(Q_{n, k})}{\frac{2^{2n}}{n+1}}\le \frac{\sum\limits_{t=1}^{n-2}\frac{2^{t}-1}{t}+3\frac{(n-2)2^{n-1}+1}{n(n-1)}}{\frac{2^{n+1}}{n+1}}.\end{array}\]
Denote by $A_{n}=\frac{\sum\limits_{t=1}^{n}\frac{2^{t}-1}{t+1}}{\frac{2^{n+1}}{n+1}}$ and $B_{n}=\frac{\sum\limits_{t=1}^{n-2}\frac{2^{t}-1}{t}+3\frac{(n-2)2^{n-1}+1}{n(n-1)}}{\frac{2^{n+1}}{n+1}}=\frac{\sum\limits_{t=1}^{n-2}\frac{2^{t}-1}{t}}{\frac{2^{n+1}}{n+1}}+3[\frac{n^{2}-n-2}{4n^{2}-4n}+\frac{n+1}{n(n-1)2^{n+1}}]$. In what follows, we will show that $\lim\limits_{n\to\infty}A_{n}=\lim\limits_{n\to\infty}B_{n}=1$.
Let $x_{n}=\sum\limits_{t=1}^{n}\frac{2^{t}-1}{t+1}=\frac{1}{2}+\frac{3}{3}+\cdots+\frac{2^{n}-1}{n+1}$ and  $y_{n}=\frac{2^{n+1}}{n+1}$. We have
\begin{equation}\label{lim-eq-1}\begin{array}{lll}
 \lim\limits_{n\to\infty}\frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}=\lim\limits_{n\to\infty}\frac{\frac{2^{n}-1}{n+1}}{\frac{(n-1)2^{n}}{n(n+1)}}=\lim\limits_{n\to\infty}\frac{n}{n-1}-\lim\limits_{n\to\infty}\frac{n}{(n-1)2^{n}}=1-0=1.\end{array}\end{equation}

By the Stolz-Ces\'{a}ro Theorem, we get\\
\[\lim\limits_{n\to\infty}A_{n}=\lim\limits_{n\to\infty}\frac{x_{n}}{y_{n}}=1.\]
Let $x_1'=x_2'=0$, and for $n\ge 3$ let $x_{n}'=\sum\limits_{t=1}^{n-2}\frac{2^{t}-1}{t}=1+\frac{3}{2}+\frac{7}{3}+\cdots+\frac{2^{n-2}-1}{n-2}$, $y_{n}'=\frac{2^{n+1}}{n+1}$. As similar as (\ref{lim-eq-1}), we have
\[\begin{array}{lll}
\lim\limits_{n\to\infty}\frac{\sum\limits_{t=1}^{n-2}\frac{2^{t}-1}{t}}{\frac{2^{n+1}}{n+1}}=\lim\limits_{n\to\infty}\frac{x_{n}'-x_{n-1}'}{y_{n}'-y_{n-1}'}=\lim\limits_{n\to\infty}\frac{n^{2}+n}{4(n^{2}-3n+2)}-\lim\limits_{n\to\infty}\frac{n(n+1)}{(n-2)(n-1)2^{n}}=\frac{1}{4}.
\end{array}\]
Thus
\[\begin{array}{ll}
\lim\limits_{n\to\infty}B_{n}&=\lim\limits_{n\to\infty}\left(\frac{\sum\limits_{t=1}^{n-2}\frac{2^{t}-1}{t}}{\frac{2^{n+1}}{n+1}}+3[\frac{n^{2}-n-2}{4n^{2}-4n}+\frac{n+1}{n(n-1)2^{n+1}}]\right)\\
&=\lim\limits_{n\to\infty}\frac{\sum\limits_{t=1}^{n-2}\frac{2^{t}-1}{t}}{\frac{2^{n+1}}{n+1}}+3\lim\limits_{n\to\infty}\frac{n^{2}-n-2}{4n^{2}-4n}+3\lim\limits_{n\to\infty}\frac{n+1}{n(n-1)2^{n+1}}\\
&=\frac{1}{4}+\frac{3}{4}-0=1
\end{array}\]
By the Squeeze Theorem, we get\\
\[\begin{array}{lll}
\lim\limits_{n\to\infty}\frac{Kf(Q_{n, k})}{\frac{2^{2n}}{n+1}}=\lim\limits_{n\to\infty}A_{n}=\lim\limits_{n\to\infty}B_{n}=1.\end{array}\]
This completes the proof.
\end{proof}

\begin{remark}
Although we get explicit formula of $Kf(Q_{n, k})$ when $k=1$ or $n-1$ in Corollary \ref{cor-4} and Corollary \ref{cor-5}, respectively. The calculation of the general representation of $Kf(Q_{n, k})$ given in Theorem \ref{thm-4} is more complex. Fortunately, Theorem \ref{thm-7} provides a simple uniform approximation function for $Kf(Q_{n, k})$ which is also independent of $k$, it means that $Kf(Q_{n, k})$ can be replaced with $\frac{2^{2n}}{n+1}$ if $n$ is large enough.
\end{remark}

redpajama_set_name: RedPajamaArXiv


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This gives us a responsibility to act on and save energy wherever we can.
Fortunately, at LCL, we have the right men and women for the job. Over the last few months we have been optimizing our cooling regulation and equipment to decrease our energy consumption without any impact on our customers. This has allowed us to cut energy consumption in the testing part of our data center by no less than 65 percent. Needless to say this makes a huge impact on the effect our data centers have on the environment.
How did we achieve this? Until now we have had a certain way to keep the temperature low in our data centers. As our customers' equipment creates a lot of heat and is sensitive to high temperatures, this is crucial to our operations. As is common practice in data centers, our server rooms have a 'cold corridor'. At LCL, these corridors have been around for year. Where we have been innovating during the past 6 months, is in temperature management. Our engineers have been experimenting with our temperature settings.
In the past we would maintain a constant temperature inside our server rooms themselves, extracting the air as it heats up and injecting cooler air. However, maintaining a constant temperature inside the cold corridor turns out to be a more efficient way of cooling, thus saving a lot of energy. The servers pull up the air from underneath the raised floor, into the cold corridor. The temperature is allowed to rise inside the room, without any effect on our equipment inside the cold corridor. This way there is less cooling to be done, hence less energy that needs to be used.
Another breakthrough was realised by modifying the speed of the fans of the cooling system outside our building. Before these would be either off or running at full speed, with no setting in between. By making the speed of the fans variable, these do not have to work at full capacity all the time, thus saving extra energy.
By using these techniques we are implementing a key part of the 'European Code of Conduct for Energy Efficiency in Data Centers'. We are the first endorser of this code in Belgium and believe very strongly in its goals. Implementing this new cooling solution has been a challenge, especially in a live environment. Realising this project gives a great deal of satisfaction. We will now work towards expanding our test area and implementing this solution in all of our data centers.


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Young playmaker looks ahead to future after returning to starting xi.
Midfielder Fergus Bell was pleased to return to our starting xi, despite Tuesday's frustrating 1-1 draw with non-league Concord Rangers in the FA Cup.
Our team will visit the Aspect Arena next week for a replay after Jordan Chiedozie's wonder-strike earned the Vanarama South outfit a replay.
"It's a frustrating time at the moment," said the playmaker. "Maybe we're not getting the rub of the green and there's a few things going against us. Obviously, if you go back a few weeks, then we were well under water with injuries too.
He continued: "It's sort of strange that I'm playing with a lot of strappings and things, but it's just nice to be back out on the pitch and not have to worry about injuries.
Looking ahead, the 23-year-old is keen to maintain his place in the starting line-up and help our team get back to winning ways – starting tomorrow afternoon when we entertain Plymouth Argyle.
"I think the intensity in training has gone up a notch [with players returning from injury] and that's because there's great competition for places," said Bell.
"The manager has got pretty hard decisions to make regarding the starting xi and that's always good for a team. It's sort of a nice headache for him to have in a way that players are now coming back to fitness.
He added: "We saw some of Plymouth's game last weekend on the coach and they just look like a real attacking threat. They've got good players across the pitch, but I do feel that we can match them. One of our strongest aspects in recent games has been that we haven't conceded too many goals.
"Our back three has been strong and Sascha [Studer, goalkeeper] has been improving as every game goes by, so it's a good test and if we come out of games like that with a win then it's only a positive, because people are probably tipping them for promotion with the run they're on, but hopefully we can come out of this with three points.
Our difficult run of matches in Sky Bet League Two continues next weekend when we head to Luton Town, who currently occupy top spot, but that is something that doesn't phase Bell.
"I think you've got to look at things in the context that you're going to play these teams at some point," added our summer signing.
"It doesn't necessarily matter whether we're on a winless run or a 10-match winning streak – the preparation changes week-by-week for the opposition that we're playing.


Please note: This glossary provides brief definitions of words but is not a legally accurate document, any omissions or incorrect statements are not the responsibility of Master Cover Insurance Services Ltd.
This is an intermediary between clients and an insurance companies. Brokers typically experts in a certain field of insurance, capable of using the insurance market to find the best cover for their clients. In the UK brokers must be approved by the FCA to operate legally.
Contractors Insurance is specially designed for businesses in the contracting industry, this policy allows you to tailor your ideal cover and includes public liability with employer's liability or contractors "all risks".
Our van insurance policies carry a 14 day cool off period. You may cancel your policy during this period and will only be charged for the time you were under cover (subject to a minimum administration charge of £20).
Our declaration stipulates the conditions that will apply to your policy, you must agree to this declaration to take out insurance.
Employers liability provides insurance cover for lawsuits	filed by injured or ill employees who take action against their employer.
The excess is typically the amount that you will have to contribute towards a claim on your insurance. Sometimes these can be set by the client, but are often fixed by the insurer. It is not normally more than a fraction of the total possible claim.
A rule that eliminates cover for certain risks, people, property classes, or locations from insurance cover on a policy. Unlikely or unusual locations are often excluded from insurance cover to allow a lower premium that covers the most likely scenarios.
An insurance type that will provide the funds to hire legal representation or meet other court costs in the event that a legal claim is made against you.
An insurance designed to pay out a set amount to the policyholder should they not be able to work as the result of an accident. A large lump payment is also normally included to be paid in the event of permanent disablement or death.
The price of an insurance policy, or certain components thereof. Typically charged annually or semi-annually.
A tax on premiums paid by the clients in addition to the cost of their insurance. IPT is typically around 5% of the policy total.
Coverage for professional damage caused by negligence or misunderstandings that can result in court cases or legal action.
Insurance for to cover a claim or payment due as a result of bodily injury or property damage caused to another person.
A quote is an estimated set of premiums based on a client's criterion for insurance coverage. Quotes are converted into policies when the cover and prices are agreed by both client and insurer/broker.
A term that may be used by insurers to describe the situation whereby the condition of an automobile or property is so extensively damaged that repair would cost more than replacement.
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When it comes to the natural health community, folks not only keep their fingers on the pulse of the latest health developments, they're also fantastic about sharing information. When a truly natural health aid comes along, you can bet there will be a good buzz about it. Such is the case with the benefits of apple cider vinegar (ACV).
Now, it should come as no surprise that apple cider vinegar is among the most popular kinds of vinegar in the nutritional and natural health worlds.
Another huge benefit of apple cider vinegar is that it only contains three calories per tablespoon. That's an incredibly low-calorie count for something you'll hopefully make a staple of your diet for a long time to come.
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For instance, a little solution of half apple cider vinegar and half water can work as an all-purpose cleaner for most kitchen surfaces. ACV can clean and deodorize the inside of appliances like your microwave or refrigerator.
You can also use straight ACV on wood cutting boards to sanitize them and make sure they stay squeaky clean.
Well, you'd think the process would be super high-tech and difficult, but it's easier than you think. For starters, you want to expose some crushed apples to yeast. This basically causes the fermentation of the sugar in the apple into alcohol. Next, add bacteria to the alcohol to further ferment it, creating acetic acid. This is the most active compound in apple cider vinegar.
If you want to go a step further and make sure your vinegar is organic and unfiltered, make sure your ACV solution contains strands of special proteins, enzymes, and friendly bacteria. These components will give your vinegar that hazy appearance.
What's the Nutrient Profile of ACV?
Well, believe it or not, despite the fact that there are good health benefits of apple cider vinegar, ACV isn't necessarily full of tons of vitamins or minerals. It does contain a significant portion of potassium, useful amino acids, and powerful antioxidants.
Also, if you'd like to find a way to naturally preserve food, you might want to try using ACV.
Can Apple Cider Vinegar Really Help Monitor Healthy Blood Sugar Levels?
Of course, when it comes to exploring apple cider vinegar benefits, there's been significant research when it comes to patients with blood sugar issues. Whether because of insulin resistance, or an inability to properly produce insulin, research suggests ACV could potentially help.
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Is it True Apple Cider Vinegar Might Lower Cholesterol and Help Your Heart?
Now, among the potential health benefits of apple cider vinegar is the chance it can help with one of the world's leading causes of death – heart health issues.
Just How Do You Use Apple Cider Vinegar?
The easiest way to experience the health benefits of apple cider vinegar is simply to start using it when you prepare your meals or cook. Try incorporating ACV into your stove-top cooking, sauces, and dressings.
Another way to get your ACV is to simply drop up to 2 tablespoons a day into a large glass of water and drink it. It's tangy, and it can be quite refreshing on a hot day. And again, you want to try to find unfiltered apple cider vinegar (make it organic when you can).
There's definitely reason to believe that regular use of apple cider vinegar could help boost your energy and have quite a few beneficial effects on your health. Though research is thin, there's more and more work being done each day to support this view.
And if you want to get started and see if apple cider vinegar is right for you, talk to your healthcare professional and add it to your diet – just don't take excessive amounts. Enjoy the health benefits of apple cider vinegar, and your body will thank you.