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Let the Young Vote
Conor Friedersdorf has some smart comments about the 10th grader who challenged Michele Bachmann to a competition over knowledge of the Constitution. I recommend it.
My reaction? If I had a magic wand, I'd give this 10th grader and her peers the right to vote — and the right to really challenge Bachmann by running against her in the next election.
First of all, as I've said before, I just don't see what is gained from making Americans ages 18-35 only partial citizens by restricting their ability to run for office. Yes, most 20 year olds shouldn't be Members of the House of Representatives, but so what? Most 50 year olds shouldn't, either! I see no reason why the electorate is supposedly able to judge the latter but not the former. Practices vary around the world, and this is one where I think the framers of the Constitution just got it wrong.
And, then, voting age. Ezra Klein had a fun little item yesterday about weighted votes in America, and he's upset that folks in Wyoming are overrepresented in the Senate, and that's fair enough. But you know who is really underrepresented? Those under 18 years of age. The more I think about it, the more I'm convinced that teenagers are perfectly capable of voting — and the more I think about it, the more I'm open to the argument for vote-from-birth.
Granted, the question becomes the meaning of voting and elections. Do we believe in elections because we believe that collective decision making yields the best decisions? Do we believe in elections because we believe that citizens should have an opportunity for meaningful political action? Or do we believe in elections because we believe that all interests should be equally represented? To the extent that the latter is the "real" reason for elections, then I think there's a strong argument against excluding even infants from the franchise (obviously, actually carried out by their parents, with families free to choose at what point the child was old enough to decide for herself).
I'd put it this way: if vote-from-birth was the norm and someone proposed eliminating it, I'm confident we would all think that was a terrible injustice. I'm not sure I'm for vote-from-birth, but I've been floating the idea here for a while now and haven't found the arguments against it very convincing at all. As far as teenage voting, well, I've made this argument before — if high school and even middle school kids are allowed to electioneer, and they are and they do, then I can't imagine why they are not allowed to vote.

redpajama_set_name: RedPajamaCommonCrawl


I have been suffering from back pain virtually all throughout my life. I thought it is of genetic origins for that my mum and my dad both have and had it all throughout my life. The pain that haunts you when you are sleeping, when you are standing or walking, when you cannot straighten out after you reached for an item on the ground is just too much to handle for a lifetime.
The worst part is that it creates a never-ending cycle of unhealthy lifestyle. Even though people tell you that by swimming or going to the gym it would be better, you just dont want to endure that horrible short term pain for a solution that may not even work. And as a result, you dont go to the gym because your back hurts because you dont go to the gym, ad infinitum.
But the Back Pain Relief4Life offers a method that will relieve you from your back pain forever, based on a fundamental physiological observation that explains why the pain developed in the first place.
Whenever you are trying to get rid of a medical condition, you will meet two kinds of professional healers. One is the doctor who will prescribe you drugs to cure your symptoms. He will try to alleviate you of your pain and will try to give you a life that looks as if you had no problems whatsoever.
But symptoms are a but a channel of communication between your mind and your body, telling you that there is a huge problem that you have to fix soon, before it turns into an even greater issue. When you try to cure symptoms, you are basically silencing someone who is trying to inform you that there is a danger and you have flee. To put it simply, it is an unwise decision.
The other kind of healer is the one who offers not instant relief by administering a drug but a way to end the cause of your symptoms. This method may be painful and may take more time than simply enduring and injection, but it will be more helpful in the long run.
Back Pain Relief4Life offers a way not to alleviate the pain but to purge the problem that causes it in the first place. It does not offer injections, drugs, painkillers that will alleviate the symptoms but will not even try to find its cause. It is an alternative solution that can be administered by a simple lifestyle change, which will completely dissipate back pain from your life in the long run.
The medical argument of Back Pain Relief4Life is that the main reason why your back hurts so much is because your muscles in your back are unevenly developed. Some of them are much larger than the others, and over the span of years or decades, this unequal redistribution of muscle mass will cause your back to literally pull apart your nerves and your spine.
The real cause of your back pain is not genetic but the fact that there is a vicious war going on in your back within your muscles and it is inflicting tremendous pressure on your nerves. Those muscles that are more developed are pushing out those that are weaker, creating injuries all around your back.
In the long run, it causes unbearable pain.
This is a physiological problem doctors never ever thought of. All they do with people who complain about back pain is to prescribe painkillers, or if the condition worsens until the pain is inhumane they inject the same drug into your back, hoping that it will solve the problem.
But it is the same foolishness as killing those who warn you of danger and expecting the danger to go away. It is madness.
As I have said before, Back Pain Relief4Life offers a solution that includes nothing more than a simple change in your habits and in your lifestyle. If you follow the guidance of the book, you will see that the pain goes away within weeks, without administering painkillers, drugs or supplements.
All you will need to do is to follow the unique training method written in the Back Pain Relief4Life guide which will help you even out the muscle distribution in your back and end the pressure the stronger muscles inflict both on the weaker ones and on your nervous system.
This exercise program was developed by a man in Serbia, who worked on to perfect a combination of movements that provides true pain relief to your back without medication. He created a whole system of back pain relief which is included in Back Pain Relief4Life. By using the system, you will be able to get rid of the haunting pain in one session and keep it away for the rest of your life.
For some people it might not work perfectly at the first try. But Back Pain Relief4Life also includes various other tips to relieve you of your back pain and return your back to its healthy anatomy. They teach you how to cure your back pain both in the short and in the long term.
The workouts of Back Pain Relief4Life are explained in a very easily understandable language with a step by step set of instructions. But if you feel that you do not understand what movements you are supposed to perform, you will be able to use the bonus targeted coaching sessions or the free one on one coaching to help you perfectly execute the prescribed workout program.
I watched as my mother and father suffered from back pains through their adult years. Every physical activity they tried to do was ruined by the curse of aching muscles and the jolts of pain constantly running through their nervous system. It renders physical activities almost impossible. Playing with your child? It is not an easy thing to do when you can barely stand up due to the constant pain. Doing the most essential daily chores? You may try, but in every second, you will feel the pain that rips through your body.
It poisons every aspect of your life from the professional to the personal. It poisons your mood and it sometimes pins you down for a whole day to the bed, finding it difficult to breathe from the pain. But for $47 only, you are given the choice to prevent such a fate from occurring, or if you are already suffering from the most agonizing back pain helps you cure yourself with a very easy exercise, never to feel its curse again.
– The guide is very cheap. At only $47, you are assured that your back pain will be forever gone.
– Avoids the use or the administration of drugs and other supplements to the individual since they will only harm the persons body rather than improving the situation.
– It is a fast pain relief, following the instruction to the dot, you will be sure to get the results that you are looking for very fast.
– Comes with some various tips that will help your back come in its perfect anatomy. Therefore, you can trust this guide to help you with every thing that you might be looking for.
– It will take some weeks before you get to see the results that you are looking for but since the guide has proven to work, it is worth waiting for the actual results.
Summary: Back Pain Relief4Life is a guide that will liberate you from the slavery of tons of drugs and supplement that claim to work and help you relief your back pain. The guide has been formulated to give the results that you are looking for using some simple techniques such as changing your everyday habits. Therefore, the guide is natural and it will help you gain the best results in a matter of weeks.


This download page is set up to provide easy access to the Triticum aestivum UK (bread wheat) genome sequence data.
The FTP download page collects datasets produced, described and used in the wheat UK genome paper.
PGSB PlantsDB hosts the wheat "UK 454 survey" data for interactive search at http://pgsb.helmholtz-muenchen.de/plant/wheat/uk454survey/searchjsp/index.jsp.
If you wish to download the Orthologous Assembly (OA) or the Low Copy Genome Assembly (LCG) in bulk files please proceed to our FTP download page (see below).


Rice, 37, said in the past, he has spent time in Western and Midwestern states experiencing different outdoor activities, even living out of a tent in Minnesota for a year. His experiences led him to pursue establishing a store in Livingston that would help people with similar passions.
"We just really want people to have fun camping," he said.
The store has everything one could need for their outdoor adventures, especially food. Many of the food options are from Montana and range from ice cream to meats and cheeses in the fridge section to dried goods and fresh produce as well. He said he has picked things that are easily packable.
An adventure is not complete without a little bit of alcohol and the business has that too. Rice said he wanted to incorporate beers that people may not find elsewhere as well as mid-range-priced wines. Beverages come bottled, boxed or in cans.
"Everything we have here is something we have tested and enjoyed," Rice said.
Rice said he has already worked with local businesses for collaborations. For example, Blue Bean Coffee Roasters came up with a "Jumping Off Point" blend and Mountain Air Sports Screen Printing in town helped produce their homemade logo for their T-shirts and sweatshirts.
Other items in the store include camping basics such as tents, axes and fire starters, as well as games, gifts and books.
The business will hold a grand opening at 11 a.m. on March 11, with opportunities for prizes. Store hours are 11 a.m. to 7 p.m. Tuesday through Thursday, 11 a.m. to 9 p.m. Friday and Saturday, and noon to 6 p.m. Sunday.
For more information, call store at 333-2191.


First project - what next?
I recently bought a 1974 CB125s from a friend who let the oil run out and seized it. This is my first project, so I'm maybe in over my head.
I've torn the little guy down, and I was hoping to get some advice on where to go next with it. I know I need to get it running, but should I take it out completely and crack open the top end? If it's seized, am I better off just putting a new Lifan engine in it and scrapping the seized one?
Any and all advice appreciated. Thanks in advance.
Re: First project - what next?
Myself, I'd crack open the motor and see what the damage is before deciding.
I have a little CB100, it was seized too.
Just pull the top end, they're not that difficult.
If it's badly pitted put a bigger sized piston, have the jug bored, should do it.
Might want to take a look at trans parts too, since it was run w/o oil.
The thing about running 4 strokes without oil and seizing them is it's normally not just a case of a new piston and hone - it's more than likely a shot cam, lifters, big/small end and crank as well.
No harm in taking what you've got apart of course - but ultimately putting in a Lifan would probably be cheaper and easier after you buy gaskets, piston, bore / hone, whatever else you'll nedd to rebuild your current lump.
There's a stack of info about the swaps on the web so it's not rocket science.
What a nice looking little bike! I hope you intend to restore it, it's too good to do anything really radical with...but then again, it's YOUR bike. I've been working on bikes for 45 years; my advice is to see if you can get it running before you even consider a tear down, a lot of these old Hondas have really low mileage and only need a little TLC to get 'em up and running (you did say "IF it's seized").
Forgot to mention the cam, valves, and such in the head, probably messed that up as well. Especially the cam.


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Home » Other Games » The World Of Path Of Exile Full Of Heroes
The World Of Path Of Exile Full Of Heroes
Path of Exile is surprising in many ways. When it comes to Diablo action RPG, it's a puzzling theme. Designers, however, enjoy the complexity, ideas and, above all, the fabulous playability.
"If that were the third Diablo, that would be something." So with this sentence I met this year really many times. I heard her from people who downloaded the Path of Exile for free and after some time they got the feeling that they had the best isometric action RPG. I finally understood that attitude, but he did not want to identify with him at once. The truth is, however, that it is a completely logical consequence of the fact that the creators were able to find the way not only for the very demanding and rigorous hard core community, but at the same time allowed for hours of simple, intuitive and, above all, entertaining and laic.
The new one, while testing publically, has received one praise after another, so when the developers turned the game to the final version, nobody was surprised to see anything out there. And that from this comparison for many people already in the beta version independent project won over the mainstream title, it was logical.
The Devil For All
So, let's go cheerfully metathy after the overthrow king of the action "RPG" of the kobylince of fables. Right from the beginning, I allowed myself to paraphrase the situation in Path of Exile, and probably always in the forefront. The game is free and nothing in the world forces you to pay. It's one of the few real-world games where the only thing you need for long hours of thrilling action is the Internet connection and the quality mouse. It is clear that even here a microtransaction model appears, but it is here as an added value.
The creators bet that the game will really have fun, so they like to buy visual effects for their armor and weapons. Offerings for action prices of all sorts of sceneries appear on the introductory screen, where you can get a half discount on a new and truly upgraded sword. Thanks to the colored letters, it all looks like a flyer from some RPG supermarket. But it is an acceptable business model that does not bother. Path of Exile is based on a proven model that we like to associate with the Diable. The truth is, however, that the creators have played so much with the details that a really interesting original has been created by the best of what appeared in the genre. Everything is, in fact, the same, but it is so different that you chuckle with blessing.
World Full Of Heroes
The basic classes are exactly just so ours finds exactly what they need, that is, a diverse sixth. Their name is different, but there is still a mix of close and remote combat, and it changes with the types of weapons. Either you crack the game, or you can do the same with magical spells and magic. You can choose a templer who is a mix of barbarian and paladin, or a duelist, who is such a guy in the uncomfortable that he loves cords. Marauder is Path of Exile as a classic barbarian and shadow then a thief. The offer is closed by a hunter and a witch. It is clear that the choice of character has a significant effect on the playing style. It is hard for a supporter of a close fight to choose a character that is perfect with bow and arrows.
On the other hand, the game offers an incredibly sophisticated character development that brings really great possibilities where your hero can go. With the new level, you will be able to open a rich tree of development where you can nicely allocate the points earned. It's just a splendor. The moment of getting a new level is a moment of endless blessing, when you go somewhere into the quiet corner of the dungeon and start thinking about how you can handle it all.
A great thing to do is to upgrade weapons. These can not only fit with gems that add specific properties, but also use leveling. The advantage is that the weapon with your warrior really "shuffle" and even make some of them a notional emotional relationship. If you go through almost half of the act with a sword and then sell it, because it really does want something better, maybe some tear will drop.
Path of Exile has a slower start. The first act takes longer than the player actually pulls. Beginning can be hard for a newcomer because understanding all the connections and melting games is not so. But no one presses you for anything. Beginning is good in this game to go in peace and quiet. The gradual declassification of the corners of the various possibilities that the game brings to you will come very soon. What can derail you at first is the absence of money. That's right, in Path of Exile there are not gold, silver, nothing.
Here, of course, you sell weapons to NPCs in cities, but you do not receive finance but items. In this case, it is a kind of fragments, which can then be composed of a scroll of identification. You can buy guns and potions from traders. A little complicated, but it is believable, because "at that time it was primarily a commodity exchange."
We Will All Meet Here
The moment when Path of Exile is really a great place for perfect experiences, it only arrives when you recognize other gaming options. Interesting diversification is the so-called leagues. It requires the creation of a new character (if you do not decide right at the beginning) and the choice of the league mentioned, which is nothing more than a higher goal of playing. You have to reach a certain level of character in a certain time. The goal is given and rewards are, for example, unique weapons.
Playing alone, without other players, is a patch for the soul who is not satisfied with the de facto forced multiplayer Diablo III. Playing with others, having the chance to get involved in someone's story and helping him with the final boss, is something that Path of Exile is just doing the right thing to do. It only depends on your taste. It is nice to create your own group of friends and play your game in more people. You can also take part in larger events organized by developers.
Story? But Who Will Meet This?
Path of Exile is still an ever-expanding work that develops both developers themselves and players. The games they create are the real reason why they are still returning to the world of Path of Exile. Nobody actually cares that the title offers a total of a dozen story.
There are two islands, one of which is dominated by some dark power that keeps order among people. On the other island there is a penal colony that you're heading with and wreck. This is how a game started, do not you remember? But it does not matter. The atmosphere of Path of Exile is working. Action and RPG elements are more sophisticated than hits such as Torchlight II and Diablo III. And just repeat that the free-to-play model claims music.
If you want to get cheapest PoE currency, U4GM always is the best choice for you. We guarantee that the activity is 100% real and effective. Have fun!
12/21/2017sky shine Other Games, Path of Exile Path Of Exile, Review
«Path Of Exile: Passive Skill Tree
Skills In Path Of Exile Take The Form Of Gems: Two Basic Types Of Skill Gems»


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About THC Pills
There's also no more distinctive odor. There will be no more hacking and coughing. All the benefits of THC without any of the drawbacks. That's fantastic news for your lungs and general health, since you won't be inhaling any unknown chemicals. Simply take a pill and go. What could be easier?
Before you start down the road of swallowing THC pills like candy, it's important to understand what's inside and why they're superior to other means of delivery. So that's what we'll go through in this post. We'll begin with the most essential information you need to know about THC pills. While marijuana can be consumed in a variety of ways, including flower, pre-rolls, waxes, shatters, and hashish, the most common is smoking it. THC pills are another alternative for some people who like to consume cannabis in other ways.
There are actually two forms of THC-infused pills. Cannabis oil capsules are the first kind. These cannabis oil pills are most often given by healthcare dispensaries and are made with raw cannabis extracts taken straight from the plant. Most patients who use weed oil pills take them to receive high doses of THC, which cancer and AIDS patients require for a therapeutic result. Today's topic will not be discussed in detail.
The second kind of THC pill, which we will review here, contains dronabinol. For the uninitiated, dronabinol is a fancy name for "THC made in a lab." It's nearly chemically identical to the THC found in weed, except it's not extracted from cannabis. Instead, chemists mix a bunch of chemicals together to generate dronabinol. And since dronabinol doesn't come from the plant, pharmaceutical companies can circumvent Schedule I restrictions to manufacture, distribute, and sell it.
Tetrahydrocannabinol, or THC for short, is the psychoactive ingredient in marijuana that makes you feel euphoric and high. THC exists in raw cannabis plants as the non-psychoactive THCA, which won't get you high until you add heat to decarboxylate it.
You can pop your bud in the oven, set in on fire, or even hold it under a magnifying glass in the hot sun for a while. Just be sure to inhale the smoke on those last two.
If you're going to use your bud in something else (like an edible or a pill), we suggest going the hot-oven route. Pretty much every other method of decarboxylation will destroy your ganja in the process. You don't want that now, do you?
Once your grass is decarboxylated, you're ready to think about making your own THC pills.
What Are THC Pills?
THC pills are made from extracted natural cannabinoids suspended in a fatty oil. Cram all that canna-goodness in a gelatin capsule, and you've got yourself a THC pill.
Even though THC pills are pretty simple in theory, there's still a lot of confusion in practice. That's because many people also refer to Marinol and Dronabinol (the legal synthetic versions of THC) as THC pills as well.
Because of the controversy surrounding THC pills, choosing the right chill pill can be a bit stressful if you're new to the cannabis game. Never fear! We'll help you understand how popping synthetic THC pills differs from smoking wonderful weed. That way, you'll understand the advantages and potential pitfalls of putting your pot in a pill vs a pipe, vape pen, or dab rig.
In the next few sections, we'll break down the scientific logic behind our preference for homemade organic marijuana THC pills over the synthetic, laboratory-made marinol and dronabinol THC pills.
We'll even wrap things up with a few tips on how to make your own custom THC chill pills at home!
The Truth About Marinol
Dronabinol is the active artificial THC-like ingredient in the prescription brand-name drug Marinol, which is given in pill form to AIDS and cancer patients for pain and nausea when all other alternatives have failed.
Marinol has proven to create a similar high, which has been shown to take effect slower but outlast the pain-relieving effects of a joint of combusted cannabis. But Marinol is not always a perfect alternative. Here's why.
Adverse Side Effects Of Marinol
Feeling high, dizziness, confusion, or sleepiness, according to the Marinol website. The National Library of Medicine (NLM) also lists weakness, stomach pain, nausea, vomiting, memory loss, and hallucinations.
About one-third of Marinol patients reportedly experience adverse Marinol side effects of drowsiness, dizziness, confusion, anxiety, changes in mood, muddled thinking, perceptual difficulties, coordination impairment, irritability, and depression from anywhere between four to six hours, according to the National Organization for the Reform of Marijuana Laws (NORML).
Marinol can give patients a marijuana overdose even more severe than the one that NY Times Columnist Maureen Dowd had when she "overdowded" by eating too much weed edible.
And in case you're wondering what "overdowd" means, it's a slang portmanteau of the words "over" (as in overdose) and "dowd" (the last name of journalist Maureen Dowd).
Some years ago, while writing a piece on the burgeoning marijuana industry in Colorado, Ms. Dowd consumed too much of a cannabis-infused candy bar — despite prior warnings — and went on a really bad trip.
She wrote about it in her column, but the event still earned her the dubious honor of having a semi-derogatory, completely mocking word "named" after her. But enough of the history lesson. Let's get back to the topic at hand.
Marinol, much like natural THCA in cannabis plants, transforms into a completely unique THC compound called 11-hydroxy-THC. This is why Marinol, much like natural THC Pills and weed edibles, is stronger than the typical Delta-9-THC compound you create from combusting or vaping cannabis.
THC pills and Marinol can take anywhere from 45 minutes to two hours to hit you and are often felt four or five times stronger compared to combusted and inhaled Delta-9-THC, just like a weed edible.
Researchers have argued that doctors have just as difficult a time prescribing adequate and safe doses of Marinol as they do getting the THC dosing of edible cannabis correct due to a lack of adequate data. This lack of data-driven research is in part due to the Drug Enforcement Agency's position on pot as a drug with no known medicinal value.
Marinol, the legal alternative to medical marijuana, is even more dangerous to drive on than a weed edible, let alone combusted cannabis (contrary to what Businessweek writes) because Marinol completely lacks the sobering anti-psychoactive qualities as well as all of the health benefits of cannabidiol or CBD.
According to the NLM, dronabinol, the active ingredient in Marinol, may also be habit-forming, leading to abnormal drug-seeking behavior with withdrawal symptoms, specifically including irritability, trouble sleeping, restlessness, hot flashes, and diarrhea if you suddenly stop taking it. Cannabis is not known to foster adverse drug-seeking behavior, unlike Marinol.
Popping THC Pills VS Smoking Pot
Based on findings from a landmark 2013 study in The Journal of Neuropsychopharmacology, Businessweek was able to conclude that legalizing cannabis in its natural smokeable form is unnecessary because it is widely (yet incorrectly) assumed that smoking cannabis is every bit as dangerous a cancerous health risk as is smoking nicotine cigarettes.
Leafly argues that THC pills are a more comfortable, convenient, and familiar option than the more traditional method of cannabis combustion. Cannabis and Nicotine to be clear, however, are not equally carcinogenic, and cannabis smoke has never been definitively linked to cancers associated with tobacco use in humans.
The stereotypes that people associate with smoking prevent many potential patients from understanding how smoking or vaporizing cannabis gives severely nauseated patients instantaneous awareness of pain relief.
This rapid onset of combusted or vaporized cannabis is the key advantage of these methods, as it provides the patient greater control to self-adjust the dose they need for pain relief by backing off and saving part of a dose for a later session or by taking additional puffs without fear of an 11-Hydroxy-THC overdose that taking too many THC pills can give you.
The major advantage of THC pills is that they are much more powerful and will last much longer than traditionally smoked or vaped pot if you can afford the patience for the 11-Hydroxy-THC to begin acting on your symptoms from your liver, giving you a groovy body high.
THC pills don't require you to learn how to smoke, they're more portable, and they don't have that same unmistakable overpowering sweet smell, thus making them much more discreet than the traditional method of combusting cannabis.
We advocate the patient's right to choose as well as experiment with their preferred method of smoking, vaporizing, dabbing, or ingesting their cannabis more so than we advocate any one method over any other. We strongly advocate utilizing a variety of cannabis consumption methods as well. We specifically suggest smoking, vaping, or dabbing for instantaneous relief and THC pills for longer-term relief.
A cancer patient suffering from acute nausea would probably want to take a THC pill right after smoking or vaporizing for an ideal combination of powerful fast-acting and long-term relief. A patient may enjoy the psychologically calming ritual of packing a pipe, also known as "packing a bowl" or perhaps rolling a joint for additional puffs as needed while waiting for the 11-Hydroxy-THC to kick in once it's been eaten.
Extra hits off of a joint, pipe, or vape pen will calm the anxious body high that 11-Hydroxy-THC can cause should the dosage in your THC pill be too strong. Extra cannabis can supplement your THC pill if it's not strong enough as well.
THC pills may be ideal for children who are not yet old enough to understand how to hit a dab rig or clear the chamber of a killer bong hit. Charlotte Figi is just such an example. Charlotte suffered from Dravet's Syndrome and experienced over 300 seizures a week. A cannabis oil extracted from a strain now known as Charlotte's Web changed all that.
Miss Figi now experiences only two or three seizures A MONTH thanks to the anticonvulsant effects of the marijuana plant.
We do recommend always using organic marijuana as your safest and healthiest option compared to other growing methods, no matter how you chose to get your THC into your endocannabinoid system!
The other major advantage of real marijuana is that it will probably cost you less to smoke it than it will to take Marinol. NORML estimates that Marinol can cost anywhere between $200 – $800 out of pocket to very sick people depending on the dosage size.
WebMD even estimates that the standard dose of dronabinol costs around $678 a month compared to $514 a month to smoke two marijuana joints a day. A dronabinol prescription can cost you as little as $15 per month if you are lucky enough to have insurance.
Still, the ironically higher costs of legal dronabinol compared to black market cannabis is why many doctors don't bother to recommend dronabinol and why many potential patients never bother to ask for or even try it, according to NORML.
Capsule Ingredients
Ground decarboxylated flower
Some THC pills contain ground-up decarboxylated flower for oral ingestion. Cannabis flower requires decarboxylation for its THCA to become active THC, which is what allows patients to tap into cannabis' benefits in a non-combusting way.
Crystalline is a powdered substance visually like table salt or sugar. Crystalline capsules contain nearly 100% pure powdered cannabinoids, often in THCA or CBDA-only formulas that allow patients to ingest their cannabis medication without the intoxicating effects associated with THC. The extraction process used to create crystalline strips cannabis of excess terpenes and other plant chemicals to leave only the desired molecule behind.
Distillate is a runny liquid that is similar to crystalline in that a distillation process is used to produce it, but its purity can run anywhere from 85% to 95% as the cannabis plant's terpenes are often reintroduced into the final product.
Distillate's cannabinoids are decarboxylated and are capable of providing users with some of the intoxicating effects associated with inhaled cannabis or ingesting edibles containing THC. Typically, distillate is devoid of a plant's fats and waxes, thereby providing patients with refined THC or CBD in precisely dosed capsules without additives.
Single Cannabinoid
Some cannabis capsules contain one variety of cannabinoid, in which case patients can find them labeled as "CBD" or "THC." These varieties do not contain any of the other medicinally beneficial cannabinoids or terpenes.
Multi-cannabinoid
Capsules that contain mixed percentages of THC and CBD work to enhance the medicinal effects and modulate the psychoactive effects to meet patients' individual needs. Capsules that include CBD can reduce users' blood pressure response to stress, among other medical benefits. CBD-infused capsules deliver relief to patients while mitigating THC's potentially adverse side effects, making the medication more tolerable to new users.
Cannabis capsules labeled as "full spectrum" include all of the cannabis plant's available biomolecules — terpenes, cannabinoids, and everything in between. Full-spectrum cannabis capsules can contain decarboxylated flower or infused-oil bases and allow patients to feel the effects of cannabis without any additives.
Nutraceutical oils
Additional nutraceuticals are often added to capsules to tailor their effects. Additives such ginkgo root, astragalus, or maca root generate energy; spirulina, chlorella, or turmeric add nutritional value; and powdered valerian root, skullcap, lemon balm, or hop flower deliver sedative effects.
Most capsule shells fall into one of three categories: gelatin, hypromellose, and pullulan. Gelatin capsules are made from animal byproducts derived from beef, while vegetarian options hypromellose and pullulan are made from non-toxic plant derivatives and fungal fermentation, respectively.
Weed Muffins
Golden Ticket Cannabis Strain


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Metallica To Debut New LP's Third Single Next Week
Posted on October 25, 2016 by philspicks
Courtesy: Blackened Records
Metallica is offering a very special treat for its fans this Halloween.
The band announced this week that it will release its new single 'Atlas, Rise' at 3p.m. EST on Oct. 31. The song is the third single to be released from the band's forthcoming album Hardwired…To Self Destruct.
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Fans can find a full list of participating stores online now here.
Hardwired…To Self Destruct is the band's first album in eight years. Its last full-length studio recording was Death Magnetic (2008). This latest effort was produced and mixed by Greg Fidelman, who also engineered and mixed Death Magnetic.
Hardwired…To Self Destruct is available for pre-order now here in a variety of platforms. It will be available in stores nationwide on Nov. 18.
More information on Hardwired…To Self Destruct is available online now along with all of the band's latest tour dates and more at:
Website: http://www.metallica.com
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text:
\section{Decomposition of the $k$-Bonacci Polyominoes}

Let ${\mathcal P}_{n,k}^{(1)}$ and  ${\mathcal P}_{n,k}^{(2)}$ denote the sets of $k$-bonacci polyominoes with $n$ columns, whose last column has height $1$ and 2, respectively. It is clear that ${\mathcal P}_{n,k}={\mathcal P}_{n,k}^{(1)} \cup {\mathcal P}_{n,k}^{(2)}$. If $P\in {\mathcal P}_{n,k}^{(1)}$, then $P$ is a unit square or $P=P'\square$, where $P' \in {\mathcal P}_{n-1,k}={\mathcal P}_{n-1,k}^{(1)} \cup {\mathcal P}_{n-1,k}^{(2)}$ $(n>1)$.  If $P\in {\mathcal P}_{n,k}^{(2)}$, then $P=P'C_j$, where $P' \in {\mathcal P}_{n-j,k}^{(1)}$ (possibly empty) and  $C_j$ is a concatenation of $j$ columns of height 2, for $1\leq j < k$.  Figure~\ref{deco} illustrates the above decomposition.

\begin{figure}[ht]
\centering
\includegraphics[scale=1]{Deco.eps}
\caption{Decomposition of a $k$-bonacci polyomino.}
\label{deco}
\end{figure}

\subsection{Area and Semiperimeter}

In this section we study the distribution of the area and semiperimeter    in ${\mathcal P}_{n,k}$.
Let $P$ be a $k$-bonacci polyomino.  We denote  by ${\texttt{area}}(P)$  the number of cells of $P$ and by
${\texttt{sper}}(P)$ the semiperimeter, that is the half of  the  perimeter of $P$ (since  the perimeter of a $k$-bonacci polyomino is always an even number).

 Define the generating function
$$F_k(x,p,q):=\sum_{n\geq 1} x^n \sum_{P\in {\mathcal P}_{n,k}}p^{{\texttt{sper}}(P)}q^{{\texttt{area}}(P)}, \quad k\geq 2,$$
where $x$ marks the length of the corresponding $k$-bonacci word, i.e., the number of cells in the bottom row of a polyomino.
Analogously, we introduce the generating functions
$$F_{k,j}(x,p,q):=\sum_{n\geq 1} x^n \sum_{P\in {\mathcal P}_{n,k}^{(j)}}p^{{\texttt{sper}}(w)}q^{{\texttt{area}}(P)},  \quad \text{for} \quad j=1, \,  2.$$
It is clear that
\begin{align}\label{GF}
F_k(x,p,q)=F_{k,1}(x,p,q)+F_{k,2}(x,p,q).
\end{align}
In Theorem~\ref{teoareaper1} we give a rational expression for the generating function $F_k(x,p,q)$.
\begin{theorem}\label{teoareaper1}
The generating function for $k$-bonacci polyominoes with respect to
the length of the corresponding word,
semiperimeter and area (marked respectively by $x$, $p$ and $q$)
is
$$F_k(x,p,q)=\frac{p^2 ((q + p q^2 ) x - (p q^3 - p^2 q^3) x^2 - p^k q^{2 k} x^k -
   p^{k+1} q^{2 k+1} x^{k+1})}{1 - (p q + p q^2) x + (p^2 q^3 - p^3 q^3) x^2 +
 p^{k+2} q^{2 k+1} x^{k+1}}.$$
\end{theorem}
\begin{proof}
From the decomposition given in Figure~\ref{deco} we have the functional equations
\begin{align*}
F_{k,1}(x,p,q)&=\underbrace{p^2qx}_{(1)} + \underbrace{pqx(F_{k,1}(x,p,q)+F_{k,2}(x,p,q))}_{(2)+(3)} \\
F_{k,2}(x,p,q)&=\underbrace{\sum_{j=1}^{k-1}p^{i+2}q^{2i}x^j}_{(4)} +\underbrace{\left(\sum_{j=1}^{k-1}p^{i+1}q^{2i}x^j\right) F_{k,1}(x,p,q)}_{(5)}\\
&=\frac{p^2(pq^2x-p^kq^{2k}x^k)}{1-pq^2x} + \frac{p(pq^2x-p^kq^{2k}x^k)}{1-pq^2x}F_{k,1}(x,p,q).
\end{align*}
To make understanding easier, cases (1) to (5) in Figure~\ref{deco} are indicated below their corresponding terms.
Solving the above system of equation and from \eqref{GF} we obtain the desired result.
\end{proof}
For example, the series expansion of the generating function $F_{3}(x,p,q)$ begins with
\begin{multline*}
(p^2 q + p^3 q^2) x + (p^3 q^2 + 2 p^4 q^3 +
    p^4 q^4) x^2 + \bm{(p^4 q^3 + 3 p^5 q^4 + 2 p^5 q^5 +
    p^6 q^5) x^3} \\
    + (p^5 q^4 + 4 p^6 q^5 + 3 p^6 q^6 + 3 p^7 q^6 +  2 p^7 q^7) x^4 + \cdots
    \end{multline*}
Figure~\ref{grafo2} shows the weights of the  $3$-bonacci polyominoes   corresponding to the bold coefficient in the above series.

\begin{figure}[ht]
\centering
\includegraphics[scale=1]{Grafo2.eps}
\caption{Weights for polyominoes in ${\mathcal P}_{3,3}$.}
\label{grafo2}
\end{figure}

\begin{corollary}\label{coroarea}
The generating functions for the total area and semi-perimeter within all of the members in ${\mathcal P}_{n, k}$ are
$$A_k(x)=\frac{3 x - 2 k x^k - 2 (2 - k) x^{k+1} + x^{2 k+1}}{(1 - 2 x + x^{k+1})^2}$$
and
$$P_k(x)=\frac{5 x - 5 x^2 - (2 + k) x^k - (1 - k) x^{k+1} + 4 x^{k+2} - x^{2k+2}}{(1 - 2 x + x^{k+1})^2},$$
respectively.
\end{corollary}
\begin{proof}
From the definition of  $F_k(x,p,q)$ we have the equalities
$$A_k(x)=\left.\frac{\partial F_k(x,1,q)}{\partial q}\right|_{q=1} \quad\text{and}\quad P_k(x)=\left.\frac{\partial F_k(x,p,1)}{\partial p}\right|_{p=1}.$$
Therefore, from Theorem~\ref{teoareaper1} we obtain the desired results.
\end{proof}
In particular, for $k = 2, 3,$ and $4$ we obtain the following generating functions:
\begin{align*}
A_2(x)&=\frac{x (3 + 2 x + x^2)}{(1 -  x -x^2)^2},\\
 A_3(x)&=\frac{x (3 + 6 x + 3 x^2 + 2 x^3 + x^4)}{(1 -  x -x^2-x^3)^2},\\
 A_4(x)&=\frac{x (3 + 6 x + 9 x^2 + 4 x^3 + 3 x^4 + 2 x^5 + x^6)}{(1 -  x -x^2-x^3-x^4)^2},\\
P_2(x)&=\frac{x (5 + x - 2 x^2 - x^3)}{(1 -  x -x^2)^2},\\
 P_3(x)&=\frac{x (5 + 5 x - 3 x^3 - 2 x^4 - x^5)}{(1 -  x -x^2-x^3)^2},\\
 P_4(x)&=\frac{x (5 + 5 x + 5 x^2 - x^3 - 4 x^4 - 3 x^5 - 2 x^6 - x^7)}{(1 -  x -x^2-x^3-x^4)^2}.
   \end{align*}

We will apply the same technique  to obtain other corollaries  throughout the article.

\subsection{Area and Semiperimeter of the Fibonacci polyominoes}

For the Fibonacci polyominoes ($k=2$) we can give some additional results.  Let $t_n(p,q)$ denote the $n$-th coefficient of  $F_2(x,p,q)$, that is,
$$t_n(p,q):= \sum_{P\in {\mathcal P}_{n,2}}p^{{\texttt{sper}}(P)}q^{{\texttt{area}}(P)}.$$

\begin{theorem}\label{fibopoly}
For all $n\geq 3$ we have
$$t_n(p,q)=pqt_{n-1}(p,q) + p^3q^3t_{n-2}(p,q),$$
with the initial values $t_1(p,q)=p^2 q + p^3 q^2$ and $t_2(p,q)=p^3 q^2 + 2 p^4 q^3$. Moreover,  for all $n\geq 1$ we have the combinatorial formula
\begin{align}\label{closedf2b}
t_n(p,q)=\sum_{i=0}^{\lfloor \frac{n+1}{2} \rfloor} \binom{n+1-i}{i}p^{n + i + 1} q^{n + i}.
\end{align}
\end{theorem}
\begin{proof}
Let $P$ be a Fibonacci polyomino with $n$ columns ($n\geq 3$).  If the last column has height 1, then the number of this kind of polyominoes is given by  $qpt_{n-1}(p,q)$. On the other hand, if the last column has height 2, then the previous column has height 1, and these polyominoes are counted by the polynomial  $p^3q^3t_{n-2}(p,q)$.  Hence the total number of Fibonacci polyominoes is given by $pqt_{n-1}(p,q) + p^3q^3t_{n-2}(p,q)$. Finally, we can use mathematical induction and the recurrence relation to prove the combinatorial identity.  This formula can be also proved by means of Zeilberger's creative telescoping method~\cite{ABZ}. Denote the summand on the right side
of the equality in \eqref{closedf2b} by $F(n,i)$, that is
$$F(n,i):= \binom{n+1-i}{i}p^{n + i + 1} q^{n + i}.$$
By the Zeilberger algorithm, $F(n,i)$ satisfies the relation
\begin{align}\label{rel1}
F(n+2,i)-pqF(n+1,i)-p^3q^3F(n,i)=G(n,i+1)-G(n,i),
\end{align}
with the certificate
$$R(n,i)=-\frac{i (2 - i + n) p^2 q^2}{(2 - 2 i + n) (3 - 2 i + n)}.$$
That is, $R(n,i)=G(n,i)/F(n,i)$ is a rational function in both variables.
If $f(n)$ denotes the right sum in the equality \eqref{closedf2b}, then summing both sides of \eqref{rel1} with respect to $i$ yields $f(n+2)-pqf(n+1)-p^3q^3f(n)=0$.  The sequences $f(n)$ and $t_n(p,q)$ satisfy the same recurrence relation and  have the same initial values, therefore these sequences coincides for all positive integers $n$.

\end{proof}

\begin{corollary}
The  total area for all Fibonacci polyominoes in ${\mathcal P}_{n,2}$ is
$$\sum_{i=0}^{\lfloor \frac{n+1}{2} \rfloor} \binom{n+1-i}{i}(n + i)=\frac{1}{5}(6nF_{n+2}+(n+2)F_n),$$
where $F_n$ is the $n$-th Fibonacci number.
\end{corollary}
\begin{proof}
From Theorem~\ref{fibopoly} the total area of the Fibonacci polyominoes is given by
$$\left.\frac{\partial t_n(1,q)}{\partial q}\right|_{q=1} =\sum_{i=0}^{\lfloor \frac{n+1}{2} \rfloor} \binom{n+1-i}{i}(n + i).$$
For the expression in terms of Fibonacci numbers we use the generating function of the total area
$$A_2(x)=\frac{x (3 + 2 x + x^2)}{(1 -  x -x^2)^2}=(3x+2x^2+x^3)\sum_{n\geq 0}c(n+2)x^n,$$
where $c(n)$ is the convolution of the Fibonacci numbers, that is
$$c(n)=\sum_{i=0}^nF_iF_{n-i}=\frac{1}{5}\left((n-1)F_n + 2nF_{n-1}\right).$$
In the last equality we use the identity (32.13) given in~\cite{Koshy}. Therefore,
$$[x^n]A_2(x)=3c(n+1)+2c(n)+c(n-1)=\frac{1}{5}(6nF_{n+2}+(n+2)F_n), \quad  n \geq 1.$$
\end{proof}

The number of Fibonacci polyominoes of area $n$ is related to the
Narayana's cows sequence ~\cite{Allou}. The
Narayana's cows sequence $b_n$ is defined by the recurrence relation  $b_{n}=b_{n-1}+b_{n-3}$ for $n\geq3$, with the initial values  $b_0=0, b_1=1$, and  $b_2=1$. The Narayana's cows sequence can be calculated with the formula
$$b_n=\sum_{i=0}^{\lfloor (n-1)/3 \rfloor} \binom{n-2i-1}{i}.$$

\begin{theorem}
The number of Fibonacci polyominoes of area $n$ is equal to
 the number of Narayana's cows $b_{n+1}$.
\end{theorem}
\begin{proof}
From Theorem~\ref{teoareaper1} the generating function of the number of Fibonacci polyominoes of a fixed area is
$$F_2(1,1,q)=\frac{q (1 + q + q^2)}{1-q-q^3}.$$
On the other hand, the generating function of the Narayana sequence is  $N(x):=\sum_{i\geq 0}b_ix^i=1/(1-x-x^3)$. Then
$N(q)=1+q+qF_2(1,1,q)$. By comparing
the $n$-th coefficient of the generating functions $N(q)$ and $F_2(1,1,q)$ we obtain  the desired result.
\end{proof}

For example,  Figure~\ref{grafo5} shows the Fibonacci polyominoes   of area 5, that is, $b_{6}=6$ polyominoes.

\begin{figure}[ht]
\centering
\includegraphics[scale=1]{Grafo5.eps}
\caption{Fibonacci polyominoes of area 5.}
\label{grafo5}
\end{figure}

\section{Number of Vertices and Edges}

The goal of this section is to enumerate the number of vertices and edges of the $k$-bonacci graphs.
Let $G$ be a $k$-bonacci graph.  We denote  by ${\textsf{ver}}(G)$ and ${\textsf{edg}}(G)$ the number of vertices and edges  of the graph $G$.  Let ${\mathcal G}_{n,k}^{(1)}$ and  ${\mathcal G}_{n,k}^{(2)}$ denote the set of $k$-bonacci graphs associated with the polyominoes in  ${\mathcal P}_{n,k}^{(1)}$ and  ${\mathcal P}_{n,k}^{(2)}$, respectively.
   Define the generating function
$$G_k(x,p,q):=\sum_{n\geq 1} x^n \sum_{G\in {\mathcal G}_{n,k}}p^{{\textsf{edg}}(G)}q^{{\textsf{ver}}(G)}, \quad k\geq 2.$$
Similarly, we have the generating functions
$$G_{k,j}(x,p,q):=\sum_{n\geq 1} x^n \sum_{G\in {\mathcal G}_{n,k}^{(j)}}p^{{\textsf{edg}}(G)}q^{{\textsf{ver}}(G)},  \quad \text{for} \quad j=1, 2.$$
In these trivariate generating functions
the variable $x$ marks the length of the corresponding $k$-bonacci word, i.e., the number of vertices in the bottom row of a graph minus one.

\begin{theorem}
The generating function for $k$-bonacci graphs with respect to
the length of the corresponding word,
the number of edges and vertices (marked respectively by $x$, $p$ and $q$)
is
$$G_k(x,p,q)=\frac{p^2 q^3((p^2 q + p^5 q^3) x - (p^7 q^4 - p^8 q^5) x^2 - p^{5k} q^{3 k} x^k -
   p^{5 k+3} q^{3 k+2} x^{k+1})}{1 - (p^3 q^2 + p^5 q^3) x + (p^8 q^5 - p^9 q^6) x^2 +
 p^{5 k+4} q^{3 k+3} x^{k+1}}.$$

\end{theorem}
\begin{proof}
From the decomposition given in Figure~\ref{deco} we have the functional equations
\begin{align*}
G_{k,1}(x,p,q)&=\underbrace{p^4q^4x}_{(1)} + \underbrace{p^3q^2x(G_{k,1}(x,p,q)+G_{k,2}(x,p,q))}_{(2)+(3)} \\
G_{k,2}(x,p,q)&=\underbrace{\sum_{j=1}^{k-1}p^{3 j + 2 j + 2}q^{3 (j + 1)}x^j}_{(4)} +\underbrace{\left(\sum_{j=1}^{k-1}p^{3 j + 2 j + 1}q^{3 j + 1}x^j\right) G_{k,1}(x,p,q)}_{(5)}\\
&=\frac{p^2 q^3 (p^5 q^3 x - p^{5 k}q^{3 k} x^k)}{1 - p^5 q^3 x} + \frac{p q (p^5 q^3 x - p^{5 k}q^{3 k} x^k)}{1 - p^5 q^3 x}G_{k,1}(x,p,q).
\end{align*}
Solving the above system of equation and from the equality $G_k(x,p,q)=G_{k,1}(x,p,q)+G_{k,2}(x,p,q)$ we obtain the desired result.
\end{proof}
For example, the series expansion of the generating function $G_{3}(x,p,q)$ begins with
\begin{multline*}
x \left(p^7 q^6+p^4 q^4\right)+x^2 \left(p^{12} q^9+2 p^{10}
   q^8+p^7 q^6\right)
   + \bm{x^3 \left(p^{16} q^{12}+2 p^{15} q^{11}+3
   p^{13} q^{10}+p^{10} q^8\right)}\\
   +x^4 \left(2 p^{21} q^{15}+3 p^{19} q^{14}+3 p^{18} q^{13}+4 p^{16} q^{12}+p^{13}
   q^{10}\right)+ \cdots
    \end{multline*}
Figure~\ref{grafo3} shows the weights of the  $3$-bonacci graphs  corresponding to the bold coefficient in the above series.

\begin{figure}[ht]
\centering
\includegraphics[scale=1]{Grafo3.eps}
\caption{Weights for the graphs in ${\mathcal G}_{3,3}$.}
\label{grafo3}
\end{figure}

\begin{corollary}
The generating functions for the total number of vertices and edges  within all of the members in ${\mathcal G}_{n, k}$ are
$$V_k(x)=\frac{10 x - 9 x^2 - 3 (1 + k) x^k - (4 - 3 k) x^{k+1} + 8 x^{k+2} -  2 x^{2k+2}}{(1 - 2 x + x^{k+1})^2}$$
and
$$E_k(x)=\frac{11 x - 5 x^2 - (2 + 5 k) x^k - (9 - 5 k) x^{k+1} + 4 x^{k+2} + 2 x^{2 k+1} - x^{2 k+2}}{(1 - 2 x + x^{k+1})^2},$$
respectively.
\end{corollary}

In particular, for $k = 2, 3$, and $4$ we obtain the following generating functions:
\begin{align*}
V_2(x)&=\frac{2 x (5 + x - 2 x^2 - x^3)}{(1 -  x -x^2)^2},\\
 V_3(x)&=\frac{x (10 + 11 x - 6 x^3 - 4 x^4 - 2 x^5)}{(1 -  x -x^2-x^3)^2},\\
 V_4(x)&=\frac{x (10 + 11 x + 12 x^2 - 2 x^3 - 8 x^4 - 6 x^5 - 4 x^6 - 2 x^7)}{(1 -  x -x^2-x^3-x^4)^2}, \\
E_2(x)&=\frac{x (11 + 5 x - x^3)}{(1 -  x -x^2)^2},\\
 E_3(x)&=\frac{x (11 + 17 x + 6 x^2 + x^3 - x^5)}{(1 -  x -x^2-x^3)^2},\\
 E_4(x)&=\frac{x (11 + 17 x + 23 x^2 + 7 x^3 + 2 x^4 + x^5 - x^7)}{(1 -  x -x^2-x^3-x^4)^2}.
   \end{align*}

Let $v_n(p,q)$ denote the $n$-th coefficient of  $G_2(x,p,q)$, that is,
$$v_n(p,q):= \sum_{G\in {\mathcal G}_{n,2}}p^{{\textsf{edg}}(G)}q^{{\textsf{ver}}(G)}.$$

\begin{theorem}\label{fibopoly2}
For all $n\geq 3$ we have
$$v_n(p,q)=p^3q^2v_{n-1}(p,q) + p^9q^6v_{n-2}(p,q),$$
with the initial values $v_1(p,q)=p^4 q^4 + p^7 q^6$ and $v_2(p,q)=p^7 q^6 + 2 p^10 q^8$. Moreover,  for all $n\geq 1$ we have the combinatorial formula
\begin{align*}
v_n(p,q)=\sum_{i=0}^{\lfloor \frac{n+1}{2} \rfloor} \binom{n+1-i}{i}p^{3 n + 1 + 3 i} q^{2 n + 2 + 2 i}.
\end{align*}
\end{theorem}

\section{Degree Sequences}

In this section we are interested in the degree sequence of a $k$-bonacci graph. Degree sequences have been well studied for Fibonacci cubes~\cite{Klav5} and Fibonacci-run graphs~\cite{OME3}.
Recall that the degree of a vertex of a graph is the number of edges that are incident to the vertex.  A $k$-bonacci graph can have only vertices of degree two, three, or four.  Let $G$ be a $k$-bonacci graph. We denote  by ${\texttt{deg}}_i(G)$ the number of vertices of degree $i$ in the graph  $G$.

We are interested in the  generating function
$$D_k(x, q_2,q_3,q_4):=\sum_{n\geq1}x^n\sum_{G\in {\mathcal G}_{n,k}}q_2^{{\texttt{deg}}_2(G)}q_3^{{\texttt{deg}}_3(G)}q_4^{{\texttt{deg}}_4(G)}, \quad k\geq 2,$$
 where $x$ marks the length of the corresponding $k$-bonacci word, i.e., the number of vertices in the bottom row of a graph minus one.
Analogously, we introduce  the generating functions
$$D_{k,j}(x, q_2,q_3,q_4):=\sum_{n\geq 1} x^n \sum_{G\in {\mathcal G}_{n,k}^{(j)}}q_2^{{\texttt{deg}}_2(G)}q_3^{{\texttt{deg}}_3(G)}q_4^{{\texttt{deg}}_4(G)}, \quad j=1,  2.$$

\begin{theorem}\label{degreegfg}
For all $k\geq 2$, the generating function $D_k(x,q_2,q_3,q_4)$ is given by
$$\frac{q_2^4 \left( (q_3^2 q_4 +q_4) x -(q_3^2
   q_4^2 - 2 q_2 q_3^2 q_4^2 +q_3^4 q_4) x^2 - q_3^{2 k} q_4^k x^k + (q_3^{2 k+2} q_4^k -2 q_2 q_3^{2 k}
   q_4^{k+1}) x^{k+1}\right)}{q_4(1 -(q_3^2 + q_4 q_3^2) x + (q_4 q_3^4 - q_2^2 q_4^2 q_3^2) x^2 + q_2^2 q_4^{k+1} q_3^{2 k}
   x^{k+1})}.$$
\end{theorem}
\begin{proof}
Let $G$ be a Fibonacci graph in ${\mathcal G}_{n,k}$.  If $n=1$, then $G$ contributes to the generating function the term $q_2^4x$. See Figure~\ref{deco2} case (1).  If $n>1$ and $G\in {\mathcal G}_{n,k}^{(1)}$, then this case contributes to the generating function the terms $$q_3^2xD_{k,1}(x,q_2,q_3,q_4) \quad \text{and} \quad q_2q_4xD_{k,2}(x,q_2,q_3,q_4),$$  as seen in Figure~\ref{deco2} cases (2) and (3). Notice that in the case (2) we have only two new vertices of degree 3.  In the case (3) we have a new vertex of degree 2 and another of degree 4. The red vertices denote the vertices of degree 2, the blue vertices denote the vertices of degree 3, and the green vertices denote the vertices of degree 4. For $n>1$ and $G\in {\mathcal G}_{n,k}^{(2)}$, this case contributes the terms   (see Figure~\ref{deco2} cases (4) and (5))
\begin{align*}
\underbrace{q_2^4\sum_{j=1}^{k-1}q_3^{2 (j - 1) + 2}q_4^{j - 1}x^j}_{(4)} +\underbrace{\left(q_2\sum_{j=1}^{k-1}q_3^{2 (j - 1) + 2}q_4^{j}x^j\right) D_{k,1}(x,q_2,q_3,q_4)}_{(5)}.
\end{align*}

\begin{figure}[ht]
\centering
\includegraphics[scale=1]{deco2.eps}
\caption{Decomposition of a $k$-bonacci graph.}
\label{deco2}
\end{figure}
Therefore, we have the functional equations
\begin{align*}
D_{k,1}(x,q_2,q_3,q_4)&=q_2^4x+ q_3^2xD_{k,1}(x,q_2,q_3,q_4) +  q_2q_4xD_{k,2}(x,q_2,q_3,q_4)\\
D_{k,2}(x,q_2,q_3,q_4)&=\frac{q_2^4 (q_3^2q_4 x - q_3^{2k}q_4^k x^k)}{q_4(1-q_3^2q_4x)} +\frac{q_2 (q_3^2q_4 x - q_3^{2k}q_4^k x^k)}{1-q_3^2q_4x}  D_{k,1}(x,q_2,q_3,q_4).
\end{align*}

Solving the above system of equation we obtain the desired result.
\end{proof}
For example, the series expansion of the generating function $D_3(x,q_2,q_3,q_4)$ begins with
\begin{multline*}
\left(q_2^4q_3^2+q_2^4\right) x+\left(2 q_2^5 q_3^2 q_4 + q_2^4 q_3^2
   + q_2^4 q_3^4 q_4 \right) x^2+\\
   \bm{\left(q_2^6 q_3^4 q_4^2 + q_2^6 q_3^2
   q_4^2 +2 q_2^5 q_3^4 q_4^2 + 2 q_2^5 q_3^4 q_4  + q_3^4 q_2^4\right) x^3} + \cdots
    \end{multline*}
Figure~\ref{grafo4} shows the weights of the  $3$-bonacci graphs  corresponding to the bold coefficient in the above series.
\begin{figure}[ht]
\centering
\includegraphics[scale=1]{Grafo4.eps}
\caption{Weights for the graphs in ${\mathcal G}_{3,3}$.}
\label{grafo4}
\end{figure}

\begin{corollary}
The generating function for the total number of vertices of degree 2 within all of the members in ${\mathcal G}_{n, k}$ is
$$D_{k}^{(2)}(x):=\frac{2 (4 x - 7 x^2 - 2 x^k + x^{k+1} + 7 x^{k+2} - x^{2 k+1} - 2 x^{2 k+2})}{(1 - 2 x + x^{k+1})^2}.$$
\end{corollary}

In particular, for $k = 2, 3$, and $4$ we obtain the following generating functions:
\begin{align*}
D_{2}^{(2)}(x)&=\frac{2 x (4 - x - 5 x^2 - 2 x^3)}{(1 -  x -x^2)^2},\\
D_{3}^{(2)}(x)&=\frac{2 x (4 + x - 4 x^2 - 8 x^3 - 5 x^4 - 2 x^5)}{(1 -  x -x^2-x^3)^2},\\
D_{4}^{(2)}(x)&=\frac{2 x (4 - 7 x - 2 x^3 + x^4 + 7 x^5 - x^8 - 2 x^9)}{(1 - 2 x + x^5)^2}.
   \end{align*}
\begin{corollary}
The generating function for the total number of vertices of degree 3 within all of the members in ${\mathcal G}_{n, k}$ is
$$D_{k}^{(3)}(x):=\frac{2 (x + x^2 - k x^k - (1 - k) x^{k+1} - 2 x^{k+2} + x^{2k+2})}{(1 - 2 x + x^{k+1})^2}.$$
\end{corollary}
In particular, for $k = 2, 3$ and $4$ we obtain the following generating functions:
\begin{align*}
D_{2}^{(3)}(x)&=\frac{2 x (1 + x + 2 x^2 + x^3)}{(1 -  x -x^2)^2},\\
D_{3}^{(3)}(x)&=\frac{2 (x + x^2 - 3 x^3 + 2 x^4 - 2 x^5 + x^8)}{(1 - 2 x + x^4 )^2},\\
D_{4}^{(3)}(x)&=\frac{2(x + x^2 - 4 x^4 + 3 x^5 - 2 x^6 + x^{10})}{(1 - 2 x + x^5)^2}.
   \end{align*}

  \begin{corollary}
The generating function for the total number of vertices of degree 4 within all of the members in ${\mathcal G}_{n, k}$ is
$$D_{k}^{(4)}(x):=\frac{3 x^2 + (1-k) x^k - (4 - k) x^{k+1} - 2 x^{k+2} + 2x^{2k+1}}{(1 - 2 x + 2 x^{k+1})^2}.$$
\end{corollary}

In particular, for $k = 2, 3$ and $4$ we obtain the following generating functions:
\begin{align*}
D_{2}^{(4)}(x)&=\frac{2 x^2 (1 + x)}{(1 -  x -x^2)^2},\\
D_{3}^{(4)}(x)&=\frac{x^2 (3 + 4 x + 4 x^2 + 2 x^3)}{(1 - x - x^2 - x^3 )^2},\\
D_{4}^{(4)}(x)&=\frac{3 x^2 - 3 x^4 - 2 x^6 + 2 x^9}{(1 - 2 x + x^5)^2}.
\end{align*}

\subsection{The degree sequence of the Fibonacci graph}
Finally, we consider the particular case of Fibonacci graphs ($k=2$).  Let $d_{n,2}(q)$ denote the $n$-th coefficient of  $D_2(x, q,1,1)$, that is,
$$d_{n,2}(q):= \sum_{G\in {\mathcal G}_{n,2}}q^{{\textsf{deg}}_2(G)}.$$
Similarly, we define the sequences
$$d_{n,3}(q):= \sum_{G\in {\mathcal G}_{n,2}}q^{{\textsf{deg}}_3(G)} \quad \text{and} \quad d_{n,4}(q):= \sum_{G\in {\mathcal G}_{n,2}}q^{{\textsf{deg}}_4(G)}.$$
\begin{theorem}\label{closeformula}
For all $n\geq 3$ we have
$$d_{n,2}(q)=d_{n-1,2}(q) + q^2d_{n-2,2}(q),$$
with the initial values $d_{1,2}(q)=2q^4$ and $d_{2,2}(q)=q^4 + 2 q^4$. Moreover,  for all $n\geq 1$ we have the combinatorial formula
\begin{align}\label{vertdeg2}
d_{n,2}(q)=\sum_{i=1}^{n} \left(\binom{n-1-\lfloor \frac{i}{2}\rfloor}{\lfloor\frac{i-1}{2}\rfloor}+  \binom{n-2-\lfloor \frac{i-1}{2}\rfloor}{\lfloor\frac{i-2}{2}\rfloor}\right)q^{i+3}.
\end{align}
\end{theorem}
\begin{proof}
Define the generating function
$D_2(x,q):=\sum_{n\geq 0} d_{n,2}(q)x^n$.
From the Theorem~\ref{degreegfg} we have the expression
$$D_2(x,q)=\frac{xq^4(2-x+2xq)}{1-x-q^2x^2}.$$
This generating function satisfies
$$D_2(x,q)-xD_2(x,q)-q^2x^2D_2(x,q)=2 q^4 x - q^4 x^2 + 2 q^5 x^2.$$
By comparing the  coefficient  of $x^n$, for  $n\geq 2$,  in the above equality, we obtain the recurrence relation for the sequence $d_{n,2}(q)$. On the other hand, the combinatorial sum \eqref{vertdeg2} is equivalent to
\begin{align}\label{closedf2}
\sum_{i\geq 0} \left(2\binom{n-i-1}{i-1}q^{2i+3}+ \binom{n-i}{i-1}q^{2i+2} + \binom{n-i-1}{i-2}q^{2i+2} \right).
\end{align}
Denote the above summand  by $F(n,i)$, that is
\begin{align*}
F(n,i):=& 2\binom{n-i-1}{i-1}q^{2i+3}+ \binom{n-i}{i-1}q^{2i+2} + \binom{n-i-1}{i-2}q^{2i+2}\\
=&\frac{(n-i-1)!\left(n(2q +1) -4 i q  + 2q-1\right)}{(i-1)!(n-2i+1)!}q^{2i+2}.
\end{align*}
By the Zeilberger algorithm, $F(n,i)$ satisfies the relation
\begin{align}\label{rel2}
F(n+2,i)-F(n+1,i)-q^2F(n,i)=G(n,i+1)-G(n,i),
\end{align}
with the certificate
$$R(n,i)=\frac{(n-i) (i-1)\left((4 i-6)q - n(2q+1)  + 1\right)}{(n+2-2 i) (n+3-2 i) \left(2 n q+n-4 i q+2 q-1\right)}.$$
If $f(n)$ denotes the combinatorial  sum in \eqref{closedf2}, then summing both sides of \eqref{rel2} with respect to $i$ yields $f(n+2)-f(n+1)-q^2f(n)=0$.  The sequences $f(n)$ and $d_{n,2}(q)$ satisfy the same recurrence relation and  have the same initial values, therefore these sequences coincides for all positive integers $n$.

\end{proof}

Notice that the  total vertices of degree 2 of all Fibonacci graphs in ${\mathcal G}_{n,2}$  is given by
$$\sum_{i=1}^{n}(i+3) \left(\binom{n-1-\lfloor \frac{i}{2}\rfloor}{\lfloor\frac{i-1}{2}\rfloor}+  \binom{n-2-\lfloor \frac{i-1}{2}\rfloor}{\lfloor\frac{i-2}{2}\rfloor}\right).$$

From a similar argument as in Theorem~\ref{closeformula} we can prove the following two theorems.

\begin{theorem}
For all $n\geq 3$ we have
$$d_{n,3}(q)=q^2(d_{n-1,3}(q) + d_{n-2,3}(q)),$$
with the initial values $d_{1,3}(q)=q^2+1$ and $d_{2,3}(q)=3q^2$. Moreover,  for all $n\geq 1$ we have the combinatorial formula
$$d_{n,3}(q)=\sum_{i=0}^{n}\left((1+q^2)q^{2(n-1-i)}\binom{n-i-1}{i} + (2q^2-q^4)q^{2(n-1-i)}\binom{n-i-2}{i}\right).$$
\end{theorem}

\begin{theorem}
For all $n\geq 3$ we have
$$d_{n,4}(q)=d_{n-1,4}(q) + q^2d_{n-2,4}(q),$$
with the initial values $d_{1,2}(q)=2q^4$ and $d_{2,2}(q)=q^4 + 2 q^4$. Moreover,  for all $n\geq 1$ we have the combinatorial formula
$$d_{n,4}(q)=\sum_{i=0}^{n+2} \left(\binom{n-1-\lfloor \frac{i-2}{2}\rfloor}{\lfloor\frac{i-3}{2}\rfloor}+  \binom{n-2-\lfloor \frac{i-3}{2}\rfloor}{\lfloor\frac{i-4}{2}\rfloor}\right)q^{i-3}.$$
\end{theorem}

Let $d_n$ denote the total number of vertices of the Fibonacci polyominoes in ${\mathcal G}_{n,2}$. In the following theorems we study the proportion between the sequences $d_{n,i}:=d_{n,i}(1)$ ($i=2, 3, 4$) and $d_n$.   Before,  we need the following result.

\begin{theorem}[Asymptotics of linear recurrences,~\cite{Flajolet2}]\label{FlaTheo}
Assume that a rational generating function $f(x)/g(x)$, with $f(x)$ and $g(x)$ relatively prime and $g(0)\neq 0$, has a unique pole $1/\beta$ of smallest modulus. Then, if the multiplicity of $1/\beta$ is $\nu$, we have
$$[x^n]\frac{f(x)}{g(x)}\sim \nu \frac{(-\beta)^\nu f(1/\beta)}{g^{(\nu)}(1/\beta)}\beta^n n^{\nu-1}.$$

\end{theorem}

\begin{theorem}\label{asymp1}
Among total degree of  vertices of all graphs in ${\mathcal G}_{n,2}$, the proportion of those that are of degree 2 is asymptotically
\[
\lim_{n\to\infty}\frac{d_{n,2}}{d_n} = \frac{7-\sqrt{5}}{22}\approx 0.21654236.
\]
\end{theorem}
\begin{proof}
The generating functions of the sequences $d_n$ and $d_{n,2}$ are rational, therefore we can use the asymptotic analysis for linear recurrences. First, note that the unique pole $1/\beta$ of the rational generating function
$$V_2(x)=\sum_{n \geq 0}d_nx^n=\frac{2 x (5 + x - 2 x^2 - x^3)}{(1 - x - x^2)^2}$$ is $\alpha=\frac{-1+\sqrt{5}}{2}$, with multiplicity 2. Therefore
$d_n\sim \frac{2(3+2\sqrt{5})}{5}\left(\frac{1+\sqrt{5}}{2}\right)^nn$. Similarly, we have $d_{n,2}\sim(\frac{1+\sqrt{5}}{2})^{n+1}n$. From these expressions we obtain the desired result.
\end{proof}

\begin{theorem}
Among total degree of  vertices of all graphs in ${\mathcal G}_{n,2}$, the proportion of those that are of degree 3 is asymptotically
\[
\lim_{n\to\infty}\frac{d_{n,3}}{d_n} = \frac{4+\sqrt{5}}{11}\approx 0.56691527.
\]
\end{theorem}
\begin{proof}
The unique pole $1/\beta$ of the rational generating function
$$\sum_{n \geq 0}d_{n,3}x^n=\frac{2 (x - x^2 + x^3 - 2 x^4 + x^6)}{(1 - 2 x + x^3)^2}$$ is $\alpha=\frac{-1+\sqrt{5}}{2}$, with multiplicity 2. Therefore
$d_{n,2}\sim \frac{2(2+\sqrt{5})}{5}(\frac{1+\sqrt{5}}{2})^nn$. Since \linebreak $d_n\sim \frac{2(3+2\sqrt{5})}{5}(\frac{1+\sqrt{5}}{2})^nn$  we obtain the desired result.
\end{proof}

\begin{theorem}
Among total degree of  vertices of all graphs in ${\mathcal G}_{n,2}$, the proportion of those that are of degree 4 is asymptotically
\[
\lim_{n\to\infty}\frac{d_{n,4}}{d_n} = \frac{7-\sqrt{5}}{22}\approx 0.216542364.
\]
\end{theorem}

\section{Number of Hamiltonian $k$-bonacci graphs}
A \emph{Hamiltonian cycle} is a cycle that visits each vertex exactly once.  Let $G$ be a $k$-bonacci graph.  We define  ${\texttt{Ham}}(G)=1$ if $G$ has a Hamiltonian cycle, and 0 otherwise.  If ${\texttt{Ham}}(G)=1$, we say that $G$ is a Hamiltonian $k$-bonacci graph.
   Define the generating function
$$H_k(x,q):=\sum_{n\geq 1} x^n \sum_{G\in {\mathcal G}_{n,k}}q^{{\texttt{Ham}}(G)}, \quad k\geq 2,$$
 where $x$ marks the length of the corresponding $k$-bonacci word, i.e., the number of vertices in the bottom row of a graph minus one.
Similarly, we have the generating functions
$$H_{k,j}(x,q):=\sum_{n\geq 1} x^n \sum_{G\in {\mathcal G}_{n,k}^{(j)}}q^{{\texttt{Ham}}(G)},  \quad \text{for} \quad j=1, 2.$$

\begin{theorem}
For all $k\geq 2$ we have
$$H_k(x,q)=\frac{x ((1 - x) x (1 - x^{2 \lfloor (k-1)/2\rfloor}) -
   q (1 + x) (1 - 2 x + x^{k+1}) (-2 + x + x^{2 \lfloor k/2\rfloor}))}{(1 - 2 x + x^{k+1}) (1 - x - 2 x^2 + x^3 + x^{2(1 +  \lfloor k/2 \rfloor})}.$$
\end{theorem}
\begin{proof}
From the decomposition given in Figure~\ref{deco} we have the functional equation
\begin{align*}
H_{k,1}(x,q)&=\underbrace{qx}_{(1)} + \underbrace{x(H_{k,1}(x,q)+H_{k,2}(x,q))}_{(2)+(3)}
\end{align*}
The polyomino given in the decomposition $(4)$ corresponds to the grid graph $P_3 \times P_i$, for $1 < i \leq k$. It is known\footnote{~See, for example the discussion on ``Math Stackexchange": \url{https://math.stackexchange.com/questions/1699203/hamilton-paths-cycles-in-grid-graphs}} that the grid graph $P_n \times P_m$ has a hamiltonian cycle if and only if at least one of $m$ or $n$ is even
or $m=n=1$. Therefore, the graph $P_3 \times P_i$ has a hamiltonian cycle if and only if $i$ is even. In this case, the generating function is given by
$$T_k(x,q)=\underbrace{q\sum_{j=1}^{\lfloor k/2 \rfloor}x^{2j-1} + \sum_{j=1}^{\lfloor (k-1)/2 \rfloor}x^{2j}}_{(4)} =\frac{x\left(q + x - x^{2\lfloor\frac{k-1}{2} \rfloor+1} -  qx^{2\lfloor\frac{k}{2} \rfloor} \right)}{1-x^2}.
$$
Therefore, we obtain the functional equation
\begin{align*}
H_{k,2}(x,q)&=\underbrace{T_k(x,q)}_{(4)} +\underbrace{\left(\sum_{j=1}^{\lfloor k/2 \rfloor}x^{2j-1}\right) H_{k,1}(x,q) +\left(\sum_{j=1}^{\lfloor(k-1)/2 \rfloor}x^{2j}\right) H_{k,1}(x,1)}_{(5)}\\
&=T_k(x,q) + \frac{x(1-x^{2\lfloor k/2 \rfloor})}{1-x^2}H_{k,1}(x,q) + \frac{x^2(1-x^{2\lfloor (k-1)/2 \rfloor})}{1-x^2}\frac{x(1-x^k)}{1-2x+x^{k+1}}.
\end{align*}
Notice that $H_{k,1}(x,1)$ is the generating function of the total number of $k$-bonacci words end in \texttt{1}.
Solving the above system of equation  we obtain the desired result.
\end{proof}

\begin{corollary}
The generating function for the total number of Hamiltonian $k$-bonacci graphs is
$$H_{k}(x):=\frac{x(1+x)(2-x-x^{2\lfloor k/2 \rfloor})}{1 - x - 2 x^2 + x^3 + x^{2 \lfloor k/2\rfloor +2}}.$$
\end{corollary}

In particular, for $2 \le k \le 7$
we obtain the following generating functions:
\begin{align*}
H_{2}(x)&=H_{3}(x)=\frac{x (2 + x)}{1 -  x -x^2},\\
H_{4}(x)&=H_{5}(x)=\frac{x (2 + x + x^2 + x^3)}{1 - x - x^2 - x^4},\\
H_{6}(x)&=H_{7}(x)=\frac{x (2 + x + x^2 + x^3 + x^4 + x^5)}{1 - x - x^2 - x^4 - x^5}.
   \end{align*}

Note that every Fibonacci graph admits a Hamiltonian cycle,
just walk along its border.
The grid $P_3 \times P_3$ does not have a Hamiltonian cycle, so
Hamiltonian 3-bonacci graphs cannot have grids $P_3 \times P_3$ as induced subgraphs. Thus, they are precisely 2-bonacci.
In general, Hamiltonian $2k$-bonacci equals
Hamiltonian $(2k+1)$-bonacci graphs, because any Hamiltonian $(2k+1)$-bonacci graph cannot
contain $P_{2k+1} \times P_3$ as
induced subgraph, and there are no $1^{2k}$ factors in the corresponding binary words.

redpajama_set_name: RedPajamaArXiv


text:
Home » Blog » The World's Most Extreme Science Laboratories
When it comes to pushing the boundaries of scientific discovery and understanding, sometimes only the most cutting-edge technologies, apparatus and environments will suffice – creating unique and extreme workspaces in the process.
To celebrate the unending spirit and determination of dedicated scientists across the globe, we'll take you on a whirlwind tour through five of the most extreme science laboratories in the world.
The Pyramid Laboratory – The Himalayas
Earning the title of the world's highest science lab, the Pyramid Laboratory in the Sagarmatha National Park of Nepal is located a staggering 16,568 feet above sea level. Found at the foot of Mount Everest, the research station is a stylish three-story pyramid-shaped laboratory housing a team of researchers studying areas as diverse as geology, climate, the environment, and human physiology.
Its primary goal? To improve the lives of the native populations of Nepal and surrounding nations and safeguard the world's fragile, high-altitude ecosystems. Those with vertigo or a fear of heights need not apply.
SNOLAB – Ontario, Canada
Roughly 2km under the Canadian province, SNOLAB is the world's deepest science laboratory – and seemingly the most likely to be used by an evil genius plotting global domination.
But for now, the decidedly non-nefarious SNOLAB is primarily focussed on the Sudbury Neutrino Observatory (SNO) experiment to study the detection of solar neutrinos – subatomic particles that have extremely small mass.
That's why the positioning of the lab underneath the Earth's surface is so important; the 2km thick layer serves to filter out cosmic radiations during research.
NOAA Aquarius Reef Base – Florida Keys
From the deepest underground lab to the deepest underwater research facility, the NOAA Aquarius Reef Base sits roughly 60 feet beneath the water's surface in the Florida Keys. Part of the Florida Keys National Marine Sanctuary, researchers have used the facility for almost 30 years as a base to study the ecology of the local reef.
And while plunging into highly pressurised waters might sound like something out of James Cameron's The Abyss, the rather homely research station includes sleeping quarters for up to six people as well as bedroom facilities and windows to watch the watery world pass by.
CERN – Geneva
Perhaps the most famous science laboratory on Earth, CERN spans more than 550 hectares across Switzerland and France and is officially the largest particle physics lab in the world.
And considering the lab is attempting to replicate the conditions of the Big Bang thanks to the Large Hadron Collider, scientific research doesn't get any bigger or more extreme than trying to uncover the secrets of the Universe.
Located 492ft below the ground, the Large Hadron Collider is the single-largest machine in the world – and CERN are looking to expand it further in the near future!
LENS-X Wind Tunnel – Buffalo, New York
The strongest gust you've been caught in is nothing compared to the LENS-X Wind Tunnel at the Calspan University of Buffalo Research Centre. Capable of creating Mach 30 speeds – 30 times the speed of sound – this wind tunnel goes hypersonic to test the most advanced aircraft, space planes and NASA's Orion space vehicle.
2.5m wide and over 30m long, LENS-X might only operate for two milliseconds at its strongest level, but that's enough to create 18m of airflow. And for companies testing out their latest designs, it's also much easier on the wallet compared to testing in the field. Whereas flight testing would usually destroy a prototype to the tune of a few million, designs can easily be tested and re-used in LENS-X for a hundred thousand dollars or so.
Svartisen Subglacial Observatory – Norway
Not to make another film reference, but this remote lab has more than a touch of John Carpenter's The Thing about it. Located in Norway? Check. Situated under 200m of ice? Check. It feels like Kurt Russell manning a helicopter is the only thing that's missing.
For the sake of humankind, that's where the comparisons fortunately end. It might take a flight, a drive and then a ferry to reach it, but this research lab is the perfect location for carrying out research into glaciers.
Under all that ice, scientists can investigate how glaciers move, how climate change might affect their surroundings, and how these huge chunks of ice can send out seismic signals.
That said, it's not exactly ideal for claustrophobic types. The extreme pressures at the bottom of the glacier cause it to get smaller and smaller, so staff have to work quickly during the six to seven days they're down there.
Mount Washington Observatory – New Hampshire
As crazy as it sounds, researchers have been braving the extreme weather of this New Hampshire-based lab since 1932.
Dedicated to developing their understanding of the natural systems that make up Earth's weather and climate, these plucky researchers maintain a weather station at the mountain's summit, and carry out a whole host of weather and climate research to get to grips with what affects conditions around us.
And they do it in winds that frequently reach 100mph. In fact, the lab was home to the second-fastest wind ever measured, with gusts of up to 231mph hitting the peak way back in 1934.
International Space Station – Outer Space
Once manned by Brit Tim Peake, the International Space Station is arguably the most extreme of all manmade science laboratories. Hurtling through space at a speed of 17,227 mph, the International Space Station calls its home 200+ miles above the Earth's surface.
Launched in 1998, the International Space Station primarily serves as a microgravity and space environment research laboratory – providing the setting for a range of experiments in biology, human biology, physics, astronomy, and meteorology.
As well as being the most extreme research lab in the world, the International Space Station is also thought to be the most expensive single item ever constructed – costing a whopping $150bn to build.
If you're looking to create a cutting-edge laboratory, or something a bit more off-the-shelf, the InterFocus team can help you realise your dream lab. For more information about how the InterFocus team can help you, visit our homepage or call us today on 01223 894833.
Why Hair Analysis Could Become an Alternative to DNA Testing
Although hair is routinely collected at the scene of a crime, these days it is considered unreliable and rarely... read more


The Best Christmas Bar Gifts On Etsy That You Can't Ignore!
Here are some amazing Christmas gifts that you'll want to offer to all your friends and at the same time wish having this in your home decor.
Here you can find a small list of christmas bar gifts, that are at the same time shipper and useful. It's true that these following are gifts, but don't forget to buy a special one just for you.
SEE ALSO: Your Favorite Mid-Century Modern Magazine Is Finally Here!
Choose your favorite drink and use this wonderful Handmade alcohol dispenser.
If you know anyone that has an industrial bar home decor, this can be a perfect gift this Christmas. There is a gift that is not too expensive and can make the day of who receives this amazing piece of handmade dispenser.
Everyone needs ideas, why not give them this: Ideas.
This Poster Cocktail can be an incredible gift for anyone who is gifted. As it can give you lots of ideas when inspiration is lacking. At the end of this article, we have a "Editor choices" that we believe this poster will fit well with any of these products.
You can always opt for a simple Bottle Opener, a simple object that can be always useful. And that can be used in a fun way in a bar, with a funny phrase.
Since we are talking about fun things, a Bar towel can be used in a fun way, and none of the guests will be indifferent. With a simple phrase like "Our Drinking Club has a Bunco problem!" can make the day different from who receives this gift.
Here are some Editor Choice, which we think you'll love.


text:
School Uniforms Persuasive
School Uniforms Are The Way To Go
School uniforms are essential towards a school's success. There are many benefits to having uniforms, yet many would still argue that uniforms hinder students' abilities to express themselves, but if uniforms are strictly enforced, students will be able to focus more in school, be less likely to indulge in gang violence, and help lower income families save money on clothes.
Background The way a person dresses is how someone is first seen and then judged. A test was made to see how people can change their judgement on a person just based off of their clothes, a group of men and women were assigned to give judgements on men and women who had Tailored suits and people who had suits you would buy…show more content…
The uniform is a symbol of school. The convenience of the uniforms are great for students as well. "While she would prefer shirts in a lighter blue, as opposed to navy, Alexandria said the uniforms made life simpler. "Yeah, it's so much easier in the morning," she said. "When I get up, I already know what I'm wearing, and it's already hung on my door." (Weiss, NYTIMES). The expression argument is great, but this expression can cause students to focus too much on their outfits instead of actually focusing on studying and learning in school. The professional look gives off the impression that the student is smart and a promising member of a society that will benefit the future. This will boost overall performance in the school due to such a positive meaning. Uniforms also have been shown to increase attendance and graduation rates. "YOUNGSTOWN, Ohio – Requiring students in urban public high schools to wear uniforms results in increased attendance, higher graduation rates and fewer student suspensions" (Campus Safety Staff). Schools with strict uniform policies have more retention rates from students which overall is great for students. Some may say that this limits expression, but this expression comes from education itself, and if students are attending school more they will learn to use this expression even further.The convenience of having something you can wear everyday without having to think about it keeps a students focus on strictly school when in…show more content…
When uniforms are in place, there is not significant , which makes students feel closer to one another bringing a more positive environment. "But some research suggests uniforms do indeed result in a more harmonious school environment, including Impact of School Uniforms on Academic Achievement and Student Behaviour, published in the U.S. in 2009" (Hassan, Toronto SUN). A more harmonious environment equates to a more peaceful workplace for students. The calmer the environment the less violence is likely to happen thus keeping students safe. Not only does this prevent bullying, but it saves families money. School clothes investments are less costly when schools have uniform policies in place. "According to a 2013 survey from the National Association of Elementary School Principals (NAESP), 77 percent of respondents estimated the average cost of school uniforms per child, per year, was $150 or less…. According to the National Retail Federation, in 2013 parents spent close to $231 per child on back-to-school clothing. In 2016 that number has gone up. Back-to-school clothing purchases are expected to be more than $235 per child. New shoes add about $130 to the tab" (Ingram theparade.com). There is a hefty price difference between the uniforms and non-uniform clothes. The benefits of Uniforms are too much. Student expression should be done during school and in their work. Looking at a parents view point, clothes at school are an expensive
The Importance Of Uniforms In Schools
Uniforms at schools make a student feel more relaxed. Parents are also impacted when schools require uniforms. Uniforms provide students with school uniqueness. However, some parents and students think that having all the students dress the same style is ridiculous. Although some people feel that uniforms are really not necessary at schools, some education experts argue that wearing a uniform can have positive results.
Dress Code Policy Analysis
This lessens differentiation and nervousness, which then results to the student being able to do his or her list of activities. To support this point, 5 Reasons Why Wearing a Proper Uniform is Important (2015) states that a dress code policy saves a lot of time and effort because the person doesn't have to choose again and again the clothes to be worn at school, thus preserving his or her time for other productive
School Uniform Advantages And Disadvantages Essay
Around the world, a lot of schools instituted the idea of students wearing uniforms so they can improve overall school performance and to reduce negative behavior. Thus, this has become a topic that has proponents that believe implementing schools uniform will enhance performance and safety between students in schools and critics that think school uniforms will cancel all the benefit that would be enjoyed. So, in order to have a better understanding on how school uniform policies affect the educational system and the students as well. We have to know the advantages and disadvantages that come from implementing a mandatory school uniform. First of all, according to oxford dictionary, "school uniforms are the special clothes worn by
Should School Uniforms Be Banned Essay
Schools can develop programs or activities that include drawing, music, dance, martial arts, theater, sculpture and even crafts. In conclusion, school uniforms benefit students in a number of ways. School uniforms are cost-efficient, promote equality among students and reduce the distractions that don't relate to school work. Even Bill Clinton, former President of the United States of America said that uniforms are a "better" choice for students. I strongly believe that school uniforms should not be banned from schools and I encourage other schools who don't require school uniforms to try as uniforms create a better learning environment for the future leaders of the
Persuasive Essay On School Uniforms
First of all, uniforms save parents heaps of money. Secondly, uniforms can diminish the economic and social barriers between students. Lastly, uniforms can provide a happier, safer environment for everyone in the school. Uniforms are cost efficient and can save parents hundreds of dollars. In 2012, families with school-age children spent an average of $688.82 on apparel and shoes back-to-school shopping.
Negative Effects Of School Uniforms
All humans have the right to freedom of speech. Is a public school dress code requirement violating that freedom of speech if it protects student human rights? Research has shown that while many people believe school uniforms to have some negative effects on teachers, students, and their parents, uniforms have more benefit than just looking nice at school. Being required to wear school uniforms prevents violence, promotes higher motivation for academic learning, and encourages more respect for everyone in the school environment. Wearing uniforms has an impact on the student's school years, so they should be required to be worn every day of school.
Should Uniforms Be Required In Schools Essay
Uniforms help students to prepare for when they leave school and may have to dress smartly or wear a uniform in their future jobs, they gain responsibility to dress smartly, and dress appropriately. It brings so many benefits for so many reasons. Parents don't even have to spend so much money in clothing for their students. If students wear non school uniforms to school, their clothing gets dirty, and over used, when students go out after school, they won't have clean clothes to go out with. With uniforms in hand, after school clothes won't be wasted, school uniforms will be only used in school.
Why Should Students Wear Uniforms To School
Another benefit of uniforms is that they will keep children more disciplined. Schools also have to deal with fights and crimes, but the rate of them will go down if students wear uniforms since fights and crimes are mainly about the designer and expensive clothes. Students should wear uniforms to school because it keeps them disciplined, focused, and they will lower the rate of violence. To begin with, students should wear uniforms to school to result in more disciplined children. In the
Should Schools Keep Dress Code Essay
A quote by Larry Wilder reveals that "68% of parents believed the uniform policy improved overall academic performance." This allows us to see that after dress codes were required in schools parents saw an increase in their kid/kids grades. Another quote by Kim Schlauch says that "Uniforms/dress codes reduce distractions, allowing students to focus more on their academics." This means that it is believed when kids follow the dress code they will be less distracted by the quality of their clothes or the clothes of others making it easier for them to work on their school work. The last quote by KIMA TV states that "The school uniforms create a better learning environment. Kids pay less attention to what they wear and can focus more on academics." This shows that kids pay more attention to schoolwork rather than what they wear which leads to an increase in academic
Importance Of School Discipline
The more it happens, the worse it gets, and it is not cool. Don't you hate when you have to call your parents in the middle of a school day and ruin their day by telling them you disrupted someone else's? It overall depends on how schools enforce discipline, but it can get ugly in some schools. Positive School Discipline is the best way to lead a school because it can lead to less suspensions and expulsions, it can help teach students better ways of acting rather than making them feel like
More about School Uniforms Persuasive


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This statement sets out BEUMER Group UK Limited's actions to understand all potential modern slavery risks related to its business and to put in place steps that are aimed at ensuring that there is no slavery or human trafficking in its own business and its supply chains. This statement relates to actions and duties during the financial year 1 January 2019 to 31 December 2019.
As part of manufacturing industry, BEUMER Group UK Limited recognises that it has a responsibility to take a robust approach to slavery and human trafficking. BEUMER Group UK Limited is absolutely committed to preventing slavery and human trafficking in its corporate activities, and to ensuring that its supply chains are free from slavery and human trafficking.
Organisational structure and supply chains
This statement covers the activities of BEUMER Group UK Limited. BEUMER Group is an international manufacturing leader in intralogistics in the fields of conveying, loading, palletising, packaging, sortation and distribution technology. Together with BEUMER Group A/S, BEUMER Group employs about 4,000 people and achieves an annual turnover of about 700 million EUR.
Responsibility for the organisation's anti-slavery initiatives is as follows
Policies: HR and the board are responsible for putting in place and reviewing policies and the process by which they were developed.
Risk assessments: Health & Safety
Investigations/due diligence: Senior Management and the board are responsible for investigations and due diligence in relation to known or suspected instances of slavery and human trafficking.
BEUMER Group UK Limited undertakes due diligence when considering taking on new suppliers, and regularly reviews its existing suppliers. The organisation's due diligence and reviews include:
mapping the supply chain broadly to assess particular product or geographical risks of modern slavery and human trafficking;
evaluating the modern slavery and human trafficking risks of each new supplier
reviewing on a regular basis all aspects of the supply chain based on the supply chain mapping;
conducting supplier audits or assessments through our own team of auditors, which have a greater degree of focus on slavery and human trafficking where general risks are identified;
creating an annual risk profile for each supplier;
taking steps to improve substandard suppliers' practices, including providing advice to suppliers through auditing and requiring them to implement action plans.
invoking sanctions against suppliers that fail to improve their performance in line with an action plan or seriously violate our supplier code of conduct, including the termination of the business relationship
BEUMER Group UK Limited organisation has reviewed its key performance indicators (KPIs) in light of the introduction of the Modern Slavery Act 2015. As a result, the organisation is continuously reviewing its existing and new supply chains.
All personnel working in supply chain within BEUMER Group UK Limited is required to complete training on modern slavery.
The organisation's modern slavery training covers:
our business's purchasing practices, which influence supply chain conditions and which should therefore be designed to prevent purchases at unrealistically low prices, the use of labour engaged on unrealistically low wages or wages below a country's national minimum wage, or the provision of products by an unrealistic deadline;
how to assess the risk of slavery and human trafficking in relation to various aspects of the business, including resources and support available;
how to identify the signs of slavery and human trafficking;
what initial steps should be taken if slavery or human trafficking is suspected;
how to escalate potential slavery or human trafficking issues to the relevant parties within the organisation;
what external help is available, for example through the Modern Slavery Helpline, and "Stronger together" initiative;
what messages, business incentives or guidance can be given to suppliers and other business partners and contractors to implement anti-slavery policies; and
what steps the organisation should take if suppliers or contractors do not implement anti-slavery policies in high-risk scenarios, including their removal from the organisation's supply chains.
Awareness-raising programme
As well as training staff, BEUMER Group UK Limited has raised awareness of modern slavery issue, by putting up posters across our premises. The policy is incorporated as part of our induction programme and on the intranet for all to access.
Board approval
This statement has been approved by the organisation's board of directors, who will review and update it annually.
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Q: Why are there parantheses in this statement containing onClickListener? What is the purpose for the the empty parentheses following onClickListener in this snippet of Android code? (I'm wondering, since onClickListener is an interface.)
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redpajama_set_name: RedPajamaStackExchange


Facebook announced on 3rd July that it has hired the team behind the startup Bloomsbury AI to cope with fake news shared on the social media platform.
The social media giant did not take the full acquisition of the company, but hired Bloomsbury AI's team in what's sometimes called an "acquihire." According to sources, Facebook plans to deploy the company's AI team, software, and tech to work on dealing with fake news and to tackle other content issues.
The news was first reported by TechCrunch on 2nd July that Facebook is in talks to acquire the AI company for up to $30 million. Facebook added on its page that the Bloomsbury team will help us grow our AI efforts in London by joining a community of strong engineering talent. We look forward to welcoming them to Facebook and we can't wait to see what we build together.
As we know Facebook is in hot waters after Cambridge Analytica scandal and a large number of users have stopped using Facebook after losing trust in the social media network by considering Facebook a medium of propagating fake content and news. In order to gain trust back, Facebook in June announced the expansion of its fact-checking program and adoption of machine learning to combat fake news.
Moreover, Facebook is also implementing machine learning technology to find duplicates of fake stories. The algorithm will downrank fake news and demote pages that are spreading sponsored hoaxes in various countries. Facebook believes the technology will help them stay accountable and track their progress.


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O Journal of Monetary Economics é um jornal académico revisado por pares que cobre pesquisas em macroeconomia e economia monetária. É publicado pela Elsevier e foi criado em outubro de 1973 por Karl Brunner e Charles I. Plosser. A partir de 2002 foi fundido com a Carnegie-Rochester Conference Series on Public Policy. A última série foi criada em 1976 e foi publicada de forma independente, originalmente pela North-Holland Publishing Company, agora uma marca da Elsevier. De acordo com o Journal Citation Reports, o jornal teve um factor de impacto de 1,726 em 2014. Desde 2014 os seus editores são Urban Jermann e Yuriy Gorodnichenko. É amplamente considerado como um dos jornais académicos de economia de maior prestígio e consistentemente classificado como um dos 10 primeiros entre todos os jornais de economia.

Revistas científicas em inglês
Revistas científicas da Elsevier
Publicações fundadas em 1973
Revistas científicas de economia

redpajama_set_name: RedPajamaWikipedia


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