NATIONAL REPORT — Almost 50 percent of retailers responding to the exclusive Convenience Store News 2017 Technology Study said they plan to add new equipment and replace technology this year, with an expected growth in capital expenditure spending of 8 percent more than in 2016.
For an analysis of this year's Technology Study results, CSNews turned to Collupy, a former retail technology leader and current executive consultant for the W. Capra Consulting Group, a leading retail technology, payments, security and strategic implementation firm.
Over the last several years, the convenience store industry has been transforming itself with the addition of stronger foodservice offerings and retailers have invested in preparation and display equipment, people, and remodeled stores. This year's CSNews Technology Study indicates the realization that systems can further help grow this aspect of the business, noted Collupy.
Foodservice ordering kiosks, both inside and outdoors, show continuing and significant growth over the past few years: more than 30 percent of survey responders indicate they've implemented or plan to implement these customer-experience technologies both in-store (12 percent in 2016 and 9 percent in 2015) and outside (6 percent in 2016 and 4 percent in 2017).
The plumbing (i.e., network and security hosting, monitoring and management) of information technology (IT) in companies continues to shift to the "cloud." Collupy explained that these technologies are often viewed as a necessity but not the best use of the time and resources of in-house IT professionals. Thus, management continues to outsource many of these activities. Sixty-four percent of respondents reported that their companies outsource IT functions. The study shows steady growth in this area, up from 50 percent in 2015 and 59 percent in 2016.
"In my discussions with retailers, I hear that the main driver for this is to 'let my people focus on systems that will bring added value to the business,'" said Collupy.
One of those added-value areas appears to be social media. Social media has become an important marketing element for c-store operators, as almost three-quarters of respondents said they are using social media applications.
For more findings, look in the October issue of Convenience Store News.


text:
'Nothing Must Happen To Nnamdi Kanu', World Igbo Congress Tells FG
The World Igbo Congress (WIC) has asked the Nigerian Government to ensure that the leader of the proscribed Indigenous Peoples of Biafra (IPOB) Nnamdi Kanu, is safe.
Efizzy reports that Kanu who jumped bail in 2017 was rearrested and brought back to Nigeria on Tuesday.
He is facing treason charges and was remanded in the custody of the Department of State Services (DSS) following an order by a Federal High Court in Abuja.
But in a statement signed and issued by its chairman, Anthony Ejiofor and spokesperson, Basil Onwukwe, the WIC urged the government to protect the rights of the secessionist leader.
The group also urged the federal government to treat the IPOB leader according to universally accepted laws.
It also called on the government to ensure that Nnamdi Kanu is treated in accordance with the rules and international conventions that guarantee that his rights and personal safety are protected.
The WIC, however, claims that the Biafran agitator was not captured in the UK and called on the British Government and the international community to investigate the circumstances surrounding Kanu's rearrest.
The group also called for the protection of the rights of Ndigbo and asked the government to ensure fairness and justice at all time.
The statement read in parts: "The UK government needs to investigate the collusion of any other country or party in this saga," the Igbo group further stated.
"The government of the US, the UN and the international community to take note of the ongoing abuse of human rights in Nigeria including this matter of abduction of Nnamdi Kanu."
"The world has taken notice of the disparity in the treatment of Boko Haram insurgents and Fulani herdsmen who have been designated as terrorists by known international agencies.
"These killers are routinely arrested, compensated, released, and oftentimes absorbed into the Nigerian security services. On the other hand, the Igbo has been subjected to massive shoot-at-sight order, arrests and abduction. We would like to emphasize that no government in the world or any international security agency has designated IPOB as a terrorist organization and Nnamdi Kanu is not on any terrorist watchlist anywhere in the world."
Published July 1, 2021 By admin@jobzify.com.ng
Nigerians should buy petrol at N256 per liter – NNPC boss, Kyari
Intersociety Reveals People Responsible For Nnamdi Kanu's Arrest

redpajama_set_name: RedPajamaCommonCrawl


The perm market continues to power on. We got 3 more new roles in last week including a Specialist Governance Administrator role for a NFP.
Fill in your details below to sign up to Rusher Roger's weekly wrap-up report on what's happening in the world of Rusher Rogers, our clients, our candidates and the recruitment market.


text:
VACCINES-for HEALTHY Dogs Only
Thread: VACCINES-for HEALTHY Dogs Only
Kris L. Christine
THE RABIES CHALLENGE FUND
PERMISSION GRANTED TO CROSS-POST THIS MESSAGE.
Vaccines are intended to be administered to HEALTHY dogs -- it is an advisory issued on vaccine labels, in veterinary literature and guidelines, as a dog's health status can have an impact on a vaccine's effectiveness and fail to elicit an immune response. Startlingly, the AAHA task force indicates (see quote below) that vaccination in a "severely immunosuppressed" dog can result in the dog acquiring the disease it is being vaccinated to prevent!
On Page 24 of the 2006 American Animal Hospital Association's Canine Vaccine Guidelines, its states under the "Sick Dogs" heading that: "As with pregnant dogs, veterinary medicine has advised against vaccination during illness, due to concerns about suboptimal seroconversion, or worse, conversion of vaccine to disease." In other words, if you vaccinate a pregnant or sick dog, not only do you run the risk of a less-than-desirable immunological response, but you run the risk of your dog contracting the disease it is being vaccinated against!
Under the "Immunological Factors Determining Vaccine Safety" column on page 16 of the American Animal Hospital Association's 2003 Canine Vaccine Guidelines it is written that: " ...an attenuated pathogen in a host which is severely immunosuppressed, or genetically more susceptible, may result in the vaccine causing the disease for which it was designed to prevent." Further on that page, under "The Immune System and Frequency of Revaccination," it cautions that: "When vaccinating an animal, the age of the animal, the animal's immune status, and interference by maternal antibodies in the development of immunity must be considered. Research has demonstrated that the presence of passively acquired maternal antibodies significantly interferes with the immune response to many canine vaccines, including CPV [parvo], CDV [distemper], CAV-2 [hepatitis] and rabies vaccines."
The Merial Imrab 3 vaccine label for dogs and cats, click here: http://merialusa.naccvp.com/prodlist...country&p=msds and then click on IMRAB 3, then on "label": "INDICATIONS: IMRAB® 3 is recommended for the vaccination of healthy cats, dogs,sheep, cattle, horses, and ferrets 12 weeks of age and older for prevention of disease due to rabies virus."
For Rabvac 3 http://wyethc.naccvp.com/view.php?pr...country&p=msds click on "label" "Rabvac 3 is a killed virus vaccine for the vaccination of healthy dogs, cats and horses against rabies."
This link http://www.knowbetterdogfood.com/dog...ccinations.php will take you to an article by Dr. Moira Drosdovech entitled Pet Vaccinations - A Time for Change, in which she states what all the veterinary medical schools and vaccine companies declare, that: "Although I have stated this in previous articles, I cannot emphasize enough that any pet receiving a vaccination should be 100% healthy. This excludes vaccinating pets with any health problems whatsoever, including diseases in "remission" such as skin diseases, cancer, thyroid problems, to name a few, and especially not those with a history of autoimmune disease. Please understand that you are not benefiting your pet at all by vaccinating while unhealthy."
Also, be aware that giving combo vaccines (multi-valent) and/or several shots at once increases the risk of adverse reactions as well as the risk that the vaccines will interfere with each other, resulting in neutralization or negation.
Combination Vaccines, Multiple Shots--on Page 16 of the 2003 AAHA Guidelines under Immunological Factors Determining Vaccine Safety, it states that: "Although increasing the number of components in a vaccine may be more convenient for the practitioner or owner, the likelihood for adverse effects may increase. Also, interference can occur among the components. Care must be taken not to administer a product containing too many vaccines simultaneously if adverse events are to be avoided and optimal immune responses are sought. "
If anyone would like copies of the American Animal Hospital Association's Canine Vaccine Guidelines, the 1992 French challenge study demonstrating that dogs were immune to a rabies challenge 5 years after vaccination, the 2003 Italian study documenting fibrosarcomas at the presumed injection sites of rabies vaccines in dogs, as well as Dr. W. Jean Dodds' papers on vaccinal adverse reactions, please e-mail me at [email protected].
Duration of Immunity to Canine Vaccines: What We Know and Don't Know, Dr. Ronald Schultz http://www.cedarbayvet.com/duration_of_immunity.htm
World Small Animal Veterinary Association 2007 Vaccine Guidelines http://www.wsava.org/SAC.htm Scroll down to Vaccine Guidelines 2007 (PDF)
The 2003 American Animal Hospital Association's Canine Vaccine Guidelines are accessible online at http://www.leerburg.com/special_report.htm .
The 2006 American Animal Hospital Association's Canine Vaccine Guidelines are downloadable in PDF format at http://www.aahanet.org/PublicDocumen...s06Revised.pdf .
Veterinarian, Dr. Robert Rogers,has an excellent presentation on veterinary vaccines at http://www.newvaccinationprotocols.com/
Founder, Co-Trustee
www.RabiesChallengeFund.org
« Oh Rhysie, if its not one thing... | VACCINATING PUPPIES--16 Weeks+ Info »


text:
module Dune::Api
  module V1
    class ProjectsController < Dune::Api::BaseController
      include PaginatedController

      has_scope :order_by
      has_scope :pg_search, as: :query
      has_scope :between_created_at,
        :between_expires_at,
        :between_online_date,
        using: %i(starts_at ends_at),
        type:  :hash

      def index
        respond_with_pagination collection
      end

      def show
        project = ::Project.find(params[:id])
        authorize project
        respond_with project, serializer: Dune::Api::ProjectSerializer
      end

      def update
        @project = Project.find(params[:id])
        authorize @project
        respond_with Project.update(params[:id], permitted_params)
      end

      def destroy
        project = Project.find(params[:id])
        authorize project

        project.push_to_trash!
        head :no_content
      end

      [:approve, :launch, :reject, :push_to_draft].each do |name|
        define_method name do
          project = Project.find(params[:id])
          authorize project

          project.send("#{name.to_s}!")
          head :no_content
        end
      end

      private

      def permitted_params
        params.permit(policy(@project || Project).permitted_attributes)[:project]
      end

      def collection
        @collection ||= begin
          if ActiveRecord::ConnectionAdapters::Column.
              value_to_boolean(params[:manageable])
            authorized_scope = policy_scope(Dune::Api::Project)
          else
            authorized_scope = Dune::Api::Project.visible
          end
          apply_scopes(scoped_by_state(authorized_scope)).without_state('deleted')
        end
      end

      def scoped_by_state(scope)
        state_scopes = params.slice(*Project.state_names).keys
        if state_scopes.any?
          scope.with_state(state_scopes)
        else
          scope
        end
      end
    end
  end
end

redpajama_set_name: RedPajamaGithub


The ScholarNet suite of products plays a very important role in private loan processes used by our school and lender customers—and we strongly believe in providing the same high level of service to all parties. Great Lakes is committed to providing lender-neutral solutions to help you expedite private loan processing.
We're contractually obligated to maintain neutrality in our presentation of lenders and loan programs and our protection of data. We do this in several ways.
Lenders make schools eligible for loan programs and write the content they display in FastChoice, setting the impression borrowers have of their organization and programs.
Schools choose to scope any of the eligible lenders and loan programs they want made available to their students in FastChoice.
All of the data used in our FastChoice and ScholarNet solutions is segregated to protect each school and lender user. Users are unable to view private and confidential business or personal data unless they are a school scoping a loan program or a lender creating a loan program for a particular school or set of schools.
Access for Great Lakes employees to help you in configuring our FastChoice and ScholarNet solutions is tightly controlled to protect confidential data for our schools, lenders, and the borrowers they serve. The Great Lakes support team only has access to the information they need in order to provide customer support that is appropriate to their role within Great Lakes.
You are responsible for updating access roles for individuals at your institution on an annual basis to protect confidential data and your use of our FastChoice and ScholarNet solutions.
What do our ScholarNet suite solutions do?
We identified a need for schools and lenders to work closely to help borrowers access private loan funds through a secure connection. This offered an opportunity for us to create tools that connect these school and lender systems.
Powered by Great Lakes—a servicer and guarantor of federal loans—our ScholarNet suite of solutions benefits from our team's experience working in a heavily regulated industry with highly sensitive information. We take a great deal of pride in remaining compliant and protecting the interests of all of our various customers, with all of our products, solutions, and services.


Israel says it arrested eight Palestinians during a raid in the town of Jenin in the north of the West Bank.
Israeli security sources said the men were militants involved in making rockets for attacks against Israel.
The raid was a rare invasion into a West Bank city by the military, which said it would carry out only essential operations after a truce in February.
It came as Israeli MPs were due to vote on holding a referendum on a pull-out from Gaza and parts of the West Bank.
Israeli Prime Minister Ariel Sharon, who already has parliament's backing for the July pullout, has said the move is a delaying tactic.
Opinion polls consistently show a majority of Israelis favour leaving Gaza, which Israel has occupied since 1967.
Unnamed Israeli military officials said the arrested men belonged to the militant group, Islamic Jihad.
However, residents told the Associated Press news agency that three of the men, arrested in a nearby village, were members of Palestinian security forces, and that the other five were not known militants.
An Israeli soldier was wounded by an explosive device during the operation.
Jenin is known as a stronghold of Palestinian militias which have sent suicide bombers to Israel.


text:
Daddygreen Jeans salesman Fentanyl Seller has been sentenced to more than seven years in prison
admin November 25, 2020, 5:14 pm 2.8k Views
Joseph Resort Gruzzo, who lives in Osaka County, was sentenced to 87 months in prison.
Three years in federal prison and probation
Due to the death of the substance controller.
In a statement released Thursday, Grasso said the Oregon Eurasian Prosecutor's Office is a "great gift."
Fentanyl Chinese drug dealers. Cartoon and Fentanyl have been linked to a deadly amount at the Wisconsin company
Ousef Richard Grasso
Richard Caruso This phrase is an important step in ending Death
Our drug shortage, Bench Brothers executive chairman
In a press release, the Internal Security Review in Seattle.
The most lethal substance is fentanyl. Close your eyes unnecessarily
The safety of the people who have to deal with it
Continuing this case is a testament to the hard work of HSI and our law
Motivating partners, daily drug production
On our roads
Crosso's report
Arrested by Payee November 2017.
He was permanently appointed by the U.S. Department of State, which sent him to Caruso.
Portland Air Transport Center. Medicine in Portland
However, the content of the second part is viewed in the clickbait found by Ziplock.
Leather Bag 1 Tests in Oregon
The laboratory confirmed that the chemical cyclopropylene, I.
Described as an add-on and list I manage
Scientists lay objects with dismay
The powder size is uniform, which improves
Caruso Lake Oswego returned home on November 21, 2017. Caruso retired
The mailboxes of his home and land seized him
Businesses work safely with other groups of employers.
The worst actions of four law firms resulted
Carvos is closed two days after the last registration
He had a piece of fentanyl. Such a style and excellent solution
Organizational reforms are needed to prevent opioid overdose
C. Our approach to preventing further conflict, says Williams, is,
The district attorney said in a statement
Good luck to all of you.
In April, Crosso pleaded guilty to more than 76,764,000 insurance claims in 2013.
Audi A4 sedan.
To America
Georgette Richard Caruso from Lake Oswego, Oregon, was sentenced to 87 months in prison.
201 inmates in federal prisons for three years on probation for the distribution of unlawful fentanyl with lethal doses.
Wisconsin.
Croso was arrested just two days after a joint effort by four law enforcement agencies
I found an internal package for Fontanel. This kind of work is necessary and requires strict law enforcement
Synthetic opioid monitoring has been abandoned and deactivated
Oregon. Thank you for the extraordinary work of everyone involved in this event.
"This trial is an important step in eliminating lethal drugs from our community," said special researcher Brad Bench.
Billing from Seattle to Seattle. Phenyl is a deadly substance. Anyone you meet will ignore security
He did not contact them. This case is evidence of the efforts of HSI and our law enforcement partners.
The daily struggle with this drug is not our direction.
According to court documents, he was a U.S. citizen on November 19, 2017. The court's decision was made. Analysts at email
verification services have found a suspicious package
Talk to cars in the United States. of. Portland Aviation Center. The package was taken to the Portland Police Department.
Medicine and grade for additional testing in a safe environment. Wear waterproof caps, a
The Internal Security Agency (HSI) was assigned to a group of high-intensity drug addicts.
We found a brilliant patent story with a heat-soaked package and comfortable powder material. The next day was a test
Oregon State Police Laboratory recognizes cyclopropyl fentanyl, opioid control program, and self-monitoring
The researchers removed the cyclopropylphenananil from the bag and replaced it with a similar inert powder. anyway,
On November 21, 2017, they took control of the delivery of inactive powder containers to Carvus House on Lake Osogo.
After a while, Caruso managed to retrieve the package from his department box. HSI agents and other working groups.
The authorities opposed the protest with all available police forces, special services and the army. "
On April 3, 2019, Crosso pleaded guilty to distributing controlled substances after his death. That's with conviction
64,764,000 were ordered and sedans of the Audi A4 were ordered.
The case is being investigated by a high-intensity working group on drug trafficking, internal security investigations,
U.S. Portland Inspection and Drug and Police. These are Scott M. Karen and
From Yulia Retaret, American Assistant to the United States. Oregon Attorney General.
The Oregon HIDTA program was established in June 1999 by the Department of National Statistics (ONDCP).
In 2015, the project expanded to Idaho and was named Oregon-Idaho.
Reservations for the Indian HIDTA area include: Oregon Kleckams, Deshutters, Douglas, Jackson, Lane, Lynn,
The statistics show the geographical location of Maleur, Marion, Mulnomah, Umatilla, and Washington, Ada, Bannok, and Canyon,
Idaho.
Drug use affects communities across the country, and drug use is a major issue in the CDC report.
Between 1999 and 2016, more than 630,000 deaths were reported, and in 2016, drug overdoses accounted for 66% of deaths.
Overdose is currently the leading cause of injury or death in the United States. Death in Oregon
Drug use increased 11% from 2013 to 2017, and in 2017, 546 people died
More From: Crimes
Babysitter jailed for 30 years for creating child pornography in custody
Dealer Fensanil Fentanyl sentenced to 13 years in prison
Two men arrested in New Jersey ordering high-end smartphones with IDs stolen from the dark internet
TrickBot has added new contact information to collect 250 million email addresses
ECh0raix for the latest ransomware is attacking QNAP NAS secretly


text:
\section{Introduction}
While mathematics and juggling have existed independently for thousands of years, it has only been in the last thirty years that the mathematics of juggling has become a subject in its own right (for a general introduction see Polster \cite{P}).  Several different approaches for describing juggling patterns have been used.  The best-known method is siteswap which gives information what to do with the ball that is in your hand at the given moment, in particular how ``high'' you should throw the ball (see \cite{BEGW}).  For theoretical purposes a more useful method is to describe patterns by the use of cards.  This was first introduced in the work of Ehrenborg and Readdy \cite{ER}, and modified by Butler, Chung, Cummings and Graham \cite {BCCG}.

These cards work by focusing on looking at the \emph{relative order} of when the balls will land should we stop juggling at a given moment. Every throw then has the effect of changing the relative ordering of the balls. But we can only effect the order of a ball that is thrown; the remaining balls will still have the remaining relative order to each other. As a consequence if there are $b$ balls there are $b+1$ different things which can happen.  Namely, we don't throw a ball (the ``$+1$'') or we throw a ball so that it will land in some order relative to the other $b-1$ balls (which can be done in $b$ ways).  The four different cards for the case $b=3$ are shown in Figure~\ref{fig:BasicCards} (in all drawings of cards the circle at the bottom indicates the hand which either does not catch the ball at that ``beat'' or catches and throws effecting the relative ordering of the ball(s); we will always think of time moving from left to right).

\begin{figure}[htb!]
\centering
\picCa \hfil \picCb \hfil \picCc \hfil \picCd
\caption{Possible juggling cards for three balls.}
\label{fig:BasicCards}
\end{figure}

The advantage of working with cards is that the cards can work independently of each other, that is the choice of card to use at a given time is not dependent on the preceding choice of cards.  In siteswap the opposite is true in that you must know all preceding throws to determine which throws are possible.

Given a set of these $n$ cards for a given $b$ we can repeat these periodically to form a pattern.  Moreover, every possible siteswap with period $n$ and at most $b+1$ balls will occur as a unique combination of these cards (see \cite{BCCG,P}).  Therefore the number of different siteswap sequences of period $n$ for exactly $b$ balls is given by
\[
(b+1)^n-b^n.
\]
If we want to find all of the juggling patterns of \emph{minimal} period $n$ and using exactly $b$ balls we can then use M\"obius inversion and divide out by the period to get
\[
\frac1n\sum_{d\mid n}\mu({\textstyle\frac{n}{d}})\big((b+1)^d-b^d\big),
\]
where $\mu$ is the M\"obius function (see \cite{BEGW}).  (We will revisit this with more detail in Section~\ref{sec:counts}.)

For as long as there has been interest in the mathematics of juggling there has been interest in extending results to multiplex juggling (where more than one ball is allowed to be caught at a time).  In Ehrenborg and Readdy they produced possible cards for multiplex juggling which were a natural generalization.  Namely multiple balls could come down at a given time and would then be redistributed appropriately.  While these cards can describe every juggling pattern there is the problem that uniqueness is lost (see Figure~\ref{fig:nonunique} for an example of two consecutive cards describing the same pattern but using different cards).  So using these cards to count multiplex juggling patterns is not straight-forward.

\begin{figure}[htb!]
\centering
\picERc \picERa \hfil \picERd \picERa
\caption{Ambiguity arising from using cards of Ehrenborg and Readdy \cite{ER}.}
\label{fig:nonunique}
\end{figure}

One approach is to distinguish the balls which come down.  This is what was done in Butler, Chung, Cummings and Graham \cite{BCCG}, an example of such a card is shown in Figure~\ref{fig:separate}.  This avoids ambiguity that might arise but does not accurately reflect multiplex juggling in practice, but rather reflects passing patterns with multiple jugglers involved each juggler catching one ball (in particular the different points that come down correspond to the different jugglers).

\begin{figure}[htb!]
\centering
\picBCCGa
\caption{An example of a card used in Butler, Chung, Cummings and Graham \cite{BCCG}.}
\label{fig:separate}
\end{figure}

In this paper we will propose a new type of card which can be used for multiplex juggling.  It solves the ambiguity problem of Ehrenbrog and Readdy and also solves the modeling problem of Butler, Chung, Cummings and Graham.  However it does come at the mild cost of having a card being dependent on the \emph{previous} card which came before.  In Section~\ref{sec:cards} we will introduce these cards, and in Section~\ref{sec:graph} we show how to use matrices associated with a weighted graph to count the number of periodic patterns of length $n$.  We then count the number of siteswap sequences and the number of juggling patterns in Section~\ref{sec:properties} and Section~\ref{sec:counts}.  In Section~\ref{sec:hand} we will consider what happens when we limit the number of balls which can be thrown. We will give some concluding remarks in Section~\ref{sec:conclusion}, including a discussion of counting crossing numbers.

Most of the enumeration techniques here are fairly standard, it is their application to counting juggling patterns that is new.  We will also see that the objects generated in the process of deriving our count seem to have independent combinatorial interest.  Moreover, while our main goal has been to enumerate juggling patterns, the cards themselves might be useful for the exploration of other combinatorial aspects of juggling.

Finally, we note that while there has some been interest in counting multiplex juggling patterns, prior to this paper there has been little success.  Butler and Graham \cite{BG} made the most progress but their focus was on counting closed walks in a state graph and were not able to efficiently enumerate all juggling patterns.

\section{Cards for multiplex juggling}\label{sec:cards}
The way that cards describe juggling patterns is through understanding the relative order of their landing times.  The ambiguity that appeared in Figure~\ref{fig:nonunique} comes from the fact that two balls are landing \emph{together} but still being kept \emph{separate} in the ordering.  Since they are separate we could order them in two ways but that does not effect the pattern.  This suggests the following simple fix: tracks no longer represent individual balls, but rather groups of balls which will land together.   So now either the ``lowest'' group doesn't land, or the lowest group lands and the balls get thrown so that they are placed in new track(s) or added to the existing tracks.

Before each throw we will have an ordered partition of the number of balls $b$ on the left, i.e., $(q_1,q_2,\ldots,q_k)$ which corresponds to the statement that were we to stop juggling we would first have $q_1$ balls land at some point; then $q_2$ balls land some time later; and so on until finally $q_k$ balls land at the end.  (Note that we do not claim that they will land one right after the other; cards are keeping track of relative ordering of when things land and not the absolute times that they will land.)  Similarly after each throw we will have another ordered partition of $b$ on the right, i.e., $(r_1,r_2,\ldots,r_\ell)$.  (The number of our parts in our two partitions need not be the same but we must have $\ell\ge k-1$.)  If anything lands then the card in the middle indicates how the $q_1$ balls get redistributed.  Examples of these cards are shown in Figure~\ref{fig:multicards} where the first card corresponds to going from $(2,1,1)$ back to $(2,1,1)$ and the second corresponds to going from $(2,2,1)$ to $(1,2,2)$.

\begin{figure}[htb!]
\centering
\begin{tabular}{c@{\qquad}c}
\picJUGA & \picJUGB\\
\begin{tikzpicture}[scale=0.4]
\fill[color=black!50!white,rounded corners] (0,0) rectangle (1,1) (0,1) rectangle (1,2);
\draw[thick,rounded corners] (0,2) rectangle (1,3) (0,1) rectangle (1,2) (0,0) rectangle (1,1) (1,0) rectangle (2,1);
\end{tikzpicture}&
\begin{tikzpicture}[scale=0.4]
\fill[color=black!50!white,rounded corners] (0,1) rectangle (1,2) (1,1) rectangle (2,2) (0,2) rectangle (1,3);
\draw[thick, rounded corners] (0,0) rectangle (1,1) (0,1) rectangle (1,2) (1,1) rectangle (2,2) (0,2) rectangle (1,3) (1,2) rectangle (2,3);
\end{tikzpicture}
\end{tabular}
\caption{Cards which can be used to model multiplex juggling.}
\label{fig:multicards}
\end{figure}

\begin{definition}
An ordered partition $(q_1,q_2,\ldots,q_k)$ can be \emph{nontrivially embedded} into an ordered partition $(r_1,r_2,\ldots,r_\ell)$ if there exists indices $1\le i_2< i_3<\cdots< i_k\le\ell$ so that $q_j\le r_{i_j}$ for $2\le j\le k$.  Note that given two ordered partitions several nontrivial embeddings are possible.  An ordered partition $(q_1,q_2,\ldots,q_k)$ can be \emph{trivially embedded} into an ordered partition $(r_1,r_2,\ldots,r_\ell)$ if and only if  $(q_1,q_2,\ldots,q_k) = (r_1,r_2,\ldots,r_\ell)$.
\end{definition}

For every nontrivial embedding of $(q_1,q_2,\ldots,q_k)$, a partition of $b$, into $(r_1,r_2,\ldots,r_\ell)$, another partition of $b$, we have a card for multiplex juggling where a throw occurred.  As an example in Figure~\ref{fig:multicards} we have also marked underneath how $(q_1,q_2,\ldots,q_k)$ embeds into $(r_1,r_2,\ldots,r_\ell)$ by drawing the partition of $(r_1,r_2,\ldots,r_\ell)$ arranged from $r_1$ on the bottom to $r_\ell$ on the top and shading where $q_2,\ldots,q_k$ sits inside the partition.  Trivial embeddings, i.e.,  $(q_1,q_2,\ldots,q_k)=(r_1,r_2,\ldots,r_\ell)$,  correspond to no throws.  All possible cards (and corresponding embeddings of partition) for multiplex juggling when $b=3$ are shown in Figure~\ref{fig:3ballcards}.

\begin{figure}[h!]
\centering
\begin{tabular}{|c|c|c|c|c|c|}
\hline \vphantom{-}&&&&&\\
\picJPaa&\picJPab&\picJPac&\picJPad&\picJPae&\picJPaf\\
\partpicA&\partpicB&\partpicC&\partpicD&\partpicE&\partpicF\\ \hline \vphantom{-}&&&&&\\
\picEXg&\picEXf&\picEXe&\picJPbd&\picEXa&\picEXb\\
\partpicG&\partpicH&\partpicI&\partpicJ&\partpicK&\partpicL\\ \hline \vphantom{-}&&&&&\\
\picEXc&\picEXd&\picEXh&\picJPcd&\picJPce&\picJPcf\\
\partpicM&\partpicN&\partpicO&\partpicP&\partpicQ&\partpicR\\ \hline \vphantom{-}&&&&&\\
\picJPda&\picJPdb&\picJPdc&\picJPdd&\picJPde&\picJPdf\\
\partpicS&\partpicT&\partpicU&\partpicV&\partpicW&\partpicX\\ \hline
\end{tabular}
\caption{All cards and corresponding embeddings of ordered partitions for $b=3$.}
\label{fig:3ballcards}
\end{figure}

We can now determine the number of cards involved by examining their interpretation using embeddings of partitions.

\begin{lemma}\label{lem:right}
The total number of ways that an ordered partition $(r_1,r_2,\ldots,r_\ell)$ can have an ordered partition embedded inside is
\[
\prod_i(r_i+1).
\]
Equivalently this is the number of cards where $(r_1,r_2,\ldots,r_\ell)$ is the ordered partition on the right.
\end{lemma}
\begin{proof}
We can think of shading the whole partition then removing parts of it as desired.  In particular for every part $r_i$ the embedding can use any of $0,1,\ldots,r_i$ in that slot.  In particular for the $i$th part there are $r_i+1$ choices and these can be made independently of each other.  Therefore there are $(r_1+1)(r_2+1)\cdots(r_\ell+1)$ possible embeddings.
\end{proof}

\begin{lemma}\label{lem:left}
The total number of ways that an ordered partition $(q_1,q_2,\ldots,q_k)$ can be embedded into an ordered partition is
\[
1+\frac{q_1+2k-2}{q_1+k-1}{q_1+k-1 \choose k-1}2^{q_1-1}.
\]
Equivalently this is the number of cards where $(q_1,q_2,\ldots,q_k)$ is the ordered partition on the left.
\end{lemma}
\begin{proof}
Suppose that a throw happens, so $q_1$ balls come down and get redistributed, possibly adding balls to existing groups to land (i.e., adding to one of the $q_i$) or creating new groups to land.  Suppose that the ordered partition we embed into has $(k-1)+\ell$ different parts; where $0\le \ell\le q_1$.  This can happen in
\[
{k-1+\ell\choose \ell}{k+q_1-2\choose q_1-\ell}=\frac{\ell+k-1}{q_1+k-1}{q_1\choose \ell}{q_1+k-1\choose k-1}
\]
different ways.  The ${k-1+\ell\choose \ell}$ divides the parts as coming from an existing part (e.g., $q_i$) or being newly created.  For the $\ell$ new parts we first need to get a contribution of $1$ coming from $q_1$ leaving $q_1-\ell$ available to distribute among the $k-1+\ell$ different parts arbitrarily which can be done in  ${k+q_1-2\choose q_1-\ell}$ ways.  Finally, we can perform some simple manipulation of binomial coefficients.

Putting this together we have that the total number of ways that $(q_1,q_2,\ldots,q_k)$ can be embedded into another ordered partition by a throw is
\begin{align*}
\sum_{\ell=0}^{q_1}\frac{\ell+k-1}{q_1+k-1}{q_1\choose \ell}{q_1+k-1\choose k-1}
&=\frac{1}{q_1+k-1}{q_1+k-1\choose k-1}\sum_{\ell=0}^{q_1}(\ell+k-1){q_1\choose \ell}\\
&=\frac{1}{q_1+k-1}{q_1+k-1\choose k-1}\bigg(\sum_{\ell=0}^{q_1}\ell{q_1\choose \ell}
+\sum_{\ell=0}^{q_1}(k-1){q_1\choose \ell}\bigg)\\
&=\frac{1}{q_1+k-1}{q_1+k-1\choose k-1}\big(q_12^{q_1-1}+(k-1)2^{q_1}\big)\\
\\
&=\frac{q_1+2k-2}{q_1+k-1}{q_1+k-1 \choose k-1}2^{q_1-1}.
\end{align*}
Combining this with the ``$+1$'' from the trivial embedding and the result follows.
\end{proof}

We can now determine the total number of cards, or equivalently the total number of embeddings possible.  Starting with $b=0$ the numbers are
\[
1,
~2,
~7,
~24,
~82,
~280,
~956,
~3264,
~11144,
~38048,
~129904,
~443520,
~1514272,\ldots.
\]
This is sequence {\tt A003480} in the OEIS \cite{OEIS} which is initiated with $a_0=1$, $a_1=2$ and $a_2=7$ and for $b\ge 3$ we have $a_b=4a_{b-1}-2a_{b-2}$.  Verifying the first few cases is straightforward (the case for $a_3=24$ is shown in Figure~\ref{fig:3ballcards}).  It remains to verify the recurrence.

\begin{theorem}\label{thm:cardrecurse}
If $a_b$ is the number of possible cards used for describing multiplex juggling with $b$ balls, then for $b\ge 3$ we have $a_b=4a_{b-1}-2a_{b-2}$.
\end{theorem}
\begin{proof}
We can count the number of cards by breaking our count up according to the ordered partition that shows up on the right side using Lemma~\ref{lem:right}.  In particular we have
\begin{equation}\label{eq:ab}
a_b=\sum_{(r_1,r_2,\ldots)\in\mathcal{R}_b}\prod_{i\ge1}(r_i+1),
\end{equation}
where $\mathcal{R}_b$ are all of the ordered partitions of $b$.

We now further break up this count by combining the ordered partitions according to the size of the first part which can be anything from $1$ to $b$.  So we have
\begin{align*}
a_b&=\sum_{j=1}^b\bigg(\sum_{(j,r_2,\ldots)\in\mathcal{R}_b}(j+1)\prod_{i\ge 2}(r_i+1)\bigg)\\
&=\sum_{j=1}^b(j+1)\bigg(\sum_{(r_2,\ldots)\in\mathcal{R}_{b-j}}\prod_{i\ge 2}(r_i+1)\bigg)\\
&=\sum_{j=1}^b(j+1)a_{b-i},
\end{align*}
where in the last step we note that we have the form of \eqref{eq:ab}.

To finish we have
\begin{align*}
a_b-2a_{b-1}+a_{b-2}&=
\sum_{j=1}^b(j+1)a_{b-j}
-2\sum_{j=1}^{b-1}(j+1)a_{b-1-j}
+\sum_{j=1}^{b-2}(j+1)a_{b-2-j}\\
&=\sum_{k=1}^b(k+1)a_{b-k}
-2\sum_{k=2}^{b}ka_{b-k}
+\sum_{k=3}^{b}(k-1)a_{b-k}\\
&=2a_{b-1}-a_{b-2}.
\end{align*}
Here we reindex the three sums to be consistent and then note that all but the first two terms will drop out.  Rearranging we conclude $a_b=4a_{b-1}-2a_{b-2}$, as desired.
\end{proof}

\section{Combining cards by taking walks in a graph}\label{sec:graph}

The advantage to using cards in order to describe juggling patterns was the ability to have our current card be chosen independently of all other cards.  With these new cards that we will use for multiplex juggling we now have to be careful in that the choice of our current card is dependent on the previous card, namely the partition on the right of the previous card must match the partition on the left of the current card.  Moreover if we are forming a pattern with period $n$ we need to make sure that our last card will also be consistent with our first card (so that we can repeat).

To help achieve this we will construct a directed multi-graph $\mathcal{G}_b$ by letting the vertices of $\mathcal{G}_b$ be the ordered partitions of $b$; and for each card we add a directed edge from the ordered partition on the left of the card to the ordered partition on the right of the card.

\begin{observation}\label{obs:bijection}
There is a one-to-one correspondence between periodic sequences using $n$ cards and closed walks of length $n$ in the graph.  In particular, to count the number of periodic sequences using $n$ cards it suffices to count the number of closed walks of length $n$.
\end{observation}

This follows by noting that each card is an edge and if two cards are used sequentially then the edges also occur sequentially in the graph, giving the correspondence between sequences of consecutive cards and walks in the graph.  Moreover the fact that we can repeat the pattern periodically indicates that we must return to the same ordered partition that we started with, so the walk is closed.

We now can use the transfer matrix method (see Stanley \cite[Ch.\ 4.7]{S}).

\begin{theorem}\label{thm:walks}
Given a directed multi-graph $\mathcal{G}$ let $A$ be the matrix with rows and columns indexed by the vertices with $A_{uv}$ equal to the number of directed arcs from $u\to v$.  Then $\trace(A^n)$ equals the number of closed walks of length $n$ in the graph.
\end{theorem}

Let $A_b$ be the matrix associated with the graph $\mathcal{G}_b$.  We have $A_0=(1)$, $A_1=(2)$,
\[
A_2=\begin{array}{r}(2)\\(1,1)\end{array}\left(\begin{array}{cc}
2 & 1 \\
1 & 3
\end{array}\right),\qquad
A_3=
\begin{array}{r}(3)\\(2,1)\\(1,2)\\(1,1,1)\end{array}\left(\begin{array}{cccc}
2 & 1 & 1 & 1 \\
1 & 3 & 2 & 3 \\
1 & 1 & 2 & 0 \\
0 & 1 & 1 & 4
\end{array}\right),
\]
and
\[
A_4=
\begin{array}{r}(4)\\(3,1)\\(1, 3)\\(2, 1, 1)\\(2, 2)\\(1, 2, 1)\\(1, 1, 2)\\(1, 1, 1, 1)\end{array}\left(\begin{array}{cccccccc}
2 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 3 & 2 & 3 & 2 & 3 & 3 & 4 \\
1 & 1 & 2 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & 4 & 1 & 3 & 3 & 6 \\
1 & 1 & 1 & 1 & 3 & 1 & 1 & 0 \\
0 & 1 & 0 & 2 & 1 & 2 & 0 & 0 \\
0 & 0 & 1 & 0 & 1 & 1 & 3 & 0 \\
0 & 0 & 0 & 1 & 0 & 1 & 1 & 5

\end{array}\right),
\]
where on the left we have marked the ordered partitions that correspond to the vertex.  For reference we also give the graphs $\mathcal{G}_0$, $\mathcal{G}_1$, $\mathcal{G}_2$, $\mathcal{G}_3$ in Figure~\ref{fig:graphs}.

\begin{figure}[htb!]
\centering
\begin{tabular}{c@{\qquad\qquad}c@{\qquad\qquad}c@{\qquad\qquad}c}
\begin{tikzpicture}
\node[draw, rectangle, rounded corners, inner sep=2pt] (a) at (0,0) {$\emptyset$};
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners] (a) -- (0.75,0)--(0.75,0.75) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $1$} --(0,0.75)--(a);
\end{tikzpicture}

&

\begin{tikzpicture}
\node[draw, rectangle, rounded corners, inner sep=2pt] (a) at (0,0) {$(1)$};
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners] (a) -- (0.75,0)--(0.75,0.75) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $2$} --(0,0.75)--(a);
\end{tikzpicture}

&

\begin{tikzpicture}
\node[draw, rectangle, rounded corners, inner sep=2pt] (a) at (0,0) {$(1,1)$};
\node[draw, rectangle, rounded corners, inner sep=2pt] (b) at (0,2) {$(2)$};
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners] (a) -- (0.75,0)--(0.75,-0.75) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $3$} --(0,-0.75)--(a);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners] (b) -- (0.75,2)--(0.75,2.75) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $2$} --(0,2.75)--(b);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners] (b) to (0.25,1) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $1$} to (a);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners] (a) to (-0.25,1) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $1$} to (b);
\end{tikzpicture}
&
\begin{tikzpicture}
\node[draw, rectangle, rounded corners, inner sep=2pt] (a) at (0,0) {$(3)$};
\node[draw, rectangle, rounded corners, inner sep=2pt] (b) at (3,0) {$(1,2)$};
\node[draw, rectangle, rounded corners, inner sep=2pt] (c) at (3,3) {$(1,1,1)$};
\node[draw, rectangle, rounded corners, inner sep=2pt] (d) at (0,3) {$(2,1)$};
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners] (a) -- (-0.75,0)--(-0.75,-0.75) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $2$} --(0,-0.75)--(a);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners] (d) -- (-0.75,3)--(-0.75,3.75) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $3$} --(0,3.75)--(d);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners] (b) -- (3.75,0)--(3.75,-0.75) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $2$} --(3,-0.75)--(b);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners] (c) -- (4,3)--(4,3.75) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $4$} --(3,3.75)--(c);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners]
    (a) to (1.5,0.3) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $1$} to (b);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners]
    (b) to (1.5,-0.3) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $1$} to (a);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners]
    (a) to (-0.3,1.5) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $1$} to (d);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners]
    (d) to (0.3,1.5) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $1$} to (a);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners]
    (a) to (1.5,1.5) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $1$} to (c);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners]
    (c) to (3,1.5) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $1$} to (b);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners]
    (d) to (1.5,3.3) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $3$} to (c);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners]
    (c) to (1.5,2.7) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $1$} to (d);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners]
    (d) to (1.5,2.1) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $2$} to (b);
\draw[-{>[scale=1.25]},thick,shorten >=2pt,shorten <=2pt, rounded corners]
    (b) to (1.5,0.9) node[circle,draw,thick, fill=white, inner sep = 1pt] {\small $1$} to (d);
\end{tikzpicture}
\\
$\mathcal{G}_0$& $\mathcal{G}_1$& $\mathcal{G}_2$& $\mathcal{G}_3$
\end{tabular}
\caption{The graphs $\mathcal{G}_0$, $\mathcal{G}_1$, $\mathcal{G}_2$ and $\mathcal{G}_3$.}
\label{fig:graphs}
\end{figure}

We note that Lemma~\ref{lem:right} can be used to determine the \emph{column} sums of $A_b$, while Lemma~\ref{lem:left} can be used to determine the \emph{row} sums of $A_b$.  Using Theorem~\ref{thm:walks} we also get the sum of all entries of the matrix.

\section{Counting multiplex siteswaps}\label{sec:properties}

We can combine Observation~\ref{obs:bijection} together with Theorem~\ref{thm:walks} to find the number of periodic sequences of $n$ cards.  For each sequence of cards we will get a siteswap pattern (recall that siteswap patterns work by recording how many beats in the future the ball(s) land which can be done by following the path along the figure formed by the cards until it goes back down).  However for a given siteswap there can be \emph{multiple} ways in which it can be represented using the cards.

The problem lies in what happens with balls that are never used.  For instance in Figure~\ref{fig:samesiteswap} we see two distinct sets of three cards which correspond to the same siteswap, the difference between them being the tracks in the unused balls.  (In siteswap notation this will be 1[112][22].)

\begin{figure}[htb!]
\centering
\picWeda \picWedb \picWedc
\hfil
\picWedd \picWede \picWedf
\caption{Two distinct sets of three cards with the same siteswap (1[112][22]).}
\label{fig:samesiteswap}
\end{figure}

Note that for normal juggling the unused balls all have the same behavior, in our language they would correspond to lying in tracks with capacity $1$.  So to count the number of siteswaps we take the difference of the ways to place $n$ cards with $b$ tracks $(b+1)^n$ and the ways to place $n$ cards with $b-1$ tracks $b^n$ (which is in bijection with the number of ways to place $n$ cards with $b$ tracks and the top track is never used).  This gives $(b+1)^n-b^n$.  We will do something similar with our cards that we are using for multiplex juggling.

\begin{theorem}\label{thm:siteswapcount}
Let $ss_b(n)$ be the number of siteswap patterns using exactly $b$ balls and with period $n$.  Then
\[
ss_b(n) = \trace(A_b^n)-\sum_{i=0}^{b-1}\trace(A_i^n).
\]
\end{theorem}
\begin{proof}
We proceed by induction on $b$.  For the base case of $b=0$ there is one card (the empty card) and so for any length $n$ there is exactly one way to position these cards.  (In siteswap notation this is 00\ldots0.)  Since $A_0=(1)$ we have $\trace(A_0^n)=1$ establishing the base case.

Now assume that it works up through $b-1$ and consider the case for $b$.  First we observe
\[
\trace(A_{b}^n) = ss_{b}(n)+\sum_{i=0}^{b-1}2^{b-i-1}ss_i(n).
\]
This follows by noting that for every siteswap which uses exactly $i$ balls we can find a sequence of cards corresponding to the ordered partitions of $i$.  We can now add $b-i$ balls to the top of each card as long as we do it consistently across the different cards.  Moreover the number of different options we have to place the $b-i$ balls equals the number of ordered partitions of $b-i$ which is $2^{b-i-1}$.

Rearranging and using our induction hypothesis we have
\begin{align*}
ss_b(n)
&=\trace(A_{b}^n)-\sum_{i=0}^{b-1}2^{b-i-1}ss_i(n)\\
&=\trace(A_{b}^n)-\sum_{i=0}^{b-1}2^{b-i-1}\bigg(\trace(A_i^n)-\sum_{j=0}^{i-1}\trace(A_j^n)\bigg)\\
&=\trace(A_{b}^n)-\sum_{i=0}^{b-1}\trace(A_i^n)\bigg(2^{b-i-1}-\sum_{k=0}^{b-i-2}2^k\bigg)\\
&=\trace(A_{b}^n)-\sum_{i=0}^{b-1}\trace(A_i^n).
\end{align*}
Establishing the result.
\end{proof}

\section{Counting multiplex juggling patterns}\label{sec:counts}
To go from counting siteswap patterns of period $n$ to counting juggling patterns of minimal period $n$ we want to do two things.  First we want to remove any pattern that has shorter period (suppose $d$ divides $n$, then any periodic sequence of cards of length $d$ can be repeated $n/d$ times to make a periodic sequence of cards of length $n$).  Second we want to divide out by $n$ since in juggling patterns there is no set start point, i.e., 1[112][22], [112][22]1, and [22]1[112] all correspond to the same juggling pattern.

For the first issue we can use M\"obius inversion (see \cite{S}).  Namely we note that if $ms_b(n)$ is the number of siteswap patterns with $b$ balls and minimal period $n$, then
\[
ss_b(n) = \sum_{d \mid n}ms_b(d).
\]
So if we let $\mu(\frac{n}{d})$ be the M\"obius function it follows
\[
ms_b(n) = \sum_{d \mid n}\mu({\textstyle\frac{n}{d}})ss_b(d).
\]
With this in hand we can now divide out by the rotational symmetry of the starting point and determine the number of juggling patterns with $b$ balls and period $n$ which we denote $jp_b(n)$.  Combining the above with Theorem~\ref{thm:siteswapcount} we get the following.

\begin{theorem}\label{thm:jpcount}
The number of juggling patterns with $b$ balls and minimal period $n$ is
\begin{equation*}
jp_b(n)=\frac1n\sum_{d\mid n}\mu({\textstyle\frac{n}{d}})\bigg(\trace(A_b^d)-\sum_{i=0}^{b-1}\trace(A_i^d)\bigg).
\end{equation*}
\end{theorem}

In Table~\ref{tab:jpcount} we give the number of minimal period juggling patterns for $b=2,3,4,5$ and period at most $15$.

\begin{table}[htb!]
\centering
\begin{tabular}{|l||r|r|r|r|}\hline
& $b=2$ & $b=3$ & $b=4$ & $b=5$\\ \hline\hline
$n=1$& $2$& $3$& $5$& $7$\\ \hline
$n=2$& $4$& $12$& $32$& $77$\\ \hline
$n=3$& $13$& $63$& $261$& $964$\\ \hline
$n=4$& $37$& $310$& $2089$& $12086$\\ \hline
$n=5$& $118$& $1618$& $17449$& $156975$\\ \hline
$n=6$& $356$& $8434$& $147807$& $2077448$\\ \hline
$n=7$& $1142$& $45142$& $1276577$& $27976399$\\ \hline
$n=8$& $3620$& $243998$& $11169023$& $381752857$\\ \hline
$n=9$& $11744$& $1336644$& $98872035$& $5267354817$\\ \hline
$n=10$& $38275$& $7392117$& $883717142$& $73358245986$\\ \hline
$n=11$& $126234$& $41247234$& $7964898829$& $1029873201879$\\ \hline
$n=12$& $418735$& $231856131$& $72305691686$& $14559160765380$\\ \hline
$n=13$& $1399610$& $1311820110$& $660528998007$& $207076019661773$\\ \hline
$n=14$& $4702499$& $7464002451$& $6067348742573$& $2961063646029819$\\ \hline
$n=15$& $15883190$& $42679372930$& $56002661734041$& $42542385162393167$\\ \hline
\end{tabular}
\caption{The number of multiplex juggling patterns of minimal period $n$ using $b$ balls.}
\label{tab:jpcount}
\end{table}

As a special case of Theorem~\ref{thm:jpcount} we have
\begin{equation}\label{eq:jp1}
jp_b(1)=\trace(A_b)-\sum_{i=0}^{b-1}\trace(A_i).
\end{equation}
We can also compute the number of period one multiplex juggling patterns directly.
\begin{theorem}\label{thm:jp1}
We have $jp_b(1)=p(b)$, where $p(b)$ is the number of unordered partitions of $b$.
\end{theorem}
\begin{proof}
In order for a card to produce a valid juggling sequence of period one with $b$ balls we must have that the (ordered) partitions on the left and right side of the card be equal.  That is $(q_1,q_2,\ldots,q_k)=(r_1,r_2,\ldots,r_k)$.  Further we must have that when the $q_1$ balls get distributed some ball gets placed into the top group.  This second requirement will force $q_k$ to ``embed'' into $r_{k-1}$ (otherwise it would have to embed into $r_k$ but the placement of at least one more ball in the top group then results in $q_k\ne r_k$).  In particular this then forces, in turn, $q_i$ to embed in $r_{i-1}$ for $i=2,\ldots,k$.  This is only possible if $q_1\le q_2\le\cdots\le q_k$.  Therefore our ordered partition on the sides of the card have the unique ordering from largest to smallest element.

Conversely, if we start with an unordered partition we can create a card as above by placing the partition on the sides of the card from largest to smallest; then all but the bottom group will shift down by one while the bottom group will then drop down and redistribute to fill differences as needed.  An example of this is shown in Figure~\ref{fig:partition} for the partition $(3,3,2,2).$
\end{proof}

\begin{figure}[htb!]
\centering
\picThu
\caption{Forming a period one juggling sequence from the partition $(3,3,2,2)$.}
\label{fig:partition}
\end{figure}

\begin{theorem}\label{thm:trace}
Let $p(b)$ denote the number of unordered partitions of $b$.  Then
\[
\trace(A_b)=p(b)+\sum_{i=0}^{b-1}2^{b-i-1}p(i).
\]
\end{theorem}
\begin{proof}
Updating equation \eqref{eq:jp1} using Theorem~\ref{thm:jp1} we have
\[
p(b) = \trace(A_b)-\sum_{i=0}^{b-1}\trace(A_i).
\]
We now proceed by induction.  We have $\trace(A_0)=1=p(0)$ (note that the sum will be empty and not contribute), establishing the base case.  Now suppose that we have established the result up through $b-1$ and consider the case for $b$.
\begin{align*}
\trace({A}_b)&=p(b)+\sum_{i=0}^{b-1}\trace(A_i)\\
&=p(b)+p(b-1)+2\sum_{i=0}^{b-2}\trace(A_i)\\
&=p(b)+p(b-1)+2p(b-2)+2^2\sum_{i=0}^{b-3}\trace(A_i)\\
&=\cdots \\
&=p(b)+p(b-1)+2p(b-2)+2^2p(b-3)+\cdots+2^{b-1}p(0)\\
&=p(b)+\sum_{i=0}^{b-1}2^{b-i-1}p(i)
\end{align*}
Establishing the result.
\end{proof}

Using this we get that the trace of $A_b$ starting with $b=0$ is
\[
1, ~2, ~5, ~11, ~24, ~50, ~104, ~212, ~431, ~870, ~1752, ~3518, ~7057, ~14138, ~28310, ~56661, \ldots.
\]
This is sequence {\tt A090764} in the OEIS \cite{OEIS}.  The recurrence that we have derived is a variant of that given in the OEIS.  We now give a different variation establishing that this is the same sequence as given in the OEIS.

\begin{theorem}
Let $\mathcal{P}(b)$ be the set of unordered partitions of $b$, and for a partition $q$ let $o(q)$ denote the number of $1$s in the partition.  Then
\[
\trace({A}_b)=\sum_{q\in \mathcal{P}(b)}2^{o(q)}.
\]
\end{theorem}

\begin{proof}
We proceed by induction on $b$.  The result is easily checked for small cases, so now assume it works up through $b-1$.  Then we have the following.
\begin{align*}
\sum_{q \in \mathcal{P}_b}2^{o(q)}&=
\sum_{\substack{q\in\mathcal{P}_b\\ o(q)=0}} 2^{o(q)}+\sum_{\substack{q\in\mathcal{P}_b \\ o(q)>0}}2^{o(q)}\\
&=\sum_{\substack{q\in\mathcal{P}_b\\o(q)=0}} 1+2\sum_{q\in\mathcal{P}_{b-1}}2^{o(q)}\\
&=\big(p(b)-p(b-1)\big)+2\trace(A_{b-1})\\
&=\big(p(b)-p(b-1)\big)+2 \bigg(p(b-1)+\sum_{i=0}^{b-2}2^{b-i-2}p(i)\bigg)\\
&=p(b)+\sum_{i=0}^{b-1}2^{b-i-1}p(i)\\
&=\trace(A_{b}).
\end{align*}
We first divide up our partitions on whether we include a $1$; then we note that there are $p(b)-p(b-1)$ partitions that do not include a $1$ and $p(b-1)$ partitions that do include a $1$ (i.e., taking a partition with $b-1$ we can append a $1$ to get a partition of $b$ which does include a $1$).  We apply the induction hypothesis, and then apply Theorem~\ref{thm:trace} once, clean up our sum, and finally apply Theorem~\ref{thm:trace} a second time.
\end{proof}

\section{Juggling patterns with hand capacity}\label{sec:hand}
One of our basic assumptions that we have employed is that we can catch and throw any number of balls at any given step.  From a practical standpoint jugglers usually limit themselves to catching and throwing at most two or three balls at a time.  The approach outlined above works just as well when we introduce a capacity constraint into how many balls can land at a given time, or equivalently how many balls can be thrown at a given time.

To do this we let $A_{b,\kappa}$ be the principal submatrix of $A_b$ by taking all rows and columns indexed by ordered partitions with all parts of size at most $\kappa$.

\begin{theorem}\label{thm:capacity}
Let $ss_{b,\kappa}(n)$ be the number of siteswap patterns using exactly $b$ balls, with period $n$ and all throws involve at most $\kappa$ balls.  Then
\begin{equation}\label{eq:capss}
ss_{b,\kappa}(n)=\trace(A_{b,\kappa}^n)-\sum_{i=\max\{0,b-\kappa\}}^{b-1}\trace(A_{i,\kappa}^n).
\end{equation}
Let $jp_{b,\kappa}(n)$ be the number of juggling patterns using exactly $b$ balls, with minimal period $n$ and all throws involve at most $\kappa$ balls.  Then
\begin{equation}\label{eq:capjp}
jp_{b,\kappa}(n)=\frac1n\sum_{d\mid n}\mu({\textstyle\frac{n}{d}})\bigg(\trace(A_{b,\kappa}^d)-\sum_{i=\max\{0,b-\kappa\}}^{b-1}\trace(A_{i,\kappa}^d)\bigg).
\end{equation}
\end{theorem}

Before beginning the proof, observe if $\kappa=1$, then this reduces to the case of normal juggling (i.e., $A_{b,1}=(b+1)$); and if $\kappa=\infty$ then this is equivalent to what we have already done.

\begin{proof}
We note that \eqref{eq:capjp} follows from \eqref{eq:capss} by applying M\"obius inversion.  So it suffices to establish \eqref{eq:capss}.

We proceed by induction on $b$. For the base case of $b=0$ there is one card (the empty card) and the capacity places no restriction on its use, and so for any length there is exactly one way to position these cards.  Since $\trace(A_0^n)=1$ this establishes the base case.

Let $r_{i,\kappa}$ be the number of ordered partitions of $i$ with each part at most $\kappa$.  By grouping based on the first part (which has size between $1$ and $\kappa$) we have
\begin{equation}\label{eq:rrecur}
r_{i,\kappa}=\sum_{j=1}^{\min\{\kappa,i\}} r_{i-j,\kappa},
\end{equation}
the $\min\{\kappa,i\}$ coming from noting that we cannot have ordered partitions with negative parts and so we need to handle the case of small $i$.

Now assume that we have established the result up through $b-1$ and consider the case for $b$.  First we observe
\[
\trace(A_{b,\kappa}^n)=ss_{b,\kappa}(n)+\sum_{i=0}^{b-1}r_{b-i,\kappa}ss_{i,\kappa}(n).
\]
This follows by noting that for every siteswap which uses exactly $i$ balls we can find a sequence of cards which uses exactly  $i$ balls.  We can now add $b-i$ balls to the top of each card as long as we do it consistently across the different cards.  Moreover the number of different options we have to place the $b-i$ balls equals the number of ordered partitions of $b-i$ with each part at most $\kappa$ which is $r_{b-i,\kappa}$.

Rearranging and using our induction hypothesis we have
\begin{align*}
ss_{b,\kappa}(n)&=
\trace(A_{b,\kappa}^n)-\sum_{i=0}^{b-1}r_{b-i,\kappa}ss_{i,\kappa}(n)\\
&=\trace(A_{b,\kappa}^n)-\sum_{i=0}^{b-1}r_{b-i,\kappa}\bigg(\trace(A_{i,\kappa}^n)-\sum_{j=\max\{0,i-\kappa\}}^{i-1}\trace(A_{j,\kappa}^n)\bigg)\\
&=\trace(A_{b,\kappa}^n)-\sum_{i=0}^{b-1}\trace(A_{i,\kappa}^n)\bigg(r_{b-i,\kappa}-\sum_{j=\max\{0,b-i-\kappa\}}^{b-i-1}r_{j,\kappa}\bigg)\\
&=\trace(A_{b,\kappa}^n)-\sum_{i=\max\{0,b-\kappa\}}^{b-1}\trace(A_{i,\kappa}^n).
\end{align*}
In going from the first to the second line we use the induction hypothesis, and in going from the second to the third line we rearrange the terms.  Finally we observe that \eqref{eq:rrecur} indicates that almost all terms will $0$ out, except for the first few initial terms which can easily be checked to be one, establishing the result.
\end{proof}

For reference we have produced the number of juggling patterns for $\kappa=2$ in Table~\ref{tab:cap2} and for $\kappa=3$ in Table~\ref{tab:cap3}.

\begin{table}[htb!]
\centering
\begin{tabular}{|l|r|r|r|r|}\hline
$\kappa=2$ & $b=2$ & $b=3$ & $b=4$ & $b=5$\\ \hline\hline
$n=1$& $2$& $2$& $3$& $3$\\ \hline
$n=2$& $4$& $9$& $18$& $30$\\ \hline
$n=3$& $13$& $47$& $134$& $314$\\ \hline
$n=4$& $37$& $224$& $950$& $3140$\\ \hline
$n=5$& $118$& $1118$& $6938$& $31886$\\ \hline
$n=6$& $356$& $5522$& $50751$& $324909$\\ \hline
$n=7$& $1142$& $27910$& $376402$& $3341566$\\ \hline
$n=8$& $3620$& $141946$& $2813824$& $34605634$\\ \hline
$n=9$& $11744$& $730544$& $21219536$& $360849352$\\
\hline
$n=10$& $38275$& $3790391$& $161190485$& $3785776259$\\
\hline
$n=11$& $126234$& $19827570$& $1232724798$&
$39941119938$\\ \hline
$n=12$& $418735$& $104422007$& $9483975303$&
$423549648963$\\ \hline
$n=13$& $1399610$& $553339258$& $73360425430$&
$4512516867634$\\ \hline
$n=14$& $4702499$& $2947940371$& $570219618745$&
$48282551418859$\\ \hline
$n=15$& $15883190$& $15780565950$& $4451677886746$&
$518633980103198$\\ \hline
\end{tabular}
\caption{The number of juggling patterns of period $n$ using $b$ balls with capacity $2$.}
\label{tab:cap2}
\end{table}

\begin{table}[htb!]
\centering
\begin{tabular}{|l|r|r|r|r|}\hline
$\kappa=3$ & $b=2$ & $b=3$ & $b=4$ & $b=5$\\ \hline\hline
$n=1$& $2$& $3$& $4$& $5$\\ \hline
$n=2$& $4$& $12$& $28$& $58$\\ \hline
$n=3$& $13$& $63$& $231$& $713$\\ \hline
$n=4$& $37$& $310$& $1840$& $8591$\\ \hline
$n=5$& $118$& $1618$& $15168$& $106073$\\ \hline
$n=6$& $356$& $8434$& $126258$& $1325570$\\ \hline
$n=7$& $1142$& $45142$& $1069002$& $16789985$\\ \hline
$n=8$& $3620$& $243998$& $9154845$& $214916096$\\ \hline
$n=9$& $11744$& $1336644$& $79252442$& $2776778019$\\
\hline
$n=10$& $38275$& $7392117$& $692290928$& $36167946945$\\
\hline
$n=11$& $126234$& $41247234$& $6095630354$&
$474470288650$\\ \hline
$n=12$& $418735$& $231856131$& $54045188641$&
$6263882726811$\\ \hline
$n=13$& $1399610$& $1311820110$& $482108239540$&
$83162406390939$\\ \hline
$n=14$& $4702499$& $7464002451$& $4323812672665$&
$1109678347266127$\\ \hline
$n=15$& $15883190$& $42679372930$& $38963338572980$&
$14873888879020290$\\ \hline
\end{tabular}
\caption{The number of juggling patterns of period $n$ using $b$ balls with capacity $3$.}
\label{tab:cap3}
\end{table}

We note that we can ask similar questions about the sum of the entries in $A_{b,\kappa}$ as well as the row and column sums as we did with $A_{b}$ (i.e., Lemma~\ref{lem:right}, Lemma~\ref{lem:left} and Theorem~\ref{thm:cardrecurse}).  However the counts are less clear, and have not appeared in the OEIS.  As an example if we count the total number of cards when $\kappa =2$ we get the following numbers, starting with $b=0$:
\[
1,~2,~7,~17,~41,~91,~195,~403,~812,~1601,~3102,~5922,~11165,~20824,~38477,\ldots.
\]
These numbers do appear to satisfy a relatively simple relationship.  In particular through $b=25$ these numbers agree with the following conjecture.

\begin{conjecture}
Let $a_b^{(2)}$ be the number of cards for multiplex juggling with $b$ balls and where each track has capacity at most $2$.  Then
\[
\sum_{b\ge 0}a_b^{(2)}x^b = \frac{1-x+x^2+x^3}{(1-x-x^2)^3}.
\]
\end{conjecture}

\section{Conclusion}\label{sec:conclusion}
By modifying the cards used for juggling, namely allowing groups of balls to be grouped together, we have found a method that works for enumerating multiplex juggling patterns.  There are still a few questions that remain, particularly in understanding what happens when we limit the number of balls that can be caught at any given time.

\bigskip

Ehrenborg and Readdy \cite{ER} introduced cards for juggling and used them to study a $q$-analog of juggling by counting crossings.  It is easy to count crossings on each card and then one simply adds up the crossings over all cards used to count the crossings of the pattern.  We note that the matrices used here can be easily adapted to this situation.  Namely for each card we count crossings (making sure to count multiplicity when balls move in groups), and then weight the card (and hence edge in the graph) by $q^k$ where $k$ is the number of crossings.  Finally we can form matrices $A_b(q)$ where we add up the weights of cards connecting ordered partitions.  As an example we have
\[
A_3(q)=
\begin{array}{r}(3)\\(2,1)\\(1,2)\\(1,1,1)\end{array}\left(\begin{array}{cccc}
2 & 1 & 1 & 1 \\
1 & q+2 & q+1 & q^2+q+1 \\
1 & q & 2 & 0 \\
0 & 1 & q & q^2+q+2
\end{array}\right).
\]
We note that Theorem~\ref{thm:siteswapcount} and Theorem~\ref{thm:capacity} can be easily modified to count the number of juggling patterns with minimal period based on the number of crossings.

\bigskip

An applied mathematical juggler might also want to add the constraint that whenever multiple balls are thrown that no two balls get thrown to the same height.  Our method can be readily adopted to this situation by simply removing any card which has two balls moved to the same track, which leads to modified graphs $\widehat{\mathcal{G}}_b$, and also modified matrices, $\widehat{A}_b$.  For example we have
\[
\widehat{A}_2=\begin{array}{r}(2)\\(1,1)\end{array}\left(\begin{array}{cc}
1 & 1 \\
1 & 3
\end{array}\right),\qquad\text{and}\quad
\widehat{A}_3=
\begin{array}{r}(3)\\(2,1)\\(1,2)\\(1,1,1)\end{array}\left(\begin{array}{cccc}
1 & 0 & 0 & 1 \\
0 & 2 & 1 & 3 \\
1 & 1 & 2 & 0 \\
0 & 1 & 1 & 4
\end{array}\right).
\]

\bigskip

The matrices $A_b$ might also have independent interest.  For example, it is easy to see that for $1\le i\le 3$ that $A_i$ is a principal submatrix of $A_{i+1}$.  This seems to continue for at least the first few cases.
Does this containment continue?  Note that this also seems to indicate a preferred ordering of the ordered partitions if we want to have (1) containment of the previous matrix in the upper left block and (2) an upper triangular matrix in the lower left block.  What is this ordering?

Things get even more interesting when we consider the characteristic polynomial of $A_b$.  If we let $P_b:=P_b(x)=\det(xI-A_b)$, then we have the following.
\[
\begin{array}{l@{\,=\,}llllllllllllll}
P_0&f_0^1\\
P_1&&f_1^1\\
P_2&&&f_2^1\\
P_3&f_0^1&&&f_3^1\\
P_4&f_0^2&f_1^1&&&f_4^1\\
P_5&f_0^5&f_1^2&f_2^1&&&f_5^1\\
P_6&f_0^9&f_1^5&f_2^2&f_3^1&&&f_6^1\\
P_7&f_0^{19}&f_1^9&f_2^5&f_3^2&f_4^1&&&f_7^1\\
P_8&f_0^{37}&f_1^{19}&f_2^9&f_3^5&f_4^2&f_5^1&&&f_8^1\\
P_9&f_0^{74}&f_2^{37}&f_2^{19}&f_3^9&f_4^5&f_5^2&f_6^1&&&f_9^1\\
P_{10}&f_0^{148}&f_1^{74}&f_2^{37}&f_3^{19}&f_4^9&f_5^5&f_6^2&f_7^1&&&f_{10}^1\\
P_{11}&f_0^{296}&f_1^{148}&f_2^{74}&f_3^{37}&f_4^{19}&f_5^9&f_6^5&f_7^2&f_8^1&&&f_{11}^1\\
P_{12}&f_0^{591}&f_1^{296}&f_2^{148}&f_3^{74}&f_4^{37}&f_5^{19}&f_6^9&f_7^5&f_8^2&f_9^1&&&f_{12}^1\\
P_{13}&f_0^{1183}&f_1^{591}&f_2^{296}&f_3^{148}&f_4^{74}&f_5^{37}&f_6^{19}&f_7^9&f_8^5&f_9^2&f_{10}^1&&&f_{13}^1
\end{array}
\]
Where the polynomials $f_i$ are given by the following.
\begin{align*}
f_0(x)&=x-1\\
f_1(x)&=x-2\\
f_2(x)&=x^2-5x+5\\
f_3(x)&=x^3 - 10x^2 + 27x -20\\
f_4(x)&=x^5 - 20x^4 + 135x^3 - 396x^2 + 518x - 245\\
f_5(x)&=x^7 - 36x^6 + \cdots - 2^5{\cdot}5{\cdot}7^2
\\
f_6(x)&=x^{11} - 65x^{10} + \cdots - 2^5 {\cdot} 3^2 {\cdot} 5^2 {\cdot} 7^3\\
f_7(x)&=x^{15} - 110x^{14} + \cdots - 2^{11} {\cdot} 3^2 {\cdot} 5^3 {\cdot} 7^4\\
f_8(x)&=x^{22} - 185x^{21} + \cdots + 2^{10} {\cdot} 3^4 {\cdot} 5^4 {\cdot} 7^6 {\cdot} 11^2\\
f_9(x)&=x^{30} - 300x^{29} + \cdots + 2^{21} {\cdot} 3^5 {\cdot} 5^6 {\cdot} 7^8 {\cdot} 11^2\\
f_{10}(x)&=x^{42} - 481x^{41} + \cdots + 2^{21} {\cdot} 3^9 {\cdot} 5^8 {\cdot} 7^{12} {\cdot} 11^3 {\cdot} 13^2
\\
f_{11}(x)&=x^{56} - 752x^{55} +\cdots + 2^{38} {\cdot} 3^{12} {\cdot} 5^{11} {\cdot} 7^{16} {\cdot} 11^4 {\cdot} 13^2\\
f_{12}(x)&=x^{77} - 1165x^{76}+\cdots - 2^{42} {\cdot} 3^{17} {\cdot} 5^{16} {\cdot} 7^{22} {\cdot} 11^9 {\cdot} 13^3\\
f_{13}(x)&=x^{101} - 1770x^{100} +\cdots - 2^{70} {\cdot} 3^{23} {\cdot} 5^{21} {\cdot} 7^{29} {\cdot} 11^{11} {\cdot} 13^4
\end{align*}

There are several interesting things that are appearing.
\begin{itemize}
\item The sequence of the exponents of $f_i$ in the $P_j$ seem to follow
\[
1,~0,~0,~1,~2,~5,~9,~19,~37,~74,~148,~296,~591,~1183,\ldots.
\]
This appears to be the sequence {\tt A178841} in the OEIS \cite{OEIS} which counts the number of pure inverting compositions of $n$.
\item The degree of the polynomials $f_i$ follow
\[
1,~1,~2,~3,~5,~7,~11,~15,~22,~30,~42,~56,~77,~101,\ldots.
\]
This appears to be the sequence {\tt A000041} in the OEIS \cite{OEIS} which counts the number of unordered partitions of $n$.
\item The second coefficients of the polynomials $f_i$ follow
\[
1,~2,~5,~10,~20,~36,~65,~110,~185,~300,~481,~752,~1165,~1770,\ldots.
\]
This appears to be the sequence {\tt A000712} in the OEIS \cite{OEIS} which counts the number of unordered partitions of $n$ into parts of $2$ kinds.
\end{itemize}

We have no explanations for any of these phenomenon, but given the nature of how the matrix is formed believe this is more than coincidence.  We look forward to more research being done into these cards and matrices.

redpajama_set_name: RedPajamaArXiv


text:
Blackwall visse a Hendre House, nei pressi di Llanrwst, nel nord del Galles dal 1833 alla sua morte.
Si interessò alla natura fin dalla sua giovinezza, in primo luogo agli uccelli e poi ai ragni, sui quali pubblicò il suo primo articolo nel 1827.

La sua opera più importante fu  Una Storia dei Ragni della Gran Bretagna e dell'Irlanda  (in 2 volumi, del 1861-1864) che menzionava 304 specie e dove dava le prime descrizioni adeguate dei ragni britannici. Solo dieci delle specie incluse erano descritte da Octavius Pickard-Cambridge e dodici dal naturalista irlandese Robert Templeton.

Corrispondenza con Charles Darwin

Blackwall scrisse quattro lettere sul tema dei ragni a Charles Darwin, il 12 febbraio 1868,
il 18 febbraio 1868, il 10 agosto 1869 ed l'8 settembre 1869. Gli originali sono conservati nel Darwin Archive della biblioteca dell'Università di Cambridge. La prima, seconda e terza lettera sono di risposta diretta a comunicazioni di Darwin, sebbene il luogo di spedizione di queste lettere, presumibilmente mandate da Blackwall, è ignoto.
Il loro argomento è, in modo ampio, la variabilità che esiste fra specie e generi di ragni.
La prima lettera comincia "io desidero che sia in mio potere dare risposte soddisfacenti alle domande contenute nella Sua lettera del 10 scorso. Ragni adulti della stessa specie differiscono in taglia e colore così notevolmente, e anche indipendentemente ad ogni aspetto delle situazioni nel quale vengono rinvenuti, che io non sono capace di assegnare una causa adeguata per questo fatto straordinario... " .

Il lavoro di Blackwall costituì un passo importante nella comparsa dell'aracnologia.
È stato uno dei primi aracnologi che si interessarono ai ragni di taglie molto piccole, in particolare quelli che appartengono ai generi Neriene e Walckenaeria (Linyphiidae).

Taxa descritti
Ha descritto e denominato, in ordine alfabetico, molti taxa, fra i quali:
 Linyphiidae Blackwall, 1859 famiglia (Araneae)
 Oecobiidae Blackwall, 1862 famiglia (Araneae)
 Salticidae Blackwall, 1865 famiglia (Araneae)
 Scytodidae Blackwall, 1864 famiglia(Araneae)
 Pycnacantha Blackwall, 1865 genere (Araneidae)

e specie:
 Alpaida grayi Blackwall, 1863 (Araneidae)
 Araeoncus humilis Blackwall, 1841 (Linyphiidae)
 Araneus hortensis Blackwall, 1859 (Araneidae)
 Baryphyma pratense Blackwall, 1861 (Linyphiidae)
 Bathyphantes gracilis Blackwall, 1841 (Linyphiidae)
 Bolyphantes luteolus Blackwall, 1833 (Linyphiidae)
 Caerostris vicina Blackwall, 1866 (Araneidae)
 Centromerita bicolor Blackwall, 1833 (Linyphiidae)
 Centromerus sylvaticus Blackwall, 1841 (Linyphiidae)
 Cnephalocotes obscurus Blackwall, 1834 (Linyphiidae)
 Dicymbium nigrum Blackwall, 1834 (Linyphiidae)
 Dicymbium tibiale Blackwall, 1836 (Linyphiidae)
 Diplocephalus cristatus Blackwall, 1833 (Linyphiidae)
 Diplocephalus picinus Blackwall, 1841 (Linyphiidae)
 Dismodicus bifrons Blackwall, 1841 (Linyphiidae)
 Epicadinus spinipes Blackwall, 1862 (Thomisidae)
 Eriophora edax Blackwall, 1863 (Araneidae)
 Eustala tristis Blackwall, 1862 (Araneidae)
 Firmicus insularis Blackwall, 1877 (Thomisidae)
 Gasteracantha frontata Blackwall, 1864 (Araneidae)
 Kilima decens Blackwall, 1866 (Araneidae)
 Misumena spinifera Blackwall, 1862 (Thomisidae)
 Misumenops gibbosus Blackwall, 1862 (Thomisidae)
 Neoscona vigilans Blackwall, 1865 (Araneidae)
 Ozyptila elegans Blackwall, 1870 (Thomisidae)
 Ozyptila trux Blackwall, 1846 (Thomisidae)
 Paraplectana thorntoni Blackwall, 1865 (Araneidae)
 Paraplectana walleri Blackwall, 1865 (Araneidae)
 Parawixia audax Blackwall, 1863 (Araneidae)
 Phrynarachne tuberosa Blackwall, 1864 (Thomisidae)
 Thomisus candidus Blackwall, 1866 (Thomisidae)
 Thomisus modestus Blackwall, 1870 (Thomisidae)
 Xysticus diversus Blackwall, 1870 (Thomisidae)
 Xysticus erraticus Blackwall, 1834 (Thomisidae)
 Xysticus luctuosus Blackwall, 1836 (Thomisidae)

Portano il suo nome
In suo onore sono state denominati molti generi e specie di ragni fra cui:
 Cozyptila blackwalli SIMON, 1895 (Thomisidae)
 Herpyllus blackwalli BRAENDEGAARD, 1966 (Thomisidae)
 Idiops blackwalli O.P.-CAMBRIDGE 1870 (Idiopidae)
 Leiobunum blackwalli MEADE 1861 (Phalangiidae)
 Salticus blackwalii CLARK, 1855 (Salticidae)
 Scotophaeus blackwalii THORELL, 1871(Gnaphosidae)

Altri progetti

Collegamenti esterni

redpajama_set_name: RedPajamaWikipedia


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What's Not: Internals aren't mobile workstation caliber-- more like Ultrabook internals in a bigger, sturdier body.
HP mobile workstations are varied animals--some are large and heavy laptops with enough power to help you design the next Chevy or BMW. Others are not quite so heavy, not so powerful but are still durable and serviceable for upgrades. Those are the kind of laptops your IT department loves--the price won't break the corporation's bank, and the IT folks can fix or replace most anything that breaks. Want more RAM? No problem! Swap the HDD or SSD? Done. The CPU and GPU might not be a CAD designer's professional dream machine, but it has enough oomph to get some drawings with a hundred parts done in AutoCAD. That's the HP ZBook 15u G3.
Despite similar names, don't confuse our ZBook 15u with the ZBook Studio (also thin and light, but $800 more expensive and more powerful) or the ZBook 15 (heavier but more powerful). The ZBook 15u has an Ultrabook CPU in a mobile workstation chassis--an interesting take on mobile workstations that HP and Lenovo have been offering for a few years. Even Dell got into the act on the gaming front with the Alienware 13, a gaming laptop that has a fairly powerful graphics card but an Ultrabook dual core CPU. Since games and 3D design programs often rely on the graphics card more than the CPU, it's a reasonable thing to try; particularly if it brings weight and thickness down, allows for a smaller charger and drops the price a little bit. The ZBook 15u however opts for the less powerful AMD FirePro W4190M 2GB DDR5 dedicated graphics that sits at the bottom of AMD's line, rather than the beefier GPUs in the competition. Is this the right way to go, especially if you want a mobile workstation that won't send you to the chiropractor or a loan shark? Read on!
The HP ZBook 15u G3u runs Windows 10 on Intel 6th generation dual core, 15 watt Ultrabook CPUs rather than the quad core 45 watt powerhouse CPUs in the non-u HP ZBook 15, Dell Precision 15 5000 series and Lenovo ThinkPad P50. It has the low end CAD oriented AMD FirePro W4190M 2GB DDR5 graphics card, and can house up to 32 gigs of RAM in two slots. It has an has SSD boot drive and a conventional 2.5" hard drive for high capacity storage. It's available with 1080p and 4K displays in two quality levels (TN and IPS). Despite its mobile workstation status, the ZBook 15u weighs just 4.18 lbs. and is 0.78" thick.
The laptop looks every bit like a ZBook rather than a consumer Ultrabook. It's beefy, built like a tank and passes MIL Spec 810G tests for humidity and shock resistance. It's a dark gray machine that's easy to grip and hold. The backlit keyboard deck is fairly rigid with just a little bit of flex in the center (you won't notice unless you're a very heavy-handed typist). The laptop weighs just 4.18 lbs., which is quite light for a workstation, but given the fairly low powered CPU and graphics card in this model, we're less impressed.
The Synaptics trackpad has dedicated right and left click buttons (nice!) as does the eraser stick pointer. It's a well behaved trackpad and the buttons are just right--not too stiff and not too mushy.
The laptop doesn't get hot on top or bottom. The low power CPU and GPU combined with plenty of internal space for heat dispersal and cooling mean you won't get sweaty hands or legs.
The bottom panel is removable for repairs and upgrades, but your IT department will die a little each time they have to do it. Most mobile workstations are fairly easy to open and the battery is removable. With the ZBook 15u, the battery is sealed inside (a concession to the slim design) and there are 11 Phillips head screws that must be removed to access the internals. One of those screws is hidden under the SD card slot plastic blank. The rest are under little rubber plugs that are best removed using a fingernail (sharp, pointy tools might tear them). These plugs vary in shape and thickness to match the contours of the laptop. Best to lay them out on the table in a rectangle to more easily remember which one goes where. Ouch. Once you open it up the internals are easy to access: two RAM slots, an M.2 SSD slot, a 2.5" drive bay, socketed Intel 8260 WiFi card and the battery.
We have the 1920 x 1080 full HD touch screen in for review. This is a TN panel with mediocre viewing angles and weak contrast that even for the business world seems a bit out of date. Color gamut isn't good, but the 280 nits (measured) brightness is adequate for use in the office or at home. The laptop's base panel represents 79% of sRGB and 50% of Adobe RGB, far from the extremely wide color gamut of HP's Dream Color displays and below the consumer laptop average in this price range of full sRGB and 75% of Adobe RGB. That means this panel isn't a good choice for designers or photo and video editors. The more expensive IPS display options are a little brighter and have wider color gamut, thankfully.
This is the sort of panel you'll have to angle forward and back to find the sweet spot for best viewing. For those who don't want a touchscreen, HP also offers the base 1080p TN "standard viewing angle" panel in non-touch form. I strongly recommend going with the 1080p or 4K "wide viewing angle" IPS displays instead for better viewing angles, contrast and color. There's no HP Dream Color wide gamut display option for the 15u--it truly is a lovely display that we enjoyed when reviewing the larger, heavier and more expensive standard HP ZBook 15.
The ZBook 15u Ultrabook workstation is a bit of an oddball: it looks and feels like a hearty, powerful workstation class laptop but it has a lowly dual core Ultrabook 15 watt CPU and low end AMD FirePro W4190M 2GB DDR5 dedicated graphics switchable with Intel HD 520 graphics. It's available with Core i5 and i7 dual core ULV CPUs, and it's not much faster or more capable than the average 13.3" Core i7 Ultrabook with low end dedicated graphics. Its appeal rests instead with the ample selection of ports, robust build quality and HP's relationship with large scale corporate buyers. I have to feel a little sorry for the folks at a big company who get the less powerful 15u version while some compatriots higher up in the engineering and CAD food chain get the more powerful ZBook Studio and ZBook 15 models.
The machine has 2 RAM slots and ships with DDR4 RAM. It can handle up to 32 gigs of RAM, which is certainly better than the average Ultrabook that's capped at 8 or 16 gigs of DDR3 RAM. It has both an M.2 SSD slot and a 2.5" drive bay, so you can have a fast boot/applications drive and a big 5400 or 7200 RPM HDD to store large files. It's available with a SATA-3 SSD or a 256 gig or 512 gig PCIe NVMe SSD that's even faster. HP calls this the HP Z Turbo Drive option, and for those who move many files daily, the extremely fast read and writes speeds are a bonus.
The HP ZBook 15u is a low power Ultrabook with a low end dedicated graphics card inside, so it should run forever, right? Not so much. HP went with a 3 cell, 46 WHr battery, which is similar in capacity to many 13.3" Ultrabooks on the market, and thus it doesn't outlast them. Several powerful quad core mobile workstations and professional 15" laptops now have very high capacity batteries to ensure stamina, and in fact some have double the battery capacity. Thus they last around 6 hours on a charge for productivity use, and the ZBook 15u does a little better at 6.5 hours.
One bright spot is the compact and light 65 watt charger. Unlike those powerful workstations with huge power bricks that weigh almost 2 pounds, the ZBook 15u's charger is similar to that of an Ultrabook.
The HP ZBook 15u is like a Hummer H3 with a Mazda Miata drivetrain--it's hard to say exactly where it fits other than as an upsell for the more powerful ZBook Studio and ZBook 15 models, or as a corporate purchase for those whose jobs don't entail heavy computational and design tasks. It's basically an Ultrabook in mobile workstation clothing, so you get the durable design, healthy port selection and serviceability of a workstation mated to the brains of a dual core 15 watt Ultrabook. The downside is your machine won't crunch through complex CAD models with many parts quickly, and thanks to a not terribly high capacity battery, runtimes aren't much better than some quad core competitors. We do strongly recommend that you opt for the wide viewing angle IPS display because the TN panel's viewing angles and contrast are hard on the eyes if not downright antiquated. The adage "you get what you pay for" applies here; the HP ZBook Studio is a much more desirable laptop than our ZBook 15u, but $800 sets them apart.
Display: 15.6", available with TN and IPS panels. 1920 x 1080 and 4K resolutions available. Intel HD 520 integrated graphics switchable with AMD FirePro W4190M 2GB DDR5 dedicated graphics. DisplayPort 1.2 with MST support and VGA port. USB-C 3.1 Gen. 1 port.
Battery: 3 cell, 46 WHr Lithium Ion rechargeable, sealed inside. 65 watt compact charger.
Performance: Intel Skylake 6th generation Core i5 and i7 dual core, 15 watt CPUs available. 8 to 32 gigs DDR4 RAM in two slots. M.2 SSD (SATA-3 and PCIe NVMe avaialble) and one 2.5" drive bay with your choice of a 5400 or 7200 RPM drive.
Size: 4.47 x 2.78 x .53 inches. Weight: 4.18 pounds.
Networking: Integrated Intel 8260AC dual band WiFi 802.11b/g/n/ac and Bluetooth 4.2. SIM card slot for 4G LTE. Gigabit Ethernet.
Expansion and Ports: 2 USB 3.0 ports, 1 USB-C 3.1 Gen. 1 port, DisplayPort, VGA, Smart Card slot, Ethernet, 3.5mm audio and SD card slot.


Healthcare is changing like never before. Nowadays, patients expect consistent quality – from medical treatment to the spatial experience which has a considerable influence on the success of therapy. At the same time, processes must become increasingly efficient.
Our sliding solutions must prove themselves in harsh everyday hospital life, and fulfil an extremely wide range of demands.
They must provide freedom from barriers, for example. This becomes possible by means of ergonomic door openings with sliding solutions, which are appreciated by wheelchair users and nurses moving hospital beds.
Sustainable and lasting products whose running costs are minimized by being low maintenance and having long service intervals help everyday processes to run smoothly.
The demand for a high degree of costeffectiveness is met by making efficient use of space. This makes it possible to have more patient and treatment rooms in a perfectly used layout.
Transparency and orientation are decisive as far as the image of a hospital is concerned. This is because a hospital must appear inviting and trustworthy.
The main focus is on people. Ones who are to remain or become healthy. And the people who are going to take care of this. Sliding applications support both of them.
Energy cost can be reduced if the right sliding solution is used outdoors.
Normal doors create unnecessary barriers and risks in hectic everyday life. Sliding doors reduce them.
Space is a scarce resource in a hospital. Sliding solutions create a high degree of space efficiency.


Lomography are huge in the analogue world starting with Holga's and Diana's then the LCA they really put analogue on the map when e'ryone was chasing 5 megapixel digital cameras.
400 ISO still not fast enough? Use Lomography Color Negative 800 35mm film for fantastic results under all lighting conditions. Quick shutters, small apertures, fast subjects and low lights call for a fast film but the last thing you want is to sacrifice your Lomographic colors and effects for a little extra speed. To all of you plastic and multi-lensers, dusk-til-dawners and hi-speed Lomographers—we've got you covered with Lomography Color Negative 800 35mm film! It guarantees amazing results even in low-lighting situations.


While I'm fairly certain that this is probably just a false positive, I figured I'd bring it to your attention that my Firefox seems to have flagged your site as a potential attack site.
Perhaps this is something you can look into resolving?
The flagging comes from Google and after doing a scan on the kimitonara subdomain, there were no problems found. Google messed up something, I bet.


There exists a prejudice against the term 'freelancer'. Many consider it a synonym for a college student who cant hold traditional office hours or a person who can't adjust to a corporate environment. As technology grows it offers unlimited options to people who for some reason are unable to step out of the door.
Blockchain on the other hand is a decentralized technology which has disrupted many industries. But, for it to function, it needs freelancers who not only know about the technology but also offer knowledge of computing languages and mathematics.
Lack of trust in a new client : Unknown clients can abscond once their work is transfer to them without paying the remittance that is due.
Lack of Marketing Skills: Many new freelancers face a crucial problem. They do not have the means nor the knowledge on how to market their services to potential clients.
Maintaining Deadlines: many a times the clients throws in outrageous deadlines which can't be met. The freelancer needs to know their strengths to know beforehand if a deadline can be met or no.
Software and Hardware: As one starts up a new service there may be a lack of required Software or Hardware in regards to your client's needs.
Lack of trust in a new client : It doesn't matter if the client is unknown. Any transaction is secured with the help of a 'Smart Contracts'. If one party renegades the transaction immediately gets noted on his Block and shall has future ramifications.
Lack of Marketing Skills: There is no need to market yourself. Once you have run a transaction successfully on the Blockchain, then your work begins to speak for itself. New clients shall see your past work every time they access the Blockchain.
Software and Hardware: No software is required as such. All you need is your brains and a supercomputing hardware program and more RAM attached to your PC. This can be acquired easily from your computer engineer.
Now that we have cleared the doubts, in regards to working as a freelancer on Blockchain Technology, let us shift our focus on to the platforms who are looking to hiring freelancers for the same.
There you have it folks. Blockchain technology has something to offer everyone. Be it in the education sector or the government. It is here to disrupt till the time it becomes the norm. Freelancers rejoice, your time has come!


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Q: akka-persistence-cassandra too many reads from metadata table We have a multi node web application that uses akka-persistence-cassandra to store incoming events. Recently we were told that our application is doing nearly 2.8mln reads per day from cassandra messages and metadata table.(combined) I understand the need to read from messages table but i am still not able to figure out why we need to read from metadata table.
Lets look at akka-persistence-cassandra api  (2.11-0.59) . This is the only read query on metadata table
def selectDeletedTo = s"""
  SELECT deleted_to FROM ${metadataTableName} WHERE
    persistence_id = ?
"""

The CassandraJournal class implements this asyncReadHighestSequenceNr :
override def asyncReadHighestSequenceNr(persistenceId: String, fromSequenceNr: Long): Future[Long] = {
    writeInProgress.get(persistenceId) match {
      case null => super.asyncReadHighestSequenceNr(persistenceId, fromSequenceNr)
      case f    => f.flatMap(_ => super.asyncReadHighestSequenceNr(persistenceId, fromSequenceNr))
    }
  }

override def asyncReadHighestSequenceNr(persistenceId: String, fromSequenceNr: Long): Future[Long] =
    asyncHighestDeletedSequenceNumber(persistenceId).flatMap { h =>
      asyncFindHighestSequenceNr(persistenceId, math.max(fromSequenceNr, h))
    }

So as i can understand from the above statement, akka  first needs to read the metadata table to get the deleted count and then use that to get the highest sequence number.
However in my application i never delete a message so metadata table is always empty . Hence math.max(fromSequenceNr, h) will always result in fromSequenceNr.
So i am thinking of overriding asyncReadHighestSequenceNr method and removing the call to read highest deleted seq number. That should save me all the extra calls to metadata table..
Any comments , ideas on how this sounds??

redpajama_set_name: RedPajamaStackExchange


Mahjong is the ultimate game of strategy and has frustrated and delighted players in equal measure ever since it first appeared on computer screens back in 1981. The original Mahjong game that originated in China doesn't hold many similarities with the computer version of the game apart from the tiles, which are the same both in style and in number. The reason for the difference is that the computer version of the game is catered to the one-player user, something which is not possible with the traditional four-player table game. Mahjong Solitaire has been made available on just about every console that has ever existed from the Macintosh (the first home of the legendary Shanghai game) to the SEGA Megadrive. Despite being deemed a game of concentration and strategy, Mahjong found a place on games consoles that were more akin to high adrenalin, fast-paced games such as Sonic the Hedgehog and Donkey Kong. Such was Mahjong's popularity that numerous versions of the solitaire based games have sold tens of millions of copies worldwide.
There are numerous different versions of Mahjong online and hundreds of different layouts, that can make the game easier of harder, the one thing that each version has in common are the rules and the objective. The aim of the game is to remove every tile from the selected layout, resulting in an empty board. This must be achieved by removing tiles in pairs – a task that is easier said than done.
The tiles are an important part of the game. Each layout will consist of 144 tiles, laid-out in various different patterns (turtle, dragon and crab are popular). There are three sets of basic tiles, numbered one to none and containing 36 tiles in each set. These sets are named bamboo, character and ball (also called dot). There are also 16 wind tiles in four styles, 12 dragon tiles in three styles and four tiles each in flower and season styles.
Tiles must be removed from the layout in pairs, for the majority of tiles this means selecting two tiles that are matching exactly, however there are a exceptions with the flower and seasons tiles which can be matched with any another tile from their particular suit. For instance, autumn can be paired with winter and an orchard with a plum. Removing the tiles is not simply a case of located a pair and selecting them, each tile selected must be "free"; this means that it should not be restricted in any way. A restricted tile will be one that is covered on either side (left or right) and covered by a tile on top, even partially covered tiles are classed as restricted.
Scoring points adds an extra dimension to the game. Points are scored by removing pairs from the layout and bonus points are awarded for removing pairs that are the same pattern and number and doing so in a row. Games can also be played against the clock, giving friends and family the chance to beat your time.


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Entertainment this Morning with Gary and Bristol Wednesday, August 24
The Olympics were good for NBC, Cole Swindell has a super power and more! Get this scoop in today's Entertainment this Morning with Gary and Bristol!
Entertainment this Morning with Gary and Bristol Wednesday, January 6
Six days a week, Gary and Bristol get up before the sun shines to bring you the latest from Nashville, New York City and Hollywood with Entertainment this Morning!
Jennifer Lawrence Isn't Happy In First 'Mockingjay Part 2' Clip
The first clip from The Hunger Games: Mockingjay - Part 2 has arrived, along with Katniss' first set of orders. In the new clip, which was released to celebrate the first advanced ticket sales for the film, Jennifer Lawrence's Katniss is grouped into her new squad. Meet the Star Squad,…
E. Oliver Whitney
Entertainment this Morning with Gary and Bristol Friday, August 21
Wake up bright and early with Gary and Bristol to get the latest from Nashville, New York City and Hollywood with Entertainment this Morning!
Entertainment this Morning with Gary and Bristol Tuesday, July 21
Six days a week Gary and Bristol get up early to bring you the latest from Nashville, New York City and Hollywood with Entertainment this Morning.
Entertainment this Morning with Gary and Bristol Thursday, July 16
Entertainment This Morning with Gary and Bristol Wednesday, April 8
Each morning Gary and Bristol bring you the latest in entertainment news from Nashville, New York City and Hollywood!
Entertainment This Morning with Gary and Bristol Tuesday, March 24
Each weekday morning on Kiss Country, we give you the latest entertainment news from Nashville, New York City and Hollywood!
'The Hunger Games: Mockingjay - Part 1' Review
'The Hunger Games: Mockingjay - Part 1' is a movie comprised almost entirely of deleted scenes. As it says right in the title, this isn't the final chapter of 'The Hunger Games' series; it's just the first half of the final chapter, and that's exactly…
Kiss Chronicles Jan 15th: What Star Admitted to Drug Use After Denying it for Years?
Every morning at 6:35 and 8:50 we let you know what's going on in Hollywood, New York and Nashville. From Lance Armstrong coming clean to Mindy McCready's heartbreak here's what making headlines with you're favorite stars:
Gary and Julie


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Rheumatologist Spire London East Hospital
by PLR | Oct 9, 2019 | Consultant Rheumatologist, Private Consultant Rheumatologist, Rheumatologist In London, Spire London East Hospital
Dr Naveen Bhadauria is a Consultant Rheumatologist at Spire London East Hospital.
Some of the principal treatments carried out by Dr Bhadauria include:
Joint pain treatment (joint injections)
Dr Bhadauria's Professional Profile
Dr Bhadauria is currently practising as a substantive NHS Consultant Rheumatologist at North Middlesex University Hospital NHS Trust.
He completed his medical degree in June 2004 at Guy's, King's and St Thomas' Medical School.
In June 2001 he completed a BSc in immunology which set the foundation for his path in to the world of rheumatology.
He entered specialty training in rheumatology and general internal medicine in 2012.
During his training, he experienced a wide variety of rheumatological problems, including the common conditions and more complex cases.
His experience in acute and general medicine has helped him to develop into a well-rounded rheumatologist who can really provide holistic care to his patients.
His current practice involves looking after patients with all aspects of general heumatology.
His specialty interest is in early inflammatory arthritis and also in the management of patients with connective tissue diseases, such as lupus or systemic sclerosis.
Other areas that he commonly manage include:
More About Dr Bhadauria's Practising
Dr Bhadauria sees on average about 50-60 patients per week in the rheumatology clinic and also on the wards.
He regularly performs joint injections in the knee, shoulder, wrist and carpal tunnel, averaging about 5-10 injections per week.
He also providing platelet-rich plasma joint injections, which is establishing itself as a highly successful form of medical technology for the treatment of joint arthritis.
Dr BhadauriaI also conducts research in his current NHS Practise, and is the research lead for his department.
He is a Principle Investigator for a clinical trial involved in looking at the benefit of two different medications in the management of Ankylosing Spondylitis (SURPASS study).
His previous research during his BSc in immunology looked at the role of heat shock proteins as possible target antigens in the development of rheumatoid arthritis.
Dr Bhadauria's Care & Treatment
An important priority for Dr Bhadauria is to maintain a high professional standard.
He regularly keeps himself up-to-date with the latest evidence and research in rheumatology so that he can provide his patients with the best possible care.
He regularly attends conferences and read the latest journals which allow him to do this.
Another important aspect to his care is that he provides consultations where he looks to make the patient feel as comfortable as possible.
This allows fully for a discussion regarding any investigation and treatment.
The patient's feelings and views are just as important as the doctors.
Dr Bhadauria's Further Medical Interests
Dr Bhadauria is the Clinical Governance lead for the rheumatology department at North Middlesex Hospital.
Clinical Interests there are Rheumatology, with special clinical interests including:
Soft Tissue Rheumatism
Make an appointment with Dr Bhadauria
Because Dr Bhadauria is based across four hospitals, you might like to contact him directly.
So, if you would like to arrange to see Dr Bhadauria, please do not hesitate to get in touch.
First of all, you could contact Medical Secretary, Daniela Ryan, by telephone on 07943 404279 or 07806 683980.
Alternatively, you can contact her via email – [email protected]


Ability Trial: Summon 12 minions with 1 or less Attack.
Promote will cause a minion to transform into a minion of your faction that costs one more.
Cards like Zyx, Jax Truesight, Jaxi, Sarlac the Eternal etc. summon more than 1 minion with 1 or less Attack and therefore count more for the trial.
Although Zeal minions like Silverguard Knight, Sol Pontiff, Second Sun but also Decorated Enlistee can have more than 1 attack when they are summoned, they will still count for the Trial because Strategos (and Auroara) proc off of a minions base attack value.
This page was last edited on 3 August 2018, at 19:51.