Keywords: Drug Resistance, Mutations, Gene Expression, Naive Bayes, Machine Learning.
of data mining and also reduces efforts in retrieving information from online resources.
not very clear but still there are some notions about it.
by increased level of expression of drug molecules.
in many of the cases this brings failure in treatment.
Khalid, Z. and Sezerman, U.
ZK Drugresist - Automatic Extraction of Drug Resistance Mutations and Expression Level Changes from Medline Abstracts.
(Chrisanthar et al., 2008; Hochhaus et al., 2011).
resistance against antibiotics (Pal and Larsoon, 2014).
from such kind of repositories is a laborious task.
analysis approach for information extraction.
of information as it has a huge collection of abstracts.
information remains a big challenge for researchers.
related information are downloaded into a database.
words making them easy to process.
Rocchio algorithm both uses bag of words approach.
summarizes all the steps involved in ZK DrugResist.
is marked as either drug or others.
the amino acid level mutations from each document.
easily be misinterpreted with the nucleotide letters.
compiled using PERL Regular Expression library.
output on the web page.
testing and k fold cross validation were employed.
resistance. This makes up to 701 documents in total.
in which mutation is mentioned but not at drug level.
classifier are listed in Table 2, 3, 4 and 5 respectively.
The graphical representation is illustrated in Figure 1.
Table 2: Classification Results of Drug Resistance vs.
Others in Training and Test Sets.
Table 3: Classification Results of Mutations vs.
Expressions in Training and Test Sets.
Table 4: Classification Results of Protein Mutations vs.
DNA Mutations in Training and Test Sets.
literature without comprising the accuracy measure.
Proceedings of RIAO, pp. 197-216.
vol.15, no. 3 ,pp. 289-96.
from literature', BMC Bioinformatics, vol.11, no.101.
PLoS ONE, vol. 3, no. 8, pp. e3062.
J Bioinform Comput Biol, vol. 5, no .6, pp.1261-1275.
Ser Workshop Genome Inform, pp. 81-90.
Assoc, vol. 5, no. 11, pp.1-11.
Med Chem, vol. 56, no. 14, pp. 5691-708.
mechanisms of drug resistanc', J Pathol, vol. 205, no.
Genome Inform, vol. 9, pp. 72-80.
epilepsy', Ann Neurol, vol. 53, no. 4, pp.469-79.
antibacterial drug resistance', Nature, vol. 406, no.


By Pioneer Staff Report on Feb 11, 2019 at 4:07 p.m.
LITTLE FALLS -- The Bemidji High School wrestling team will enter into the Section 8-3A Team Tournament in Little Falls on Saturday, Feb. 16, as the No. 2 seed.
The Lumberjacks, ranked No. 8 in Class 3-A, will match up with the winner of No. 7 Little Falls and No. 10 Rocori in the quarterfinals at 12 p.m.
BHS was bested by only Willmar -- ranked No. 6 in 3-A -- when seedings were announced on Monday. The Cardinals will take on the winner of No. 8 Alexandria and No. 9 Moorhead.
No. 3 Brainerd will face No. 6 St. Cloud Tech, with the victor advancing to square off against the winner of the Bemidji match. A meeting between No. 4 Sartell-St. Stephen and No. 5 Sauk Rapids-Rice rounds out the rest of the field.
The section semifinals will begin at 2:30 p.m. on Saturday, and the finals will follow at 4:30 p.m.


When I saw that the Pantone Colour Of The Year for 2019 was Living Coral I was so excited! I think that's such a perfect colour and I love it! I also thought the message behind the colour was a really great addition this year, you can read more about that here.
This post contains affiliate links or ads. Inspired By Rossland Optical sponsored this post. You can follow them on their Website, Instagram, Facebook and Twitter.
I think this is the perfect shade to incorporate into makeup looks though. I'm excited to see what I can pull together this year to work with that colour! So this week I wanted to share my thoughts on a great sheer, coral lipstick – the Tarte Amazonian Butter Lipstick in Coral Blossom.
I love lipstick but I also love pairing them with different styles and colours of glasses! Wearing makeup and glasses can be intimidating but I swear it's a lot of fun when you figure out what works for you and the glasses that you wear and love! I'll be partnering with my favourite local Optical store Inspired By Rossland Optical to bring you some posts surrounding the topic of Makeup and Eyewear! I was able to go in and pick out a few pairs of glasses that I thought would match really well with some of the coral lipsticks that I have in my collection. So there will be another post coming out later this month with another glasses and lipstick pairing!
I picked out this stunning Philosopheyes frame (PHP-980 C002) to match my favourite sheer coral lipstick. I love that the sheerness of the lipstick helps to balance out the boldness of the frames.
The packaging of this lipstick is so fun! It has a paper like texture to it and I find that is fairly unique when it comes to lipstick bullets.
It has a minty feeling to it that just gives a quick tingle to the lips. I find that to be refreshing and I don't mind it since it is such a short sensation.
It is a very sheer and glossy formula. But it still manages to pack a punch when it comes to the colour. I do recommend exfoliating your lips before applying this lipstick. If you have any dry patches on your lips it can stick to those and it doesn't always look the greatest when it does that. I always seem to have that problem when it comes to sheer lipsticks so I'm definitely used to exfoliating before application if I want everything to look smooth and seamless!
This isn't extremely long wearing since it is a sheer formula. I find that it's super easy to apply throughout the day though so I don't necessarily mind that it fades quickly.
Overall I love wearing this lipstick shade! It's the perfect shade to pair with a bold, black frame! Are you someone who loves switching up your glasses if you wear them? I love having a few different pairs and deciding which look I'm going for every morning!


text:
Q: Fetching the SFDC data on to sencha using equi join I am using the Work orders and cases from SFDC. In my js, i have established a relation that work order "belongs to" cases. Then in my PHP file i am making the query using equi join case id as the foreign key in work orders to retrive the case number. But on sfdc side , in work orders the look up has been mentioned on the case id and not on the case number. the query fails and i am not able to fetch the required data.
This is my query. And the Work Looks ups to case i.e One case can have multiple work orders.
SELECT Id, FS_Account_Name__c, FS_Asset_Number__c, FS_Case_Number__c,
       Name, FS_Comment__c,FS_Defect_Type__c,FS_Description__c, WO_Status__c,CaseNumber
from FS_Work_Order__c,Case
where FS_Work_Order__c.FS_Case_Number__c = Case.Id
LIMIT 100

A: SOQL does not support joins with multiple objects in the FROM clause. Instead, you'll want to traverse the object relationships from either the parent or the child. Take a look at the API Doc about Relationship Queries. I'm making a guess of your relationship names (they can be found in the describeSObject() results), but the queries should look something like this:
Parent Relationship Query
SELECT FS_Case_Number__r.Id,
       FS_Case_Number__r.FS_Account_Name__c,
       FS_Case_Number__r.FS_Asset_Number__c,
       FS_Case_Number__r.FS_Case_Number__c,
       FS_Case_Number__r.Name,
       FS_Case_Number__r.FS_Comment__c,
       FS_Case_Number__r.FS_Defect_Type__c,
       FS_Case_Number__r.FS_Description__c,
       WO_Status__c,
       FS_Case_Number__r.CaseNumber
FROM FS_Work_Order__c

Child Relationship Query
SELECT Id,
       FS_Account_Name__c,
       FS_Asset_Number__c,
       FS_Case_Number__c,
       Name,
       FS_Comment__c,
       FS_Defect_Type__c,
       FS_Description__c,
       (SELECT WO_Status__c FROM FS_Work_Order__r),
       CaseNumber
FROM Case

redpajama_set_name: RedPajamaStackExchange


text:
\section{Introduction}

\qquad Recently, Fibonacci and Lucas numbers have investigated very largely
and authors tried to developed and give some directions to mathematical
calculations using these type of special numbers [2,9,16,17]. One of these
directions goes through to the \textit{Tribonacci }and the \textit
Tribonacci-Lucas numbers. }In fact Tribonacci numbers have been firstly
defined by M. Feinberg in 1963 and then some important properties over this
numbers have been created by [5,8,10,13,19]. On the other hand,
Tribonacci-Lucas numbers have been introduced and investigated by author in
[4]. In addition, there exists another mathematical term, namely to be
\textit{incomplete}, on Fibonacci, Lucas and Tribonacci numbers. As a brief
background, the incomplete Fibonacci, Lucas and Tribonacci numbers were
introduced by authours [6,11,12,18], and further the generating functions of
these numbers were presented by authours. Moreover, in [14,15], it is
defined and examined recurrence relations of the incomplete Fibonacci, Lucas
$p$-numbers and $p$-polynomials . We may also refer [3,18] for further
studies about to have incompleteness of special numbers.

For $n\geq 2$, it is known that while the Tribonacci sequence $\left\{
T_{n}\right\} _{n\in
\mathbb{N}
}$ is defined by
\begin{equation}
T_{n+1}=T_{n}+T_{n-1}+T_{n-2}\ \ \ (T_{0}=0,\ T_{1}=T_{2}=1),  \label{1.1}
\end{equation
the Tribonacci-Lucas sequence $\left\{ K_{n}\right\} _{n\in
\mathbb{N}
}$ is defined by

\begin{equation}
K_{n+1}=K_{n}+K_{n-1}+K_{n-2}\ \ (K_{0}=3,\ K_{1}=1,\ K_{2}=3).  \label{1.2}
\end{equation
There is also well known that each of the Tribonacci and Tribonaccci-Lucas
numbers is actually a linear combination of $\alpha ^{n}$, $\beta ^{n}$ and
\gamma ^{n}.$ In other words,
\begin{equation}
\left.
\begin{array}{c}
T_{n}=\frac{\alpha ^{n+1}}{(\alpha -\beta )(\alpha -\gamma )}+\frac{\beta
^{n+1}}{(\beta -\alpha )(\beta -\gamma )}+\frac{\gamma ^{n+1}}{(\gamma
-\alpha )(\gamma -\beta )} \\
\text{and} \\
K_{n}=\alpha ^{n}+\beta ^{n}+\gamma ^{n}
\end{array
\right\}  \label{1.3}
\end{equation
where $\alpha ,\beta $ and $\gamma $ are roots of the characteristic
equations of (\ref{1.1}) and (\ref{1.2})\ such that
\begin{eqnarray*}
\alpha &=&\frac{1+\sqrt[3]{19+3\sqrt{33}}+\sqrt[3]{19-3\sqrt{33}}}{3},\
\beta =\frac{1+w\sqrt[3]{19+3\sqrt{33}}+w^{2}\sqrt[3]{19-3\sqrt{33}}}{3}, \\
\ \gamma &=&\frac{1+w^{2}\sqrt[3]{19+3\sqrt{33}}+w\sqrt[3]{19-3\sqrt{33}}}{3
,\ \ w=\frac{-1+i\sqrt{3}}{2}.
\end{eqnarray*
Meanwhile we note that equations in (\ref{1.3}) are called the Binet
formulas for Tribonacci and Tribonacci-Lucas numbers, respectively.

Moreover, authors defined a large class \ of polynomials by Fibonacci and
Tribonacci numbers [1, 7]. Such polynomials are called Fibonacci polynomials
and Tribonacci polynomials \cite{9}, respectively. In 1973, Hoggatt and
Bicknell \cite{7} introduced Tribonacci polynomials. The Tribonacci
polynomials $T_{n}\left( x\right) $ are defined by the recurrence relatio
\begin{equation*}
T_{n+3}\left( x\right) =x^{2}T_{n+2}\left( x\right) +xT_{n+1}\left( x\right)
+T_{n}\ \left( x\right) ,
\end{equation*
where $T_{0}\left( x\right) =0,\ T_{1}\left( x\right) =1,\ T_{2}\left(
x\right) =x^{2}.$

On the other hand, in [12], incomplete Tribonacci polynomials are defined b
\begin{equation}
T_{n}^{\left( s\right) }\left( x\right)
=\sum\limits_{i=0}^{s}\sum\limits_{j=0}^{i}\dbinom{i}{j}\dbinom{n-i-j-1}{i
x^{2n-3\left( i+j\right) -2},  \label{1.4}
\end{equation
where $0\leq s\leq \left\lfloor \frac{n-1}{2}\right\rfloor $. In here, for
x=1$, it is obtained incomplete Tribonacci numbers. The recurrence relation
of these polynomials i
\begin{equation}
\begin{tabular}{ll}
$\ T_{n+3}^{\left( s\right) }\left( x\right) =$ & $x^{2}T_{n+2}^{\left(
s\right) }\left( x\right) +xT_{n+1}^{\left( s\right) }\left( x\right)
+T_{n}^{\left( s\right) }\left( x\right) -x\sum\limits_{j=0}^{s}\dbinom{s}{j
\dbinom{n-s-j}{s}x^{2n-3\left( s+j\right) }\ \ $ \\
& $\ -\sum\limits_{j=0}^{s}\dbinom{s}{j}\dbinom{n-s-j-1}{s}x^{2n-3\left(
s+j\right) -2}
\end{tabular}
\label{1.5}
\end{equation
$\ \ $

Also the generating function of incomplete Tribonacci numbers i
\begin{equation}
Q_{s}\left( z\right) =\frac{T_{2s+1}+z\left( T_{2s+2}-T_{2s+1}\right)
+z^{2}\left( T_{2s+3}-T_{2s+2}-T_{2s+1}-2\right) -g\left( z\right) }
1-z-z^{2}-z^{3}},  \label{1.6}
\end{equation
where $g\left( z\right) =\left( z^{2}+z^{3}\right) \frac{\left( 1+z\right)
^{s}}{\left( 1-z\right) ^{s+1}}$ and $T_{n}$ is $n$-th Tribonacci number.

In the light of the above paragraph, the main goal of this paper is to
improve the Tribonacci-Lucas numbers\ with a different viewpoint . In order
to do that we first define Tribonacci-Lucas polynomials and then by
presenting Tribonacci-Lucas numbers and polynomials as a binomial sum\textit
, }we define\textit{\ }the incomplete Tribonacci-Lucas numbers and
polynomials.

After that we find the generating function of incomplete Tribonacci
polynomials which is given as the open problem in [12]. Also, we obtain some
properties and generating functions of incomplete Tribonacci-Lucas numbers
and polynomials.

\section{Tribonacci-Lucas polynomials and pascal-like triangle}

\qquad In the following table, we give the pascal-like triangle of
Tribonacci-Lucas numbers and each element of this table is defined in
similar way as in the tribonacci triangle.

\begin{eqnarray*}
&
\begin{tabular}[t]{l|lllllll}
$n\backslash i$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $\cdots $ \\ \hline
$0$ & $3$ &  &  &  &  &  &  \\
$1$ & $1$ & $2$ &  &  &  &  &  \\
$2$ & $1$ & $6$ & $2$ &  &  &  &  \\
$3$ & $1$ & $8$ & $10$ & $2$ &  &  &  \\
$4$ & $1$ & $10$ & $24$ & $14$ & $2$ &  &  \\
$5$ & $1$ & $12$ & $42$ & $48$ & $18$ & $2$ &  \\
$\vdots $ &  &  &  & $\vdots $ &  &  &
\end{tabular}
\\
&&\text{\textit{Table1}. Tribonacci-Lucas triangle}
\end{eqnarray*}

\qquad Let $B\left( n,i\right) $ be the element in the $n$-th row and $i$-th
column of the Tribonacci-Lucas triangle. By using the triangle, we have

\begin{equation}
B\left( n+1,i\right) =B\left( n,i\right) +B\left( n,i-1\right) +B\left(
n-1,i-1\right) ,  \label{2.1}
\end{equation
where $B\left( n,0\right) =1,\ B\left( n,n\right) =2$ for $n\in
\mathbb{Z}
^{+}.$

By using the Table 1, we have the Tribonacci-Lucas numbers as binomial sum
\begin{equation*}
K_{n}=\sum\limits_{i=0}^{\left\lfloor \frac{n}{2}\right\rfloor }B\left(
n-i,i\right) .
\end{equation*
In here, the sum of elements on the rising diagonal lines in the
Tribonacci-Lucas triangle is the Tribonacci-Lucas number $K_{n}$.
Furthermore, we write
\begin{equation}
K_{n}=\sum\limits_{i=0}^{\left\lfloor \frac{n}{2}\right\rfloor
}\sum\limits_{j=0}^{i}\frac{n}{n-i-j}\dbinom{i}{j}\dbinom{n-i-j}{i},\ \ \ (\
n>i+j)  \label{2.2}
\end{equation
since these coefficients hold the relation
\begin{equation*}
\left\{
\begin{array}{c}
B\left( n,i\right) =\sum\limits_{j=0}^{i}\frac{n+i}{n-j}\binom{i}{j}\binom
n-j}{i},\ \ \ (n>i) \\
B(n,i)=2,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (n=i
\end{array
\right. .
\end{equation*
\qquad

Now, we introduce the Tribonacci-Lucas polynomia
\begin{equation*}
K_{n+3}\left( x\right) =x^{2}K_{n+2}\left( x\right) +xK_{n+1}\left( x\right)
+K_{n}\ \left( x\right) ,\ \
\end{equation*
where $K_{0}\left( x\right) =3,\ K_{1}\left( x\right) =x^{2},\ K_{2}\left(
x\right) =x^{4}+2x.$ Note that $K_{n}\left( 1\right) =K_{n},$ $n\in
\mathbb{N}
$. \ It is given a few Tribonacci-Lucas polynomials in the following
\begin{equation*}
\begin{tabular}{ll}
$K_{0}\left( x\right) =3,$ & $K_{4}\left( x\right) =x^{8}+4x^{5}+6x^{2},$ \\
$K_{1}\left( x\right) =x^{2},$ & $K_{5}\left( x\right)
=x^{10}+5x^{7}+10x^{4}+5x,$ \\
$K_{2}\left( x\right) =x^{4}+2x,$ & $K_{6}\left( x\right)
=x^{12}+6x^{9}+15x^{6}+14x^{3}+3,$ \\
$K_{3}\left( x\right) =x^{6}+3x^{3}+3,$ & $K_{7}\left( x\right)
=x^{14}+7x^{11}+21x^{8}+28x^{5}+14x^{2}.
\end{tabular
\end{equation*
\qquad \qquad \qquad\ \ \qquad

In similarly with the Tribonacci-Lucas triangle, we define the
Tribonacci-Lucas polynomials triangle
\begin{eqnarray*}
&
\begin{tabular}[t]{l|lllllll}
$n\backslash i$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $\cdots $ \\ \hline
$0$ & $3$ &  &  &  &  &  &  \\
$1$ & $x^{2}$ & $2x$ &  &  &  &  &  \\
$2$ & $x^{4}$ & $3x^{3}+3$ & $2x^{2}$ &  &  &  &  \\
$3$ & $x^{6}$ & $4x^{5}+4x^{2}$ & $5x^{4}+5x$ & $2x^{3}$ &  &  &  \\
$4$ & $x^{8}$ & $5x^{7}+5x^{4}$ & $9x^{6}+12x^{3}+3$ & $7x^{5}+7x^{2}$ &
2x^{4}$ &  &  \\
$5$ & $x^{10}$ & $6x^{9}+6x^{6}$ & $14x^{8}+21x^{5}+7x^{2}$ &
16x^{7}+24x^{4}+8x$ & $9x^{6}+9x^{3}$ & $2x^{5}$ &  \\
$\vdots $ &  &  &  & $\vdots $ &  &  &
\end{tabular}
\\
&&\text{\textit{Table2}. Tribonacci-Lucas polynomials triangle}
\end{eqnarray*
\qquad

Let $B\left( n,i\right) \left( x\right) $ be the element in the $n$-th row
and $i$-th column of the Tribonacci-Lucas polynomials triangle. By using the
triangle, we have

\begin{equation}
B\left( n+1,i\right) \left( x\right) =x^{2}B\left( n,i\right) \left(
x\right) +xB\left( n,i-1\right) \left( x\right) +B\left( n-1,i-1\right)
\left( x\right) ,  \label{2.3}
\end{equation
where $B\left( n,0\right) \left( x\right) =x^{2n},\ B\left( n,n\right)
\left( x\right) =2x^{n}$ for $n\in
\mathbb{Z}
^{+}.$

By using the Table 2, we have
\begin{equation*}
K_{n}\left( x\right) =\sum\limits_{i=0}^{\left\lfloor \frac{n}{2
\right\rfloor }B\left( n-i,i\right) \left( x\right) .
\end{equation*
In here, the sum of elements on the rising diagonal lines in the Table 2 is
the Tribonacci-Lucas polynomials $K_{n}\left( x\right) .$Furthermore, we
write
\begin{equation}
K_{n}\left( x\right) =\sum\limits_{i=0}^{\left\lfloor \frac{n}{2
\right\rfloor }\sum\limits_{j=0}^{i}\frac{n}{n-i-j}\dbinom{i}{j}\dbinom{n-i-
}{i}x^{2n-3i-3j},\ \ \ \ \ (n>i+j)  \label{2.4}
\end{equation
since these coefficients satisfy the relatio
\begin{equation*}
\left\{
\begin{array}{c}
B\left( n,i\right) \left( x\right) =\sum\limits_{j=0}^{i}\frac{n+i}{n-j
\binom{i}{j}\binom{n-j}{i}x^{2n-i-3j},\ \ \ \ \ \ (n>i) \\
B(n,i)(x)=2x^{n},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
(n=i
\end{array
\right. .
\end{equation*
\qquad

Thus, by considering equations (\ref{2.2}) and (\ref{2.4}), we introduce the
incomplete Tribonacci-Lucas numbers and incomplete Tribonacci-Lucas
polynomials.

Now, we get new recurrence relations, some properties and generating
functions of incomplete Tribonacci-Lucas numbers and polynomials.

\section{The incomplete Tribonacci-Lucas polynomials and Tribonacci-Lucas
numbers}

\begin{definition}
The incomplete Tribonacci-Lucas polynomials $K_{n}^{\left( s\right) }\left(
x\right) $ are defined b
\begin{eqnarray}
K_{n}^{\left( s\right) }\left( x\right) &=&\sum\limits_{i=0}^{s}B\left(
n-i,i\right) \left( x\right)  \label{3.1} \\
&=&\sum\limits_{i=0}^{s}\sum\limits_{j=0}^{i}\frac{n}{n-i-j}\dbinom{i}{j
\dbinom{n-i-j}{i}x^{2n-3\left( i+j\right) },  \notag
\end{eqnarray
where $0\leq s\leq \left\lfloor \frac{n}{2}\right\rfloor $ for $n\in
\mathbb{Z}
^{+}.$
\end{definition}

In Definition 1, for $x=1$, we define the incomplete Tribonacci-Lucas
numbers, that is, $K_{n}^{\left( s\right) }\left( 1\right) =K_{n}\left(
s\right) .$

To reveal the importance of this subject, we can express the relationships
as in the following:

\begin{itemize}
\item $K_{n}^{\left( \left\lfloor \frac{n}{2}\right\rfloor \right) }\left(
x\right) =K_{n}\left( x\right) $ $\ $(the relationship between incomplete
Tribonacci-Lucas polynomials and Tribonacci-Lucas polynomials),

\item $K_{n}\left( \left\lfloor \frac{n}{2}\right\rfloor \right) =K_{n}$
(the relationship between incomplete Tribonacci-Lucas numbers and
Tribonacci-Lucas numbers).
\end{itemize}

From Definition 1, we have a few incomplete Tribonacci-Lucas polynomials as

\begin{eqnarray*}
&
\begin{tabular}[t]{l|lllll}
$n\backslash s$ & $0$ & $1$ & $2$ & $3$ & $\cdots $ \\ \hline
$1$ & $x^{2}$ &  &  &  &  \\
$2$ & $x^{4}$ & $x^{4}+2x$ &  &  &  \\
$3$ & $x^{6}$ & $x^{6}+3x^{3}+3$ &  &  &  \\
$4$ & $x^{8}$ & $x^{8}+4x^{5}+4x^{2}$ & $x^{8}+4x^{5}+6x^{2}$ &  &  \\
$5$ & $x^{10}$ & $x^{10}+5x^{7}+5x^{4}$ & $x^{10}+5x^{7}+10x^{4}+5x$ &  &
\\
$6$ & $x^{12}$ & $x^{12}+6x^{9}+6x^{6}$ & $x^{12}+6x^{9}+15x^{6}+12x^{3}+3$
& $x^{12}+6x^{9}+15x^{6}+14x^{3}+3$ &  \\
$\vdots $ &  &  & $\vdots $ &  &
\end{tabular}
\\
&&\text{\textit{Table3}. Incomplete Tribonacci-Lucas polynomials }
\end{eqnarray*}

By taking account of Table 3, we can write

\begin{equation}
K_{n}^{\left( 0\right) }\left( x\right) =x^{2n},  \label{3.3}
\end{equation
\begin{equation}
K_{n}^{\left( 1\right) }\left( x\right) =x^{2n}+nx^{2n-3}+nx^{2n-6},\ \
n\geq 3  \label{3.4}
\end{equation}

\begin{equation}
K_{n}^{\left( \left\lfloor \frac{n}{2}\right\rfloor \right) }\left( x\right)
=K_{n}\left( x\right) ,  \label{3.5}
\end{equation}

\begin{equation}
K_{n}^{\left( \left\lfloor \frac{n-2}{2}\right\rfloor \right) }\left(
x\right) =\left\{
\begin{array}{c}
\ \ \ K_{n}\left( x\right) -2x^{\frac{n}{2}},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ n\geq 2,\ even \\
K_{n}\left( x\right) -\left( nx^{\frac{n+3}{2}}+nx^{\frac{n-3}{2}}\right) ,\
\ \ \ \ \ \ \ n\geq 2,\ od
\end{array
.\right.  \label{3.6}
\end{equation}

\begin{proposition}
For $\ 0\leq s\leq \left\lfloor \frac{n}{2}\right\rfloor $ and $n\in
\mathbb{Z}
^{+}$, we have the following recurrence relations;
\end{proposition}

\begin{enumerate}
\item[i)] \textit{The homogeneous recurrence relation of the incomplete
Tribonacci-Lucas polynomials }$K_{n}^{\left( s\right) }\left( x\right)
\textit{\ is
\begin{equation}
K_{n+3}^{\left( s+1\right) }\left( x\right) =x^{2}K_{n+2}^{\left( s+1\right)
}\left( x\right) +xK_{n+1}^{\left( s\right) }\left( x\right) +K_{n}^{\left(
s\right) }\left( x\right) .  \label{3.7}
\end{equation}

\item[ii)] \textit{The non-homogeneous recurrence relation of the incomplete
Tribonacci-Lucas polynomials }$K_{n}^{\left( s\right) }\left( x\right)
\textit{\ is
\begin{eqnarray}
K_{n+3}^{\left( s\right) }\left( x\right) &=&x^{2}K_{n+2}^{\left( s\right)
}\left( x\right) +xK_{n+1}^{\left( s\right) }\left( x\right) +K_{n}^{\left(
s\right) }\left( x\right)  \label{3.8} \\
&&-xB\left( n+1-s,s\right) \left( x\right) -B\left( n-s,s\right) \left(
x\right) .  \notag
\end{eqnarray}
\end{enumerate}

\begin{proof}

\begin{description}
\item[i)] From Definition 1, let us label $x^{2}K_{n+2}^{\left( s+1\right)
}\left( x\right) +xK_{n+1}^{\left( s\right) }\left( x\right) +K_{n}^{\left(
s\right) }\left( x\right) $ by $RHS$. Actually we writ
\begin{eqnarray*}
RHS &=&x^{2}\sum\limits_{i=0}^{s+1}B\left( n+2-i,i\right) \left( x\right)
+x\sum\limits_{i=0}^{s}B\left( n+1-i,i\right) \left( x\right)
+\sum\limits_{i=0}^{s}B\left( n-i,i\right) \left( x\right) \\
&=&x^{2}\sum\limits_{i=0}^{s+1}B\left( n+2-i,i\right) \left( x\right)
+x\sum\limits_{i=1}^{s+1}B\left( n+2-i,i-1\right) \left( x\right)
+\sum\limits_{i=1}^{s+1}B\left( n+1-i,i-1\right) \left( x\right) \\
&=&\sum\limits_{i=0}^{s+1}\left( x^{2}B\left( n+2-i,i\right) \left( x\right)
+xB\left( n+2-i,i-1\right) \left( x\right) +B\left( n+1-i,i-1\right) \left(
x\right) \right) \\
&&-xB\left( n+2,-1\right) \left( x\right) -B\left( n+1,-1\right) \left(
x\right) .
\end{eqnarray*
Then, by considering $\dbinom{n}{-1}=0$ and the equation (\ref{2.3}), we
finally hav
\begin{equation*}
RHS=\sum\limits_{i=0}^{s+1}B\left( n+3-i,i\right) \left( x\right)
=K_{n+3}^{\left( s+1\right) }\left( x\right)
\end{equation*
as required.

\item[ii)] By considering the equations (\ref{2.3}), (\ref{3.7}) and
Definition 1, we hav
\begin{eqnarray*}
\sum\limits_{i=0}^{s+1}B\left( n+3-i,i\right) \left( x\right)
&=&x^{2}\sum\limits_{i=0}^{s+1}B\left( n+2-i,i\right) \left( x\right)
+x\sum\limits_{i=0}^{s}B\left( n+1-i,i\right) \left( x\right)
+\sum\limits_{i=0}^{s}B\left( n-i,i\right) \left( x\right) \\
\sum\limits_{i=0}^{s}B\left( n+3-i,i\right) \left( x\right)
&=&x^{2}\sum\limits_{i=0}^{s}B\left( n+2-i,i\right) \left( x\right)
+x\sum\limits_{i=0}^{s}B\left( n+1-i,i\right) \left( x\right)
+\sum\limits_{i=0}^{s}B\left( n-i,i\right) \left( x\right) \\
&&-B\left( n+2-s,s+1\right) +x^{2}B\left( n+1-s,s+1\right) \\
K_{n+3}^{\left( s\right) }\left( x\right) &=&x^{2}K_{n+2}^{\left( s\right)
}\left( x\right) +xK_{n+1}^{\left( s\right) }\left( x\right) +K_{n}^{\left(
s\right) }\left( x\right) -xB\left( n+1-s,s\right) -B\left( n-s,s\right) .
\end{eqnarray*}
\end{description}
\end{proof}

By using Table 3, for $x=1$, we have incomplete Tribonacci-Lucas numbers in
the following Table 4
\begin{eqnarray*}
&
\begin{tabular}[t]{l|lllll}
$n\backslash s$ & $0$ & $1$ & $2$ & $3$ & $\cdots $ \\ \hline
$1$ & $1$ &  &  &  &  \\
$2$ & $1$ & $3$ &  &  &  \\
$3$ & $1$ & $7$ &  &  &  \\
$4$ & $1$ & $9$ & $11$ &  &  \\
$5$ & $1$ & $11$ & $21$ &  &  \\
$6$ & $1$ & $13$ & $37$ & $39$ &  \\
$\vdots $ &  &  & $\vdots $ &  &
\end{tabular}
\\
&&\text{\textit{Table4}. Incomplete Tribonacci-Lucas numbers }
\end{eqnarray*}

\begin{corollary}
For $\ 0\leq s\leq \left\lfloor \frac{n}{2}\right\rfloor $ and $n\in
\mathbb{Z}
^{+},$ we have the following recurrence relations;
\end{corollary}

\begin{enumerate}
\item[i)] \textit{The homogeneous recurrence relation of the incomplete
Tribonacci-Lucas numbers }$K_{n}\left( s\right) $\textit{\ is
\begin{equation}
K_{n+3}\left( s+1\right) =K_{n+2}\left( s+1\right) +K_{n+1}\left( s\right)
+K_{n}\left( s\right) .  \label{3.9}
\end{equation}

\item[ii)] \textit{The non-homogeneous recurrence relation of the incomplete
Tribonacci-Lucas numbers }$K_{n}\left( s\right) $\textit{\ is
\begin{eqnarray}
K_{n+3}\left( s\right) &=&K_{n+2}\left( s\right) +K_{n+1}\left( s\right)
+K_{n}\left( s\right)  \label{3.10} \\
&&-B\left( n+1-s,s\right) -B\left( n-s,s\right) .  \notag
\end{eqnarray}
\end{enumerate}

\begin{proposition}
The relation between of incomplete Tribonacci polynomials $T_{n}^{\left(
s\right) }\left( x\right) $ and incomplete Tribonacci-Lucas polynomials
K_{n}^{\left( s\right) }\left( x\right) $ i
\begin{equation*}
K_{n}^{\left( s\right) }\left( x\right) =T_{n+1}^{\left( s\right) }\left(
x\right) +xT_{n-1}^{\left( s-1\right) }\left( x\right) +2T_{n-2}^{\left(
s-1\right) }\left( x\right) ,
\end{equation*
where $1\leq s\leq \left\lfloor \frac{n-1}{2}\right\rfloor $ and $n>2.$
\end{proposition}

\begin{proof}
Proof of its can easily do by using Definition 1 and the equation (\ref{1.4
).
\end{proof}

\begin{corollary}
The relation between of incomplete Tribonacci numbers $T_{n}\left( s\right) $
and incomplete Tribonacci-Lucas numbers $K_{n}\left( s\right) $ i
\begin{equation*}
K_{n}\left( s\right) =T_{n+1}\left( s\right) +T_{n-1}\left( s-1\right)
+2T_{n-2}\left( s-1\right) ,
\end{equation*
where $1\leq s\leq \left\lfloor \frac{n-1}{2}\right\rfloor $ and $n>2.$
\end{corollary}

\begin{theorem}
For $n,\ h\geq 1$ and $0\leq s\leq \left\lfloor \frac{n}{2}\right\rfloor ,$
the sum of incomplete Tribonacci-Lucas numbers i
\begin{equation}
\sum_{i=0}^{h-1}K_{n+i}\left( s\right) =\frac{1}{2}\left( K_{n+h+2}\left(
s+1\right) -K_{n+2}\left( s+1\right) +K_{n}\left( s\right) -K_{n+h}\left(
s\right) \right) .  \label{3.11}
\end{equation}
\end{theorem}

\begin{proof}
Let us use the principle of mathematical induction on $h$ to prove (\re
{3.11}). While, for $h=1$, it is easy to see tha
\begin{equation*}
K_{n}\left( s\right) =\frac{1}{2}\left( K_{n+3}\left( s+1\right)
-K_{n+2}\left( s+1\right) +K_{n}\left( s\right) -K_{n+1}\left( s\right)
\right) .
\end{equation*
As the usual next step of inductions, let us assume that it is true for all
positive integers $h.$ That is
\begin{equation*}
\sum_{i=0}^{h-1}K_{n+i}\left( s\right) =\frac{1}{2}\left( K_{n+h+2}\left(
s+1\right) -K_{n+2}\left( s+1\right) +K_{n}\left( s\right) -K_{n+h}\left(
s\right) \right) .
\end{equation*
Therefore, we have to show that it is true for $h+1.$ In other words$,$ we
need to chec
\begin{equation*}
\sum_{i=0}^{h}K_{n+i}\left( s\right) =\frac{1}{2}\left( K_{n+h+3}\left(
s+1\right) -K_{n+2}\left( s+1\right) +K_{n}\left( s\right) -K_{n+h+1}\left(
s\right) \right) .
\end{equation*
Hence, we can writ
\begin{eqnarray*}
\sum_{i=0}^{h}K_{n+i}\left( s\right) &=&\sum_{i=0}^{h-1}K_{n+i}\left(
s\right) +K_{n+h}\left( s\right) \\
&=&\frac{1}{2}\left( K_{n+h+2}\left( s+1\right) -K_{n+2}\left( s+1\right)
+K_{n}\left( s\right) -K_{n+h}\left( s\right) \right) +K_{n+h}\left( s\right)
\\
&=&\frac{1}{2}\left( K_{n+h+2}\left( s+1\right) -K_{n+2}\left( s+1\right)
+K_{n}\left( s\right) +K_{n+h}\left( s\right) \right) \\
&=&\frac{1}{2}\left( K_{n+h+3}\left( s+1\right) -K_{n+2}\left( s+1\right)
+K_{n}\left( s\right) -K_{n+h+1}\left( s\right) \right) .
\end{eqnarray*}
\end{proof}

The following proposition give the sum of incomplete Tribonacci-Lucas
poynomials, that is, sum of the $n$-th row of the Table 3.

\begin{proposition}
For $l=\left\lfloor \frac{n}{2}\right\rfloor ,$ we have the equalit
\begin{equation}
\sum_{s=0}^{l}K_{n}^{\left( s\right) }\left( x\right) =\left( l+1\right)
K_{n}\left( x\right) -\sum\limits_{i=0}^{l}\sum\limits_{j=0}^{i}\frac{in}
n-i-j}\dbinom{i}{j}\dbinom{n-i-j}{i}x^{2n-3\left( i+j\right) }.  \label{3.12}
\end{equation}
\end{proposition}

\begin{proof}
From Definition 1, we hav
\begin{equation*}
K_{n}^{\left( s\right) }\left( x\right) =\sum\limits_{i=0}^{s}B\left(
n-i,i\right) \left( x\right) .
\end{equation*
Then, we can writ
\begin{eqnarray*}
\sum_{s=0}^{l}K_{n}^{\left( s\right) }\left( x\right)
&=&\sum\limits_{s=0}^{l}\sum\limits_{i=0}^{s}B\left( n-i,i\right) \left(
x\right) \\
&=&\left( l+1\right) B\left( n,0\right) \left( x\right) +lB\left(
n-1,1\right) \left( x\right) +\cdots +B\left( n-l,l\right) \left( x\right) \\
&=&\sum\limits_{i=0}^{l}\left( l+1-i\right) B\left( n-i,i\right) \left(
x\right) \\
&=&\sum\limits_{i=0}^{l}\left( l+1\right) B\left( n-i,i\right) \left(
x\right) -\sum\limits_{i=0}^{l}iB\left( n-i,i\right) \left( x\right) \\
&=&\left( l+1\right) K_{n}^{\left( l\right) }\left( x\right)
-\sum\limits_{i=0}^{l}\sum\limits_{j=0}^{i}\frac{in}{n-i-j}\dbinom{i}{j
\dbinom{n-i-j}{i}x^{2n-3\left( i+j\right) }.
\end{eqnarray*
Since $l$ is $\left\lfloor \frac{n}{2}\right\rfloor ,$ we obtai
\begin{equation*}
\sum_{s=0}^{l}K_{n}^{\left( s\right) }\left( x\right) =\left( l+1\right)
K_{n}\left( x\right) -\sum\limits_{i=0}^{l}\sum\limits_{j=0}^{i}\frac{in}
n-i-j}\dbinom{i}{j}\dbinom{n-i-j}{i}x^{2n-3\left( i+j\right) }.
\end{equation*}
\end{proof}

The following corollary shows the sum of the $n$-th row of the Table 4. It
is obtained from (\ref{3.12}) with $x=1.$

\begin{corollary}
For $l=\left\lfloor \frac{n}{2}\right\rfloor ,$ we have the following
equality
\begin{equation}
\sum_{s=0}^{l}K_{n}\left( s\right) =\left( l+1\right)
K_{n}-\sum\limits_{i=0}^{l}\sum\limits_{j=0}^{i}\frac{in}{n-i-j}\dbinom{i}{j
\dbinom{n-i-j}{i}.  \label{3.13}
\end{equation}
\end{corollary}

\section{Generating functions of the incomplete Tribonacci and
Tribonacci-Lucas polynomials and numbers}

\begin{lemma}
Let $\left\{ S_{n}\right\} _{n=0}^{\infty }$ be a complex sequence
satisfying the non-homogeneous third-order recurrence relation. Then we hav
\begin{equation*}
S_{n}=aS_{n-1}+bS_{n-2}+cS_{n-3}+r_{n},
\end{equation*
where $a,b,c\in
\mathbb{C}
,~n\geq 3~$and $r_{n}
\mathbb{N}
\rightarrow
\mathbb{C}
$ is a sequence. Hence the generating function $U\left( x\right) $ of $S_{n}$
is given b
\begin{equation}
U\left( x\right) =\frac{S_{0}-r_{0}+x\left( S_{1}-aS_{0}-r_{1}\right)
+x^{2}\left( S_{2}-aS_{1}-bS_{0}-r_{2}\right) +G\left( x\right) }
1-ax-bx^{2}-cx^{3}},  \label{4.1}
\end{equation
where $G\left( x\right) $ denotes the generating function of $r_{n}$.
\end{lemma}

\begin{proof}
Let $U(x)$ and $G(x)$ be two generating functions for complex sequences
S_{n}$ and $r_{n}$, respectively, wher
\begin{equation}
U(x)=S_{0}+S_{1}x+S_{2}x^{2}+S_{3}x^{3}+\ldots +S_{n}x^{n}+\ldots ,
\label{4.2}
\end{equation
\begin{equation}
G\left( x\right) =r_{0}+r_{1}x+r_{2}x^{2}+r_{3}x^{3}+\ldots
+r_{n}x^{n}+\ldots .  \label{4.3}
\end{equation
If $U(x)$ given in (\ref{4.2}) multiply with $ax,$ $bx^{2}$ and $cx^{3},$
respectively, then we ge
\begin{equation}
\left.
\begin{array}{c}
axU\left( x\right) =aS_{0}x+aS_{1}x^{2}+aS_{2}x^{3}+aS_{3}x^{4}+\ldots
+aS_{n}x^{n+1}+\ldots \\
bx^{2}U\left( x\right)
=bS_{0}x^{2}+bS_{1}x^{3}+bS_{2}x^{4}+bS_{3}x^{5}+\ldots +bS_{n}x^{n+2}+\ldots
\\
cx^{3}U\left( x\right)
=cS_{0}x^{3}+cS_{1}x^{4}+cS_{2}x^{5}+cS_{3}x^{6}+\ldots +cS_{n}x^{n+3}+\ldot
\end{array
\right\} .  \label{4.4}
\end{equation
Consequently, by subtracting (\ref{4.3}) and (\ref{4.4}) from (\ref{4.2}),
it is obtained the equatio
\begin{equation*}
U\left( x\right) =\frac{S_{0}-r_{0}+x\left( S_{1}-aS_{0}-r_{1}\right)
+x^{2}\left( S_{2}-aS_{1}-bS_{0}-r_{2}\right) +G\left( x\right) }
1-ax-bx^{2}-cx^{3}}
\end{equation*
which completes the proof of the Lemma.\medskip
\end{proof}

Now, we examine the problem which is given for incomplete Tribonacci
polynomials in [15] .

\begin{theorem}
The generating function of the incomplete Tribonacci polynomials
T_{n}^{\left( s\right) }\left( x\right) $ is given b
\begin{equation*}
Q_{s}\left( x,z\right) =\sum_{i=0}^{\infty }T_{i}^{\left( s\right) }\left(
x\right) z^{i}=z^{2s+1}U_{s}\left( x,z\right) ,
\end{equation*
where $U_{s}\left( x,z\right) =\frac{T_{2s+1}\left( x\right) +z\left(
T_{2s+2}\left( x\right) -x^{2}T_{2s+1}\left( x\right) \right) +z^{2}\left(
T_{2s}\left( x\right) -2x^{s+1}\right) -\left( xz^{2}+z^{3}\right) \frac
\left( x+z\right) ^{s}}{\left( 1-x^{2}z\right) ^{s+1}}}{\left(
1-x^{2}z-xz^{2}-z^{3}\right) }$.
\end{theorem}

\begin{proof}
Let $s$ be fixed positive integer. By using the equations (\ref{1.4}) and
\ref{1.5}), we hav
\begin{eqnarray*}
T_{n}^{\left( s\right) }\left( x\right) &=&0\ \ \ \ \ \ \left( 0\leq
n<2s+1\right) , \\
T_{2s+1}^{\left( s\right) }\left( x\right) &=&T_{2s+1}\left( x\right) , \\
T_{2s+2}^{\left( s\right) }\left( x\right) &=&T_{2s+2}\left( x\right) , \\
T_{2s+3}^{\left( s\right) }\left( x\right) &=&T_{2s+3}\left( x\right)
-x^{s+1},
\end{eqnarray*
and
\begin{eqnarray}
T_{n}^{\left( s\right) }\left( x\right) &=&x^{2}T_{n-1}^{\left( s\right)
}\left( x\right) +xT_{n-2}^{\left( s\right) }\left( x\right)
+T_{n-3}^{\left( s\right) }\left( x\right) -\sum\limits_{j=0}^{s}\dbinom{s}{
}\dbinom{n-3-s-j}{s}x^{2n-5-3\left( s+j\right) }  \label{4.5} \\
&&-\sum\limits_{j=0}^{s}\dbinom{s}{j}\dbinom{n-4-s-j}{s}x^{2n-8-3\left(
s+j\right) },  \notag
\end{eqnarray
where $\ n\geq 4+2s.$ Also, we replace $S_{0},S_{1},...,S_{n}$ by
T_{2s+1}^{\left( s\right) }\left( x\right) ,~T_{2s+2}^{\left( s\right)
}\left( x\right) ,...,T_{n+2s+1}^{\left( s\right) }\left( x\right) $,
respectively. Assume that $r_{0}=r_{1}=0,\ r_{2}=x^{s+1}$ and
\begin{equation*}
r_{n}=\sum\limits_{j=0}^{s}\dbinom{s}{j}\dbinom{n-2+s-j}{s
x^{2n-3+s-3j}+\sum\limits_{j=0}^{s}\dbinom{s}{j}\dbinom{n-3+s-j}{s
x^{2n-6+s-3j}.
\end{equation*
Furthermore, by considering [20, page 127]$,$ the generating function
G\left( x,z\right) $ of the $\left\{ r_{n}\right\} \ $is
\begin{equation*}
G\left( x,z\right) =\left( xz^{2}+z^{3}\right) \frac{\left( x+z\right) ^{s}}
\left( 1-x^{2}z\right) ^{s+1}}.
\end{equation*
Therefore, by using Lemma 9, the generating function $U_{s}\left( x,z\right)
$ of the sequence $S_{n}$ is
\begin{eqnarray*}
U_{s}\left( x,z\right) \left( 1-x^{2}z-xz^{2}-z^{3}\right) +G\left(
x,z\right) &=&T_{2s+1}\left( x\right) +z\left( T_{2s+2}\left( x\right)
-x^{2}T_{2s+1}\left( x\right) \right) \\
&&+z^{2}\left( T_{2s}\left( x\right) -2x^{s+1}\right) .
\end{eqnarray*
Eventually, we conclude that $Q_{s}\left( x,z\right) =z^{2s+1}U_{s}\left(
x,z\right) $ as required.
\end{proof}

In the above theorem, if we take $x=1$, it is obtained the generating
function of the Tribonacci numbers in [12].

\begin{corollary}
The generating function of the Tribonacci numbers $T_{n}\left( s\right) $ is
given b
\begin{equation*}
Q_{s}\left( z\right) =\sum_{i=0}^{\infty }T_{i}\left( s\right)
z^{i}=z^{2s+1}U_{s}\left( z\right) ,
\end{equation*
where $U_{s}\left( z\right) =\frac{T_{2s+1}+z\left( T_{2s+2}-T_{2s+1}\right)
+z^{2}\left( T_{2s}-2\right) -\left( z^{2}+z^{3}\right) \frac{\left(
1+z\right) ^{s}}{\left( 1-z\right) ^{s+1}}}{\left( 1-z-z^{2}-z^{3}\right) }$.
\end{corollary}

\begin{theorem}
The generating function of the incomplete Tribonacci-Lucas polynomials
K_{n}^{\left( s\right) }\left( x\right) $ is given b
\begin{equation*}
W_{s}\left( x,z\right) =\sum_{n=0}^{\infty }K_{n}^{\left( s\right) }\left(
x\right) z^{n}=z^{-1}Q_{s}\left( x,z\right) +\left( xz+2z^{2}\right)
Q_{s-1}\left( x,z\right) ,
\end{equation*
where $s>1$ and $Q_{s}\left( x,z\right) $ is the generating function of
incomplete Tribonacci polynomials$.$
\end{theorem}

\begin{proof}
Let $W_{s}\left( x,z\right) $ be generating function of the incomplete
Tribonacci-Lucas polynomials, that is $W_{s}\left( x,z\right)
=\sum_{n=0}^{\infty }K_{n}^{\left( s\right) }\left( x\right) z^{n}.$

By using Proposition 4, Theorem 10 and the property of sum , we definitely
hav
\begin{eqnarray*}
\sum_{n=0}^{\infty }K_{n}^{\left( s\right) }\left( x\right) z^{n}
&=&\sum_{n=0}^{\infty }\left( T_{n+1}^{\left( s\right) }\left( x\right)
+xT_{n-1}^{\left( s-1\right) }\left( x\right) +2T_{n-2}^{\left( s-1\right)
}\left( x\right) \right) z^{n} \\
&=&z^{-1}Q_{s}\left( x,z\right) +\left( xz+2z^{2}\right) Q_{s-1}\left(
x,z\right) .
\end{eqnarray*}
\end{proof}

For $x=1$ in Theorem 12, we can present the generating function of
incomplete Tribonacci-Lucas numbers.

\begin{corollary}
The generating function of the incomplete Tribonacci-Lucas numbers
K_{n}\left( s\right) $ is given b
\begin{equation*}
W_{s}\left( z\right) =\sum_{n=0}^{\infty }K_{n}\left( s\right)
z^{n}=z^{-1}Q_{s}\left( z\right) +\left( z+2z^{2}\right) Q_{s-1}\left(
z\right) ,
\end{equation*
where $s>1$ and $Q_{s}\left( z\right) $ is the generating function of
incomplete Tribonacci numbers$.$
\end{corollary}

redpajama_set_name: RedPajamaArXiv


text:
Home>Mathematics
Babylonian Mathematics
By: Dyana Hepburn
Dyana Hepburn's image for:
"Babylonian Mathematics"
Caption: Babylonian Numerals
Image by: mathlair
© CC BY http://mathlair.allfunandgames.ca/babylonian.php
As merchants plied the Mediterranean Sea they wrote down the cargo they carried in their boats. This was the beginning of the numerical system that eventually developed into the Roman, Indian, and European system we use today. It was not the brainchild of scholars but of practical businessmen attempting to keep track of their commodities. Clay tablets found show pictures of cows, denoted by horns, and sheep, long ovals. Each animal was drawn, but after awhile they learned to save time by putting a dot or a line above an animal's head to denote a quantity of more than one.
They inscribed cuneiform slashes and semi-circles with a reed that grew prevalently in the Euphrates-Tigris Rivers area (probably first used by the Sumerians who lived in the southwesterly region of what became Mesopotamia after the end of the Sumerian civilization). It was cut at an oblique angle and easily made a wedge shaped slash or curve in a moist clay tablet.
Around 2000 BC the Babylonians invaded Sumer and took in most of their numerical system, eliminating the numbers between 1 (wedge) and 10 (hook) which they wrote as a wedge and a hook. Two would be 2 wedges, etc. up to 10, then 11 would be a wedge and a hook up to 59 which was written as 5 hooks and 4 wedges. They used a sexagesimal system with a base 60. To represent 60, a wedge was placed in the 60's place. They invented the positional system and the zero to be used as a placeholder by using two oblique wedges. Babylon is not credited with inventing the zero because they only used it as a placeholder (but only later, 200 years after the Persian invasion). India is credited with inventing the zero concept because it was the first to use it as a number.
Tablets from the Old Babylonian period 1900-1600 BC reveal that they knew of the Pythagorean Theorem-finding the hypotenuse and legs of the right triangle-but did not prove it. It was not until 560-460 BC that the Greek mathematician Pythagoras of Samos provided the proof for it.
Babylonian mathematics was much more advanced than that of Egypt, whose main nongeometrical contribution was star clocks. Babylon unfortunately, rather than having a great reputation for its mathematical advancement, is thought of as a superstitious culture because of its religious, mystical beliefs which limited its thinking, especially in the field of medicine (in which Egypt excelled), relying on spirits and other worldly powers to heal the body.
There was some use of symbols, but not much. Like the Egyptians their algebra was essentially rhetorical (they discussed and solved problems without writing them down symbolically). The procedures used to solve problems were taught through examples and no reasons or explanations were given. Also, like the Egyptians, they recognized only positive rational numbers, although they did find approximate solutions to problems which had no exact rational solution.
"Their sexagesimal system facilitated their developing algebra: they had a general procedure equivalent to solving quadratic equations, although they recognized only one root and that had to be positive. In effect, they had the quadratic formula. They also dealt with the equivalent of systems of two equations in two unknowns and a few equivalent to solving equations of higher degree." (History of Algebra, 2004).
Around 1700 BC the Babylonians found the approximation 1; 24; 51; 10 in their base 60 notation for the square root of 2, which works out to 1.414212963 in decimal notation. However, they likely did not realize that 2 was a different kind of number-an irrational number. Another problem they had was that there was no decimal point, sometimes resulting in ambiguous numbers.
Fragmentary, scattered texts from the Seleucid Era (310 BC-75 AD) and the Old Babylonian Period (1900-1600 BC) do not give enough information to verify that the Babylonian astronomers used the heliocentric model in their calculations, but they made some very precise synodic (luni-solar) observations and calculations which have led some to believe that they did actually use it in their calculations. "One of the fundamental units of time and motion applied in Babylonian astronomy appears to have been the mean synodic month of 29;31,50,8,20 days." (Harris, 2004).
John Harris writes in "Babylonian Planetary Theory and the Heliocentric Concept": "The complete synodic arc and synodic period for Mars still provide the correct motion for Earth in degrees per day. The standard unit of time in all cases is the Babylonian year of 12;22,8 mean synodic months treated as the time required for Earth to complete one sidereal revolution of 360 degrees.
"Given the undoubted awareness of accurate sidereal periods for the superior planets, implicit sidereal periods for the inferior planets, accurate sidereal, synodic, draconic, and anomalistic months, and varying velocity functions for the planets, sun, and moon-all readily understood in terms of a cohesive framework-it seems reasonable to conclude that the Babylonians almost certainly possessed a well-developed, fictive heliocentric planetary model by at least 250 BCE, and quite possibly much earlier."
Harris also says that from the calculations of the Babylonians of mean sidereal revolutions and planetary mean synodic arcs, it seems that they might have used three different models to calculate the luni-solar, inferior and superior planets, and star motions, using both the geocentric model for the planetary and sidereal periods and the Earth as heliocentric for the superior planets.
The Babylonians worked with simple methodology, some likening it to counting; their tools were simple but they achieved a superiority in their time in mathematics which has left behind a legacy attesting to the greatness that their civilization once attained.
Britannica.com, 32 vols. 2002.
Harris, John. "Babylonian Planetary Theory and the Heliocentric Concept."
"Time and Tide." 2003, 2004
http://www.spirasolaris.ca/sbb2c.html
"The History of Algebra." "I Only Have Eyes for Algebra." 2004.
http://www.ucs.louisiana.edu/~sxw8045/history.htm
Yolkowski, James. "Babylonian Mathematics." Math Lair. 2002, 2004.
http://www.stormloader.com/aiy/babylonian.html
More about this author: Dyana Hepburn
Understanding the Branch of Mathematics Trigonometry Including Sine Cosine and Law of Sines
Mathematics Mode Median and mean Averages
Eulers Life and Contribution to Mathematics
Mathematics History Group Theory Arithmetic Algebra
Mathematics and Political Science

redpajama_set_name: RedPajamaCommonCrawl


The Knicks and Nets have both been a disaster this season and enter Thursday's meeting with a combined record of 8-26. It's shocking but that's the reality. How are we looking to place our NBA picks when these two collide Thursday night?
Remember the preseason NBA conversations? The Miami Heat were supposed to be atop the Eastern Conference while the Chicago Bulls and Indiana Pacers were expected to fight for the second and third seeds. Then, it would be up to the New York Knicks and New Jersey Nets to duke it out for fourth and fifth in the conference. As it turns out, that thinking was way off.
The Nets are very beaten up right now. If you're going to take the Nets with your sports picks, you should note that they're without Paul Pierce right now and are very likely to be without Deron Williams (doubtful) as well. That leaves the team without their floor general and one of their most complete scorers.
The New York Knicks have endured a long losing streak but they've played some good teams in the process. This is more than a winnable game for them and you're getting them as a dog on the NBA odds.
The Knicks can't even win a game these days. When you look at New York, they've lost nine in a row and probably would struggle to beat the New York Liberty these days – let alone another NBA team. Their only wins this season have come against Milwaukee, Charlotte and Atlanta. Those aren't exactly NBA championship contenders. The Knicks are awful and furthermore, they are in complete disarray. Head coach Mike Woodson is on the hot seat as it looks like the team has tuned him out.
As for the Nets, they're not much better but at least they have injuries to blame. They haven't been healthy since the start of the season and that's hurt them.
This is a toss-up if you've ever seen one. The Nets are without Williams and Pierce, the Knicks are without Tyson Chandler. The Nets have issues on the sidelines as they just removed Lawrence Frank from the equation while the Knicks have coaching problems of their own with Woodson being heavily scrutinized. The Knicks are only a nine-game losing streak while the Nets have lost 11 of 14.
At the end of the day, I'll take a shot with the Knicks with my NBA picks. The main reason is they've been off since Sunday and that extra time should allow them to regroup.


Two sizes: 34mm and 38mm case diameter. The model on this page is 5'7" and she is wearing size 34mm.
Minimalist, yes, but far from simple. Refined detailing – including the domed sapphire glass, lacquered varnish dial and brushed case – make this watch stand out from the rest. This watch can be easily dressed up or down, making it appropriate for everyday use.
Elegant, feminine proportions. Perfect for small and medium wrists. This is by far the most popular size among women. The model on this page is wearing this size. Great for larger wrists or women who like a larger watch face.
Generally fits wrists with circumferences between 13.6 and 18.3cm. Generally fits wrists with circumferences between 13.6 and 19.1cm.
I love how classic the watch is. Definitely recommended.


text:
Bertrand Georges, Ph.D.
Dr. Georges has served as chief technology officer for Altimmune since March 2015. He has nearly two decades of experience in product formulation and manufacturing, preclinical and early-stage clinical development, with a focus on T cell and peptide-based vaccines. Prior to joining Altimmune, Dr. Georges co-founded Immune Targeting Systems (ITS) Limited and served as the chief technology officer from 2004 until its acquisition by Altimmune in March 2015. Prior to ITS, Dr. Georges served as the of head of immunology at SEDAC-Therapeutics from 1999 until 2003. He received his Ph.D. in Molecular Immunology from the Pasteur Institute. Dr. Georges has co-authored 20 publications and is the co-inventor of 13 patent families.


text:
Q: How do sites like hubpages.com put content on subdomains without DNS? I am assuming that there is not someone at hubpages.com adding ServerAlias manually to make subdomains work.
I want to do something similar with Wordpress but I have a hard time grasping the concept of how it works.
From my understanding you need DNS for subdomains to work properly and you need to add a vhost.
I searched stackoverflow for "fake subdomain" creation, but usually the answer is "vhost". Is there a way then to auto-create vhosts for a nonexistent subdomain?

A: The DNS part is done using a wildcard DNS record:
*.hubpages.com.     900 IN  A   66.211.109.24

The server part is probably done using an automated system to create vhosts when users sign up.

A: This is done with a wildcard DNS entry to get the domains to reach the webserver.
Once at the webserver, there's a few techniques you can use with Apache
mod_vhost_alias
mod_vhost_alias allows you dynamically map the virtual hostname onto a document root.
This is a good technique when you want to serve different codebases for different domains, and you have the means to dynamically configure them. I've used this technique to automatically deploy git branches of code by having a custom 404 handler deploy them on demand.
fallthrough to default VirtualHost
Alternatively, rely on the fact that an 'unknown' hostname will fall through to the first defined vhost, then you simply have the application on that host self-configure itself for the supplied domain. I used to run pastebin.com, and this is how I did it there.
This technique is good when the codebase for each domain is identical, only the configuration needs to change.


text:
Makena Compoc. Professor Sturdevant. 5/5/2017. Why Did
1170 WordsMay 6, 20175 Pages
Makena Compoc Professor Sturdevant 5/5/2017 Why Did We Need the Constitution? Taking a look at the events that led up to the American Revolution, one could assume that eventually the 13 colonies were going to rebel against Britain. Events that led to the American Revolution include the Stamp Act and the Townshend Act, which eventually led to the Boston Massacre and the Boston Tea Party. After all of these events, the 13 colonies decided that they wanted to declare their freedom from Britain. Thus, sparked the American Revolution lasting six years, and after this came the Declaration of Independence, the Articles of Confederation, and the drafting of the U.S. Constitution. Before the Stamp Act was introduced by Britain in 1765, there…show more content…
The French destroyed the British fort and forced them to surrender. Meanwhile, Washington and an Indian named Tanaghrisson invaded a French camp in Pennsylvania. The leader of the camp, Ensign Joseph de Jumonville explained that the French were on their way to warn Britian about invading the so-called "French territory" in the Ohio River Valley. In the midst of negotiating, Tanaghrisson killed Jumonville. Preparing to defend his men from the French, Washington build a fort but was forced to surrender by a French/Indian ally. Dinwiddie heard of the news, and reached out to the bordering colonies, only receiving a response from North Carolina who said they would only fight within their own colony. The colonies reached out to Britain for help, and France reached out to the other European countries for help, and thus began the Seven Years War. The Seven Years War lasts until 1763, when the Treaty of Paris is signed, giving Britain control of Canada and all of the territory west of the 13 colonies. It was during this war when the colonists lived in fear of towns being raided by the British and French, but also any conflicts or battles that may arise between the Indians. Towards the end of the war was when colonists began to finally realize how far away they were from Britain. "In
More about Makena Compoc. Professor Sturdevant. 5/5/2017. Why Did
The Court Of School District
The Issues Surrounding The Amazon Rainforest
Mariah Miller. Professor Fandel. Eng 105. 5 May 2017. Global
The Art And Beauty : Leonardo 's Benci And Renaissance Portraits Of Women
Apple 's Line Of Smartphones
The Pathophysiology Of Depression And Depression


text:
Sinéad Burke: 'Fashion Has the Power to Change How Society Views People.' | Podcasts
To subscribe to the BoF Podcast, please follow this link.
Sinéad Burke refuses to be excluded, despite fashion's poor record on welcoming people with disabilities.
In a wide-ranging interview, featured in the third episode of The BoF Show, Burke reminisces on her fashion journey — from calling out the industry for entrenched behaviours at BoF VOICES in 2017; to advising luxury brands as founder and chief executive of consultancy "Tilt the Lens".
Here, we share the full interview with Burke exclusively on The BoF Podcast.
Watch the third episode of The BoF Show, "Belonging: The Business Case for Diversity in Fashion"


This summer has been a crucial time for me and my executive committee. Personally, I am working a summer job, relaxing with family and friends at home, and occasionally taking a weekend roadtrip. However, even though I have been away from State College I still cannot keep my mind off of it. I am exteremly excited for all the growth and development that the chapter will experience this fall.
My EC and I have used this summer to evaluate our performance as the officers of Alpha Delta and how we can further develop the chapter with our influence and decisions. We have been in contact and further planned out goals with each of our chairmen this summer so that we can hit the ground running in August. At the top of our list is the Rush Process and how we can make Fall '16's rush the most successful to date.
We are fully confident that Brother Dalton Kramer (head rush chairman) and his committee will be able to bring in a strong group of young men that will not let our colors fade for want of wearers. This summer our chapter was also able to add a new member, Brother John White, former GM of the Psi Chapter will be Alpha Delta's new Assistant Alumni Advisor.
With his past experiences in dealing with houses and finances, Brother White is truly a great addition to the amazing team that surrounds our chapter. My EC and every brother looks forward to this fall to re-connect and continue our goal of becoming the best fraternity at Penn State!


In addition to advanced treatment and our personalized, comprehensive approach to dental care, we also wish to make your visits as comfortable and convenient as possible. We've provided a link below to fill out your forms before arriving for your appointment. You may have also received an email with this same link.
We participate with most insurance plans and several Medicaid options. Give our office a call for details about your particular plan. Ensuring you get the most out of your benefits is a top priority for us. Any large out-of-pocket expenses can be handled more affordable. Care Credit, a third-party financing company, provides our patients with flexible payment plans for amounts over $200.
Before your first appointment, print and fill out the new patient forms, bring them with you, and save some waiting room time.


text:
/content/basf/www/my/en/who-we-are/organization
/content/basf/www/my/en/who-we-are/organization/management
/content/basf/www/my/en/who-we-are/organization/management/supervisory-board
Board of Executive Directors
Direction, Corporate Management & Control
Competence profile
The term of office of the Supervisory Board commenced following the Annual Shareholders' Meeting on May 3, 2019, in which the shareholder representatives on the Supervisory Board were elected. It terminates upon conclusion of the Annual Shareholders' Meeting which resolves on the discharge of members of the Supervisory Board for the fourth complete financial year after the term of office commenced; this is the Annual Shareholders' Meeting in 2024.
In accordance with the Statutes, the Supervisory Board of BASF SE comprises twelve members.
The term of office of the Supervisory Board commenced following the Annual Shareholders' Meeting on May 3, 2019, in which the shareholder representatives on the Supervisory Board were elected. In accordance with the applicable article of the Statutes as of the date of election, it terminates upon conclusion of the Annual Shareholders' Meeting that resolves on the discharge of members of the Supervisory Board for the fourth complete business year after the term of office commenced; this is the Annual Shareholders' Meeting on April 25, 2024.
Dr. Jürgen Hambrecht resigned his Supervisory Board mandate as of the conclusion of the Annual Shareholders' Meeting on June 18, 2020. The Annual Shareholders' Meeting on June 18, 2020, appointed Dr. Kurt Bock to the Supervisory Board as his successor, who was elected as the new Chairman of the Supervisory Board in the subsequent Supervisory Board meeting. The Supervisory Board member Dr. Alexander C. Karp resigned from the Supervisory Board at the end of the Supervisory Board meeting on July 22, 2020. The Ludwigshafen local court (Amtsgericht) appointed Liming Chen as a substitute member effective October 8, 2020. The shareholders confirmed the appointment of Liming Chen through the election of the Supervisory Board at the Annual Shareholders' Meeting on April 29, 2021. Accordingly, the Supervisory Board comprises the following members:
December 17, 2021 - Alessandra Genco and Stefan Asenkerschbaumer to be proposed for election to the Supervisory Board of BASF SE
Members of the Supervisory Board of BASF SE
Dr. Kurt Bock
Chairman of the Supervisory Board of BASF SE
Residence: Heidelberg, Germany | Nationality: German
Kurt Bock was born in Rahden, Eastern Westphalia, Germany in 1958. He studied business administration in Münster and Cologne and at Pennsylvania State University. In 1985, he earned his doctorate (Dr. rer. pol.) from the University of Bonn. In the same year, he began his career at BASF in finance. From 2003 to 2011, he was Chief Financial Officer of BASF SE and, since 2007, additionally Chairman and Chief Executive Officer of BASF Corporation in the United States. He was Chairman of the Board of Executive Directors of BASF from 2011 to 2018.
Bock has been a member of the BASF Supervisory Board since June 18, 2020, and his mandate terminates upon conclusion of the Annual Shareholders' Meeting in 2024.
Memberships of statutory supervisory boards in Germany:
Fuchs Petrolub SE (listed company), Chairman
BMW (listed company), Member
Memberships in comparable domestic and foreign supervisory bodies of commercial enterprises:
Prof. Dr. Thomas Carell
Professor for Organic Chemistry at Ludwig Maximilian University Munich
Place of residence: Munich, Germany | Nationality: German
Prof. Dr. Thomas Carell was born in Herford, Germany in 1966 and studied chemistry at the University of Münster, Germany. He received his doctorate in organic chemistry from the University of Heidelberg, Germany in 1993. After postdoctoral studies at Massachusetts Institute of Technology, Cambridge he obtained his habilitation from the Swiss Federal Institute of Technology Zurich, Switzerland in 1998. Carell has been a professor for organic chemistry at Ludwig Maximilian University in Munich, Germany since 2004.
Carell has been a member of the BASF Supervisory Board since May 3, 2019, and his mandate terminates upon conclusion of the Annual Shareholders' Meeting in 2024.
Memberships in comparable domestic and foreign supervisory bodies of commercial enterprise:
Dame Alison J. Carnwath DBE
Senior Advisor Evercore Partners
Residence: Exeter, UK | Nationality: British
Dame Alison J. Carnwath was born in Derby, UK in 1953 and studied Economics and German at the University of Reading/UK and University of Munich/Germany. Since 2000, she has been Senior Advisor at Lexicon Partners Limited, later Evercore Partners.
Carnwath has been a member of the BASF Supervisory Board since May 2, 2014, and her mandate terminates upon conclusion of the Annual Shareholders' Meeting in 2024.
Zurich Insurance Group AG (listed company), independent, non-executive member of the Board of Directors
Zürich Versicherungs-Gesellschaft AG (subsidiary of Zurich Insurance Group AG) (non-listed company), independent, non-executive member of the Board of Directors
PACCAR Inc. (listed company), independent member of the Board of Directors
Coller Capital Ltd. (non-listed company), non-executive member of the Board of Directors
Broadwell Capital Limited (non-listed company), non-executive member of the Board of Directors
Euro Garages Limited (non-listed company), non-executive member of the Board of Directors
Liming Chen
Chairman IBM Greater China Group
Residence: Beijing/China | Nationality: Singaporean
Liming Chen was born in 1960 in Xinjiang/China. He graduated from Shihezi University in Xinjiang in 1982 with a Bachelor of Science. In 1989, he obtained his master's degree in food science from Cornell University in Ithaca, New York. Between 1994 and 2008, Chen held various leadership roles at the chemical company Condea (since 2001: Sasol) in Singapore and China. From 2008 to 2015, he served as President of BP China and Chairman of BP (China) Holding Company. Chen is Chairman of IBM Greater China Group, based in Beijing, China, a position he has held since 2015.
Chen has been a member of the BASF Supervisory Board since October 8, 2020, and his mandate terminates upon conclusion of the Annual Shareholders' Meeting in 2024.
IBM China Investment Company Ltd. (non-listed company), chair (intragroup membership)
IBM (China) Company Ltd. (non-listed company), chair (intragroup membership)
IBM Global Services (DaLian) Company Limited (non-listed company), chair (intragroup membership)
IBM Solution and Services (ShenZhen) Company Ltd. (non-listed company), chair (intragroup membership)
IBM Financing and Leasing Company Ltd. (non-listed company), chair (intragroup membership)
IBM Factoring (China) Company Ltd. (non-listed company), chair (intragroup membership)
Inspur Power Commercial Systems Company Ltd. (non-listed company), chair (intragroup membership)
IBM Taiwan Corporation (non-listed company), director (intragroup membership)
Tatjana Diether
Member of the Works Council of BASF SE, Ludwigshafen Site and of the BASF Works Council Europe
Residence: Limburgerhof, Germany | Nationality: German
Tatjana Diether was born in Worms, Germany in 1975. She joined BASF Aktiengesellschaft (since January 14, 2008, BASF SE) as an apprentice for technical drawing in 1991. Following a vocational commercial training at a private business school, she was an instructor for vocational training at BASF.
Diether has been a member of the Works Council of BASF SE since 2003 and has been a full-time member of the Works Council since 2008. Diether was elected to the BASF Works Council Europe in 2017.
Diether has been a member of the BASF Supervisory Board since May 4, 2018, and her mandate terminates upon conclusion of the Annual Shareholders' Meeting in 2024.
Franz Fehrenbach
Vice Chairman of the Supervisory Board of BASF SE
Chairman of the Supervisory Board of Robert Bosch GmbH
Residence: Stuttgart, Germany | Nationality: German
Franz Fehrenbach was born in Kenzingen, Germany in 1949 and studied industrial engineering at the University of Karlsruhe, Germany. From 1999, he was a member of the Board of Management of Robert Bosch GmbH, from 2003 until 2012 Chairman of the Board of Management. Since 2012, he has been Chairman of the Supervisory Board of Robert Bosch GmbH and Managing Partner of Robert Bosch Industrietreuhand KG (RBIK).
Fehrenbach has been a member of the BASF Supervisory Board since January 14, 2008, and his mandate terminates upon conclusion of the Annual Shareholders' Meeting in 2024.
Robert Bosch GmbH (non-listed company), Chairman
Stihl AG (Stihl Holding AG & Co. KG group company) (non-listed company), Deputy Chairman
Stihl Holding AG & Co. KG (non-listed company), Member of the Advisory Board
Linde plc (listed company), Member of the Board of Directors
Waldemar Helber
Deputy Chairman of the Works Council of BASF SE, Ludwigshafen Site
Residence: Otterbach, Germany | Nationality: German
Waldemar Helber was born in Weilerbach, Germany in 1957. He trained as a power current technician at BASF AG (since January 14, 2008 BASF SE) from 1972 to 1976. He has been a member of the Works Council of BASF AG, Ludwigshafen Site, since 2003 and a full-time trade union delegate since 2007. Since 2010, Helber has been a member of the BASF Works Council Europe.
Helber has been a member of the BASF Supervisory Board since April 29, 2016, and his mandate terminates upon conclusion of the Annual Shareholders' Meeting in 2024.
Sinischa Horvat
Chairman of the Works Council of BASF SE, Ludwigshafen Site; and Chairman of BASF's Joint Works Council and of the BASF Works Council Europe
Sinischa Horvat was born in Pirmasens, Germany in 1976 and trained as a process control technician at BASF Aktiengesellschaft (since January 14, 2008 BASF SE) from 1993 to 1997.
After his studies of business administration, he worked in the site information management at the Ludwigshafen site from 2002 to 2007.
He joined the Works Council of BASF Aktiengesellschaft in 2002 and has been a full-time member of the Works Council at the Ludwigshafen site since 2007. Horvat joined the BASF Europa Betriebsrat (BASF Works Council Europe) in 2013.
Horvat has been a member of the BASF Supervisory Board since May 12, 2017, and his mandate terminates upon conclusion of the Annual Shareholders' Meeting in 2024.
Anke Schäferkordt
Supervisory Board Member
Residence: Cologne, Germany | Nationality: German
Anke Schäferkordt was born in Lemgo, Germany in 1962 and studied business administration at the University of Paderborn, Germany. From 2005 to 2018, she was CEO of RTL Television and Mediengruppe RTL Deutschland GmbH. From 2012 to 2017, she was Co-CEO of RTL Group S.A. and from 2012 to 2018, she was a member of the Executive Board of Bertelsmann SE & Co. KGaA.
Schäferkordt has been a member of the BASF Supervisory Board since December 17, 2010, and her mandate terminates upon conclusion of the Annual Shareholders' Meeting in 2024.
Serviceplan Group Management SE (non-listed company), Partner with unlimited liability of Serviceplan Group SE & Co. KG, member
Bayerische Motoren Werke Aktiengesellschaft (listed company), member
Wayfair Inc. (listed company), non-executive director
Denise Schellemans
Full-Time Trade Union Delegate
Residence: Brecht, Belgium | Nationality: Belgian
Denise Schellemans was born in Niel, Belgium in 1961. She trained as medical-technical assistant at the University of the Province of Antwerp. From 1982 to 1985, she worked in cancer research at the University of Antwerp. Schellemans has worked in the medical service at BASF Antwerpen from 1985 to 1994. In 1994, she was named a full-time trade union delegate. Since 1995, she has been a member of the local Works Council and in 1999 was named vice president of the BASF Euro Dialog (since April 2008 BASF Works Council Europe).
Schellemans has been a member of the BASF Supervisory Board since January 14, 2008, and her mandate terminates upon conclusion of the Annual Shareholders' Meeting in 2024.
Roland Strasser
Regional Manager of the Rhineland-Palatinate/Saarland branch of IG BCE
Residence. Riedstadt, Germany | Nationality: German
Roland Strasser was born in Frankfurt am Main, Germany in 1975. From 1991 to 1994, he was an apprentice in industrial mechanics at Hoechst AG in Frankfurt. Following his vocational training, he worked in various business areas at Hoechst AG. Following the spin-off of the pharmaceuticals business, he transferred to the newly founded Hoechst Marion Roussel which then became part of Aventis Pharma in 1999 with the merger of Hoechst and Rhône-Poulenc.
Roland Strasser has been the Trade Union Secretary of the Mining, Chemical and Energy Industries Union (IG BCE) since 2001 and regional manager of the Rhineland-Palatinate/Saarland branch of IG BCE since November 2017.
Strasser has been a member of the BASF Supervisory Board since May 4, 2018, and his mandate terminates upon conclusion of the Annual Shareholders' Meeting in 2024.
AbbVie Komplementär GmbH (non-listed company), Member
V & B Fliesen GmbH (non-listed company), Member
Villeroy & Boch AG (listed company), Member
Michael Vassiliadis
Chairman of the Mining, Chemical and Energy Industries Union
Residence: Hanover, Germany | Nationality: German
Michael Vassiliadis was born in Essen, Germany in 1964 and trained as a chemical lab technician at Bayer AG in Dormagen, Germany. In 1986, he began his full-time union work at IG CPK (since 1997 IG BCE). Since March 2004, he has been a member of the Managing Board of IG BCE. In October 2009, he was elected Chairman of IG BCE.
Vassiliadis has been a member of the BASF Supervisory Board since August 1, 2004, and his mandate terminates upon conclusion of the Annual Shareholders' Meeting in 2024.
Steag GmbH (non-listed company), Member
RAG AG (non-listed company), Vice chairman
Henkel AG & Co KGaA (listed company), Member
Vivawest GmbH (non-listed company), Member


Strike a high strung and high profile look with the blue and white designer wholesale flannel shirts.Significantly high quality flannel shirt with endearing base and well styled finesse. Carefully blended range of designer flannel shirts like this collection now available at factory prices for bulk buyers. Flannel Clothing manufacturer offers fabulous facilities to bulk buyers.


text:
At ASCB 2015: CRISPR/Cas9 + HPSC = human PKD lab model
CRISPR turns human pluripotent stem cells into PKD model
American Society for Cell Biology
SAN DIEGO, CA -- CRISPR/Cas9 is hot. News of the revolutionary gene editing technique that is already shaking up bioscience has finally reached the news media and the public. Now comes a first rate example of how CRISPR is changing the pace of biomedical research by linking up with another cutting edge technology -- human pluripotent stem cells (hPSCs). Benjamin S. Freedman, now at the University of Washington, and his colleagues in Joseph Bonventre's lab at Harvard Medical School, have used CRISPR/Cas9 to guide hPSCs into becoming a human cell-based lab model system for polycystic kidney disease (PKD). The most common inherited kidney disorder, affecting one in 500 Americans, PKD currently is not curable and, without long-term dialysis or kidney transplant, can be fatal. Freedman will speak about the PKD model at ASCB 2015 in San Diego on Monday, Dec. 14, 2015, following its earlier publication in Nature Communications.
PKD's hallmark is the formation of damaging, balloon-like cysts in kidney tubules. In the early 2000s, cell biologists linked cyst formation to gene mutations that affect the primary cilia, hair-like projections from cells that seem to act as sensory antennae. These fundamental discoveries were made in non-human organisms such as the algae, Chlamydomonas reinhardtii, and the zebrafish, Danio rerio. But the exact disease mechanism in humans is still not well understood in part because there hasn't been a good human model of PKD in kidney cells.
CRISPR gave Freedman et al. a more precise tool to remodel the hPSC genome to include PKD mutations in the disease-linked genes, PKD1 and PKD2. The researchers then used a 3-D cell culture system to coax their mutant and healthy hPSCs down the differentiation pathway into becoming kidney progenitor cells and finally the proximal tubule cells found in kidney nephrons. In the mutant tubule cells, they observed the formation of large, translucent cyst-like structures but not in their healthy controls. These observations and others have convinced the researchers that their CRISPR/Cas9 and hPSC system produces a stable, biologically accurate human model for a common genetic disease where new understanding and new therapies are desperately needed.
Modeling Polycystic Kidney Disease Cystogenesis with Genome?Modified Human Pluripotent Stem Cells
B.S. Freedman1,2, T.I. Steinman3, J. Zhou2, J.V. Bonventre2
1Department of Medicine, Division of Nephrology, University of Washington, Seattle, WA
2Department of Medicine, Renal Division, Brigham and Women's Hospital/Harvard Medical School, Boston, MA,
3Department of Medicine, Division of Nephrology, Beth Israel Deaconess Medical Center/Harvard Medical School, Boston, MA
Contact Author:
Benjamin S. Freedman
benof@uw.edu
At ASCB 2015
Author presents:
Microsymposium 8: The Role of the Cytoskeleton in Disease and Repair
Monday, Dec. 14, 2015
Microsymposia Room 2
Journal citation: http://www.nature.com/ncomms/2015/151023/ncomms9715/abs/ncomms9715.html
John Fleischman
jfleischman@ascb.org
@ASCBiology
http://www.ascb.org


Author: Frewen, Thomas A.; Couzin, Iain D.; Kolpas, Allison; Moehlis, Jeff; Coifman, Ronald; Kevrekidis, Ioannis G.
The coarse-grained, computer-assisted analysis of models of collective dynamics in animal groups involves (a) identifying appropriate observables that best describe the state of these complex systems and (b) characterizing the dynamics of such observables. We devise "equation-free" simulation protocols for the analysis of a prototypical individual-based model of collective group dynamics. Our approach allows the extraction of information at the macroscopic level via parsimonious usage of the detailed, "microscopic" computational model. Identification of meaningful coarse observables ("reduction coordinates") is critical to the success of such an approach, and we use a recently-developed dimensionality-reduction approach (diffusion maps) to detect good observables based on data generated by local model simulation bursts. This approach can be more generally applicable to the study of coherent behavior in a broad class of collective systems (e.g., collective cell migration).


text:
Home » Sports » Booming Fort St. John comes with billion dollar price tag
Booming Fort St. John comes with billion dollar price tag
by Energeticcity.ca Staff June 9, 2015 Updated April 19, 2022
Fort St. John could see as many as 84,000 residents by 2065, and will require nearly $1 billion in new infrastructure by then to service that growth.
Those are some key findings Urban Systems will deliver to city council at a committee of the whole meeting this afternoon from its 50 Year Growth Study final report.
The report forecasts a number population growth scenarios and infrastructure needs up to 2065, and will serve as a key document for Fort St. John as it juggles its spending priorities.
That study predicts a number of population scenarios, from a slow population growth to 46,000 residents by 2065 if the city grows at rates between one and 2.2 per cent a year.
If the city grows any faster, sustaining average annual growth in the 2.2 to 3 per cent range, it could see as many as 84,000 new residents by 2065.
The report settles on more modest growth in the range of 2 to 3 per cent over the next several decades, with population more likely to settle around 62,000 people.
Fort St. John saw the second highest growth in British Columbia in 2014 at 4.7 per cent. The city is anticipating rapid and exponential growth with construction of the $8.8-billion Site C dam, along with billions of dollars more in proposed coastal LNG plants that will feed off gas from the region.
Servicing that growth won't come cheap. The cost of new water supplies, sewers, roads, parks, fire halls, and police stations is estimated by Urban Systems at $846 million.
Transportation needs the greatest
Not surprisingly, the estimated cost of new transportation infrastructure takes up a sizeable chunk of that total at $247.1 million, or 30 per cent.
The report identifies a number of priorities, from bypass road upgrades and several four-laning road projects in the neighbourhood of $115 million. The city will need to spend roughly $132 million for new major roads and intersections. Some of those costs wouldn't be born entirely by the city, the report notes, with developers and other sources of funding playing a part.
The city will need anywhere from 700 to 2,600 hectares of new land to accommodate its growth, according to the report, more than 50 per cent of which would be allocated for residential and road use.
New civic buildings would cost at least $140M
When it comes to municipal facilities, Fort St. John will need to build itself a new city hall, two new fire stations, a new RCMP building, along with a host of other city facilities.
The report pegs a two-phase city hall construction at $33 million to eventually complement an increased staff of nearly 140 people. The first phase of a new city hall would be required by 2025, with the second phase needed by 2040.
Two new fire halls would need to be built to a tune of $24 million a piece, with the first needed by 2025, and the second needed by 2035. Report estimates staff increases of 48 firefighters.
Meanwhile, a new RCMP detachment is estimated at $34 million. Fort St. John RCMP will need 98 new city officers by 2065, with another 25 needed for regional policing.
The city will also need a new public works building by 2040 to the tune of $17.5 million, but will need to make about $2.6 million in upgrades to its existing building to handle new growth by 2017, the report states.
The city would also need a new bylaw services building, estimated at $4 million, with eight new officers brought on board. The report also budgeted $2.3 million for a new visitor centre.
New parks, trails, sports fields galore
The report projects a need for 160 hectares of new parkland, with the creation of 22 new parks across the city — including 17 new neighbourhood pars, four community parks, and only a single, new "destination" park. The report notes Fort St. John is currently limited in supply of neighbourhood parks.
About 50 kilometres of new trails would need to be built to link new developments and loop off the city's existing trail system.
The report also projects a need for 27 new sports fields, half a dozen new indoor ice sheets along with two outdoor hockey rinks, a second curling facility, and another indoor soccer pitch.
The city will likely need to begin planning for a second pool at some point, the report notes. It would also benefit from a community centre that could accommodate a gymnasium and multi-purpose rooms for a variety of sports and events.
The report pegs new parkland costs at $32 million, with new indoor and outdoor rec facilities estimated around $120 million.
The report also projected the city will need to spend another $120 million on sanitary and storm sewers.
Fort St. John will also need to develop a new water supply, and it has apparently reached a tentative agreement with BC Hydro to withdraw water from the reservoir created by Site C for future city needs.
More on that story, and how much the city will need to spend to ease its water shortfall, can be read here.