text:
\section{Introduction}
\label{S-I}

Random walks with long-memory arose naturally in applied mathematics and statistical physics. One of them is the so-called
elephant random walk (ERW) introduced by Sch\"utz and Trimper \cite{Schutz04} in the early 2000s, in order to investigate how long-range memory
affects the random walk and induces a crossover from a diffusive to superdiffusive behavior. The ERW is a one-dimensional discrete-time random
walk on $\mathbb{Z}$, which has a complete memory of its entire history. It was referred to as the ERW in allusion to the famous saying that elephants
can always remember where they have been before.
\ \vspace{2ex} \par
The ERW starts at the origin at time zero,
$S_0=0$. At time $n=1$, the elephant moves to the right with probability $q$ and to the left with probability
$1-q$ where $q$ lies between $0$ and $1$, which means that $S_1=X_1$ where $X_1$ has a Rademacher $\mathcal{R}(q)$ distribution.
Then,  at any time $n+1 \geq 1$, the elephant chooses uniformly at random an integer $k$ among the previous
times $1,\ldots,n$ and it moves exactly in the same direction as that of time $k$ with probability $p$ or in the opposite direction
with probability $1-p$. In other words,
\begin{equation}
\label{STEPS}
   X_{n+1} = \left \{ \begin{array}{ccc}
    +X_{k} &\text{ with probability } & p, \vspace{2ex}\\
    -X_{k} &\text{ with probability } & 1-p.
   \end{array} \right.
\end{equation}
Therefore, the position of the ERW at time $n+1$ is given by
\begin{equation}
\label{POSERW}
S_{n+1}=S_{n}+X_{n+1}.
\end{equation}
The asymptotic behavior of the ERW is closely related to the value of the probability $p \in [0,1]$ called the memory of the ERW.
The elephant random walk is said to be subcritical if $0 \leq p < 3/4$, critical if $p=3/4$ and supercritical if $3/4< p \leq 1$. In this paper, we shall focus
our attention on the supercritical case where the memory parameter $3/4< p \leq 1$. It has been shown recently by three different approaches \cite{Baur16},
\cite{Bercu18}, \cite{Coletti17} that
\begin{equation}
\label{ASCVG}
 \lim_{n \rightarrow \infty} \frac{S_n}{n^{2p-1}}=L \hspace{1cm} \text{a.s.}
\end{equation}
where $L$ is a non-degenerate random variable. However, to the best of our knowledge, the distribution
of $L$ is far from being known. The only thing we know about the distribution of $L$ is that it is not Gaussian \cite{Bercu18}
Baur and Bertoin \cite{Baur16} established a very successful connection
between the ERW and generalized P\'olya urns  \cite{Chauvin11}, \cite{Pemantle07}. Thanks to this connection and two functional limit theorems
for multitype branching processes due to Janson \cite{Janson04}, it is possible to compute all the moments of $L$, see also \cite{Chauvin11}.
The martingale approach of Bercu \cite{Bercu18} allows us to compute the moments of $L$ in two different ways.
On the one hand, it is proven in \cite{Bercu18} that the almost sure convergence \eqref{ASCVG}
also holds in $\mathbb{L}^4$, which means that
\begin{equation}
\label{L4CVG}
 \lim_{n \rightarrow \infty} \mathbb{E}\Bigl[ \Bigl| \frac{S_n}{n^{2p-1}} -L \Bigr|^4 \Bigr]=0.
\end{equation}
Consequently, in order to compute the first four moments of $L$, it is only necessary to calculate
the first four moments of $S_n$ and to make use of  Lebesgue's dominated convergence theorem.
On the other hand, denote
\begin{equation}
\label{DEFL}
 L_n = \frac{a_n S_n}{\Gamma(a+1)}
\end{equation}
where $a=2p-1$ and the sequence $(a_n)$ is given by $a_1=1$ and for all $n\geq 2$,
\begin{equation}
\label{DEFAN}
a_n =  \prod_{k=1}^{n-1}\Bigl(\frac{k+2p-1}{k}\Bigr)^{-1} = \frac{\Gamma(n)\Gamma(a+1)}{ \Gamma(n+a)} .
\end{equation}
We already saw in \eqref{L4CVG} that
\begin{equation}
\lim_{n \rightarrow \infty}L_n=L \hspace{1cm} \text{a.s.}
\end{equation}
Therefore, our strategy is to calculate the moments of $L$ in two different ways.
We were happily surprised to find that this approach is the source of new hypergeometric identities.
For example, we deduce from the double calculation of $\mathbb{E}[L^2]$ that for all $a>1/2$,
\begin{equation*}
\frac{1}{(2a-1)\Gamma(2a)}=\frac{2a}{(2a-1)\Gamma(a+1)^2}\left(1 -\Bigl(\frac{a}{a+1}\Bigr)^2{}_{3}F_2 \Bigl( \begin{matrix}
{1,1,2a+1}\\
{a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)\right)
\end{equation*}
leading to
\begin{equation*}
a^2 {}_{3}F_2 \Bigl( \begin{matrix}
{1,1,2a+1}\\
{a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)=(a+1)^2-\frac{(\Gamma(a+2))^2}{\Gamma(2a+1)}
\end{equation*}
where $\!{}_{p}F_q$ stands for the hypergeometric function defined for all $z \in \mathbb{C}$ by
\begin{equation}
\label{HGpq}
{}_{p}F_q \Bigl( \begin{matrix}
{a_1,\ldots,a_p}\\
{b_1,\ldots,b_q}\end{matrix} \Bigl|
{\displaystyle z}\Bigr)
=\sum_{n=0}^{\infty}
\frac{(a_1)_n\,\cdots\,(a_p)_n}
{(b_1)_n\,\cdots\,(b_q)_n\, n!} z^n.
\end{equation}
Here, $(a)_n$ is the rising Pochhammer symbol given by $(a)_0=1$ and, for all $n\geq 1$,
$(a)_n=a(a+1)\cdots(a+n-1)$. Via the double calculation of $\mathbb{E}[L^d]$ for $d=2,3,4$,
we will obtain new hypergeometric identities associated to the function ${}_{d+1}F_d$.
\ \vspace{2ex} \par
A wide range of literature is available on hypergeometric identities.
We refer the reader to the very complete survey of Zudilin \cite{Zudilin} as well as to the more recent contributions
of Grondin \cite{Grondin} and Milgram \cite{Milgram}.
Our work is in the spirit of the classical Watson theorem, see the summation formulas in Lavoie \cite{Lavoie} and
Maier \cite{Maier}.
Krattenthaler and Srivinasa Rao \cite{Kratten} develop integral methods to prove many identities.
To our best knowledge, our identities are not one of them.
\ \vspace{2ex} \par
Our main tool is the very successful method of telescoping sums that has been widely used by Zeilberger \cite{Zeilberger}
and implemented in computer algebra systems in \cite{PWZ}. We would like to point out
that our identities were not directly recognized by Maple or Mathematica.
\ \vspace{2ex} \par
The paper is organized as follows.
The main results of the paper are given in Section \ref{S-MR}. In Section \ref{S-M2}, we compute in two different ways the first order moments
of the random variable $L$. It allows us to prove our new hypergeometric identities for the functions ${}_{3}F_2$, ${}_{4}F_3$ and ${}_{5}F_4$
evaluated at $z=1$.
Section \ref{S-Md} is devoted to the proof of the general case involving the higher order moments of $L$.
A short conclusion and perspectives are postponed to Section \ref{S-CP}.

\section{Main results}
\label{S-MR}

In order to simplify the presentation of our main results, denote $a=2p-1$ and $b=2q-1$.  We recall from \cite{Bercu18} that the first four moments of $L$ are given by
\begin{eqnarray}
\label{MOML1}
\mathbb{E}[L]  &=& \frac{b}{\Gamma(a+1)},\\
\label{MOML2}
\mathbb{E}[L^2] &=& \frac{1}{(2a-1)\Gamma(2a)}, \\
\label{MOML3}
\mathbb{E}[L^3] &=& \frac{b(a+1)}{a(2a-1)\Gamma(3a)}, \\
\label{MOML4}
\mathbb{E}[L^4] &=& \frac{6(2a^2+2a-1)}{(4a-1)(2a-1)^2\Gamma(4a)}.
\end{eqnarray}

\noindent
By the calculation of the moments of $L$ in two different ways, we obtain the following  hypergeometric identities.
\begin{thm}
\label{T-NHI-MOM}
For any $a>1/2$, we have
\begin{equation}
\label{T-NHI1}
\frac{1}{(2a-1)\Gamma(2a)}=
\frac{2a}{(2a-1)\Gamma(a+1)^2}\left(1 -\Bigl(\frac{a}{a+1}\Bigr)^2{}_{3}F_2 \biggl( \begin{matrix}
{1,1,2a+1} \\
{a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)\right).
\end{equation}
\begin{eqnarray}
\frac{(a+1)}{a(2a-1)\Gamma(3a)} & = & \frac{3(a+1)}{(2a-1)\Gamma(a+1)^3}\left(1- \frac{3a^2}{(a+1)^3}
{}_{4}F_3 \Bigl( \begin{matrix}
{1,1,2,3a+1} \nonumber\\
{a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)\right. \\
& - &
\left.\frac{a^3}{(a+1)^3}
{}_{4}F_3 \Bigl( \begin{matrix}
{1,1,1,3a+1} \\
{a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)\right).
\label{T-NHI2}
\end{eqnarray}
\begin{eqnarray}
\frac{6(2a^2+2a-1)}{(4a-1)(2a-1)^2\Gamma(4a)}& = & \frac{24a(2a^2+2a-1)}{(4a-1)(2a-1)^2\Gamma(a+1)^4}\biggl(1
\nonumber \\
& - &
\left.
\frac{6a^2}{(a+1)^4}
{}_{5}F_4 \Bigl( \begin{matrix}
{1,1,2,2,4a+1} \nonumber\\
{a+2,a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)\right. \\
& - &
\left.\frac{4a^3}{(a+1)^4}
{}_{5}F_4 \Bigl( \begin{matrix}
{1,1,1,2,4a+1} \nonumber\\
{a+2,a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)\right. \\
& - &
\left.
\frac{a^4}{(a+1)^4}
{}_{5}F_4 \Bigl( \begin{matrix}
{1,1,1,1,4a+1} \\
{a+2,a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)
\right).
\label{T-NHI3}
\end{eqnarray}
\end{thm}

\begin{rem}
Another identity coming from the calculation of $\mathbb{E}[L^2]$ is given, for any $a>1/2$, by
\begin{eqnarray}
\label{T-NHI4}
\frac{1}{(2a-1)\Gamma(2a)} & = & \frac{1}{\Gamma(a+1)^2}
\!{\ }_{3}F_2 \Bigl( \begin{matrix}
{1,1,1} \\
{a+1,a+1}\end{matrix} \Bigl|
1 \Bigr)\\
   +\frac{ a^2 }{(2a-1)\Gamma(a+2)^2} &  & \hspace{-1cm} \left(
{ }_{3}F_2 \Bigl( \begin{matrix}
{1,1,2}\\
{a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)
- 2a\,
{ }_{3}F_2 \Bigl( \begin{matrix}
{1,1,2a+1}\\
{a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)\right). \nonumber
\end{eqnarray}
\end{rem}

\noindent
A more general hypergeometric identity is as follows.

\begin{thm}
\label{T-NHI-GEN}
For any $a>1/2$ and for all $d \geq 2$, we have
\begin{equation}
\label{T-NHIGEN}
\sum_{k=0}^{d-2} \binom{k}{d}a^{d-k}
{}_{d+1}F_d \Bigl( \begin{matrix}
{1_{d-k},2_k, ad+1} \\
{a+2,a+2,\ldots,a+2}\end{matrix} \Bigl|
1 \Bigr)
=(a+1)^d - \frac{\Gamma(a+2)^d}{\Gamma(ad+1)}
\end{equation}
where $1_{d-k}=1,\dots,1$ of length $d-k$ and $2_{k}=2,\dots , 2$ of length $k$.
\end{thm}

\section{Proofs of the first order moment results}
\label{S-M2}

We shall now proceed to the proof of Theorem \ref{T-NHI-MOM}. Our proof relies on the Lebesgue's dominated convergence theorem
together with a telescoping argument. We already saw that the sequence
$(L_n)$ satisfies
$$
\sup_{n \geq 1} \mathbb{E}[L_n^4] < \infty.
$$
Hence, the dominated convergence theorem implies that for any integer $d \leq 4$,
\begin{equation}
\label{LEBLN}
\lim_{n\rightarrow \infty} \mathbb{E}[L_n^d]=\mathbb{E}[L^d].
\end{equation}
where the limiting random variable $L$ is given by \eqref{DEFL}. Therefore, we have
\begin{eqnarray}
 \mathbb{E}[L^d]\Gamma(a+1)^d- \mathbb{E}[X_1^d] & = & \lim_{n\rightarrow \infty} \mathbb{E}[L_n^d]\Gamma(a+1)^d- \mathbb{E}[X_1^d],\nonumber \\
 & & \nonumber \\
 & = &

 \lim_{n\rightarrow \infty} \mathbb{E}[(a_nS_n)^d]- \mathbb{E}[(a_1S_1)^d]\nonumber\\
 & = & \lim_{n\rightarrow \infty} \sum_{k=2}^n(\mathbb{E}[(a_kS_k)^d]- \mathbb{E}[(a_{k-1}S_{k-1})^d])\nonumber \\
 & = & \sum_{k=2}^\infty\mathbb{E}[(a_kS_k)^d]- \mathbb{E}[(a_{k-1}S_{k-1})^d].
 \label{TEL}
\end{eqnarray}
We shall now proceed to the proof of \eqref{T-NHI1} by the computation of the second moment of the random variable $L$.
We have from identity (B.19) in \cite{Bercu18} that
\begin{equation}
\label{MOMSN2}
\mathbb{E}[S_n^2]=\frac{n}{2a -1}\left( \frac{\Gamma(n+2a)}{\Gamma(n+1)\Gamma(2a)} -1\right).
\end{equation}
Consequently, we obtain from \eqref{DEFAN} together with \eqref{TEL} and \eqref{MOMSN2} that
\begin{align*}
 \mathbb{E}[L^2] & \Gamma(a+1)^2 -1  =\sum_{n=2}^\infty (a_n^2\mathbb{E}[S_n^2]-a_{n-1}^2\mathbb{E}[S_{n-1}^2])\\

 = &\frac{\Gamma(a+1)^2 }{2a-1} \sum_{n=2}^\infty\left[\frac{1}{\Gamma(2a)}\left(\frac{\Gamma(n)\Gamma(n+2a)}{\Gamma(n+a)^2}-\frac{\Gamma(n-1)\Gamma(n-1+2a)}{\Gamma(n-1+a)^2}\right) \right.\\
 &\hskip2.5cm -\left. \frac{\Gamma(n)\Gamma(n+1)}{\Gamma(n+a)^2}+\frac{\Gamma(n-1)\Gamma(n)}{\Gamma(n-1+a)^2}\right].
 \end{align*}
However, we have by the cancellation technique that
\begin{equation*}
\sum_{n=2}^\infty\left[ \frac{\Gamma(n)\Gamma(n+1)}{\Gamma(n+a)^2}-\frac{\Gamma(n-1)\Gamma(n)}{\Gamma(n-1+a)^2}\right]=-\frac{\Gamma(1)\Gamma(2)}{\Gamma(a+1)^2}=-\frac{1}{\Gamma(a+1)^2}
\end{equation*}
Therefore, we deduce from the previous calculation that
\begin{align*}
 \mathbb{E}[L^2]-\frac{1}{\Gamma(a+1)^2}
  = &\frac{-a^2}{(2a-1)\Gamma(2a)}\sum_{n=0}^\infty\frac{\Gamma(n+1)\Gamma(n+2a+1)}{\Gamma(n+2+a)^2} +\frac{1}{(2a-1)\Gamma(a+1)^2}\\
   = &\frac{-a^2}{(2a-1)\Gamma(2a)}
   {}_{3}F_2 \Bigl(\begin{matrix}
{1,1,2a+1} \\
{a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)\frac{\Gamma(2a+1)}{\Gamma(a+2)^2}+\frac{1}{(2a-1)\Gamma(a+1)^2}
 \end{align*}
which clearly leads to \eqref{T-NHI1}.
We are now in position to prove \eqref{T-NHI2} by the calculation of the third moment of $L$. It follows from identity (B.21) in \cite{Bercu18} that
 \begin{equation}
\mathbb{E}[S_n^3] = \frac{b}{(2a -1)\Gamma(n)}\left(
\frac{3(a +1)\Gamma(n+3a)}{\Gamma(3a + 1)}
-\frac{\Gamma(n+a)}{\Gamma(a+1)}(3n+a +1)
\right).
\label{MOMSN3}
\end{equation}
Hence, we have from \eqref{TEL} and \eqref{MOMSN3} that
\begin{align*}
 \mathbb{E}[L^3] & \Gamma(a+1)^3 -b =\sum_{n=2}^\infty (a_n^3\mathbb{E}[S_n^3]-a_{n-1}^3\mathbb{E}[S_{n-1}^3])\\
= &\frac{b\Gamma(a\!+\!1)^3}{2a-1} \sum_{n=2}^\infty\left[\frac{3(a+1)}{\Gamma(3a+1)}\!\left(\frac{\Gamma(n)^3}{\Gamma(n\!+\!a)^3}\frac{\Gamma(n\!+\!3a)}{\Gamma(n)}\!-\!\frac{\Gamma(n\!-\!1)^3}{\Gamma(n\!-\!1\!+\!a)^3}\frac{\Gamma(n\!-\!1\!+\!3a)}{\Gamma(n-1)}\right)\right.\nonumber\\
 & -\frac{3}{\Gamma(a+1)}\left(\frac{\Gamma(n)^3}{\Gamma(n+a)^3}\frac{\Gamma(n+1+a)}{\Gamma(n)}-\frac{\Gamma(n-1)^3}{\Gamma(n-1+a)^3}\frac{\Gamma(n+a)}{\Gamma(n-1)}\right)\nonumber\\
 & \left.+\frac{2a-1}{\Gamma(a+1)}\left(\frac{\Gamma(n)^3}{\Gamma(n+a)^3}\frac{\Gamma(n+a)}{\Gamma(n)}-\frac{\Gamma(n-1)^3}{\Gamma(n-1+a)^3}\frac{\Gamma(n-1+a)}{\Gamma(n-1)}\right)\right].
 \end{align*}
On one hand,
\begin{align*}
 & \sum_{n=2}^\infty\frac{3(a+1)}{\Gamma(3a+1)}\left(\frac{\Gamma(n)^3}{\Gamma(n+a)^3}\frac{\Gamma(n+3a)}{\Gamma(n)}-\frac{\Gamma(n-1)^3}{\Gamma(n-1+a)^3}\frac{\Gamma(n-1+3a)}{\Gamma(n-1)}\right)\\
& =\frac{3(a+1)}{\Gamma(3a+1)}\sum_{n=2}^\infty\frac{\Gamma(n-1)^2\Gamma(n-1+3a)}{\Gamma(n+a)^3}\left(-3(n-1)a^2-a^3\right)\\
 & =\frac{3(a+1)}{\Gamma(3a\!+\!1)}\!\left[-3a^2\sum_{n=0}^\infty\frac{\Gamma(n\!+\!1)\Gamma(n\!+\!2)\Gamma(n\!+\!1\!+\!3a)}{\Gamma(n+2+a)^3}\!-\!a^3\sum_{n=0}^\infty\frac{\Gamma(n\!+\!1)^2\Gamma(n\!+\!1\!+\!3a)}{\Gamma(n+2+a)^3}\right]\\

 & =\frac{3(a+1)}{\Gamma(a+2)^3}\left[-3a^2
{}_{4}F_3 \Bigl( \begin{matrix}
{1,1,2,3a+1} \nonumber\\
{a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)
 -
a^3
{}_{4}F_3 \Bigl( \begin{matrix}
{1,1,1,3a+1} \\
{a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)
\right].
  \end{align*}
On the other hand,  once again by the cancellation argument, we clearly have
\begin{equation*}
\sum_{n=2}^\infty\left(\frac{\Gamma(n)^3}{\Gamma(n+a)^3}\frac{\Gamma(n+1+a)}{\Gamma(n)}-\frac{\Gamma(n-1)^3}{\Gamma(n-1+a)^3}\frac{\Gamma(n+a)}{\Gamma(n-1)}\right)=-\frac{\Gamma(a+2)}{\Gamma(a+1)^3}=-\frac{a+1}{\Gamma(a+1)^2}
\end{equation*}
and
\begin{equation*}
\sum_{n=2}^\infty \left(\frac{\Gamma(n)^3}{\Gamma(n+a)^3}\frac{\Gamma(n+a)}{\Gamma(n)}-\frac{\Gamma(n-1)^3}{\Gamma(n-1+a)^3}\frac{\Gamma(n-1+a)}{\Gamma(n-1)}\right)=-\frac{1}{\Gamma(a+1)^2}.
\end{equation*}
Putting together those three contributions, we obtain that
\begin{align}
 \mathbb{E}[L^3] & -\frac{b}{\Gamma(a+1)^3} = \frac{3b(a+1) }{(2a-1)\Gamma(a+2)^3}\left(-3a^2{}_{4}F_3 \Bigl( \begin{matrix}
{1,1,2,3a+1} \nonumber\\
{a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr) \right. \\
& \left. -a^3
{}_{4}F_3 \Bigl( \begin{matrix}
{1,1,1,3a+1} \\
{a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)
\right)+\frac{b}{2a-1}\left(\frac{3(a+1)}{\Gamma(a+1)^3}-\frac{(2a-1)}{\Gamma(a+1)^3}\right)\nonumber
 \end{align}
which can be rewritten as
 \begin{align}
\hskip0.5cm \mathbb{E}[L^3] & = \frac{3b(a+1) }{(2a-1)\Gamma(a+2)^3}\left(-3a^2{}_{4}F_3 \Bigl( \begin{matrix}
{1,1,2,3a+1} \\
{a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)\right.\nonumber \\\
& \left. -a^3
{}_{4}F_3 \Bigl( \begin{matrix}
{1,1,1,3a+1} \\
{a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)
\right)+\frac{3b(a+1)}{(2a-1)\Gamma(a+1)^3} \nonumber
\end{align}
which is exactly what we wanted to prove.
Now we are going to establish \eqref{T-NHI3} by the calculation of the fourth moment of $L$. It follows from identity (B.22)  in \cite{Bercu18} that
\begin{align}
\mathbb{E}[S_n^4] = &\frac{1}{(2a-1)^2\Gamma(n)} \left(\frac{\Gamma(n+4a)}{(4a-1)\Gamma(4a)}6(2a^2+2a-1))-2(3n+2+2a)\frac{\Gamma(n+2a)}{\Gamma(2a)}\right.\nonumber\\
 &\left.+\frac{4a^2-12a+5}{(4a-1)}\Gamma(n+1)+3\Gamma(n+2)\right).
 \label{MOMSN4}
\end{align}
Using \eqref{TEL}, we get  that
\begin{equation*}
\mathbb{E}[L^4] \Gamma(a+1)^4-1= \sum_{n=2}^\infty (a_n^4\mathbb{E}[S_n^4]-a_{n-1}^4\mathbb{E}[S_{n-1}^4])
\end{equation*}
\begin{align}
 & = \frac{\Gamma(a\!+\!1)^4}{(2a\!-\!1)^2}\sum_{n=2}^\infty\left[\frac{6(2a^2\!+\!2a\!-\!1)}{(4a-1)\Gamma(4a)}\left(\!\frac{\Gamma(n)^4}{\Gamma(n\!+\!a)^4}\frac{\Gamma(n\!+\!4a)}{\Gamma(n)} \!-\!\frac{\Gamma(n\!-\!1)^4}{\Gamma(n\!-\!1\!+\!a)^4}\frac{\Gamma(n\!-\!1\!+\!4a)}{\Gamma(n-1)}\!\right)\right.\nonumber\\
 &\hskip0.5cm -\frac{6}{\Gamma(2a)}\left(\frac{\Gamma(n)^4}{\Gamma(n+a)^4}\frac{\Gamma(n+1+2a)}{\Gamma(n)}-\frac{\Gamma(n-1)^4}{\Gamma(n-1+a)^4}\frac{\Gamma(n+2a)}{\Gamma(n-1)}\right)\nonumber\\
 &\hskip0.5cm  \left.+\frac{4(2a-1)}{\Gamma(2a)}\left(\frac{\Gamma(n)^4}{\Gamma(n+a)^4}\frac{\Gamma(n+2a)}{\Gamma(n)}-\frac{\Gamma(n-1)^4}{\Gamma(n-1+a)^4}\frac{\Gamma(n-1+2a)}{\Gamma(n-1)}\right)\right.\nonumber\\
 &\hskip0.5cm  \left.+\frac{4a^2-12a+5}{(4a-1)}\left(\frac{\Gamma(n)^4}{\Gamma(n+a)^4}\frac{\Gamma(n+1)}{\Gamma(n)}-\frac{\Gamma(n-1)^4}{\Gamma(n-1+a)^4}\frac{\Gamma(n)}{\Gamma(n-1)}\right)\right.\nonumber\\
 &\hskip0.5cm  \left.+3\left(\frac{\Gamma(n)^4}{\Gamma(n+a)^4}\frac{\Gamma(n+2)}{\Gamma(n)}-\frac{\Gamma(n-1)^4}{\Gamma(n-1+a)^4}\frac{\Gamma(n+1)}{\Gamma(n-1)}\right)\right].\nonumber
 \end{align}
For the first part of the series, we have
\begin{align*}
& \!\sum_{n=2}^\infty\!\frac{6(2a^2\!+\!2a\!-\!1)}{(4a-1)\Gamma(4a)}\left(\frac{\Gamma(n)^4}{\Gamma(n\!+\!a)^4}\frac{\Gamma(n\!+\!4a)}{\Gamma(n)}\!-\!\frac{\Gamma(n\!-\!1)^4}{\Gamma(n\!-\!1\!+\!a)^4}\frac{\Gamma(n\!-\!1\!+\!4a)}{\Gamma(n-1)}\!\right)\\
& =\frac{6(2a^2+2a-1)}{(4a-1)\Gamma(4a)}\left[-6a^2\sum_{n=0}^\infty\frac{\Gamma(n+1)\Gamma(n+2)^2\Gamma(n+1+4a)}{\Gamma(n+2+a)^4}\right.\\
&\left.-4a^3\sum_{n=0}^\infty\frac{\Gamma(n+1)^2\Gamma(n+2)\Gamma(n+1+4a)}{\Gamma(n+2+a)^4} -a^4 \sum_{n=0}^\infty\frac{\Gamma(n+1)^3\Gamma(n+1+4a)}{\Gamma(n+2+a)^4}\right]\\
 & =\frac{24(2a^2+2a-1)a}{(4a-1)\Gamma(a+2)^4}\left[-6a^2{}_{5}F_4 \Bigl( \begin{matrix}
{1,1,2,2,4a+1} \nonumber\\
{a+2,a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr) \right. \\
 &  \left.
 -4a^3 {}_{5}F_4 \Bigl( \begin{matrix}
{1,1,1,2,4a+1} \nonumber\\
{a+2,a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)
 -a^4 {}_{5}F_4 \Bigl( \begin{matrix}
{1,1,1,1,4a+1} \\
{a+2,a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr) \right].
  \end{align*}
Using a well-known cancellation technique on the terms of the second, the third, the fourth and the fifth part of the series,
we obtain that
\begin{align*}
& \sum_{n=2}^\infty\left(-\frac{\Gamma(n)^4}{\Gamma(n+a)^4}\frac{\Gamma(n+1+2a)}{\Gamma(n)}+\frac{\Gamma(n-1)^4}{\Gamma(n-1+a)^4}\frac{\Gamma(n+2a)}{\Gamma(n-1)}\right)
=\frac{\Gamma(2a+2)}{\Gamma(a+1)^4},\\
& \sum_{n=2}^\infty\left(\frac{\Gamma(n)^4}{\Gamma(n+a)^4}\frac{\Gamma(n+2a)}{\Gamma(n)}-\frac{\Gamma(n-1)^4}{\Gamma(n-1+a)^4}\frac{\Gamma(n-1+2a)}{\Gamma(n-1)}\right)=-\frac{\Gamma(2a+1)}{\Gamma(a+1)^4},\\
& \sum_{n=2}^\infty\left(\frac{\Gamma(n)^4}{\Gamma(n+a)^4}\frac{\Gamma(n+1)}{\Gamma(n)}-\frac{\Gamma(n-1)^4}{\Gamma(n-1+a)^4}\frac{\Gamma(n)}{\Gamma(n-1)}\right)
=-\frac{1}{\Gamma(a+1)^4},\\
& \sum_{n=2}^\infty\left(\frac{\Gamma(n)^4}{\Gamma(n+a)^4}\frac{\Gamma(n+2)}{\Gamma(n)}-\frac{\Gamma(n-1)^4}{\Gamma(n-1+a)^4}\frac{\Gamma(n+1)}{\Gamma(n-1)}\right)
=-\frac{2}{\Gamma(1+a)^4}.
\end{align*}
Putting together all these contributions, we find that
\begin{eqnarray}
\mathbb{E}[L^4] & = & \frac{24a(2a^2+2a-1)}{(4a-1)(2a-1)^2\Gamma(a+1)^4}\biggl(1 -\frac{6a^2}{(a+1)^4}
{}_{5}F_4 \Bigl( \begin{matrix}
{1,1,2,2,4a+1} \nonumber\\
{a+2,a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)\\
& & -
\frac{4a^3}{(a+1)^4}
{}_{5}F_4 \Bigl( \begin{matrix}
{1,1,1,2,4a+1} \\
{a+2,a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)\nonumber \\
&  & - \frac{a^4}{(a+1)^4}
{}_{5}F_4 \Bigl( \begin{matrix}
{1,1,1,1,4a+1} \\
{a+2,a+2,a+2,a+2}\end{matrix} \Bigl|
1 \Bigr) \biggl).\nonumber
\end{eqnarray}
In all our computations, $a=2p-1$ was a real number in $]1/2,1]$ but
the hypergeometric identitities remain valid for any $a>1/2$ by analytic continuation.
\hfill $\videbox$\\

\section{Proof of the higher order moment results}
\label{S-Md}
The computation of the first order moments of the random variable $L$ has led us to different identities involving the functions
${}_{3}F_2$, ${}_{4}F_3$ and ${}_{5}F_4$ evaluated at $z=1$.
It is natural to expect some more general identity coming from the higher order moments of $L$.
This will be the core of the proof of Theorem \ref{T-NHI-GEN}.
\ \vspace{2ex}\par
First of all, in order to understand well how the elephant moves, denote by $(\mathcal{F}_n)$ the increasing sequence of $\sigma$-algebras,
$\mathcal{F}_n=\sigma(X_1,\ldots,X_n)$. It is straightforward to see from \eqref{STEPS} that the
conditional distribution $\mathcal{L}(X_{n+1}|\mathcal{F}_n)=\mathcal{R}(p_n)$, which means that
\begin{equation}
\label{NSTEPS}
   X_{n+1} = \left \{ \begin{array}{ccc}
    1 &\text{ with probability } & p_n \vspace{2ex}\\
    -1 &\text{ with probability } & 1-p_n
   \end{array} \right.
\end{equation}
where
$$
p_n=\frac{1}{2}\Bigl(1+a\,\frac{S_n}{n}\Bigr).
$$
As a matter of fact, for any time $n \geq 1$, we have from \eqref{STEPS} that
\begin{equation*}
\mathbb{E}[X_{n+1} | \mathcal{F}_{n}] =a\frac{S_n}{n} \hspace{1cm}\mbox{a.s.}
\end{equation*}
Consequently, for all $d\ge 1$, we deduce $\mathbb{E}[X_{n+1}^d | \mathcal{F}_{n}]=1$ if $d$ is even, while if $d$ is odd
\begin{equation*}
\mathbb{E}[X_{n+1}^d | \mathcal{F}_{n}] = a\frac{S_n}{n} \hspace{1cm}\mbox{a.s.}
\end{equation*}
On the one hand, if $d$ is even, $d=2d'+2\ge 1$, we obtain from \eqref{POSERW} that
\begin{eqnarray}
 \mathbb{E}[S_{n+1}^{d} | \mathcal{F}_{n}] & = &  \mathbb{E}[(S_n+X_{n+1})^{d}| \mathcal{F}_{n}]=\sum_{k=0}^{d}\binom{d}{k}S_n^{d-k} \mathbb{E}[X_{n+1}^k | \mathcal{F}_{n}], \nonumber \\
 & = & \sum_{k=0}^{d'+1}\binom{d}{2k}S_n^{d-2k} + \sum_{k=0}^{d'}\binom{d}{2k+1}S_n^{d-2k-1}a\frac{S_n}{n}, \nonumber \\
 & = &  1+ \sum_{k=0}^{d'} \left(\binom{d}{2k}+\frac{a}{n}\binom{d}{2k+1}\right)S_n^{d-2k}.
 \label{MCDSNE}
 \end{eqnarray}
On the other hand, if $d$ is odd, $d=2d'+1\ge 1$, we find from \eqref{POSERW} that
\begin{eqnarray}
 \mathbb{E}[S_{n+1}^{d} | \mathcal{F}_{n}] & = &  \mathbb{E}[(S_n+X_{n+1})^{d}| \mathcal{F}_{n}]=\sum_{k=0}^{d}\binom{d}{k}S_n^{d-k} \mathbb{E}[X_{n+1}^k | \mathcal{F}_{n}], \nonumber\\
 & = & \sum_{k=0}^{d'}\binom{d}{2k}S_n^{d-2k} + \sum_{k=0}^{d'}\binom{d}{2k+1}S_n^{d-2k-1}a\frac{S_n}{n},\nonumber\\
 & = &  \sum_{k=0}^{d'} \left(\binom{d}{2k}+\frac{a}{n}\binom{d}{2k+1}\right)S_n^{d-2k}.
 \label{MCDSN0}
 \end{eqnarray}
Identities \eqref{MCDSNE} and \eqref{MCDSN0} can be rewritten in a single form as follows. For $d$ even, $d=2d'+2$
or for $d$ odd, $d=2d'+1$, we have
\begin{equation}
\label{CONDMOMSND}
 \mathbb{E}[S_{n+1}^{d} | \mathcal{F}_{n}] = \sum_{k=0}^{d'} A_{k,d,n} S_n^{d-2k}+
 \mathds{1}_{d \text{ is even}}
 \end{equation}
where
$$
A_{k,d,n}=\binom{d}{2k}+\frac{a}{n}\binom{d}{2k+1}
$$
and
$$
\mathds{1}_{d \text{ is even}}=
\left \{ \begin{array}{cc}
    1 &\text{ if } d  \text{ is even, } \vspace{2ex}\\
    0 &\text{ if } d \text{ is odd. }
   \end{array} \right.
$$
By taking the expectation on both sides of \eqref{CONDMOMSND}, we deduce that
\begin{equation}
\label{MOMSND}
 \mathbb{E}[S_{n+1}^{d}] = \sum_{k=0}^{d'} A_{k,d,n} \mathbb{E}[S_n^{d-2k}]+
 \mathds{1}_{d \text{ is even}}
 \end{equation}
which leads for all $n \geq 1$, to
\begin{equation}
\label{INDUC}
Z_{n+1}=A_n Z_n + B_n
\end{equation}
where  $Z_n=\mathbb{E}[S_n^{d}]$, $A_n=A_{0,d,n}$ and
$$
B_n= \sum_{i=1}^{d'} A_{i,d,n} \mathbb{E}[S_n^{d-2i}]+
\mathds{1}_{d \text{ is even}}.
$$
It is not hard to see that \eqref{INDUC} implies that
\begin{equation}
\label{NINDUC}
Z_{n+1} = Z_1\prod_{k=1}^{n} A_k + \sum_{k=1}^{n} B_k \prod_{\ell=k+1}^{n} A_\ell.
\end{equation}
Consequently, we deduce from \eqref{MOMSND} and \eqref{NINDUC} that
\begin{equation*}
\mathbb{E}[S_{n+1}^{d}]= \mathbb{E}[X_{1}^{d}]\prod_{k=1}^nA_{0,d,k}+\mathds{1}_{d \text{ is even}}\sum_{k=1}^{n} \!\prod_{\ell=k+1}^n \! A_{0,d,\ell}+
\sum_{i=1}^{d'} \sum_{k=1}^{n} A_{i,d,k}\!\!\prod_{\ell=k+1}^n\!\!A_{0,d,\ell}\mathbb{E}[S_k^{d-2i}].
\end{equation*}
Replacing $A_{k,d,n}$ in the above identity, we obtain that
\begin{align}
\label{CALCSND}
 \mathbb{E}[S_{n}^{d}] & =\frac{\Gamma(n+ad)}{\Gamma(n)} \left(\frac{1}{\Gamma(ad+1)}\mathbb{E}[X_{1}^{d}]+\mathds{1}_{d \text{ is even}}\sum_{k=1}^{n-1} \frac{\Gamma(k+1)}{\Gamma(k+1+ad)}\right. \\
  + \sum_{i=1}^{d'} \sum_{k=1}^{n-1} &\left. \frac{\Gamma(d+1)}{\Gamma(d\!-\!2i\!+\!1)\Gamma(2i\!+\!2)}\left[2i\left(1\!-\!\frac{a}{k}\right)\!+\!1\!+\!\frac{ad}{k}\right]
  \frac{\Gamma(k+1)}{\Gamma(k\!+\!ad\!+\!1)}\mathbb{E}[S_k^{d-2i}]\right). \nonumber
 \end{align}
Furthermore, we have from \eqref{DEFAN} that
\begin{equation}
\lim_{n\rightarrow \infty} \frac{a_n^d \Gamma(n+ad)}{\Gamma(n)}=\Gamma(a+1)^d.
\label{LIMAND}
\end{equation}
It clearly follows from \eqref{CALCSND}, \eqref{LIMAND} and by induction on $d$ that for all $d \geq 2$ even,
\begin{equation}
\label{SUPSNDEVEN}
\sup_{n \geq 1} a_n^d \mathbb{E}[S_n^d] < \infty.
\end{equation}
Hence, we deduce from \eqref{SUPSNDEVEN} together with H\"{o}lder's inequality for all $d \geq 1$ odd,
\begin{equation*}
\label{SUPSNDODD}
\sup_{n \geq 1} a_n^d \mathbb{E}[|S_n|^d] < \infty,
\end{equation*}
which ensures that for all integer $d \geq 1$, the sequence $(L_n)$ is bounded in $\mathbb{L}^d$,
\begin{equation}
\label{SUPLND}
\sup_{n \geq 1} \mathbb{E}[|L_n|^d] < \infty.
\end{equation}
Consequently, the Lebesgue's dominated convergence theorem implies that for all integer $d \geq 1$,
\begin{equation}
\label{LIMLD}
\lim_{n\rightarrow \infty} \mathbb{E}[L_n^d]=\mathbb{E}[L^d].
\end{equation}
We are now in position to prove \eqref{T-NHIGEN}. We already saw from \eqref{TEL} that
\begin{equation}
 \mathbb{E}[L^d]\Gamma(a+1)^d- \mathbb{E}[X_1^d]  =  \sum_{k=2}^\infty\mathbb{E}[(a_kS_k)^d]- \mathbb{E}[(a_{k-1}S_{k-1})^d].
\end{equation}
On the one hand, we obtain from \eqref{DEFAN} and \eqref{CALCSND} that
\begin{align*}
 & \frac{a_n^d \Gamma(n+ad)}{\Gamma(n)} -  \frac{a_{n-1}^d\Gamma(n-1+ad)}{\Gamma(n-1)},\\
 & =  \frac{\Gamma(n)^{d-1}\Gamma(a+1)^{d} }{\Gamma(n+a)^{d}} \Gamma(n+ad) -
 \frac{\Gamma(n-1)^{d-1}\Gamma(a+1)^{d}}{\Gamma(n-1+a)^{d}}\Gamma(n-1+ad), \\
& =  \frac{\Gamma(a+1)^d \Gamma(n-1)^{d-1}}{\Gamma(n+a)^d}\Gamma(n-1+ad)\Bigl((n-1)^{d-1}(n-1+ad)-(n-1+a)^d\Bigr),\\
& = - \frac{\Gamma(a+1)^d\Gamma(n-1)^{d-1}}{\Gamma(n+a)^d}\Gamma(n-1+ad)\sum_{k=0}^{d-2}\binom{d}{k}(n-1)^ka^{d-k}.
\end{align*}
Now, by summing on $n$, we obtain that
\begin{eqnarray}
 & & \sum_{n=2}^{\infty}\Bigl(\frac{a_n^d\Gamma(n+ad)}{\Gamma(n)\Gamma(ad+1)} -  \frac{a_{n-1}^d\Gamma(n-1+ad)}{\Gamma(n-1)\Gamma(ad+1)}\Bigr)
\nonumber\\
 & = & -\frac{\Gamma(a+1)^d}{\Gamma(ad+1)}\sum_{k=0}^{d-2}\binom{d}{k}a^{d-k}\sum_{n=2}^{\infty}\frac{\Gamma(n-1)^{d-1-k}\Gamma(n)^{k}}{\Gamma(n+a)^d}\Gamma(n-1+ad), \nonumber\\
 & = & -\frac{1}{(a+1)^d}\sum_{k=0}^{d-2} \binom{k}{d}a^{d-k}
{}_{d+1}F_d \Bigl( \begin{matrix}
{1_{d-k},2_k, ad+1} \\
{a+2,a+2,\ldots,a+2}\end{matrix} \Bigl|
1 \Bigr)
\label{GENHYP1}
\end{eqnarray}
where $1_{d-k}=1,\dots,1$ of length $d-k$ and $2_{k}=2,\dots , 2$ of length $k$.
On the other hand, as we have a telescoping series,
\begin{equation}
\sum_{n=2}^{\infty} \Bigl( \frac{a_n^d \Gamma(n+ad)}{\Gamma(n)\Gamma(ad+1)} -  \frac{a_{n-1}^d\Gamma(n-1+ad)}{\Gamma(n-1)\Gamma(ad+1)} \Bigr) =\frac{\Gamma(a+1)^{d}}{ \Gamma(ad+1)}-1.
\label{GENHYP2}
\end{equation}
Finally, \eqref{GENHYP1} together with \eqref{GENHYP2} immediately lead to \eqref{T-NHIGEN}, which completes the proof
of Theorem \ref{T-NHI-GEN}.
\hfill $\videbox$\\

\vspace{-3ex}
\section{Conclusion and perspectives}
\label{S-CP}

In this paper, we have computed the moments of the position of the ERW is two different ways.
It led us to find new hypergeometric identities. However, even if we know all the moments of $L$,
its distribution is far from being known. Baur and Bertoin \cite{Baur16} established a very successful connection
between the ERW and generalized P\'olya urns. Unfortunately, the results of Chauvin and Pouyanne \cite{Chauvin11}
as well as the induction formulas of Janson \cite{Janson04} seem useless to go deeper into the knowledge
of the distribution of $L$. It would be very interesting to specify what is the exact distribution of $L$.
\ \vspace{2ex} \par
Another source of hypergeometric identities, coming from statistical physics, is the elephant random walk with stops \cite{Cressoni13}.
The ERW by stops starts at the origin at time zero, $S_0=0$. At time $n=1$, the elephant moves to the right with probability $s$ and to the left with probability $1-s$ where $s$ lies between $0$ and $1$. Then,  at any time $n+1 \geq 1$, the elephant chooses uniformly at
random an integer $k$ among the previous times $1,\ldots,n$ and we have
\begin{equation*}
   X_{n+1} = \left \{ \begin{array}{ccc}
    +X_{k} &\text{ with probability } & p, \vspace{1ex}\\
    -X_{k} &\text{ with probability } & q, \vspace{1ex}\\
    0 &\text{ with probability } & r,
   \end{array} \right.
\end{equation*}
where $p+q+r=1$. In the special case $r=0$, we find again the standard ERW.
By the same procedure as before, we obtain the following identity, which can be seen as a generalization of \eqref{T-NHI4}.
For any $a>0$ and $b>0$ such that $2a>b$,
\begin{eqnarray}
\label{T-NHI5}
\frac{1}{(2a-b)\Gamma(2a)} & = & \frac{1}{\Gamma(a+1)^2}
\!{\ }_{3}F_2 \Bigl( \begin{matrix}
{1,1,b} \\
{a+1,a+1}\end{matrix} \Bigl|
1 \Bigr) \\
   +\frac{ a^2 }{(2a-b)\Gamma(a+2)^2} &  & \hspace{-1cm} \left( b\,
{ }_{3}F_2 \Bigl( \begin{matrix}
{1,1,b+1}\\
{a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)
- 2a\,
{ }_{3}F_2 \Bigl( \begin{matrix}
{1,1,2a+1}\\
{a+2,a+2}\end{matrix} \Bigl|
1 \Bigr)\right). \nonumber
\end{eqnarray}

\section*{Acknowledgements}
The authors would like to thank Alin Bostan, Henri Cohen and Wadim Zudilin for fruitful discussions
on a preliminary version of this paper.
\bibliographystyle{acm}

redpajama_set_name: RedPajamaArXiv


I am sure anybody who has stayed in Bangalore understands what "by 2" is. Be it coffee, be it juice or soups, Bangaloreans love "by 2". We are also going "by 2" to extend our support to Earth Hour initiative. We will be shutting down half of our servers during earth hour. I am sure our programmers have written good enough code to still serve your requests fast enough.


text:
The Academy of Veterinary Technicians in Diagnostic Imaging is dedicated to enabling the accredited veterinary technician the opportunity to advance their knowledge and skills to achieve their goal and receive the recognition title: Veterinary Technician Specialist in the field of Veterinary Diagnostic Imaging (VTS-DI).
The VTS-DI specialty certification is conferred for a period of five (5) years, beginning on the first day of the month that the examination is passed and ending on the same date five (5) years later.
NAVTA has officially announced the diagnostic imaging specialty as having received provisional recognition on August, 8 2018.
Refer to the sections to the right to learn more on the certification process:
2022 EXAMINATION
CREDENTIALS REQUIREMENTS THAT EACH APPLICANT MUST SUBMIT
PRIOR TO BE DECLARED ELIGIBLE FOR EXAMINATION:
Pre-Application:
Be a veterinary technician/nurse who has graduated from an American Veterinary Medical Association (AVMA) or Canadian Veterinary Medical Association (CVMA) accredited education program in Veterinary Technology or is credentialed to practice (if available in applicant's state) as a veterinary technician/nurse by a recognized credentialing body in the United States, Canada, or other country;
Show evidence of a passing score on the Veterinary Technician National Exam (VTNE) or State licensing exam.
Be subject to the voting on of specific candidates for unrecognized programs or foreign credentialing by the Board of Regents.
Five years (a minimum of 10,000 hrs.) of work experience or its equivalent in the field of veterinary medicine with 75% of that work experience (7,500 hrs.) dedicated in the field of Diagnostic Imaging, clinical or research based. All work experience must be completed within five to seven years prior to the application.
Continuing Education:
A minimum of forty (40) continuing education (CE) hours related to veterinary diagnostic imaging and advanced imaging modalities. The continuing education must be completed within the last five to seven years immediately prior to submitting the application.
At least ten (10) hours of the CE must be completed within the year of submission.
The CE must be RACE-approved or its equivalent, and/or received from an accredited veterinary technician school, school of veterinary medicine, or national/state/local conferences with approved CE in diagnostic imaging methods and/or use of advanced imaging modalities.
Proof of attendance is required. Applicants must provide scanned copies provided by the organization to show CE content, relevance and be given by an approved speaker (see AVTDI Bylaws for more specific information on CE requirements).
The CE should be listed in the provided template format for review by the Credentials Approval Committee (CE documentation template is provided in the Pre-Application Packet).
Letters of Recommendation:
Provide two (2) letters of recommendation (refer to Articles and Bylaws for approved references), submitted separately and confidentially to the AVTDI.
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All Pre-Application requirements are to be submitted directly on the
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Deadline for packet submission for this application term is on October 1, 2021 at 12:00 AM (E.S.T)
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Please note: Submission of the AVTDI Final Application is only applicable to those applicants who have been accepted through the AVTDI Pre-Application process.
Completion of the Veterinary Diagnostic Imaging Skills Form:
The skills form documents those skills that have not been simply performed but mastered by the candidate and are necessary to practice as a veterinary technician at an advanced level in the field of diagnostic imaging.
The skills form may be completed within the last five (5) years immediately prior to submitting the application;
Section A (General Radiology) and Section C (Supplemental Skills/Nursing and Pharmacology) may not be duplicated.
Skills in Section B (Advanced Imaging and Contrast Examinations) may be duplicated only if those skills are demonstrated using a different modality.
Refer to the AVTDI Articles and Bylaws for approved signers for skills form and for additional information on completing this form.
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Case record logs are to be maintained for one (1) year immediately preceding the submission of the application. The cases shall reflect the mastery of general radiology knowledge and advanced imaging skills required to be applied for imaging of multiple species of veterinary patients;
A minimum of forty-five (45) cases must be recorded, not to exceed sixty (60);
Recorded cases should have a minimum of two (2) species documented;
A minimum of one (1) skill from the veterinary diagnostic imaging skills form should be cross referenced in each case log.
Detailed Case Reports:
Six (6) detailed case reports original to the applicant are to be maintained for one (1) year immediately preceding the submission of the pre-application.
All case reports must include their corresponding case log number.
Recorded cases should have a minimum of two (2) species.
New for 2022 Applicants:
All imaging cases must be anonymized and submitted in a digital format;
Two (2) to five (5) images representative of the imaging study in each case report will be submitted with the final application and written case reports.
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text:
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A vocational education curriculum for those students interested in enhancing their career skills in a specific field. Students only focus on major courses without taking any general education classes. In order to obtain a certificate, students are required to complete a minimum of 18 units or more depending on the major.
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text:
What Is Volcanic Glass?
B. Leslie Baird
After a volcanic eruption, some high viscosity molten lava can cool into what is known as volcanic glass.
Volcanic glass is that which is naturally formed when high viscosity molten lava cools. The glass is created because this lava contains less water and more crystal material. Its chemical composition is similar to that of granite, which is composed of alkali feldspar and quartz. The materials in the magma, the water content, and speed of cooling all contribute to a variety of natural glasses.
Pumice igneous rocks contain volcanic ash, which has very small amounts of volcanic glass.
Magma can crystallize while cooling. The higher the viscosity of the lava, the more likely it will be to turn into glass. Igneous rock is also produced from the cooling of lava, and often contains crystal particles. Much of Earth's crust is formed from igneous rock material.
Rapid vaporization of rhyolitic or felsic lava produces the volcanic glass called obsidian. While obsidian can be formed with a variety of crystals, it has high silica content. This glass is usually black in color, but the presence of hematite can create brown, green, and red shades as well. Mount Helka in Iceland is one of the volcanoes known for producing obsidian. While it is a glass, it is also considered a natural gemstone.
Volcanic glass can form when lava cools.
Tachylyte is an igneous rock that is glassy in appearance. It is low in silica and formed when non-acidic lava cools very quickly. This glassy rock is commonly black with thin brown sections. Magnetite granules can be found in tachylyte as well. This type of volcanic glass can be found in the Hawaiian Islands and the Inner Hebrides of Scotland.
Pele's tears are small pieces of lava that have cooled quickly. These pieces become teardrop-shaped and black in color. Often found in Hawaii, the glass was named for the Hawaiian goddess of volcanoes. Apache tears are another type of black volcanic glass. These tear-shaped pieces of obsidian can be translucent when held to light, and sometimes include shades of brown or red.
Perlite is a volcanic rock, and can be used in planting mixtures to aid in aeration.
Pumicite is volcanic ash and is very abrasive. It contains very small particles of volcanic glass. The ash can be used for polishing and grinding. Perlite is a volcanic rock that can be used as an insulator in building construction. It can also be found in potting soils, where it is used to improve drainage and aeration.
Spontaneous changes, devitrification, occur in volcanic glass that can make it become stone-like in appearance. These changes frequently begin in areas where larger crystals or cracks are present in the glass. Over time, the glass will become crystals of feldspar, tridymite, and quartz. This process occurs over a long period of time — up to millions of years — but most volcanic glass found is less than 65 million years old.
What Is a Volcanic Rock?
What are Some Types of Lava Formations?
What is Magma?
What is Obsidian?
What is Glass?
What is Igneous Rock?
By: R. Sueswit Apriliant
By: Patricia Hofmeester
By: Vibe Images
By: monamakela.com


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State's Attorney's Office
Montgomery County, MD
John McCarthy, State's Attorney
Prosecution Units
District Court Division
Felony Division l
Felony Division ll
Felony Division lll
Gang Division
Juvenile Court Division
Special Prosecutions Division
Special Victims Division
Citizen Complaint Bureau
Cold Case Review
Collateral Revision
CPU Bond Division
Information Systems Support
Interagency Truancy
IT Division
Mental Health Courts
Police Academy Training Program
Truancy Prevention
Victims/Witness Home
Abused Persons Program
Protective & Peace Orders
Victim Assistance Sexual Assault Program (VASAP)
Victim Information & Notification Everyday (VINE)
Teen Court
eDiscovery Manuals
Interagency Truancy Review Board
Department of Juvenile Justice, Intake Department: 301.279.1580
Assistant State's Attorney, George Simms: 240.777.7383
Our State's Attorney John McCarthy strongly believes education prevents crime. With this in mind he has tasked a number of Assistant State's Attorneys with the responsibilities of monitoring Montgomery County schools and their truancy abatement programs. Mr. McCarthy believes that an interagency approach is crucial in helping our children become adults.
Attending school is a normal and important part of the growing up process. When a student is repeatedly absent without an obvious reason, it is a warning of serious difficulties. Research data shows that students who become truant and eventually drop out of school put themselves at a long-term disadvantage for becoming productive citizens.
Parents bear primary responsibility for ensuring that their child regularly attends school. It is the school's role to monitor student attendance and provide meaningful educational incentives that promote good attendance. Many factors can impact school attendance. If school attendance is an issue for your child, please contact your child's teacher or school counselor for assistance. Another resource is the pupil personnel worker assigned to your child's school. Contact your child's school for the name and contact information for your pupil personnel worker.
Because truancy is a community issue, Montgomery County has an interagency workgroup to address truancy and dropout prevention. The following agencies are members of this workgroup:
Montgomery County Department of Health and Human Services-Child Welfare Services and Health Services
Maryland Department of Juvenile Services
Montgomery County, Maryland State's Attorney's Office
Montgomery County Department of Police, Family Services Division
Montgomery County Housing Opportunities Commission
Each agency has agreed to participate in an overall plan to respond to the habitually truant student. This plan includes school-based and Department of Student Services interventions, a referral process for cases resistant to intervention, an interagency Truancy Review Board, and a system for referral to court. The plan applies to all students between five and sixteen years of age. Legal action applies to each person who has legal custody or care and control of a child who is 5 years of age or older and under 16.
The Montgomery County State's Attorney's Office works with the Montgomery County Public Schools, the Department of Juvenile Justice, the Montgomery County Police Department, and the Montgomery County Department of Health and Human Services to combat truancy in our public schools.
The objective of this process is to promote school attendance. As we have observed, children who are not in school often become potential victims of criminal activity or actively commit crimes.
Those children who have a certain number of unexcused absences are identified by the school and the children and their parents are worked with by the school community to address the reasons for the truancy and improve the child's attendance at school. Those children who attendance does not improve are referred to the next stage of the process which is the Interagency Truancy Review Board.
The parents and the child are called in for a meeting to find out why the child is remaining truant from school. The Board generally enters into an attendance plan with the parent and child. Those cases in which the child remains truant are then referred to the Montgomery County State's Attorney's Office for possible prosecution.
If you have any questions regarding problems with your child's attendance at school, you may contact the following:
The Pupil Personnel Worker at your child's school;
Department of Juvenile Justice, Intake Department - (301) 279-1580;
State's Attorney's Office, Assistant State's Attorney, George Simms (240) 777-7383
If a teacher, administrator, parent, neighbor, or friend believes that they know of a child who is truant and in need of assistance please contact:
Mr. Steven Neff
Director of Pupil Personnel and Attendance Services
Steve_Neff@mcpsmd.org
Contact the Truancy Prevention Program Staff


I hope you've been enjoying looking at the images that I've taken. I've been fortunate to have been given the opportunity to photograph so many amazing people! Life goes by quickly and photography helps to preserve all the great memories.
Photography can do great things. It can bring comfort to a loved one far away, it can bring confidence to a shy teenager, it can remind you of things and events that you had forgotten, it can make you laugh, and it can bring you joy. But most of all it can remind you that you are a loving, valuable, amazing person!
Clients hire me because I am dependable, prompt, easy going, and I get the job done. I enjoy meeting people and helping them to photograph at their best. I love hearing their stories and getting to know them. The best compliment is when my clients recommend me to others.
I would love to share in your awesomeness! Contact me today! I am located in Little Rock, Arkansas.


Clemson, S.C. – Clemson's men's and women's basketball teams will hold public scrimmages at Littlejohn Coliseum Saturday. The Lady Tigers will scrimmage at 10:00 AM, while the men will scrimmage at 12:30 PM. The scrimmages are free and open to the public.
Van Hilderbrand, Clemson ticket manager, announced that Clemson fans will be able to purchase basketball season tickets at the coliseum beginning at 11:00 AM. The ticket office has place balloons in seats that are still available for purchase. The purchase opportunity will be available until 1:30 PM.


This is the online version of Sky and Telescope, the astronomy magazine for the amateur and enthusiast community. Materials include news articles, observing tips, information on products and equipment, how-to advice, and other features. Resources for observing include highlights of currently visible objects, a weekly calendar of celestial events, suggestions for observing projects, an interactive sky chart and observing tools, and other items. Other resources include a glossary of astronomical terms, software, blogs, a photo gallery, and a search tool for locating clubs and astronomical organizations.


text:
Eugene Weekly : Music : 2.26.09
Archive by EW-staffPosted on 02/24/2012
Move and Groove
Some bands just know how to make you move. They effortlessly combine their talents to create a sound that turns each song into a musical dance partner who first shows off a few moves, then beckons you onto the dance floor and finally sets you to grooving in ways you're not likely to soon forget. Basin & Range is one of those bands, and each member supplies the necessary musical bones, flesh and sinews to create the ultimate dance partner on their new EP, Release Party.
Not content to simply make groovy jazz or nasty funk, the band combines both of these styles with R&B and rock to make for one heck of a dance party. "Muggy" is a perfectly funky jam replete with sax and guitar solos that shimmer and echo in assorted directions. This song is emblematic of the entire album's presentation, as is the head-bobbing, hand-clapping stomper "Parade." It's a good time from start to finish. The album's five tracks are similar only in that they get you on your feet and moving before you know what's hit you. Apart from that, these songs are diverse and interesting, and you never get the same thing twice. If you love to dance, this is a show you will be glad you went to. Basin & Range play a CD release show at 9:30 pm Friday, Feb. 27, at the Oak Street Speakeasy. 21+. Free.
— Brian Palmer
Whose Clues?
It's the kind of listing that makes a certain kind of music fan sit right up and hit Google: Sam Bond's is bringing us The Clues, a band featuring former Arcade Fire member Brendan Reed and former Unicorns fellow Alden Penner. Last fall, Pitchfork.com posted one song, "Perfect Fit," from The Clues' forthcoming self-titled record (due on Constellation Records in May). It's more Unicorns than AF, at least at first; it's a little playful, with goofy vocals and a major shift in tone near the end. A video of the band at Pop Montreal shows them going from a spare vocals/guitar song to one with quirky whistling and the occasional vocal harmony — and a tiny bit of epic swelling and swooning. The Constellation site describes the record as "idiosyncratic and enigmatic, but exuberantly infectious and approachable," and a wee sample of early recordings (on www.villavillanola.com) certainly backs up that claim. The Clues, Fiction and Leo London play at 8:30 pm Sunday, March 1, at Sam Bond's. 21+. $8. — Molly Templeton
Synth Jam
STS9's stoned electronics feel sort of like raving in Jell-O. The electronic answer to jam bands, this instrumental outfit's been making its unusual blend of drum 'n' bass, rock and roll and straight-up funk for more than 10 years, first as Sector 9, but after copyright conflicts with a longboard company, as Sound Tribe Sector 9 (or STS9). The hippie-esque name belies the hypnotic nature of the music, in which thumping drums give way to smooth, repetitive guitar licks; it's kinda like what might happen if My Bloody Valentine started making dance music. Their latest record, Peaceblaster, explores the dichotomy between the two words its title mushes together, sleek, seamless live instrumentals bleeding into each other as pounding percussion adds a sense of tension and urgency. Live, these guys know how to keep a party going: expect a two-hour long set complete with remixes and improvised alterations to old songs. STS9 and Pnuma PA Set play at 8 pm Thursday, March 5, at the McDonald Theatre. $22.50 adv., $25 door. — Sara Brickner
Demon Chaser
With contemporary Irish music, there's always a sense of the past meeting the future, and no more so than when a band takes a traditional song and gives it a new life. Circled by Hounds have done a great job with the song "A Man You Don't Meet Everyday." The arrangement they've chosen is understated, with guitar strumming behind the lilting flute of new member Hanz Araki. Sung by Kathryn Claire, it's reassuring and friendly, a lovely gift to give their fans as a hidden track on their new CD, Howl No Demon Louder.
Those who enjoyed the zest of Eugene band Toad in the Hole will be glad to know the remains of that band soldier on. Now based in Portland, the band that was a three-piece on 2006's Chasing Our Tales has expanded to a quintet.
About half the tracks on Howl are instrumental and are underpinned by the rhythm of guest banjoist Ezra Holbrook and accordionist Johnny B. Connolly. Each band member's playing is wonderful, with Claire and Sarah Dee's weaving fiddles turning cartwheels in the tune "Open Road." Strong percussion is an anchor throughout, as is Claire's warm and openhearted voice. Though most of the songs CBH chose for Howl are traditionals, one tune in particular stands out: "The Tinkerman," an original written by Claire and guitarist Matthew Hayward-McDonald. The song grows from a plaintive fiddle soaring over a waltzing rhythm to a driving, percussive, unstoppered force that swirls on itself until it comes to rest again.
While CBH's earlier CD had an energy to it that stemmed from the sheer joy of playing, the group now seems to have matured into a deeper palette. When everyone plays together, they become one smooth ensemble, a band with the confidence to appreciate the past while navigating new terrain and making each song their own. Circled by Hounds celebrate the release of their new CD at 5 pm Saturday, Feb. 28, at Tsunami Books (all ages) and with Casey Neill and the Norway Rats at 9:30 pm Saturday, Feb. 28, at Sam Bond's Garage (21+, $7). — Vanessa Salvia
Are You In Or Are You Out?
Archive 10 years ago
If five beach bums tried to surf their way to the classic California rock sound (The Doors, The Byrds, etc.) but instead got lost in a cloud of pot smoke and found themselves at a goth house party, you would end up with something like The Growlers. The Growlers are from Orange County; they like a little reverb on their guitars; there's a little bit of dark soul in their sound; and they do look and act baked much of the time, but that's where the similarities to those other bands end. Continue reading →
Little Shop of Horrors at Pleasant Hill
Just like the herbaceous monstrosity at the heart of this campy musical romp, the cult appeal of Little Shop of Horrors just keeps on growing. But if you've merely seen the movie starring Steve Martin and Rick Moranis, you only know half the story. Continue reading →
30 Years of PIELC
ArchiveNews 11 years ago
Tell your friends "I'm going to spend the weekend at a law conference" and they'll figure you are in for a really horrible couple days. But when it comes to the UO's Public Interest Environmental Law Conference (PIELC), attendees are actually in for some fun and excitement. Continue reading →
Tags: PIELC / Public Interest Environmental Law Conference


and respected,they were the custodians of our traditional cultures and beliefs.
The words and decision of the king is law and everyone must obey.
are still a handful of powerful kings (traditional rulers) that still command the respect of thousands of people across Nigeria.
This post contains the list of the top 10 most powerful kings in Nigeria 2019 in no particular order.
The Emir of Kano is one of the most powerful and respected King in Nigeria.The title commands respect from millions of muslims in Nigeria.
The current Emir of Kano is Mallam Muhammad Sanusi II,he was formerly the governor of the Central Bank of Nigeria. His predecessor Alhaji Dr. Ado Bayero ruled for 51 years before his death.
The title of Oyo is a much respected title among the yorubas. The Alaafin Of Oyo is regarded as one of the most powerful kings in Nigeria.
The ruling Alaafin of Oyo is Lamidi Olayiwola III,who is the successor of AlaafinGbadegesi Ladigbolu II. In ancient times the Alaafin Of Oyo was seen as a supreme leader by the yoruba people.
The title of the Sultan Of Sokoto has been held in high esteem,he is seen as the spiritual leader to millions of Muslims in Nigeria.
This position was respected dating back to the era of the British colonization.
The current Sultan is Muhammadu Sa'ad Abubakar IV,he was the successor of his brother Muhammadu Maccido who lost his life in a plane crash.
Ooni of Ile-Ife is also one of the most powerful kings and title in Nigeria.
The current Ooni of Ile-Ife is Oba Ogunwusi,while the previous Ooni of Ile-Ife is Ooni Okunade Adele Sijuwade. The late Ooni Okunade Adele Sijuwade was one of the richest kings in Nigeria before his death.
According to rumors,the Ooni of Ile ife has a strong role in determining who becomes the governor of Osun state.
The Dein of Agbor made history as one of the youngest kings to ascend to the throne after the death of his father.
The current Dein of Agbor is called Benjamin Ikenchukwu Keagborekuzi.
The Oba of Benin is also one of the most powerful kings in Nigeria. The current Oba of Benin is Oba Ehinede who succeded his late father Omo N'Oba N'Edo Uku Akpolokpolo Erediauwa.
The title of Oba of Benin dates back to the ancient Kingdom of Benin before the invasion of the colonial masters. The oba of Benin also plays an important role in issues regarding to politics in the state.
The Oba of Lagos is the traditional ruler and ceremonial head of one of the largest cities in Africa.
The current Oba of Lagos is his Royal Highness Oba Rilwan Babatunde Osuolale Aremu Akiolu,he is also referred to as the Eleko of Eko.
The Oba of Lagos is powerful and also very much involved in government politics of the state.Lagos is the most populous city in Nigeria.
The Olu of Warri is the traditional ruler and king of the Warri people. The current Olu of Warri is Godwin Toju Emiko who is a lawyer by profession.
He is the King of the Itsekiri people.
The Obi of Onitsha is a much revered and respected title. The current Obi of Onitsha is Igwe Nnayelugo Alfred Nnaemeka Achebe.
He is the traditional ruler and Obi of Onitsha people. The king is rated as one of the most powerful and richest kings in Nigeria.
The Olu Of Ibadan is a very prominent title held by Oba Saliu Adetunji.
Olubadan means Lord of Ibadan.The Title was once held by late Oba Samuel Odulana Odugade before he passed on at the age of 101.
Ibadan is an ancient Kingdom and the title of Olubadan of Ibadan has been respected and revered for generations.


Both my co-writer Cameron and I have recently penned screeds on a worrisome trend we've seen in Personal Finance blogs – the hyperbolic obsession with "Debt Zero". "Debt Zero" is, of course, the idea that above all else Debt Must Be Paid Down.
The first clue that something is wrong with that story is the inclusion of the word "must". There are very few absolutes in life – the snarky amongst you will acknowledge death and taxes – and in this situation is no different. You see, it's naive to assume that things come for free. Every single thing you do is a trade off.


An ogling Joe Calderone returned to the stage to present Britney Spears with the Michael Jackson Video Vanguard Award. The two together then introduced Beyoncé, who, before starting to sing "Love on Top," confirmed pregnancy rumors by saying "I want you to feel the love that's growing inside of me." She ended the song by rubbing her belly to reveal her baby bump as a proud and expectant father Jay-Z received congratulations from the audience.
The girls from the "Jersey Shore" presented the "Best Female Video Award" to Lady Gaga and, in an intergenerational twist, was joined by Academy Award®-winning actress Cloris Leachman, who joked that the audience shouldn't confuse her with Betty White.
Presenters at the"2011 MTV Video Music Awards" included Katy Perry, Miley Cirus, Paul Rudd, Shaun White, Will Ferrell, Jack Black, Zoe Saldana, Kevin Hart and Victoria Justice join Jonah Hill, Rick Ross, Seth Rogen, Selena Gomez, Kim Kardashian, Odd Future, Nicki Minaj, Taylor Lautner, Drake, Joe Jonas and Jared Leto.
Official sponsors of the 2011 MTV Video Music Awards are 5® Gum, COVERGIRL, Dove® go fresh Deodorant, Kia Motors America, PacSun, Sharpie®, State Farm®, Taco Bell® and Verizon.


Jesus Michael Martinez (JMM Sales & Consulting, Ltd.), who represents Cyria Group, Inc. and serves as the company's Vice President of Sales and Marketing, was recently elected by his peers to the C.A.R.E. (Cooperative Association of Resort Exchangers) Board of Directors at the recent Spring Conference in Virginia Beach, VA.
Cyria Group is a marketing and sales support company focused on the fulfillment of timeshare trade-ins with complete commitment to retention on behalf of its resort partners and developers. Cyria's mission aligns with the core values of C.A.R.E., to utilize inventory and generate revenue while continuing to offer expanded exchange and fulfillment options.
Mr. Martinez began attending C.A.R.E. conferences in 2013. One year later, Jesus was honored with the prestigious Rookie of the Year Award. This award is presented to a first-time attendee selected by the Board of Directors for outstanding contributions to the success of C.A.R.E. through participation on committees, panel discussions, or in any capacity that benefits conference attendees or the organization as a whole.
"C.A.R.E., the Cooperative Association of Resort Exchangers, provides a networking platform and travel industry educational opportunities. We off our Member base the opportunity to exchange inventory among themselves and to utilize inventory and generate revenue through wholesale rentals. In addition, the C.A.R.E. Association hosts an Annual Conference every year for Members to attend and share best practices while having the opportunity to network", explains Mr. Martinez.
Jesus Martinez's role as a new board member will focus on advising the C.A.R.E. committees for MarketPlace, Strategic Planning, and New Members. Mr. Martinez has worked in the vacation ownership industry for the past seven years and is currently the President and Owner of JMM Sales & Consulting, Ltd., as well as the V.P. of Cyria Group, Inc. Additionally, Jesus represents ARDA, CVOA, and AMDETUR member companies in good standing.
Cyria Group exclusively offers timeshare consultation services to developers, timeshare owners, and vacation ownership resorts. With a proven track record, Cyria Group specifically supports ethical solutions to help owners who need to be relieved of their interests by giving them access industry-supported programs.


Last month I showed another view of this sculpture, Spirit Catcher, that sits on the waterfront in Barrie, Ontario. Here is another look at it.
The following song is sung to the Fisher and Water Spirit.
I love your photo of the sculpture - with the birds flying by it... beautiful!
and I LOVE that pow wow video! I'm going to put it on my FB page! I keep trying to catch the hoop dancers when they come to Flagstaff, but I'm always out of town... I'll keep hoping!
Mary MacADNski: We have a few pow-wows here every year but I have not attended one in a long time. Summer is soooo busy. I love the music. I like this 'spirit catcher' shot too.
Great shots Mary. Its an impressive sculpture.
Mary MacADNski: It moves too. The wing tips show the air capabilities.
Mary MacADNski: I think they are on this city waterfront. There were gulls everywhere.
The birds add something to it..
Mary MacADNski: The statue moves too so this one shows best how moving things can look still.
The birds really are right at home with this beauty, Mary.
Mary MacADNski: The waterfront was filled with gulls. It was cool that I caught them there.
That looks rather quirky and I do think b&w has the edge.
Mary MacADNski: The statue moves too so I see it as dancing with the birds, yes, quirky.
Mary MacADNski: All are moving together.
B&W is excellent. It is also great that you caught it at the moment when all birds had the catcher as background - as if they are caught spirits.
Mary MacADNski: There were gulls everywhere on this city waterfront. Too many in some cases, like on the beach.
It is certainly very unusual.
Mary MacADNski: It moves too so together they are moving.
Like Lisl, I initially thought the birds were part of it - very effective!
Mary MacADNski: I love visiting the relatives and going to this park over the years.
This is wonderful and you did a great job with the crop of the Spirit catcher. I think I like the coloured one, the bird add to the picture, Mary.
Mary MacADNski: I prefer the coloured one for a change but only for a few reasons. I don't think there any more of these in the Unassigned bin.
Is this a sculpture by Alexander Calder? Some parts of it like sharp knives, but the birds don't get hurt.
Mary MacADNski: The sculptor is Ron Baird. The like tells more about the work.
I thought at first that the birds were part of it. It looks to be a huge sculpture.
Mary MacADNski: It is huge and sits on the waterfront. I have a link there if you want more info.
Works best in mono I think, Mary - wonderful with the gulls.
Mary MacADNski: It should be done in greyscale rather than black and white. It would look good in an office in the city it resides in.
I love the movement you've captured in these images, Mary. A bit magical.
Mary MacADNski: The statue move too so I like the concept. Thank you.
Mary MacADNski: They move with the Spirit Catcher's wings - all flying together.
Incredible! Love those birds...gives scale to the sculpture.
Mary MacADNski: They are all moving, the birds and the wings of the statue.


Alright it's practically Valentine's Day again, that time of year where our thoughts run more in the romantic direction. As a consequence, the ads flaunting his or her admiration because "He went to Jared's" (Kay's Jewelry, Macy's jewelry counter, or whatever jewelry store strikes your liking) will show you just how much your loved one really cares for you (Ahh). Or does this sound familiar, "Hallmark Cards say "Make Every Heart Happy" (have you priced greeting cards lately? It's expensive making hearts happy). Ok, indulge me one more time because I'm on a roll, Kohl's Department Store says, "Gifts From The Heart, It's Love At First Save". From your heart to their pockets, now that's what I call love. Something else that's a little unsettling to me is that we celebrate love officially only one day a year, but don't we love every day? It's much like Black History Month that we celebrate once a year; for those of us that are black aren't we black year round? Why the labels on everything, can't we just be?


DWR stands for Durable Water Repellancy -- although it's not durable enough. It's what causes water to bead up and helps the GoreTex do its thing.
And when the drysuit is 10 years old and you have tryied everything and is still lets water through the fabric. Is it dead then?
Any tips to restore breathability? I find my drysuit drier when I immerse myself (roll, practice rescues) frequently than on days where the outer fabric stays (mostly) dry.
Leave it soaking in fresh water?
Patrik: A ten-year-old drysuit that has been used extensively and been exposed to lots of UV may, in fact, be dead, but check with the manufacturer.
Sean: Without knowing what your drysuit is made of, this is hard to answer. Is it possible that you're simply sweating less on days when you immerse yourself more?


Brendan Fehr's TV directorial debut aired yesterday on Lifetime. If you missed the premiere … or you want to watch it again, please head on over to Lifetime.com and watch the short online. (Subscribers only via TV provider). "The Ugly Christmas Sweater" is available until January 7th, 2018.


Don't cry. It's going to be okay. The cows behind my house only eat tulips. Delbert's annoyed. Bud refused to answer questions. I'm not like Dannie. She appears to have a lot of friends. You shouldn't try to do that by yourself. I'd like to know the phone number of the nearest American Express office. You should start as early as you can. If you ever need someone to talk to, I'm here.
Let's get wild. Deborah tied the balloon to Jerome's wrist. We shouldn't tell them anything. Francois has changed his email address recently. Kim and Andrea got divorced, just like you predicted. She shall come here. What made him so angry then, I wonder? I'm staying at Van's house. Not everyone has as much money as you. I'm not much of a sailor.
Ritchey is now in the kitchen. Brooke was visibly disappointed. You need to improve your people skills. He felt a pain in his back. Did you hire them? Danielle is a very talented man. I'm not really that thirsty. You're angry now. Mayo doesn't express himself well. He's got more books than all the others put together.
Portuguese is a Romance language. Micheal sweated. Get me some coffee. The investment firm tricked customers into buying worthless stock. Every night I go to bed at ten. He is the best of us all. I'll tell them later. Please tell Bjorne to stop smoking. I dried the sink with a dishrag and now it's dry. The registration deadline is noon Monday.
One can't do lots of things at the same time. I'd like to go to Boston again someday. Is that still my room? Is it accessible? Laurie picked up his coffee cup and started drinking out of it. My uncle lives in an apartment. Norma has a black belt in karate. After drinking all night, Bob was dead to the world. We're just joking around. The House voted thirty-three times.
We go to work by subway. That's perfectly understandable. I thought you might actually do it. Bjorne and Ritalynne sat opposite each other. Bertrand apologized on Monday. I faxed Lori a copy of the contract. I want you to know that Ping's important to me. Here's where it gets tricky. I don't remember getting paid for the work. No one will believe us.
I'm sorry, but I have no more time. He traveled west by train. I helped Leon fill out the application form. Jared is in a great deal of pain. The Germans were very friendly. I like studying wild flowers. That market has been rapidly expanding. Allan works on Tuesday and Friday nights at the Konbini. Thank you very much for informing me that the shipment will be delayed. What do I tell Kanthan?
We all want Diana to be happy. I have one with me. I almost called Arlene again today. Shuvra didn't want to take any chances. Have you heard about her? I was mad. You have told another lie. Another patient has died. Let me explain the rules. I don't remember the combination.
The Atlantic Ocean is very big. Some days before the test, he put the wind up him properly. She did so out of curiosity. Let's share a cab. They were injured. The play is over. Last week I saw a German movie. I can't find my red hat anywhere. I won't try to persuade them. Jianyun yelled at me for no reason.
I hope this treaty will contribute to peace in the world. Horses have three gaits: Walking, trotting and galloping. She's a young student. Don't teach me how to live! Nothing was changed. Here's your drink, Dawson. Tolerant is an engineer. Where should I go to buy art books and catalogs? It's Manuel who needs convincing. Barbra and Dawson ate snails at the restaurant last night.


text:
PHILOSOPHY >
Lord of Elves And Eldils
Richard Purtill
A fascinating look at the fantasy and philosophy of C.S. Lewis and J.R.R Tolkien. The two men were friends and fellow professors at Oxford, renowned Christian thinkers who both "found it necessary to create for the purposes of their fiction other worlds?not utopias or dystopias, but different worlds."
"The great importance of [Lewis and Tolkien] is that they have succeeded in restating certain traditional values?in a way that they make an imaginative appeal to a very wide audience, young and old, traditionalist and non-traditionalist."
?Richard Purtill, Author, J.R.R. Tolkien: Myth, Morality, and Religion
About Richard Purtill
PHI022000 (PHILOSOPHY / Religious)
Doors in the Walls of the World
"There are more things in heaven and earth, Horatio, than are dreamt of in your philosophy." — Hamlet After William Shakespeare's Horatio sees the ghost of Hamlet's father, and scarcely believes his own eyes, Hamlet tells him that there is more to reality than he can know or imagine, including ghosts...
by Sri, Edward
"Don't be so judgmental!" "Why are Christians so intolerant?" "Why can't we just coexist?" In an age in which preference has replaced morality, many people find it difficult to speak the truth, afraid of the reactions they will receive if they say something is right or wrong...
Ignatius Bible (RSV), 2nd Edition Large Print - Hardcover
by Press, Ignatius
The Revised Standard Version of the Bible is acclaimed by many as the clearest, most accurate and most beautiful modern translation of the Bible. This newly designed and typeset 2nd edition of the popular Ignatius RSV Catholic Bible is a contemporary English translation that revises archaic language of the first edition, but avoids dumbing-down the text...
Ignatius Bible (RSV), 2nd Edition Large Print - Softcover
Ignatius Bible (RSV), 2nd Edition Large Print - Leather
Called to Be the Children of God
by Olson, Carl E.
This book gathers fourteen Catholic scholars to present, examine, and explain the often misunderstood process of "deification". The fifteen chapters show what "becoming God" meant for the early Church, for St. Thomas Aquinas and the greatest Dominicans, and for St. Francis and the early Franciscans...
After the Natural Law
by Hill, John Lawrence
The "natural law" worldview developed over the course of almost two thousand years beginning with Plato and Aristotle and culminating with St. Thomas Aquinas in the thirteenth century...
A special 2-DVD collector's edition of the acclaimed epic film on St. Augustine of Hippo. Filmed in Europe, the first full-length feature movie on Augustine, it uses a historic backdrop to tell the true story of one of the Church's most beloved and well-known Saints. Its message of sin, conversion, redemption is as timely today as it was in the 5th century of Augustine...
The Ninth Day
With The Ninth Day, acclaimed film director Volker Schlondorff creates a moral thriller that "succeeds in illuminating an almost unimaginably dark story" (The New York Times). Inspired by the true story of a dissident priest's temporary furlough from Dachau, The Ninth Day transforms an ethical crisis into a nail biting showdown and unspeakable horror into transcendent redemption...
A Deeper Vision
by Royal, Robert
In this wide-ranging and ambitious volume, Robert Royal, a prominent participant for many years in debates about religion and contemporary life, offers a comprehensive and balanced appraisal of the Catholic intellectual tradition in the twentieth century...
Life for Life
Auschwitz, 1941. One of the prisoners, Jan, escapes from the German concentration camp while working at a gravel pit. Thanks to the help of good-hearted people he finds shelter. There Jan hears tragic news about ten random inmates sentenced to death by starvation by the Nazis as a punishment for his escape. One among the convicts is Fr...
The Wife of Pilate and Other Stories
These three novellas from the acclaimed German writer Gertrud von le Fort, newly translated for the first time into English for this volume, are from her later works of historical fiction, in which she displays her mastery as a dramatist of ideas...
Who Designed the Designer?
The "New Atheists" are pulling no punches...
Finding True Happiness
by Spitzer, Fr. Robert J.
One of the hottest topics in contemporary culture is happiness. The explanation for this current fixation seems to lie in the contrary phenomenon—unhappiness. Despite the fact that we have tremendous access to every imaginable form of entertainment, we experience a pervading sense of insecurity, emptiness, and malaise amid sporadic peak experiences...
A Time of Renewal
by Francis, Mother Mary
Mother Mary Francis, abbess of a Poor Clare Monastery for over forty years, left an enduring legacy in her writings and in the conferences she gave to her spiritual daughters. In this work she presents beautiful meditations on the liturgical season of Lent, revealing the treasures of the liturgy to Christians in all walks of life...
On Human Life
by Montin, Giovanni Battista
On Human Life, also known by the Latin name Humanae Vitae, made headlines worldwide. Everybody talked about the document when it was issued in 1968 but few people actually read. It is perhaps the most controversial document in modern Church history...
The Hope of the Family
by Muller, Gerhard
"Today, the family is in crisis--it is in crisis worldwide", Pope Francis has said. "Young people don't want to get married, they don't get married, or they live together. Marriage is in crisis, and so the family is in crisis...


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Q: Linq to Sql Chained Intersect I can't figure out why this isn't intersecting all of the items in the loop, just the last 2.  I think it has something to do with IQueryable
var outerquery = db.Employees.Where(x => x.Name = "Smith").Select(x => x.EmployeeID);

foreach(var name in nameList){
var innerQuery = db.Employees.Where(x => x.Name = name).Select(x => x.EmployeeID);
outerquery = outerquery.Intersect(innerQuery);
}

return outerquery.ToList();

EDIT -
A more concrete example.  The table has approx 35 million records.
The table has ID, ConceptID, Word.  Words can have multiple ConceptIDs & there is 1 word per record.  I was to intersect a search string 'shoulder pain chronic' and get all the ConceptIDs that share those 3 words.  It should return:
Concept1234 - shoulder
Concept1234 - pain
Concept1234 - chronic

What I am getting (just the last 2):
Concept1234 - pain
Concept1234 - chronic

Doing an OR on 35 million records is rough even with this monster server I have & an intersect is the only way to do it in less than a second.
What I am trying to generate with LINQ to SQL (Entity Framework) is this -
SELECT ConceptID FROM WordTable WHERE Word = 'shoulder'
INTERSECT
SELECT ConceptID FROM WordTable WHERE Word = 'pain'
INTERSECT
SELECT ConceptID FROM WordTable WHERE Word = 'chronic'

A: You have outerquery inside the foreach loop which gets replaced in each iteration of the loop and you lose previous data.

redpajama_set_name: RedPajamaStackExchange


text:
Outlapsed: A stunning aerial journey into the Outback
BBFF News
BBFF2016, Byron Bay, Film Festival, Media Release
This mind-blowing montage of the Australian Outback is brought to life by the Byron Bay-based photographer and documentary filmmaker Evan Malcolm. Evan has taken his eye for the still image and used it to capture truly awe-inspiring moments in time, bringing life and character to the Australian landscape. Using a medley of aerial drone footage
And the Circus Leaves Town
And the Circus Leaves Town by director Mete Souzer is a Turkish delight. It opens with a murdered man, his head smashed against a rock. A stranger removes a letter stuffed in his mouth. We are not told what it is about or who has murdered him. The townspeople of the nearby village seem unlikely
BBFF2016, Byron Bay, Media Release
This animated adaption of the Japanese children's book of the same name comes from the academy-award nominated team behind the short film The Dam Keeper – Robert Kondo and Daisuke 'Dice' Tsutsumi. The film has won no less than seven Best Animation awards from film festivals all over the world and rightly so. It is
Ncube8 – Free Virtual Reality Workshops Announced
bbff
BBFF2016, Virtual
The Byron Bay International Film Festival is this year poised to boldly go beyond the "reality" of the silver screen, with an exploration of the potential of virtual reality (VR) in an artist's palette. The organisers are offering an artist-oriented program, called Ncube8, to "open new dimensions" to talented visual creatives keen to take their


To answer an incoming call press the Green phone key on the R25 / R25T.
To end a call press the Red phone key on the R25 / R25T.
How do I lock / unlock the Aspera R25T?
To lock & unlock the Aspera R25 / R25T press the menu (Top left key) button followed by the star * key which is located under the 7 Key on your keypad.
To access the menu on the Aspera R25 / R25T press the top left key or the centre key.
The Aspera R25 / R25T takes a Micro SD card, A Micro SD card adds storage to the phone. The Aspera R25 / R25T is compatible with up to 32GB Micro SD card.
What is the difference between the Aspera R25/ R25T?
The R25 does not have the Vodafone and Optus 3G 900MHz network, meaning the phone will not work on either the Vodafone network or the Optus Network. However, The Aspera R25T has the Vodafone and Optus 3G 900MHz network and has the Telstra 3G network, meaning it will work on any Australia 3G network.


text:
Charlotte McClure
2021 Statistics Articles Videos
College: Virginia
Class: Freshman
Conference: Atlantic Coast
Hometown: Ridgeland, SC
TopDrawerSoccer Ranking
Top 100 Ranking: N/A
Conference Top 20 Ranking:: N/A - Atlantic Coast
* Conference Game
Sa PCT
vs Richmond 0 0 0 0 0 0 0 0 0 0 0.0
@ West Virginia 0 0 0 0 0 0 0 0 0 0 0.0
@ George Mason 0 0 0 0 0 0 0 0 0 0 0.0
vs George Washington 0 0 0 0 0 0 0 0 0 0 0.0
vs Santa Clara 0 0 0 0 0 0 0 0 0 0 0.0
vs James Madison 0 0 0 0 0 0 0 0 0 0 0.0
vs Penn State 0 0 0 0 0 0 0 0 0 0 0.0
vs Oklahoma 0 0 0 0 0 0 0 0 0 0 0.0
@ Wake Forest* 0 0 0 0 0 0 0 0 0 0 0.0
vs Duke* 0 0 0 0 0 0 0 0 0 0 0.0
@ NC State* 0 0 0 0 0 0 0 0 0 0 0.0
@ North Carolina* 0 0 0 0 0 0 0 0 0 0 0.0
@ Boston College* 0 0 0 0 0 0 0 0 0 0 0.0
@ Syracuse* 0 0 0 0 0 0 0 0 0 0 0.0
vs Notre Dame* 0 0 0 0 0 0 0 0 0 0 0.0
vs Louisville* 0 0 0 0 0 0 0 0 0 0 0.0
vs Miami (FL)* 0 0 0 0 0 0 0 0 0 0 0.0
@ Florida State* 0 0 0 0 0 0 0 0 0 0 0.0
vs Clemson 0 0 0 0 0 0 0 0 0 0 0.0
vs Florida State 0 0 0 0 0 0 0 0 0 0 0.0
vs High Point 0 0 0 0 0 0 0 0 0 0 0.0
vs Wisconsin-Milwaukee 0 0 0 0 0 0 0 0 0 0 0.0
vs BYU 0 0 0 0 0 0 0 0 0 0 0.0
Please note that all stats are submitted by each individual college. Please contact that college's Sports Information Department directly regarding any mistakes or discrepancies.


Esta es la fuuncion que comienza a tomar audio desde la Kinect, gobernada desde el programa del usuario, nuestro danzante.
Starts capturing audio from the device. The data can be read using the returned stream.
readStaleThreshold: If there are no reads to the stream for longer than this threshold the DMO discards any buffered audio. This prevents stale data from being returned in scenarios such as speech recognition and dialog systems, when the consumption of audio samples may stop for a while.
Higher values for this threshold (e.g.: multiple seconds) are advisable if clients notice dropped audio packets, such as is more likely in high CPU-load scenarios.
Pass TimeSpan.MaxValue to avoid hitting this threshold.
A stream of audio samples, in 16khz, 16-bit PCM format.
If the stream data is not read and the buffered audio exceeds the size specified in readStaleThreshold, the oldest samples are discarded.
If the data is not read for more than 500ms, buffered data is discarded to prevent returning old audio.
Y vea otra tarea: la Grabación en audio de performances, ahora en video.


text:
Judge 'sad' to see 21-year-old's army hopes dashed after drug conviction
Christine Cunningham
Kiaron Bishop was jailed for three years and four months for possession of class A with intent to supply. Picture: Norfolk Constabulary - Credit: Norfolk Constabulary
A judge said it was sad to see a young man have his hopes dashed of joining the Army after he was given a suspended jail term for a drug offence in Norwich.
Nicholas Fletcher, 21, had come to the UK from Granada to join the army but got involved in a drug enterprise when his cousin Sherese Fletcher, 21, who was living in Southwell Road, Norwich, recruited him to help sell Class A drugs, Norwich Crown Court heard.
Although Nicholas Fletcher never actually got round to selling any drugs before police raided the address in December, last year, the court heard it would mean the end to any hope of an army career.
Nicholas Fletcher, of no fixed address, and Sherese Fletcher admitted conspiracy to supply Class A drugs and Kiaron Bishop, 22, admitted possession of Class A drugs with intent to supply.
Brian Reece, prosecuting, said when police raided the address they found wraps of heroin and cocaine in the bathroom and £2,600 in cash in a bag of clothing.
He said that text messages on phones seized showed that Sherese had recruited her cousin to sell drugs, although she had not actually sold any drugs herself.
Recorder Alastair Wilson jailed Bishop for three years, four months and imposed a 22 month sentence suspended for two years on Sherese Fletcher and ordered her to do 150 hours community service.
He imposed an 18 month sentence suspended for two years on Nicholas Fletcher and ordered him to do 100 hours community service.
He told him: 'It is very sad to hear you came here from Granada hoping to join the British Army and now as a result of this conviction I am sure it will be impossible.'
Sam Parham, for Nicholas Fletcher, said that he had never actually got round to selling any drugs but was still learning how it might work: 'His involvement was limited.'
He said he had only come to UK four weeks previously with hopes of joining the army and said: 'That ambition is now hopeless.'
Philip Farr, for Sherese Fletcher said that her involvement was limited and she had never dealt any drugs: 'She has simply helped out in recruiting one other person.'
Richard Wood, for Bishop, said he was acting on instructions by others higher up for only a modest benefit.


text:
Coronavirus: Why it's important and safe to donate blood
Scott Procter
Richmond Palladium-Item
RICHMOND, Ind. — The impact of the novel coronavirus has reared its head in the realm of blood donations, causing The American Red Cross and Reid Health to declare that they're in the midst of a blood shortage.
As of Tuesday, nearly 4,000 Red Cross blood drives have been canceled across the country due to COVID-19 concerns resulting in more than 100,000 fewer blood donations, Duchess Adjei, the communication director of Red Cross' Indiana Region, said.
"It's a really severe impact and we are officially in a severe blood shortage," Adjei said. "Already the flu season earlier this year had an impact on that and now with this uncertain pandemic, it really is shifting a lot of things for a lot of different people and businesses."
Donating blood will fulfill a critical need
Adjei encourages any healthy person who feels normal to donate blood and help fulfill a "critical need." Concerns of being asymptomatic while being a carrier of the virus due to a lack of testing nationwide or becoming infected as a result of donation are ones Red Cross is aware of. However, Adjei said there is no evidence that shows an individual can contract the coronavirus, or any other respiratory virus, through a blood transfusion.
Ultimately, Adjei said if you feel healthy and comfortable, you should feel safe donating blood.
Red Cross and Reid Health take extra steps to keep blood drives sanitary
Additional safety protocols have been put in place at Red Cross blood drives in the form of routinely sanitizing donor-touched areas, the use and changing of gloves with each donor, using sterile collection sets for every donation and more.
Adjei acknowledged how blood donations might go against the encouraged idea of "social distancing" but she compared the service to a grocery store or pharmacy in that blood drives are an essential component to ensuring the health of the community.
Coronavirus coverage:Richmond-related news about early pandemic concerns
"The Red Cross will continue to hold blood drives during this unprecedented time to help meet patients' needs," Adjei said. "Social distancing does not have to mean social disengagement."
As Red Cross deals with a blood shortage throughout the state and nationwide, Reid Health is experiencing one of its own.
Rapidly occurring community safeguards in response to the spread of the coronavirus nearly prompted Richmond's Reid Health to move Tuesday's St. Patrick's Day blood drive to a mobile unit.
Reid CEO Craig Kinyon instead decided that "this was not the time to limit blood collections" and even donated platelets of his own at Tuesday's blood drive.
"There's a critical blood shortage and we need the blood," Kinyon said in a press release. "If the people that need the blood aren't helping get the blood, then something is wrong."
Similar to Red Cross, Reid Health has altered protocol to ensure safety of donors, patients and staff. People now enter the hospital from one door where they are greeted by staff members in surgical masks. They are then asked a series of questions regarding their travel, health and possible contact with anyone diagnosed with COVID-19.
"They go from the checkpoint, directly to the elevator to donate here and leave the same way," director of laboratory services Chuck McGill said in a press release. "It limits their exposure to the hospital, and the hospital exposure to them with social distancing."
Wapahani coach tests positive for COVID-19 Ball State University coronavirus cases continue to climb, doubling over the weekend How New Castle football not only beat Tindley on Saturday, but was able to play at all United Way's 2020 Day of Caring goes virtual
Reid totaled 173 donors with 133 whole blood donations to go along with three double red cell donations and two platelet donations on Tuesday for 164 percent of its collection goal.
"Some want to say if we err on the side of safety, nothing will happen," Kinyon said. "Then nothing would happen, we wouldn't have blood, and we'd start cancelling surgeries."
Scott Procter is the sports and education reporter at the Palladium-Item. Follow him on Twitter @SprocterPI or contact him at sprocter@gannett.com.


text:
Nada mejor que esto
14 fechas, ninguna en España, pero al fin, 19 años después de su última visita a nuestro continente, ¡¡John Mellencamp, en Europa!! No better than this: ni un título del Real Madrid esta temporada puede ser mejor que esto!
For the first time in 19 years, John Mellencamp is set to tour the UK and Europe with dates set for late June and July 2011. The No Better Than This tour is designed as a uniquely formatted «evening with,» and brings Mellencamp's music to theatre-sized venues, a rare opportunity to experience one of America's most iconic songwriters and performers in an intimate setting, highlighting all phases of his artistry.
The new No Better Than This Tour dates will take in Denmark, Germany, Netherlands, Ireland, UK, France, Italy, Switzerland and tickets for the UK dates go on sale Friday 18th March at 9AM viawww.gigtsandtours.com and 24 hr cc hotline 0844 811 0051.
The tour will feature both classic material associated with the Rock and Roll Hall of Fame inductee's 35+ year long career as well as new music from No Better Than This, released last year by Rounder Records and produced by T Bone Burnett. The album has garnered great critical acclaim with Q stating, «Veteran blue-collar songsmith goes back to basics… with Mellencamp relishing his role as grizzled troubadour steeped in Rootsy traditions of America's rural south.»
It was recorded in three historic locations in the American South: Savannah's First African Baptist Church, Sun Studio in Memphis and Room 414 of San Antonio's Gunter Hotel where Robert Johnson recorded in 1936. The sessions were documented on a 55-year-old mono tape recorder using just one vintage microphone at each location. Rolling Stone's review of the album noted, «Mellencamp has never sounded looser or easier on a record,» suggesting the artist was «channeling spirits and stepping into period styles. They fit him perfectly.» The Americana Music Association recently honored Mellencamp with its Lifetime Achievement Award for Songwriting, presented by Rosanne Cash.
As a reflection of the diversity of styles and material that have been a hallmark of Mellencamp's eclectic career, the concerts are formatted into three segments. These showcase John Mellencamp in an acoustic context, fronting a small combo recalling blues/country and rockabilly roots and with his full rock band. Musicians backing Mellencamp include Michael Wanchic (guitar), Andy York (guitar), Miriam Sturm (violin), John Gunnell (electric and upright bass), Dane Clark (drums and percussion) and Troye Kinnett (accordion/keyboards).
Of special note is the fact that the tour's «opening act,» is a documentary film by Kurt Markus entitled It's About You. Shot on Super8 film over the course of 2009's Bob Dylan-John Mellencamp-Willie Nelson tour of minor league baseball stadiums, it chronicles the creation of No Better Than This while serving as a cinematic 'think piece' on the state of the country at the end of the first decade of the 21st Century. The film had its theatrical premiere at this year's SXSW® Film Festival in Austin, Texas.
John Mellencamp No Better Than This Tour,
24th Copenhagen, DEN – Vega
25th Berlin, GER – TempoDrom
26th The Hague, NLD – Forum Theatre
28th Dublin, IRL – Canal Theatre
1st Manchester GBR, – O2 Apollo
2nd London, GBR – HMV Hammersmith Apollo
3rd TBA
5th Paris, FRA – Olympia
6th Munich, GER – Circus Krone
7th Stuttgart, GER – Beethovensall
9th Vigevano, ITL – Castello Sforzesco
10th Rome, ITL – The Cavea
12th Udine, ITL – Castello
13th Zurich, CHE – Live at Sunset
Publicado en Conciertos, MusicaEtiquetado como Europe, John Mellencamp, No better than this
Anterior Waters-Gilmour
Siguiente Se busca
the beautiful loser dice:
Yo lo ví en Febrero en el Radio City Music Hall de Nueva York y me imaginé que haría gira europea porque está muy en forma y se encuentra muy a gusto con este disco.
Ya he sacado mis entradas para Londres.
pues ahi nos veremos, loser 😉
DAWE dice:
Ahí tas quedao solo!!!
Que pinta el fútbol en esto!!!
Ale! Que te aproveche!
londres!!!!! y queda la fecha del día 3! a ver si hay suerte y cae el doblete!!! 🙂
Si se pasara por BCN no dudaria en ir, perono soy tan fan como para escaparme a Europa.
Corazon de rock & roll


If you could wake up tomorrow and find that one thing had changed about the world, what would you have that one thing be?.
oh goody…a great blog to follow! I, too, am still learning wordpress.com. I am confused…do you NOT have to put "wordpress.com" when you make your blog? I thought you had to do "susiartquilt.wordpress.com" , not "susiartquilt.com ???
Duh, Im a little slow!