text:
\section{Introduction}\label{sec.intro}

Random walks on lattices, such as the face-centered cubic lattice, are an
important concept in various applications in physics, chemistry, ecology,
economics, and computer science, when lattice vibration problems (phonons),
diffusion models, luminescence, Markov processes and other random processes
are studied.
A fundamental object to investigate is the probability generating function of
a lattice, called the lattice Green's function (LGF).  For example, the return
probability of a lattice can be expressed in terms of the LGF. The LGFs of
three-dimensional lattices have been computed and analyzed
in~\cite{Joyce98,Guttmann09}, for higher-dimensional lattices
see~\cite{Broadhurst09,Guttmann10}. We present a completely different approach
to LGFs that is based on computer algebra techniques, with which we are not
only able to confirm independently the previously known results, but also go
beyond. We believe that this methodology can be applied successfully to many
other, yet unsolved problems of similar flavor, and therefore should be
popularized in the community.

The paper is organized as follows. In Section~\ref{sec.intro} we explain the
general setting in which we work and introduce basic notions, such as Bravais
lattice (Section~\ref{sec.bravais}), random walk (Section~\ref{sec.random}),
and lattice Green's function (Section~\ref{sec.lgf}); in
Section~\ref{sec.diffeq} an approach to LGFs via differential equations is
motivated. We derive the integral representation of the LGF and show how a
differential equation is connected to it. Section~\ref{sec.em} is dedicated to
different methods how to compute the ODE of a LGF in a nonrigorous way:
Section~\ref{sec.intrep} reviews the method for computing Taylor coefficients
used in~\cite{Broadhurst09,Guttmann10} and Sections~\ref{sec.count}
and~\ref{sec.multi} count random walks in order to obtain sufficient data to
construct the ODE. The main contribution of our work is contained in
Section~\ref{sec.ca}.  The method of creative telescoping is described in
Section~\ref{sec.ct}; it enables us to compute the desired ODEs in a
mathematical rigorous way, including correctness certificates. Applying this
method to the LGF of the fcc lattice confirms the results
of~\cite{Broadhurst09,Guttmann10} in dimensions four and five, and yields an
ODE for the LGF of the six-dimensional fcc lattice that was not known
previously (see Section~\ref{sec.results}).

\subsection{Bravais Lattices}\label{sec.bravais}

We consider lattices in $\mathbbm R^d$ that are given as infinite
sets of points
\[
  \bigg\{\sum_{i=1}^d n_i{\bm a}_i: n_1,\dots,n_d\in\mathbbm Z\bigg\} \subseteq\mathbbm R^d
\]
for some linearly independent vectors ${\bm a}_1,\dots,{\bm a}_d\in\mathbbm R^d$
(throughout this paper, vectors are denoted by bold letters).  In
three dimensions such lattices are called \emph{Bravais lattices}.
The simplest instance of such a lattice is obtained by choosing
${\bm a}_i={\bm e}_i$, the $i$-th unit vector; the result is the integer
lattice~$\mathbbm Z^d$ which is also called the \emph{square lattice} (for
$d=2$), or the \emph{cubic lattice} (for $d=3$), or the
\emph{hypercubic lattice} (for $d\geq 4$).

The \emph{face-centered (hyper-) cubic lattice} (fcc lattice) is
obtained from the (hyper-) cubic lattice by adding the center point of
each (two-dimensional) face to the set of lattice points. In two
dimensions this operation is trivial: the faces of the square lattice
$\mathbbm Z^2$ are all unit squares with corners
$(m,n),(m+1,n),(m+1,n+1),(m,n+1)$ for integers $m,n\in\mathbbm Z$. Their
center points are $\mathbbm Z^2+(\frac12,\frac12)$ which together with $\mathbbm Z^2$
again yields a square lattice, more precisely a copy of $\mathbbm Z^2$ which
is rotated by 45 degrees and shrunk by a factor of $\sqrt{2}$. The
situation becomes more interesting in higher dimensions. For example,
in three dimensions there are 6 faces of the unit cube, and their
center points together with all integral translates have to be
included.  It is not difficult to see that the three-dimensional fcc
lattice consists of four copies of $\mathbbm Z^3$, namely
\[
  \textstyle\mathbbm Z^3
    \cup\left(\mathbbm Z^3+\left(\frac12,\frac12,0\right)\right)
    \cup\left(\mathbbm Z^3+\left(\frac12,0,\frac12\right)\right)
    \cup\left(\mathbbm Z^3+\left(0,\frac12,\frac12\right)\right).
\]
Similarly the fcc lattice in four dimensions consists of 7 copies of
$\mathbbm Z^4$, and in general the $d$-dimensional fcc lattice is composed of
$1+\binom{d}{2}$ translated copies of $\mathbbm Z^d$.

The study of Bravais lattices was inspired by crystallography in as much as
the atomic structure of crystals forms such regular lattices. While
the cubic lattice is quite rarely found in nature (e.g., in polonium)
due to its small \emph{atomic packing factor} (the proportion of space
that is filled when a sphere of maximal radius is put on each lattice
point, in a way that these spheres do not overlap), the fcc lattice is
more often encountered, for example, in aluminium, copper, silver, and
gold.  The atomic packing factor of the fcc lattice is
$\sqrt{2}\pi/6$, the highest possible value as was shown by
Hales in his famous proof of the Kepler conjecture~\cite{Hales05}.

\subsection{Random Walks}\label{sec.random}
For the sake of simplicity, the fcc lattice as introduced in the previous
section, is stretched by a factor of~$2$ in all coordinate directions so that
all lattice points have integral coordinates. This convention is kept
throughout the paper as it does not change the relevant quantities that we are
interested in (e.g., the return probability, see below).

The aim of this paper is to study random walks on the fcc lattice in
several dimensions. We consider walks that allow only steps to the
nearest neighbors of a point (with respect to the Euclidean metric).
Furthermore it is assumed that all steps are taken with the same
probability. For example, consider a point $(k,m,n)$ in the
three-dimensional cubic lattice $(2\mathbbm Z)^3$. It is the common corner
point of 8 cubes.  The nearest neighbors in the 3D fcc lattice are
then the center points of some of those faces which have $(k,m,n)$ as a
corner point. Note that they all have distance $\sqrt{2}$ whereas the
other corner and face-center points are farther away (their distance
is $\geq2$) and hence not reachable in a single step. Thus the number
of possible steps is $8\cdot 3/2=12$ (number of adjacent cubes times
the number of adjacent faces per cube, divided by two since each face
belongs to two cubes).  The same situation is encountered at the
center point of some face and hence every point in the 3D fcc lattice
has exactly 12 nearest neighbors; this number is called the
\emph{coordination number} of the lattice.

The above considerations can be generalized to arbitrary dimensions in
a straight-forward manner; one finds that the set of permitted steps
in the $d$-dimensional fcc lattice is given by
\begin{equation}\label{eq.steps}
  \textstyle\left\{(s_1,\dots,s_d)\in\{0,-1,1\}^d:|s_1|+\dots+|s_d|=2\right\}
\end{equation}
and thus its coordination number is $4\binom{d}{2}$.

\subsection{Lattice Green's Function}\label{sec.lgf}
Let $p_n({\bm x})$ denote the probability that a random walk which started
at the origin~${\bm 0}$ ends at point~${\bm x}$ after $n$~steps.  Note that in
our setting of unrestricted walks, $c^np_n({\bm x})$ is an integer and
gives the total number of walks that end at location~${\bm x}$ after
$n$~steps, where $c$ is the coordination number of the lattice.

In order to achieve information about random walks on the fcc lattice,
the following multivariate generating function is introduced:
\begin{equation}\label{eq.gfx}
  P({\bm x};z) = \sum_{n=0}^\infty p_n({\bm x})z^n.
\end{equation}
This function is called the \emph{lattice Green's function} (LGF).
By defining the \emph{structure function}
$\lambda({\bm k})=\lambda(k_1,\dots,k_d)$ of a lattice to be the discrete
Fourier transform
\[
  \lambda({\bm k}) = \sum_{{\bm x}\in\mathbbm R^d} p_1({\bm x})e^{i{\bm x}\cdot{\bm k}}
\]
of the single-step probability function~$p_1({\bm x})$, the generating
function~\eqref{eq.gfx} can be expressed as the $d$-dimensional integral
\[
  P({\bm x};z) = \frac{1}{\pi^d}\int_0^\pi\dots\int_0^\pi
    \frac{e^{i{\bm x}\cdot{\bm k}}}{1-z\lambda({\bm k})}\rmd k_1\dots\rmd k_d.
\]
We shall be interested in walks which return to the origin and which
we therefore call \emph{excursions}. The LGF for excursions is given
by
\begin{equation}\label{eq.lgf}
  P({\bm 0};z) = \sum_{n=0}^\infty p_n({\bm 0})z^n =
            \frac{1}{\pi^d}\int_0^\pi\dots\int_0^\pi
              \frac{\rmd k_1\dots\rmd k_d}{1-z\lambda({\bm k})}.
\end{equation}
In the following, we will only refer to this special instance when talking
about LGFs.  This function allows one to calculate the \emph{return
  probability}~$R$, sometimes also referred to as the \emph{P\'{o}lya number},
of the lattice.  It signifies the probability that a random walk that started
at the origin will eventually return to the origin. It can be computed via the
formula
\begin{equation}
  R = 1-\frac{1}{P({\bm 0};1)} = 1-\frac{1}{\sum_{n=0}^{\infty} p_n({\bm 0})}.
\end{equation}
\begin{example}\label{ex.sq}
Consider the square lattice~$\mathbbm Z^2$ which admits the steps $(-1,0)$,
$(1,0)$, $(0,-1)$, and $(0,1)$. Its structure function is
\[
  \textstyle\lambda(k_1,k_2) = \frac14\left(e^{-ik_1}+e^{ik_1}+e^{-ik_2}+e^{ik_2}\right)
                   = \frac12\left(\cos k_1+\cos k_2\right).
\]
and therefore its LGF is (see, e.g., \cite{Guttmann10})
\[
  P(0,0;z) = \frac{1}{\pi^2} \int_0^\pi\int_0^\pi\frac{\rmd k_1\rmd k_2}
               {1-\frac{z}{2}\left(\cos k_1+\cos k_2\right)}
           = \frac{2}{\pi}\mathbf{K}(z^2)
\]
where $\mathbf{K}(z)$ is the complete elliptic integral of the first
kind. The fact that the above integral diverges for $z=1$ immediately
implies that the return probability $R=1$; in other words that every
random walk will eventually return to the origin, a result that was
already proven in 1921 by P\'{o}lya~\cite{Polya21}.
\exampleEnd
\end{example}

\begin{example}\label{ex.fcc2}
We have already remarked that the two-dimensional fcc lattice is
nothing else but a rotated and stretched version of the square
lattice.  Nevertheless let's have a look at the LGF when the step set
$\{(-1,-1),(-1,1),(1,-1),(1,1)\}$ is taken. Its structure function is
\begin{eqnarray*}
  \lambda(k_1,k_2)
    & = & \textstyle\frac14\big(e^{-i(k_1+k_2)}+e^{-i(k_1-k_2)}+e^{i(k_1-k_2)}+e^{i(k_1+k_2)}\big)\\
    & = & \textstyle\frac12\big(\cos(k_1+k_2)+\cos(k_1-k_2)\big) = \cos k_1\cos k_2,
\end{eqnarray*}
using the well-known angle-sum identity
$\cos(x \pm y) = \cos x\cos y \mp \sin x\sin y$.
Although the structure function differs from that in
Example~\ref{ex.sq} the LGF is the same:
\begin{equation}\label{eq.lgf2d}
  P(0,0,z) = \frac{1}{\pi^2} \int_0^\pi\int_0^\pi\frac{\rmd k_1\rmd k_2}{1-z\cos k_1\cos k_2}
           = \frac{2}{\pi}\mathbf{K}(z^2)
\end{equation}
as shown in Equation (6) of~\cite{Guttmann10}.  Note also that the
different distances between nearest-neighboring lattice points---$1$
in Example~\ref{ex.sq} and $\sqrt{2}$ in Example~\ref{ex.fcc2}---carry
no weight since only excursions (and not walks with arbitrary end
points) are investigated.
\exampleEnd
\end{example}

It is now an easy exercise to compute the structure function~$\lambda({\bm k})$
for the $d$-dimensional fcc lattice:
\[
  \lambda({\bm k}) = \binom{d}{2}^{-1}\sum_{m=1}^d\sum_{n=m+1}^d \cos k_m\cos k_n.
\]
The LGF is then given as the $d$-fold integral~\eqref{eq.lgf}
and the return probability can be computed by integrating over
$1/(1-\lambda({\bm k}))$

\subsection{The Differential Equation Detour}\label{sec.diffeq}
The return probability in the fcc lattice in three dimensions was
first computed by Watson~\cite{Watson39} as one of the three
integrals which were later named after him, and which give the return
probabilities in different three-dimensional lattices. These
probabilities can be expressed in terms of algebraic numbers, $\pi$,
and values of the Gamma function at rational arguments. For example, the
probability of returning to the origin in the 3D fcc lattice is given
by
\[
  1 - \frac{16\sqrt[3]{4}\pi^4}{9\big(\Gamma(\frac13)\big)^6}.
\]
For the three-dimensional fcc lattice, Joyce expressed the lattice Green's
function also in terms of complete elliptic integrals~\cite{Joyce98}, but the
expression is fairly complicated and for the higher-dimensionsal fcc lattices
no such evaluation is known at all.  Similarly we don't know of any
closed-form representation of the return probabilities in higher dimensions.

Instead we will derive differential equations for the corresponding
LGFs. Although less explicit than the previously mentioned closed-form
results, such an implicit representation of the LGF provides considerable
insight. It allows one to compute the number of excursions efficiently
for any fixed number of steps, as well as the return
probability with very high precision (see Section~\ref{sec.ca}). But
also the differential equations themselves reveal very interesting
properties that are worth investigation.

To motivate our approach and to illuminate the origin of
Equation~\eqref{eq.lgf}, consider an arbitrary lattice in~$\mathbbm Z^d$ with
some finite set~$S\subset\mathbbm Z^d$ of permitted steps. Then clearly the
probability function~$p_n({\bm x})$ satisfies the constant-coefficient
recurrence
\begin{equation}\label{eq.trivrec}
  p_{n+1}({\bm x}) = \frac{1}{|S|}\sum_{{\bm s}\in S}p_n({\bm x}-{\bm s}).
\end{equation}
Let~$F({\bm y};z)$ denote the multivariate generating function
\[
  F({\bm y};z)=\sum_{n=0}^\infty\sum_{{\bm x}\in\mathbbm Z^d}p_n({\bm x}){\bm y}^{{\bm x}}z^n.
\]
Multiplying both sides of~\eqref{eq.trivrec} by~${\bm y}^{{\bm x}}z^n$
and summing with respect to $n$ and ${\bm x}$ gives
\begin{eqnarray*}
\sum_{n=0}^{\infty}\sum_{{\bm x}\in\mathbbm Z^d}p_{n+1}({\bm x}){\bm y}^{{\bm x}}z^n & = &
  \frac{1}{|S|}\sum_{n=0}^{\infty}\sum_{{\bm x}\in\mathbbm Z^d}\sum_{{\bm s}\in S}p_n({\bm x}-{\bm s}){\bm y}^{{\bm x}}z^n\\
\frac{1}{z}\sum_{n=1}^{\infty}\sum_{{\bm x}\in\mathbbm Z^d}p_n({\bm x}){\bm y}^{{\bm x}}z^n & = &
  \frac{1}{|S|}\sum_{{\bm s}\in S}\sum_{n=0}^{\infty}\sum_{{\bm x}\in\mathbbm Z^d}p_n({\bm x}){\bm y}^{{\bm x}+{\bm s}}z^n\\
\frac{1}{z}\left(F({\bm y};z)-1\right) & = &
  \frac{1}{|S|}\sum_{{\bm s}\in S}{\bm y}^{{\bm s}}F({\bm y};z)
\end{eqnarray*}
Thus we obtain
\[
  F({\bm y};z) = \frac{1}{1-\frac{z}{|S|}\sum_{{\bm s}\in S}{\bm y}^{{\bm s}}}
\]
and the LGF $P({\bm 0};z)$ is nothing
else but the constant term $\langle{\bm y}^{{\bm 0}}\rangle F({\bm y};z)$. A differential
equation for this expression can be derived from an operator of the form
\begin{equation}\label{eq.ctmult}
  A(z,\D{z}) + \D{y_1}B_1 + \dots + \D{y_d}B_d
\end{equation}
that annihilates the expression $F({\bm y};z)/(y_1\cdots y_d)$.
Here the symbol~$\D{x}$ denotes the partial derivative w.r.t.~$x$
and the $B_j$ are differential operators that may involve
$y_1,\dots,y_d,z$ as well as $\D{y_1},\dots,\D{y_d},\D{z}$.
The fact that~$A$ may only depend on $z$ and $\D{z}$ is crucial
and therefore explicitly indicated. In Section~\ref{sec.ca}
we will discuss how to find such an operator. From
\[
  \big\langle y_1^{-1}\cdots y_d^{-1}\big\rangle A(z,\D{z})\frac{F({\bm y},z)}{y_1\cdots y_d}\> +\>
  \sum_{j=1}^d \big\langle y_1^{-1}\cdots y_d^{-1}\big\rangle \D{y_j}B_j\frac{F({\bm y};z)}{y_1\cdots y_d} = 0
\]
and the fact that the coefficient of $y^{-1}$ in an expression of
the form $\D{y}\sum_{n=-\infty}^\infty a_ny^n$ is always zero, it follows
that $A\big(\langle{\bm y}^{{\bm 0}}\rangle F({\bm y};z)\big)=A\big(P({\bm 0};z)\big)=0$.
Also in Section~\ref{sec.ca} we will demonstrate how the
operator~\eqref{eq.ctmult} is used to derive a differential equation
for the $d$-fold integral
\[
  \int\cdots\int \frac{d{\bm y}}{(y_1\cdots y_d)(1-\frac{z}{|S|}\sum_{{\bm s}\in S}{\bm y}^{{\bm s}})} =
  \int\cdots\int \frac{d{\bm k}}{1-z\lambda({\bm k})}.
\]

\section{An Experimental Mathematics Approach}\label{sec.em}
This section presents some results that were obtained in a
non-rigorous way using the method of guessing~\cite{Kauers09}.
That is, for finding a linear differential equation
\[
  c_m(x)f^{(m)}(x)+\dots+c_1(x)f'(x)+c_0(x)f(x)=0
\]
for a certain function~$f(x)$, one computes the first terms of the Taylor
expansion of~$f(x)$ and then makes an ansatz with undetermined polynomial
coefficients~$c_j(x)$. If the resulting linear system is overdetermined (i.e.,
if sufficiently many Taylor coefficients were used) but still admits a
nontrivial solution, then the detected ODE is very likely to be correct.
Another strategy to gain confidence in the result, is to test it with further
Taylor coefficients, that were not used in the computation. However, this
method can never produce a rigorous proof of the result and there always
remains a (very small) probability that the guess is wrong. For this reason,
any result obtained in this fashion (e.g., the ODEs presented
in~\cite{Guttmann09,Broadhurst09}) is termed a \emph{conjecture}.

\subsection{Starting from the Integral Representation}\label{sec.intrep}
In this section we briefly recapitulate some previous work done by Broadhurst
and Guttmann, who used the integral representation~\eqref{eq.lgf} of the LGF
as a starting point.

Guttmann computed a differential equation for the LGF of the four-dimensional
fcc lattice~\cite{Guttmann09} (see also Theorem~\ref{thm.fcc4}): for this
purpose, the four-fold integral was rewritten as a double integral whose
integrand was expanded as a power series.  Term-by-term integration yielded a
truncated Taylor expansion of the LGF which allowed him to apply the method of
guessing.

Recently, Broadhurst had obtained an ODE for the LGF of the five-dimensional
fcc lattice~\cite{Broadhurst09} (see also Theorem~\ref{thm.fcc5}), a result
that required several days of PARI calculations. Broadhurst's strategy
consisted in expanding the integrand in~\eqref{eq.lgf} as a geometric series
$\sum_{n=0}^\infty \lambda({\bm k})^nz^n$ and in expanding $\lambda({\bm k})^n$ using
the multinomial theorem (which gives a $(m-1)$-fold sum if $m$ is the number
of summands in $\lambda({\bm k})$). The inner terms can now be integrated using
Wallis' formula
\[
  \int_0^\pi \cos(x)^{2n}\rmd x = \frac{\pi}{4^n}\binom{2n}{n}.
\]
The structure function of the 5D fcc lattice consists of
$10$~summands.  Thus the computation of the $n$-th Taylor coefficient
of~$P({\bm 0};z)$ requires the evaluation of a $9$-fold sum, or in other
words, has complexity $O(n^9)$.

\subsection{Counting the walks}\label{sec.count}
A different way to crank out as many Taylor coefficients of the LGF as
necessary is to explicitly count all possible excursions with a certain number
of steps. Let $a_n({\bm x})$ be the number of walks from~${\bm 0}$ to~${\bm x}$ with
$n$~steps and let $c$ denote the coordination number of the lattice, then the
lattice Green's function $P({\bm 0};z)=\sum_{n=0}^\infty a_n({\bm 0})(z/c)^n$, as we
have already remarked earlier. The values of the $(d+1)$-dimensional sequence
($d$~again denotes the dimensionality of the lattice) can be computed with the
recurrence~\eqref{eq.trivrec}. To~obtain the first $n$ Taylor coefficients
hence requires one to fill the $(d+1)$-dimensional array
$\big(a_m({\bm x})\big)_{0\leq m,x_1,\dots,x_d<n}$ with values (by symmetry it
suffices to consider the first octant only, which again by symmetry can be
restricted to the wedge $x_1\geq x_2\geq\dots\geq x_d$). Still, this has
complexity~$O(n^{d+1})$.  Further optimizations consist in cutting off the
regions where the sequence can be predicted to be zero (e.g. when $x_j>n$ for
some~$j$), and to truncate the $x_j$-coordinates at $n/2$ (since we are
interested in excursions, points too far from the origin~${\bm 0}$ are not
relevant). Although the complexity is better than before, computing the full
array of values can be quite an effort. Again, in the example of the 5D fcc
lattice, about $115$ Taylor coefficients are necessary to recover the
recurrence for $a_n({\bm 0})$ (which then gives rise to the differential equation
of~$P({\bm 0};z)$), and hence the full array contains about $115\cdot
58^5/5!\approx 6.3\cdot 10^8$ values! Fortunately we can do better.

\subsection{Multi-Step guessing}\label{sec.multi}
How can the recurrence for $a_n({\bm 0})$ be computed without calculating
all the values of the multivariate sequence~$a_n({\bm x})$ in the
box~$[0,n]^{d+1}$ (or some slightly optimized version of it)?  In the
previous section, we first computed lots of data, then threw away most
of it, and did a single guessing step.  But the guessing can be done
in several steps which we call \emph{multi-step guessing}.  The method
is illustrated on the 5D fcc example. As before, we start with the
recurrence~\eqref{eq.trivrec} to crank out a moderate number of values
for the six-dimensional sequence~$a_n(x_1,\dots,x_5)$, namely in the
box~$[0,15]^6$, which takes about 30 seconds only. From this array, we
pick the values of~$a_n(x_1,x_2,x_3,0,0)$ which constitute a
four-dimensional sequence that we denote with~$b_n(x_1,x_2,x_3)$. The
data is now used to guess recurrences for this new function~$b$. One
of these recurrences is
\[
  \begin{array}{l}
  (n+1)b_n(x_1,x_2+3,x_3+1) - (n+1)b_n(x_1,x_2+1,x_3+3) +{}\\
  (n+1)b_n(x_1+1,x_2,x_3+3) - (n+1)b_n(x_1+1,x_2+3,x_3) +{}\\
  (n+1)b_n(x_1+1,x_2+3,x_3+4) - (n+1)b_n(x_1+1,x_2+4,x_3+3) -{}\\
  (n+1)b_n(x_1+3,x_2,x_3+1) + (n+1)b_n(x_1+3,x_2+1,x_3) -{}\\
  (n+1)b_n(x_1+3,x_2+1,x_3+4) + (n+1)b_n(x_1+3,x_2+4,x_3+1) +{}\\
  (n+1)b_n(x_1+4,x_2+1,x_3+3) - (n+1)b_n(x_1+4,x_2+3,x_3+1) +{}\\
  (x_2+2)b_{n+1}(x_1+1,x_2+2,x_3+3) - (x_3+2)b_{n+1}(x_1+1,x_2+3,x_3+2) -{}\\
  (x_1+2)b_{n+1}(x_1+2,x_2+1,x_3+3) + (x_1+2)b_{n+1}(x_1+2,x_2+3,x_3+1) +{}\\
  (x_3+2)b_{n+1}(x_1+3,x_2+1,x_3+2) - (x_2+2)b_{n+1}(x_1+3,x_2+2,x_3+1) = 0
  \end{array}
\]
which has the disadvantage that it does not allow us to compute the
values $b_n(0,0,0)$ since the leading coefficient vanishes;
unfortunately this phenomenon occurs frequently in this context. An
additional recurrence that does not suffer from this handicap is much
larger and therefore not reproduced here. Anyway, guessing these
recurrences can be done in less than a minute.

Now these recurrences can be used to compute more values for the
sequence $b_n(x_1,x_2,x_3)$ (in 30 seconds one can now go up to
$n=30$) which in turn are used to~guess recurrences
for~$b_n(x_1,x_2,0)$.  These latter recurrences allow to compute
$a_n({\bm 0})=b_n(0,0,0)$ for $0\leq n\leq 115$ in about $2.5$
minutes. Voil\`{a}! The whole computation takes less than 5 minutes on
a modest laptop. Of course it is a matter of trial and error to
determine how many coordinates are set to~$0$ in each step (in the
above example, we did~$2$ in the first step, $1$ in the second, and
again~$2$ in the third step).

\section{A Computer Algebra Approach}\label{sec.ca}
Again, we want to emphasize that the results presented in the previous
section are certainly nice, but lack mathematical rigor. To
achieve ultimate confidence in their correctness we have to apply a
different method. One possible such method is called \emph{creative
  telescoping}, a short introduction of which is given in the
following section.  After that we are able to state our results in
the form of theorems.

\subsection{Creative Telescoping}\label{sec.ct}
This method has been popularized by Zeilberger in his seminal
paper~\cite{Zeilberger90}.  Since then it has been applied to
innumerable identities involving hypergeometric summations, multisums,
integrals of special functions, and various other kinds of problems.
The basic idea is very simple and we illustrate it on the example
of a definite integral $F(z)=\int_a^b f(x,z)\rmd x$. The main step in the
algorithm consists in finding a partial differential equation for
$f(x,z)$ that can be written in the form
\begin{equation}\label{eq.ct}
  \big(A(z,\D{z}) + \D{x}B(x,z,\D{x},\D{z})\big)(f(x,z)) = 0
\end{equation}
where the \emph{telescoper} $A\in\mathbbm C(z)\langle\D{z}\rangle$ and the \emph{delta
  part} $B\in\mathbbm C(x,z)\langle\D{x},\D{z}\rangle$ are differential operators,
with the previously introduced notation of $\D{x}$ being the partial
derivative w.r.t.~$x$. By $\mathbbm C(z)\langle\D{z}\rangle$ we denote the
non-commutative Ore algebra that can be viewed as a polynomial ring in the
``variable''~$\D{z}$ with rational function coefficients in~$\mathbbm C(z)$.  The
result of the algorithm is a (possibly) inhomogeneous linear ODE for the
integral~$F(z)$ that is obtained by integrating Equation~\eqref{eq.ct}:
\[
  A\big(F(z)\big) + \Big[B\big(f(x,z)\big)\Big]_{x=a}^{x=b}=0.
\]
In applications one frequently encounters the situation that the second part
vanishes, yielding a homogeneous ODE. This is because many integrals that
occur in practice, have \emph{natural boundaries}. With our study of LGFs, we
are in a similar situation: in Section~\ref{sec.diffeq} it was shown that the
telescoper of~\eqref{eq.ctmult} automatically annihilates the LGF.
Nevertheless, in some of the present cases we did do the additional (but
superfluous) check that the differential equation is indeed homogeneous by
plugging in the boundaries of the integral, and got confirmation.

If this method is applied to a one-dimensional integral with
hyperexponential integrand (i.e., its logarithmic derivative
is a rational function), then it is called the Almkvist-Zeilberger
algorithm. Its summation counterpart is the celebrated Zeilberger
algorithm for hypergeometric summation. For our purposes we have
to generalize the input class for the integrand to the so-called
$\partial$-finite holonomic functions: a function~$f(x_1,\dots,x_d)$
is called $\partial$-finite if for each~$x_i$ there exists a linear ODE
for~$f$ with respect to~$x_i$. If in addition $f$ is holonomic
(the definition of this notion is somewhat technical and is omitted
here) then the existence of creative telescoping operators like
\eqref{eq.ctmult} or \eqref{eq.ct} is guaranteed.

The first algorithm to compute~\eqref{eq.ct} for general $\partial$-finite functions
(our examples fall into this class, too) was proposed in~\cite{Chyzak00}. It
can deal with single integrations only and thus has to be applied iteratively
for multiple integrals. Its main drawback is its complexity that makes it
impossible to apply it to the problems discussed in this
paper. In~\cite{Koutschan10c} we have developed a different approach to
compute~\eqref{eq.ct} which is much better suited for large examples involving
multidimensional integrals. In addition, it can directly deal with multiple
integrals by computing operators of the form~\eqref{eq.ctmult}, but in the
present context it turned out to be more efficient to do the integrations step
by step. Both algorithms are implemented in our Mathematica package
\texttt{HolonomicFunctions}~\cite{Koutschan10b}, and a detailed introduction
into the topic is given in~\cite{Koutschan09}. The following example
demonstrates how this method is applied to the previously discussed
two-dimensional lattice.
\begin{example}
Looking at the integrand of Equation~\eqref{eq.lgf2d} one realizes
that it is not $\partial$-finite since no linear ODE with respect to~$k_1$
can be found (and analogously for~$k_2$). But by means of the simple
substitutions $\cos k_1\to x_1$ and $\cos k_2\to x_2$
we can overcome this trouble: the integral now reads
\begin{equation}\label{eq.int2d}
  P(z) = \frac{1}{\pi^2} \int_{-1}^1\int_{-1}^1\frac{\rmd x_1\rmd x_2}{(1-z x_1 x_2)\sqrt{1-x_1^2}\sqrt{1-x_2^2}}.
\end{equation}
Let $f(x_1,x_2,z)$ denote the above integrand; it is easily verified
that it is a $\partial$-finite function. The three ODEs w.r.t. $x_1$, $x_2$,
and $z$ are given by the operators
\begin{eqnarray*}
  G_1 & = & (x_1x_2z-1)D_{\!z}+x_1x_2,\\
  G_2 & = & (x_2^2-1)(x_1x_2z-1)D_{\!x_2}+(2x_1x_2^2z-x_1z-x_2),\\
  G_3 & = & (x_1^2-1)(x_1x_2z-1)D_{\!x_1}+(2x_1^2x_2z-x_1-x_2z),
\end{eqnarray*}
so that $G_i\big(f(x_1,x_2,z)\big)=0$ for $i=1,2,3$. In this example,
it is an easy exercise to check that the creative telescoping
operator
\begin{equation}\label{eq.ct2d}
  z(z^2-1)D_{\!z}^2+(3z^2-1)D_{\!z}+z\>+\>\D{x_1}\frac{x_2-x_1^2x_2}{x_1x_2z-1}\>+\>\D{x_2}\frac{x_2z-x_2^3z}{x_1x_2z-1}
\end{equation}
annihilates the integrand~$f$. Indeed, it can be written as a
linear combination
\[
  \left(\frac{z(z^2-1)}{x_1x_2z-1}D_{\!z}+\frac{x_1x_2z(z^2+1)-3z^2+1}{(x_1x_2z-1)^2}\right)G_1
  -\frac{x_2}{(x_1x_2z-1)^2}(zG_2+G_3)
\]
of the previously computed operators. It follows that
the double integral~\eqref{eq.int2d} satisfies the ODE
\[
  z(z^2-1)P''(z)+(3z^2-1)P'(z)+zP(z) = 0
\]
whose solution is the elliptic integral~$\mathbf{K}(z^2)$.

Alternatively, the two integrations can be performed in two steps (the
strategy that will be applied to the higher-dimensional fcc lattices).
In the first step (integration w.r.t.~$x_1$) the following two
creative telescoping operators are found:
\begin{eqnarray*}
  &&(x_2^2z^2-1)D_{\!z}+x_2^2z\>+\>\D{x_1}(x_1^2-1)x_2\\
  &&(x_2^2-1)(x_2^2z^2-1)D_{\!x_2}+x_2(2x_2^2z^2-z^2-1)\>+\>\D{x_1}(x_1^2-1)(x_2^2-1)z.
\end{eqnarray*}
They certify that the integral $\int_{-1}^1 f(x_1,x_2,z)\rmd x_1$ is annihilated
by $(x_2^2z^2-1)D_{\!z}+x_2^2z$ and $(x_2^2-1)(x_2^2z^2-1)D_{\!x_2}+x_2(2x_2^2z^2-z^2-1)$.
Next the operator
\[
  z(z^2-1)D_{\!z}^2+(3z^2-1)D_{\!z}+z\,-\,\D{x_2}\frac{x_2z(x_2^2-1)}{(x_2^2z^2-1)}
\]
which is a linear combination of the previous ones,
again reveals the same ODE for the double integral.
\exampleEnd
\end{example}

\subsection{Results}\label{sec.results}
Using the above methodology and software, we have computed
differential equations for the LGFs of the fcc lattices in four, five,
and six dimensions, and rigorously proved their correctness.
Additionally, this allows the computation of the return probabilities in
the respective lattices up to very high precision.

\begin{theorem}\label{thm.fcc4}
The lattice Green's function of the four-dimensional face-centered cubic lattice
\[
  P(z) = \frac{1}{\pi^4} \int_0^\pi\int_0^\pi\int_0^\pi\int_0^\pi\frac{\rmd k_1\rmd k_2\rmd k_3\rmd k_4}
  {1-\frac{z}{6}\big(\cos k_1\cos k_2+\cos k_1\cos k_3+\dots+\cos k_3\cos k_4\big)}
\]
satisfies the differential equation
\[
  \begin{array}{l}
  (z-1) (z+2) (z+3) (z+6) (z+8) (3 z+4)^2 z^3 P^{(4)}(z) + {}\\
  2 (3 z+4) (21 z^6+356 z^5+2079 z^4+4920 z^3+3676 z^2-2304 z-3456) z^2 P^{(3)}(z) + {}\\
  6 (81z^7+1286 z^6+7432 z^5+19898 z^4+25286 z^3+11080 z^2\!-5248 z-5376) z P''(z) + {}\!\!\!\!\!\!\\
  12 (45 z^7+604 z^6+2939 z^5+6734 z^4+7633 z^3+3716 z^2+224z-384) P'(z) + {}\\
  12 (9 z^5+98 z^4+382 z^3+702 z^2+632 z+256) z P(z) = 0.
  \end{array}
\]
\end{theorem}
\begin{proof}
Here we give only an outline of the proof. The calculations in extenso
are provided as a Mathematica notebook in the electronic supplementary
material~\cite{Koutschan11a} (to be downloaded from
http:/$\!$/www.koutschan.de/data/fcc/).

The substitutions $\cos k_j\to x_j$ transform the integrand of the four-fold
integral to
\begin{equation}\label{eq.intg}
  f(x_1,\dots,x_4,z) =
  \frac{1}{\big(1-\frac{z}{6}(x_1x_2+x_1x_3+\dots+x_3x_4)\big)\cdot
           \prod_{j=1}^4\sqrt{1-x_j^2}}.
\end{equation}
This expression is $\partial$-finite and thus a Gr\"obner basis of the
zero-dimensional annihilating left ideal can be computed
(\texttt{ann0} in the notebook). Next, operators
$A_j(x_2,x_3,x_4,z,\D{x_2},\D{x_3},\D{x_4},\D{z})$ and
$B_j(x_1,x_2,x_3,x_4,z,\D{x_1},\D{x_2},\D{x_3},\D{x_4},\D{z})$
for $1\leq j\leq 4$ are computed, such that $A_j+\D{x_1}B_j$ is an
element in the left ideal generated by~\texttt{ann0}. This fact can be
easily tested by reducing it with the Gr\"obner basis: the remainder
being~$0$ answers the membership question in an affirmative way. In
the notebook, the $A_j$'s are collected in the variable \texttt{ann0},
and the $B_j$'s in the variable \texttt{delta1}. We conclude that
$A_1$, $A_2$, $A_3$, and $A_4$ generate an annihilating left ideal for
the integral $\int_0^\pi f(x_1,x_2,x_3,x_4,z)\rmd x_1$. In a similar
fashion, the integrations with respect to $x_2$, $x_3$, and $x_4$ are
performed, yielding a single ODE in $z$ that annihilates~$P(z)$.
\end{proof}
Note that this theorem confirms the conjectured result given
in~\cite{Guttmann09}.  Guttmann also observed that the differential equation
given in Theorem~\ref{thm.fcc4} has \emph{maximal unipotent monodromy} (MUM),
i.e., its indicial equation is of the form $\lambda^n$ and hence has only $0$
as a root, and additionally satisfies the \emph{Calabi-Yau condition}. Many
LGFs of other lattices fall into this class, too, and therefore this fact may
not seem too surprising.

In order to receive the return probability in the four-dimensional fcc
lattice, holonomic closure properties are applied to compute a differential
equation for
\[
  \frac{P(z)}{1-z}=\sum_{n=0}^\infty \bigg(\sum_{k=0}^n p_k({\bm 0})\bigg)z^n,
\]
which in turn gives a recurrence for $f(n)=\sum_{k=0}^n p_k({\bm 0})$:
\[
  \begin{array}{l}
  (n+2) (n+3)^2 (n+4) (35 n^2+420 n+1252) f(n) + {}\\
  (n+3) (n+4) (595 n^4+11375 n^3+79874 n^2+244384n+276024) f(n+1) + {}\\
  3 (n+4) (1015 n^5+24780 n^4+240253 n^3+1156976 n^2 + {}\\
  \qquad 2769392 n+2638272) f(n+2) + {}\\
  (3325 n^6+107100 n^5+1427695n^4+10080600 n^3+39767416 n^2 + {}\\
  \qquad 83134488 n+71984160) f(n+3) - {}\\
  4 (2065 n^6+62580 n^5+788848 n^4+5295615 n^3+19973086 n^2 + {}\\
  \qquad 40139838 n+33590844) f(n+4) - {}\\
  12 (735 n^6+25200 n^5+359282 n^4+2725632 n^3+11601091 n^2 + {}\\
  \qquad 26259960 n+24690708) f(n+5) + {}\\
  288 (35 n^2+350 n+867) (n+6)^4 f(n+6) = 0.
  \end{array}
\]
The initial values
\[
  f(0)=1,\quad f(1)=1,\quad f(2)=\frac{25}{24},\quad f(3)=\frac{19}{18},\quad f(4)=\frac{1637}{1536},\quad f(5)=\frac{549}{512}
\]
are easily (of course, not by hand!) computed by counting the number of excursions
of length up to~$5$. For the return probability
\[
  R = 1 - \left(\lim_{n\to\infty} f(n)\right)^{-1}
\]
we need to evaluate the limit of the sequence~$f(n)$. This can be done very
accurately when knowing the asymptotics of the sequence. We apply the method
described in~\cite{WimpZeilberger85}, which has been implemented in
Mathematica~\cite{Kauers11a}, and obtain the following basis of asymptotic
solutions:
\begin{eqnarray*}
  s_1(n) & = & \frac{1}{n^2}\left(-\frac{1}{2}\right)^n
    \left(1-\frac{5}{6n}+\frac{67}{24n^2}+\frac{1459}{144n^3}+O\Big(\frac{1}{n^4}\Big)\right),\\
  s_2(n) & = & \frac{1}{n^2}\left(-\frac{1}{3}\right)^n
    \left(1-\frac{5}{2n}+\frac{51}{8n^2}-\frac{143}{8n^3}+O\Big(\frac{1}{n^4}\Big)\right),\\
  s_3(n) & = & \frac{1}{n^2}\left(-\frac{1}{6}\right)^n
    \left(1-\frac{45}{14n}+\frac{4633}{392n^2}-\frac{112407}{5488n^3}+O\Big(\frac{1}{n^4}\Big)\right),\\
  s_4(n) & = & \frac{1}{n^2}\left(-\frac{1}{8}\right)^n
    \left(1-\frac{52}{9n}+\frac{812}{27n^2}-\frac{45820}{243n^3}+O\Big(\frac{1}{n^4}\Big)\right),\\
\end{eqnarray*}
\begin{eqnarray*}
  s_5(n) & = & \frac{1}{n}\left(1-\frac{1}{n}+\frac{7}{9n^2}-\frac{7}{18n^3}+O\Big(\frac{1}{n^4}\Big)\right),
    \qquad\qquad\qquad\quad\\
  s_6(n) & = & 1.
\end{eqnarray*}
Obviously, the first four solutions do not significantly contribute
as they tend to~$0$ very rapidly. Thus by taking into account $s_5(n)$
and $s_6(n)$ only, and computing the asymptotic expansion to a higher order
(e.g., $30$), allows us to obtain (at least) 100 correct digits of the limit.

\begin{corollary}\label{cor.d4}
The LGF of the four-dimensional fcc lattice, evaluated at $z=1$ is
\[
  P(1)\approx 1.10584379792120476018299547088585107443954623663875285836499,
\]
and therefore the return probability is
\[
  R\approx 0.09571315417256289673531676490121018570070881963801735768774.
\]
\end{corollary}
\noindent
Note: Corollaries~\ref{cor.d4} and~\ref{cor.d5} have been confirmed
independently by employing the certified numerics implemented in the Maple
package \texttt{NumGfun}~\cite{Mezzarobba10}.

\begin{theorem}\label{thm.fcc5}
The lattice Green's function of the five-dimensional fcc lattice
\[
  P(z) = \frac{1}{\pi^5} \int_0^\pi\cdots\int_0^\pi\frac{\rmd k_1\rmd k_2\rmd k_3\rmd k_4\rmd k_5}
  {1-\frac{z}{10}\big(\cos k_1\cos k_2+\cos k_1\cos k_3+\dots+\cos k_4\cos k_5\big)}
\]
satisfies the differential equation
\[
  \begin{array}{l}
  16 (z-5) (z-1) (z+5)^2 (z+10) (z+15) (3 z+5) (15678 z^6+144776 z^5+{}\\
  \quad 449735 z^4+ 933650 z^3-1053375 z^2+3465000 z-675000) z^4 P^{(6)}(z)+{}\\
  8 (z+5) (3057210 z^{12}+97471734 z^{11}+1048560285 z^{10}+3939663705 z^9-{}\\
  \quad 4878146975 z^8-87265479875 z^7-304623830625 z^6-266627903125 z^5+{}\\
  \quad 254876515625 z^4-1289447109375 z^3-503550000000 z^2+{}\\
  \quad 1774828125000 z- 354375000000) z^3 P^{(5)}(z)+{}\\
  10 (27279720 z^{13}+923795772 z^{12}+11725276842 z^{11}+68439921540 z^{10}+{}\\
  \quad 148313757125 z^9-382134335775 z^8-3351125770500 z^7-7801785421250 z^6-{}\\
  \quad 3779011321875 z^5-7716298734375 z^4-39702348750000 z^3+{}\\
  \quad 3393646875000 z^2+ 23905125000000 z-5568750000000) z^2 P^{(4)}(z)+{}\\
  5 (255864960 z^{13}+7892060544 z^{12}+92744995638 z^{11}+524857986060 z^{10}+{}\\
  \quad 1350059072325 z^9\!-465440555100 z^8\!-13545524756500 z^7\!-26918293320000 z^6-{}\\
  \quad 3649915059375 z^5-77498059625000 z^4-190176960000000 z^3+{}\\
  \quad 40530375000000 z^2+45343125000000 z-13162500000000) z P^{(3)}(z)+{}\\
  5 (496679040 z^{13}+13819981248 z^{12}+149186684934 z^{11}+810956145330 z^{10}+{}\\
  \quad 2287368823475 z^9+1646226060075 z^8-8282515456375 z^7-6199228765625 z^6+{}\\
  \quad 13367806743750 z^5-110925736437500 z^4-133825053750000 z^3+{}\\
  \quad 44457862500000 z^2+ 5055750000000 z-3240000000000) P''(z)+{}\\
  10 (167064768 z^{12}+4143853440 z^{11}+40678130502 z^{10}+209673119160 z^9+{}\\
  \quad 607021304825 z^8+689643286650 z^7-135661728250 z^6+3711617481250 z^5+{}\\
  \quad 2664478321875 z^4-21210430812500 z^3-7268326875000 z^2+{}\\
  \quad 4816462500000 z- 189000000000) P'(z)+{}\\
  30 (7525440 z^{11}+163913184 z^{10}+1443544710 z^9+6925739310 z^8+{}\\
  \quad 19123388575 z^7+21336230625 z^6+36477006875 z^5+187923165625 z^4-{}\\
  \quad 55567000000 z^3-346865625000 z^2+84037500000 z+27000000000) P(z) = 0.
  \vspace{-1mm}
  \end{array}
\]
\end{theorem}
\begin{proof}
The proof is very analogous to that of Theorem~\ref{thm.fcc4} and is given
in detail in the supplementary material~\cite{Koutschan11a}.
\end{proof}
Again, we are happy to report that our proof confirms the conjectured ODE
of~\cite{Broadhurst09}.  Remarkably enough, the indicial equation of the
differential equation presented in Theorem~\ref{thm.fcc5} is
$\lambda^5(\lambda-1)$ and hence the ODE lacks MUM. For the same reason it is
not a Calabi-Yau differential equation.

\begin{corollary}\label{cor.d5}
The LGF of the five-dimensional fcc lattice, evaluated at $z=1$ is
\[
  P(1)\approx 1.04885235135491485162956376369999275945402550465206640313845,
\]
and therefore the return probability is
\[
  R\approx 0.04657695746384802419337442059480329107640239774632112930532.
\]
\end{corollary}

\begin{theorem}\label{thm.fcc6}
The lattice Green's function of the six-dimensional face-centered cubic lattice
\[
  P(z) = \frac{1}{\pi^6} \int_0^\pi\cdots\int_0^\pi
  \frac{\rmd k_1\rmd k_2\rmd k_3\rmd k_4\rmd k_5\rmd k_6}
  {1-\frac{z}{15}\big(\cos k_1\cos k_2+\cos k_1\cos k_3+\dots+\cos k_5\cos k_6\big)}
\]
satisfies a differential equation of order~$8$ and with polynomials
coefficients of degree~$43$. Its leading coefficient is
\[
  \begin{array}{l}
  z^6 (z-3) (z-1) (z+4) (z+5) (z+9) (z+15)^2 (z+24) (2 z+3) (2 z+15)\\
  \times (4 z+15) (7 z+60) q(z)
  \end{array}
\]
where $q(z)$ stands for a certain irreducible
polynomial of degree~$25$, and its indicial equation is
$\lambda^6(\lambda-1)^2$.  The full equation is too long to be printed
here, but can be found in~\cite{Koutschan11a}.
\end{theorem}
\begin{proof}
The proof is very analogous to that of Theorem~\ref{thm.fcc4} and is given
in detail in the supplementary material~\cite{Koutschan11a}.
\end{proof}
As in the five-dimensional fcc lattice, the differential equation
of Theorem~\ref{thm.fcc6} lacks MUM and therefore is not Calabi-Yau.

\begin{corollary}
The LGF of the six-dimensional fcc lattice, evaluated at $z=1$ is
\[
  P(1)\approx 1.02774910062749883985936367927396850209243990900114872425172,
\]
and therefore the return probability is
\[
  R\approx 0.02699987828795612426936417542619638021612262676239501413843.
\]
\end{corollary}

We want to conclude this section with an overview of our results
concerning the return probabilities, which reveals an interesting
dependence on the dimension of the lattice:
\begin{center}
\begin{tabular}{|c|l|}\hline
Dimension & Return Probability \\ \hline
2 & 1 \\
3 & 0.256318236504649 \\
4 & 0.095713154172563 \\
5 & 0.046576957463848 \\
6 & 0.026999878287956 \\ \hline
\end{tabular}
\end{center}

\section{Outlook}
While the calculations for Theorem~\ref{thm.fcc4} and
Theorem~\ref{thm.fcc5} are performed in a few minutes respectively hours,
it was a major effort of several days to compute the certificates that
prove Theorem~\ref{thm.fcc6}; they are several hundred MegaBytes in
size. With the methods described in this paper and with the current
hardware, it is completely out of the question to attack the fcc
lattice in seven dimensions.  An interesting question is whether the
pattern that showed up in dimensions four to six continues. This would
suggest a differential equation of order~$10$ with indicial equation
$\lambda^7(\lambda-1)^3$. But who knows?

For the three corollaries we computed the approximations for the return
probabilities with more than one hundred valid digits.  But we have no clue
what their exact values are. Banderier evaluated these numbers up to
several thousand digits~\cite{Banderier11}, but also he was unable to identify
the closed forms. So we leave these questions open, as a challenge for future
research.

\subsection*{Acknowledgments}
Anthony Guttmann aroused my interest in computing LGFs during his talk
at the 65th S\'{e}minaire Lotharingien de Combinatoire in Strobl,
Austria.  I would like to thank him for his encouragement and advice.
I am grateful to my colleague Manuel Kauers for interesting
discussions and for pointing me to the trick of multi-step guessing.
This research was carried out while the author was employed at the
Research Institute for Symbolic Computation (RISC), Linz, Austria.
The author was supported by the Austrian Science Fund (FWF): P20162-N18.

\bibliographystyle{unsrt}

redpajama_set_name: RedPajamaArXiv


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The device should help her keep up with assignments while the school building is closed during the COVID-19 pandemic.
"We went on spring break and school is done," Zolah's mom Janell Williams explained. "To have something she could go pick up from school that was hers and she has that ownership over, I feel like kind of brought that familiarity of school to our house."
Zolah's younger brother, Charlie, received an iPad Monday to help him tackle his first grade assignments. It means they no longer have to share the one family computer with their parents. And the Williams wouldn't have been able to buy a laptop and iPad for Zolah and Charlie on their own.
A donation from SchoolSmartKC helped Scuola Vita Nuova buy the devices. In fact, SchoolSmartKC donated more than $1.5 million to Kansas City Public Schools plus charter schools and organizations within KCPS's boundaries this month, allowing those groups to purchase more than 2,500 laptops or tablets and more than 4,300 WiFi hotspots.
SchoolSmartKC has existed for about four years. It collects donations from groups like the Hall Family Foundation, Walton Family Foundation and Ewing Marion Kauffman Foundation to fund programs within schools to close the achievement gap among Kansas City's urban core.
SchoolSmartKC said schools like SVN Charter were making a lot of progress toward closing the achievement gap then COVID-19 and the corresponding stay-at-home orders flipped the educational model upside down.
So the organization pivoted. It had a pot of money with which it was going to fund specific programming, instead it granted that $1.5 million to organizations to buy devices and lend them to students for this semester.
"Our schools were doing great work, they were already kind of addressing these issues, but they needed a jolt. They needed some real support to kind of accelerate that given the unique complexities of COVID," explained Awais Sufi, the CEO of SchoolSmartKC.
Sufi said by closing the achievement gap students are more likely to stay in school, get good jobs and break the cycle of poverty. He said giving kids access to a computer and internet now could help keep them on a successful track.
Students have to give the devices back to their schools at the end of the semester. But the silver lining is technology isn't going anywhere. Now these schools have devices they can incorporate into the curriculum next school year.
Johnson County, KS
COVID-19 Video Update

redpajama_set_name: RedPajamaCommonCrawl


The consortium awards TEMA+ graduates with double or multiple degrees (depending on the student's mobility track), all accredited on national level, as well as a diploma supplement, allowing them to enter into the professions in the area of European Heritage in local, regional, national and territorial administration, in urban planning and research institutions, related to heritage and identity.
The Erasmus Mundus (Erasmus+) programme offers a scholarship to excellent Programme and Partner Country students that consists of a contribution to the students' participation costs (including the tuition fees,for Erasmus Mundus scholarships, and June 17, 2018 for self-paying students. library and laboratory costs, full insurance coverage and any other mandatory costs related to the students' participation in the programme); a contribution to student travel and installation costs; and a monthly subsistence allowance of a 1000€ for the entire duration of the study programme.


Blogposts can be about everything that crosses my mind. Get inspired by my favorite quotes, beautiful images and more.
Fashion is not about the current trends or expensive items. It's about combining & creating. Get inspired! Reading books is something I love to do. Find my reviews here.
Poetry, something so close to music. I can totally lose myself in a poem. I also share poems written by myself. Music plays a part in everyone's life. There's nothing I love more than music. I do covers of songs.


text:
Where does India Stand in the Mobile Gaming Market
Early 90's when India had just opened up her economy, it was the formative days of economic liberation. Today, having come so far, India is no more a growing economy but a robust economy big and huge enough in the Asian sub-continent to impact the world business.
Take the case of the Indian gaming industry. In the last couple of years, India has emerged as a major gaming hub after China with enormous opportunitiesthat make up for a potentially huge market. The availability of cheaper but powerful smartphones, increased internet penetration, improved mobile and internet services and better digital payment options have all added to the exponential growth of the gaming market in India.
Take a look at these parameters to understand the current position of India in the growing mobile market and believe it or not, we are a powerful force simply hard to be ignored!
The gaming industry in India was estimated at $543 million in 2016 and is expected to grow to $801 million by 2022. In 2016, India accounted to 0.55 percent of the global gaming industry and this further estimated to grow in the next five years. Increased penetration of smartphones, greater adoption of freemium games are the factors aiding the growth of the industry.
Games download
According to App Annie, an online analytics firm, in 2016 India became the fifth biggest gaming market in the world. It is expected to surpass Brazil and Russia too who stand in position three and four respectively. Compared to 2014, in the year 2016 the games download had doubled with Google Play clocking in 13 times more downloads in comparison with iOS. It is mixed bag of games download in India. With card games being one of the popular categories, rummy game download too witnessed an increase in number.
Mobile gamers
Yet another factor exhibiting significant growth is the number of online gamers in the country. According to the CII-TechSci Research report, at 198 million in 2015, the mobile gamers are expected to grow substantially to 628 million by 2020. This would further see a boost in the numbers with an increase to 1.16 billion by 2030. The shift from console games to mobile games, ease of playing on mobile phones and better wireless connectivity have added immensely to the growth of these numbers. The popular games among mobile gamers have been Poker, Teen Patti and rummy games.
7 Myths About BYOD Debunked


Need Help From Mod Isailix!
Forums » Towny Server Chat / Discussions » Need Help From Mod Isailix!
Hey buddy I need your help with something in towny for my grand resurgence on this server!! Its a build you did for someone else on the server before the reset. I have been waiting for a while but haven't seen you on. I missed the one day you came on which was prob for a second and you haven't been on since. This is a very specific build and its quite difficult for me to do off hand. But its not very complicated...just technical.
If you guys can reach him outside the game that would be a great help! I just want to know if he will at least lend me a hand!! Thankyou very much!
Hey man, I'd be glad to help you out, but as of right now, I have internet that almost doesn't function. I can't load any webpage right now, much less get on the server. It is the main reason I've been inactive this year and I hope to get it fixed soon.
I just talked with our ISP earlier today and we got most of the issue worked out. When this all blows over, I'll gladly help.
thats a shame to hear I mean why do we even pay for these blasted things when they dont even work half the time. Its why I got the lower end for a cell phone service I just use whatsapp for the most part lolol. Yeah ltr it took me two years to open my town and beofre the reset and then a month later you reset the server...so I want to get it ready in a quarter of that time! So I will need your help within 4 months! you got time. I just wanted to know the specifics of what you need and the cost.
Its the prismarine build with 4 circles and two towers (tower per circle) with a middle wooden center piece (third circle) with a fourth circle off to the side in between the two towers.
See Example... It was a rather large build several plots in diameter. I hope this gives you enough information to do it exactly the same because I simply loved it!
The items needed are Prismarine, Prismarine Brick, Dark Prismarine, Quartz, Glow Stone, Leaf Blocks. The amount, money and time needed I dont know so I need to organize that with you when you have the ability to do so thankyou.
There's a warp called /prisshop where you can buy a stack of shards for only 5 bucks. It autofills as well so I'd start from there. I know that there were stone slabs too, but that shouldn't be hard to get.
EDIT (12:53PM EST): I have started building the structure on a singleplayer game, I'll let you know everything we'll need as far as resources later. Tomorrow most likely.
if you could build it to the way you left it off the template that would be great because it was perfect for what I was doing with it. Like it was more than perfect it was a god send. For right now I am just going to work on filling a few dub chest in preparation.
If possible, I'd like you to add me on discord.
That should've worked, try it again and if it doesn't work then you can send yours and I'll add you.


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The "Triplets" Situation at a New School
The first day of middle school, the triplets were asked to stand up and tell a little about themselves. My daughter got up there: "My name is Carly Fox. I'm a triplet. I hate being a triplet." She was heading back to her seat when she turned around. "Oh, and I also hate my last name." Way to go on the positive self-talk, Carly. I looked up the number for the school psychologist and added it to my phone contacts so I'd know it when it came up.
Maybe I should have been one of those mothers who verbally assaults their kid by yelling:"Good Job!" every three seconds from birth through masters degree.
I told Carly-- I guess I can still call her "Carly". She didn't publicly denounce her first name. Maybe she just ran out of time. She does give me wrinkled nose face when I mention that my first name choice for her was "Samantha" but apparently Samantha Fox was a porno star and so daddy voted it down. (Funny how none of the women I tell that to ever heard of her but all the men smile knowingly like I just sent them down some smut-filled memory lane.)
So, I told Carly to feel free to revert "Fox" back to my husband's original family name: "Fuchs". And being from New York, I had no trouble giving her several specific examples of how that could go terribly awry at every middle school in the nation.
I also enumerated for her several of my girlfriends with whom I'd grown up, who lamented their last names for years and then married into far worse ones. (I don't dare get into them here. They might be reading. And you know who you are.)
As for the triplet thing, I told her that we hadn't really entertained the thought of reducing the number of embryos, and that now it was kind of late: The cut-off to make that decision was third grade.
While I dislike what she said up there for twenty-two or three reasons, I totally understand. When triplets attend a new camp or school or anything, it's different than one child. When you introduce triplets, it's like the circus side show has come to town. Everybody's staring and trying not to look shocked.
It's funny how students and campers seem to get over the novelty and forget all about it a lot faster than teachers and counselors. The girls don't look thaaat much alike: They have different colored eyes and one wears glasses. And not only don't they dress alike, you can tell just by looking at them that they're whole dressing methodology couldn't be more different: One tries on six different outfits in her room, running into the bathroom each time to look in the bathroom mirror, then runs downstairs to get the whole effect in the full-length mirror. She's practicing for the track team, without even knowing it. The other daughter puts on a t-shirt and leggings at night and calls it pajamas. Then in the morning: "Hayley! It's time to get up and get dressed! She rolls over in bed, eyes still closed: "I am dressed!"
And yet at least one teacher will ask all year: "Are you Carly or are you Hayley?" But okay. Teachers have a lot of kids to remember. Much worse, the last thing tweens want every time they're coming down the hallway is to be announced:
"Here come the Fox triplets!" like they're one big kid.
Then- and this is where her hatred for our last name comes into play- one teacher last year used to yell out:
"What does the Fox say?!" Yeah it was hilarious the first, second, thirty-fourth, fifty-second... okay it was never hilarious.
They also get: "Who let the Foxes out? Who? Whooo?!"
I keep trying to remind the kids that these people mean no harm. Still, as mother of this skulk of Foxes, I'm always tempted to get nose to nose with the wiseacre and say: "Who let the idiot out? Who? Whooo?"
Thanks a lot for stopping by. I hope you were able to add a few laughs to your day. If you'd like more laughs at my/your kids' expense, please sign onto my monthly newsletter and/or check-out my eBook, specifically for parents at this time of year. Available on all Amazons-- Now available at the Kindle Library too.
newsletter: http://laughingisconceivable.com (at top)
eBook: https://www.amazon.com//dp/B07261ST2P/
Tagged: back-to-school, dads, de-stress, funny, humor, kids, Laugh, mind/body, moms, multiples, parenting, parents, triplets
Newer PostWhy I Couldn't Wouldn't Shouldn't Be an IVF Nurse
Older PostMiddle School Prep- Triplets Style


Interested parties are invited to make submissions to the First Principles Review of Defence.
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publish whole or part of the submission in other Defence products, including the First Principles Review of Defence, with attribution to the author.
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Subject to the above conditions, submissions with the authors' express consent to publish will be uploaded to the Defence website during the consultation process and will be removed at a time deemed appropriate by Defence.


This article comes from The Old Stone Bank History of Rhode Island , Vol. III, pages 140-142, published by the Providence Institution for Savings, 1939. Transcribed by Beth Hurd for non-commercial use only.
University Hall was only ten years old when it was used as a military hospital in 1780.
Detail of illustration from The Early History of Brown University by Dr. Reuben Aldridge Guild (1864).
It was the year 1780 and Christmas time in Providence, where not many more than twenty-five hundred war-weary souls were preparing for the age-old holiday observance with just a little more enthusiasm than they had been able to arouse for, what seemed to them, a long, long time. And, it had been a long time since the startling news of the fateful clash of arms at Lexington had disrupted the ordinary course of things in all Rhode Island. April 19, 1775, then seemed like a day in the dim past, only faintly recalled as the time when it all began. Too many things had happened, had occurred in rapid succession, for any one hour, day, or event to stand out in any great degree of prominence then, for the crushing burdens of war still rested heavily upon the shoulders of Rhode Islanders and of all those in America who, for more than five years, had resolutely sought liberty through armed resistance.
Long forgotten by the people of Providence in 1780 were the quiet times of ordered existence. No longer were the laborious but tranquil pursuits of agriculture, and the methodical tasks of trade and commerce familiar to the comparatively few men who remained at home; no longer were household duties for the women routines of regular habits and practices. Providence like many American communities was a military camp; talk of war filled the air; machinery of war cluttered the streets; no family could escape or avoid the destructive, disconcerting effects of a bitter struggle that still raged on native soil. For many months since '75 Providence had been near to the actual fighting, hundreds of soldiers had tramped through its streets to battle, and on their way back to home or hospital. Young men, old men, many had smelled the smoke of battle, been shaken by the sudden burst of bombardment. While old wounds and new wounds were healed under the patient care of frightened, hopeful wives and mothers—cherished notes from the front were pressed against anxious breasts. In all wars man-made instruments of destruction shatter the minds and bodies of humans far out of the range of their death-dealing missiles. Battlefields encompass vastly more than the serried ranks and drawn up reserves of fighting armies.
But, it was now Christmas in Providence and it was the year 1780 when something had happened that gave renewed hope to those who sought liberty through strife. In the summer six thousand men had come from far-off France to help the American colonists in their cause, and this formidable force of eager allies had the good fortune to complete a long and stormy ocean voyage in Rhode Island waters. Leaving their ships in Newport harbor, Count de Rochambeau's troops were promptly quartered in the old seaport at the lower end of the Bay. Many were ill, however, after the crossing, and had to be provided immediately with hospital care.
The ill French soldiers who remained in Newport quartered in the Old State House and in the Congregational Church; some were sent to Bristol; and about three hundred and forty were transported to Providence and placed in University Hall, on College Hill, Brown's original structure, which was ordered to be turned over to the French by the Council of War. Brown University's precious pile of perfect Colonial design was a hospital for sick, lonesome native French warriors when the snow began to fall lightly late in the afternoon of December 24, 1780.
For many weeks the Frenchmen who had escaped illness and who could actively prepare for service at the front with Washington's army had virtually spellbound both Newport and Providence with their dazzling uniforms and courtly manners. People took their new guests as a form of wine, responding miraculously to their invigorating presence. Rhode Islanders, bruised and battered by every blow known to war, began to laugh again in the renewed gaiety of social life, and turned with a new heart to build up a bustling business and trade. The best mansions were thrown open to the French officers, and many homesteads, especially in Newport, were allotted to them as quarters. Brilliant parties and balls sprang into immediate vogue, and the two chief seaports of Rhode Island outdid themselves in courtesy and hospitality. Newport witnessed most of the social renaissance but Providence and other communities were not far behind in the whirl of entertainment. Nor were the Frenchmen one whit behind. Perhaps they had heard that a reputation for foppish ways and broad views about life preceded their arrival in this country; at any rate, they deliberately set out to shatter the truth of such unfair rumors. Certainly, by acting from the beginning with the most perfect decorum and grace, they dispelled whatever apprehensions American people may have had. Rather, the sight of their brilliant regalia, their gleaming swords, cockades, buttoned boots, the white uniforms of the Deux-Ponts Regiment, the green and white of the Saintonge, the black and red of the Bourbonnais, and the rose facings of the Soissonais [sic] with white and rose pluming surmounting grenadier caps, soon caused many a feminine heart to flutter, perhaps to feel that war really wasn't so terrible after all.
But the belles of Rhode Island were not alone in feeling the delightful quickening of the pulse and the excited beating of the heart. Charming as the Frenchmen had found America, they were unanimous in their praise of its beautiful women. Refined, attractive, gallant, and always considerate, they vied among themselves for the honor of paying tribute to their fair Colonial hostesses and partners at gay social functions. Many a French heart was left behind on the shores of Narragansett Bay at parting time, and many a romance, friendship, and broken heart can never be recalled. But, we are thinking particularly now of Christmas Eve, in 1780, when, in the midst of the joy and merriment that a chance meeting of the Old World and the new brought to these shores, three hundred or more strangers rested quietly in the dimly-lighted halls and chambers of old University Hall on College Hill.
Somewhere that afternoon a group of French officers had met intending to celebrate the approaching Christmas Eve. They were thinking and talking of Paris and of little villages in their faraway homeland—they pictured in their minds the familiar scenes of Christmas back home. They wondered what the holiday would be like in Puritanical New England where, for many years, old world customs had been put aside for more austere, conservative rites and forms of observance. But, several rounds of Madeira wine, and perhaps a few hot toddies which the cold November days in Rhode Island had taught these strangers to relish, soon dispelled all somber thoughts and longings for home. Songs soon replaced thoughts and words, and when candle lights began to flicker in the gray of twilight, the merry spirit of an old time Christmas spread to every corner of the tavern room, not far from the Market House. The tunes they sang were familiar to all, but the words sounded strange to the Providence folks who peered through the steaming panes of the tavern windows and who crowded into the room to join the happy Frenchmen. Greetings were exchanged, well wishes extended, toasts drunk to General Washington, to Count de Rochambeau, to Lafayette, to the pretty girls of Providence, to Rhode Island, to Christmas.
Then some one suggested that a serenade be given the sick Frenchmen on the Hill; and hardly had the suggestion been offered when an impromptu procession was formed in front of the tavern. Two musicians with a flute and a viola appeared from somewhere, volunteer serenaders appeared from everywhere. Headed by a popular tavern keeper in the town who linked arms with a stalwart Grenadier from the Royal Deux Ponts, the singing, laughing, arm-waving file, becoming longer and longer by the minute, circled Market Square and then tramped through the mud and slush up the hill leading to the Military Hospital long since vacated by President Manning's prospective lawyers, doctors, and preachers.
There were no Van Winkle gates to pass through then, no sidewalks for hand clapping spectators to stand upon and watch the fun. College Hill that night was just a sloping, treelined cowpath, but no College victory parade along that famous thoroughfare ever had more genuine enthusiasm, heartfelt spirit than did the procession of jolly carollers on the eve of Christmas in 1780.
Standing beneath the great elms that were probably the grandparents of the stately trees now surrounding Brown's original college building, Frenchmen, Englishmen, officers, soldiers, tavern-keepers, boys, pretty girls, and aged patriarchs waited for the signal, and then, as the snow fell lightly upon cockades, bonnets, fur caps, and beavers, they sang of Christmas. It was an old tune, a French tune, but both English and foreign words seemed to fit the rhythm and meter of the melody. Words did not matter, it was the spirit of the singing, and the three hundred or more lonesome, bodily-ill strangers behind those hoary walls heard the voices; they knew it was Christmas in Providence, Christmas everywhere. Tears came to eyes, prayers were muttered, heads fell back upon rough straw pallets, but a bond of union, of sympathy and understanding, had joined all in a common spirit of peace on earth, goodwill to men. And, as the last strains of "Lift up your gates, ye Princes, and let the child be born," or some other old European carol, reached the ears of those who listened within, some one ran to the steps of the ancient edifice and shouted loudly, "Merry Christmas to all, Merry Christmas to all," and the answer came back from every hall, from every room, "Joyeux Noel, Joyeux Noel." It was a Merry Christmas for all in Providence, in 1780.
Return to Old Stone Bank History of Rhode Island index.
John Williams Haley (1897-1963), former vice president of the Narragansett Brewing Company, was best known for his weekly radio program, "The Rhode Island Historian," which ran from 1927 to about 1953 on WJAR. Several hundred of his radio scripts were published in pamphlet form by the Providence Institute for Savings ("The Old Stone Bank"), and many were later reprinted in the four-volume Old Stone Bank History of Rhode Island.
Although sung to an old French melody, the lyrics of the Christmas carol "A Day, A Day of Glory," from which the line "Lift up your gates, ye Princes, and let the child be born" comes, weren't written until the nineteenth century. John M. Neale (1818-1866) based his lyrics on Psalm 24:7.


text:
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Daily Current affairs is one of the most important topics to score marks in the Mains Exam of any Banking Exam. This will also act as a score booster in the Mains Exam. We are providing Current Affairs Quiz regularly in our Bankersdaily and you can also attend the Daily Current Affairs in the actual exam format which will also enhance your time. Check the Current Affairs Quiz of September 05, 2018.
Q.1) Japan and which of the following Indian state has signed agreements to promote bilateral cooperation in culture, business and interaction between the academia?
Q.2) Pranab Mukherjee has inaugurated Smart gram Yojana Projects in which of the following state?
Q.3) Which of the following plans aims to build 100 new airports across India in the next 10-15 years with an investment of USD 60 billion?
Q.4) Which of the following state government plans to provide free mobile phones to women of BPL families?
Q.5) Name the India coffee field force app launched by Suresh Prabhu?
Q.6) Name the powerful typhoon that has slammed into western Japan recently?
Q.7) Name the Indo-Kazakhstan Joint Army Exercise that will be conducted in Otar region, Kazakhstan?
Q.8) The National Bank for Agriculture and Rural Development has sanctioned an amount of how much crore under the Rural Infrastructure Development Fund (RIDF) to West Bengal?
Q.9) Where is the headquarters of NATO located?
Q.10) Who was elected as the president of Pakistan?
Q.11) Who will become the 46th Chief Justice of India?
Q.12) Om Prakash Mitharwal is related to which of the following sport?
Q.13) When is the International Day of Charity observed?
Q.14) What is the currency of Kazakhstan?
Q.15) Who is the founder and CEO of Amazon?


The greatest living performer of all-time celebrated her 35th birthday yesterday - Yes, I did say all time, as male or female it really makes no difference. Who even compares? They don't call her the Queen for nothing.
With that being said, in honour of the Queen's 35th birthday yesterday, here are ten of her her most important life lessons.
This is the most valuable advice anyone can ever give you. Bey teaches us that mistakes are an integral part of the growth process. Mistakes should be seen as lessons and are something we should use to inspire us to carry on instead of instilling fear. You know what they say right? If you can't beat fear, then just do it scared. Beyoncé is a prime example of that.
Everyone knows Beyoncé is a proud feminist and humanist. In a world where inequality is so rife in numerous aspects, it's important to identify areas such as sexism and racism that need adequate support and representation. I am also a feminist (liberal and intersectional) and Bey is one of the people who inspired me to be vocal about the cause. If you can't support your own gender and things, events and causes that affect you and others like you, then who can you support?
One of the things I love most about Beyoncé is her profound humility. I have never met a more humble, selfless and generous soul than her, and despite all of her accolades, achievements and successes, she continues to remain down to earth as ever. Bey teaches me that true humility is not necessarily thinking less of yourself, but thinking of yourself, less. That's the essence of what makes us human and is the only pathway to success. This is something I pride myself upon and will always live my life by. Bey may have a song about an Ego - but she certainly doesn't harbour one herself.
"You wake up, Flawless," - Remember?
Bey teaches us that although pain is a strange thing at times, it's necessary because it makes us stronger. "They say true love's the greatest weapon, to end the war caused by pain," she sings in All Night, in one of the most telling verses from Lemonade. But why is pain so important? Because most importantly of all, it has an end. The battle will always be won, even when it looks like it won't be.
Is there anyone more feared than the individual who learns to seek validation from no one, but the reflection they see in the mirror? Let's just ask Bey shall we? This is an on-going process for me, but Bey's lessons about self-acceptance cannot be under-looked.
Just says it all - No elaboration necessary. If at first you don't succeed, then just dust yourself off and try again. Take it. Be it. Own It. After all that's what Bey does. Well, I mean, she's Beyoncé, she's the Queen. I strongly doubt there's anything she can't do, let's be realistic here.
Self-worth and self-love are to of the most important self-evaluation's anybody can ever make. Bey teaches me that no matter what happens, you should always know your worth. Never settle for anything less than you deserve. You're a queen, so you deserve to be treated like one.
Out of all of the hundreds of lessons Bey has taught me, this is the one I always remember. How many of us have ever been in that situation when we've been hurt by someone we care about? Whether that's family, friends or a significant other, we often want nothing more than to "hurt" them back, but by doing so we'd create a seemingly never-ending cycle. Bey teaches us that by letting go, we're not saying what happened was okay - we're just making the choice not to be ruled by it.
When you are served lemons, such as hurt, pain, anger, denial, apathy or emptiness, you make lemonade instead. Accountability, forgiveness, hope, redemption, love… The list will always go on.
Lemonade is available to download on iTunes.


According to the Kaiser Family Foundation, California averaged around 7.8 million Medicaid/CHIP enrollees prior to the ACA; this number has since shot up about 58%, to 12.3 million people.
One of the chief arguments in favor of the ACA has always been that uninsured people tend to not get treated for ailments/injuries at all (for obvious reasons), and then end up going to the emergency room when that gangrenous foot leaves them no choice. Not only does this end up costing much more to treat, the hospital is also stuck footing the bill (no pun intended) since the uninsured also tend not to have any money to pay out of pocket either (if they did, the odds are that they'd have some form of insurance, after all).
The idea was that by expanding Medicaid and offering financial assistance to help people buy private insurance policies, the load on emergency rooms would be lightened in two ways: A reduction in the actual number of visits (as people get covered, hopefully they'll use that coverage to make a regular doctor's appointment to hopefully prevent emergencies from happening in the first place); and a reduction in the number of case of hospitals being stiffed for the bill.
The only hard numbers shown are the one from Q4 2015,but the earlier numbers can be estimated pretty closely. The obvious starting point here is Q4 2013...the last quarter before the ACA's Medicaid expansion (and exchanges for private policies) went into effect.
California had about 38.43 million residents in 2013, which went up to around 39.14 million as of July 2015, or 1.8%. Let's figure around 2% even over that two year period. This suggests that privately insured enrollees haven't been impacted one way or the other by the ACA, but slightly impacts the other categories.
Medicare enrollment in California went up from 5.302 million to 5.644 million over those 2 years, or 6.5%, so I'm not sure what accounts for the 21% increase in ER visits. Then again, the ACA's impact on Medicare is less direct for the most part (aside from fixing the Part D "donut hole"), so I'm not even gonna try to touch that one.
That leaves Medicaid, Uninsured and "Other Coverage". I'm assuming the "Other" drop is mainly due to people being transferred over to Medicaid from other state-run programs; assuming this is the case, that cancels out around 98,000 of the Medicaid increase.
That's a decrease of 174,000 uninsured ER patients...and assuming the vast bulk of these folks were low income/poor, that likely means a good 170K or so fewer uncompensated ER visits, which is a good thing.
In terms of total ER visits, when you combine all 5 categories that's an increase from 2,645,000 to 3,024,000, or 14.3% overall. Again, that 2% population increase accounts for some of this, leaving around an adjusted 12% increase (or around 320,000).
On the surface, this makes it sound like the ACA is only achieving one of it's two ER goals: Reducing the number of uncompensated visits. On the other hand, it's entirely possible that a lot of those 320,000 additional ER visits were people who have been suffering from nasty ailments for many years but who didn't dare even go into th emergency room to have them treated.
Assuming CA's numbers are representative nationally, I'd call this 1.5 out of 2 points.


Indian wear is always exciting to don but often begs the question - what beauty look to pair with it? While accessorizing with jhumkas and slipping on juttis comes easily, figuring out what makeup look to pair with traditional Indian wear can leave most beauty lovers flummoxed. In times like those, we're thankful we've got the stars of Bollywood to turn to. Since traditional gear is their second skin, they know how to work a beauty look with their lehengas and sarees to perfection.
* When you look as lovely as Shraddha, pulling off a minimal makeup look comes easy. With a pastel toned salwar kameez, Shraddha Kapoor opts for kohl-rimmed eyes, defined brows, blush pink lips and her trademark peaches and cream complexion making this perfect for a day at work.
* Giving us tips on going glamorous with Indian wear, Jacqueline opts for shimmery champagne eye makeup with full brows, lush lashes and thickly lined eyes. Topping it off is a glossy plum lip; that's all you need to play showstopper at the next Indian wedding you head to.
* No one pulls off a regal traditional look quite like Kareena can. With a flawlessly covered face as her base, she wears bronze smoked-out eye makeup with filled-in brows, rosy cheeks and a pastel pink lip. If you're going traditional for your next family function, Kareena's makeup look is your answer.
* The ever-lovely Alia looks absolutely radiant in her low-key Indian look. Alia's dewy skin is highlighted to perfection with a flush of colour on the cheeks. With it, she wears kohl-lined eyes, defined eyelashes, naturally filled brows and poppy pink lips, making it understated enough for a puja function.
* Vaani shows us that glamorous brown eye makeup is just made for Indian wear. Her face is carefully contoured, brows filled in and lips finished off with a glossy neutral coat of colour to create a dramatic look that's not OTT, perfect for a friend's shaadi.


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Ebang International Holdings Inc
EBON CLASS ACTION NOTICE: The Law Offices of Frank R. Cruz Files Securities Fraud Lawsuit Against Ebang International Holdings, Inc. (Businesswire)
The Law Offices of Frank R. Cruz Files Securities Fraud Lawsuit Against Ebang International Holdings, Inc. (EBON)
The Law Offices of Frank R. Cruz announces that it has filed a class action lawsuit in the United States District Court for the Southern District of New York captioned Zaker v. Ebang International Holdings Inc., et al., (Case No. 1:21-cv-03060) on behalf of persons and entities that purchased or otherwise acquired Ebang International Holdings, Inc. ("Ebang" or the "Company") (NASDAQ: EBON) securities between June 26, 2020 and April 5, 2021, inclusive (the "Class Period"). Plaintiff pursues claims under Sections 10(b) and 20(a) of the Securities Exchange Act of 1934 (the "Exchange Act").
Investors are hereby notified that they have until 60 days from this notice to move the Court to serve as lead plaintiff in this action.
If you are a shareholder who suffered a loss, click here to participate.
On April 6, 2021, before the market opened, Hindenburg Research published a report alleging, among other things, that Ebang is directing proceeds from its IPO last year into a "series of opaque deals with insiders and questionable counterparties." According to the report, Ebang raised $21 million in November 2020, claiming the proceeds would go "primarily for development," and that instead the funds were directed to repay related-party loans to a relative of the Ebang's Chief Executive Officer, Dong Hu. The report also noted that Ebang's earlier efforts to go public on the Hong Kong Stock Exchange had failed due to widespread media coverage of a sales inflation scheme with Yindou, a Chinese peer-to-peer online lending platform that defrauded 20,000 retail investors in 2018, with $655 million "vanish[ing] into thin air."
On this news, the Company's share price fell $0.82, or approximately 13%, to close at $5.53 per share on April 6, 2021, on unusually heavy trading volume.
On April 6, 2021, after the market closed, Ebang issued a statement stating that, though it believed the report "contain[ed] many errors, unsupported speculations and inaccurate interpretations of events," the "Board, together with its Audit Committee, intends to further review and examine the allegations and misinformation therein and will take whatever necessary and appropriate actions may be required to protect the interest of its shareholders."
On this news, the Company's share price fell $0.12, or 2.17%, to close at $5.41 per share on April 7, 2021. The stock price continued to decline over the next trading session by $0.38, or 7%, to close at $5.03 per share on April 8, 2021, on unusually heavy trading volume.
Throughout the Class Period, Defendants made materially false and/or misleading statements, as well as failed to disclose material adverse facts about the Company's business, operations, and prospects. Specifically, Defendants failed to disclose to investors: (1) that the proceeds from Ebang's public offerings had been directed to an low yield, long term bonds to an underwriter and to related parties rather than used to develop the Company's operations; (2) that Ebang's sales were declining and the Company had inflated reported sales, including through the sale of defective units; (3) that Ebang's attempts to go public in Hong Kong had failed due to allegations of embezzling investor funds and inflated sales figures; (4) that Ebang's purported cryptocurrency exchange was merely the purchase of an out-of-the-box crypto exchange; and (5) that, as a result of the foregoing, Defendants' positive statements about the Company's business, operations, and prospects were materially misleading and/or lacked a reasonable basis.
Follow us for updates on Twitter: twitter.com/FRC_LAW.
If you purchased Ebang securities during the Class Period, you may move the Court no later than 60 days from this notice to ask the Court to appoint you as lead plaintiff. To be a member of the Class you need not take any action at this time; you may retain counsel of your choice or take no action and remain an absent member of the Class. If you purchased Ebang securities, have information or would like to learn more about these claims, or have any questions concerning this announcement or your rights or interests with respect to these matters, please contact Frank R. Cruz, of The Law Offices of Frank R. Cruz, 1999 Avenue of the Stars, Suite 1100, Los Angeles, California 90067 at 310-914-5007, by email to info@frankcruzlaw.com, or visit our website at www.frankcruzlaw.com. If you inquire by email please include your mailing address, telephone number, and number of shares purchased.
This press release may be considered Attorney Advertising in some jurisdictions under the applicable law and ethical rules.
The Law Offices of Frank R. Cruz, Los Angeles
Frank R. Cruz, 310-914-5007
fcruz@frankcruzlaw.com
www.frankcruzlaw.com
Copyright Business Wire 2021
Ebang International Ho... - I dag
Ebang International Ho...


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For women-friendly bank, SBI merger is bittersweet newshttps://indianexpress.com/article/business/banking-and-finance/for-women-friendly-bank-sbi-merger-is-bittersweet-news-4594557/
For women-friendly bank, SBI merger is bittersweet news
The bank was established by the UPA government on November 19, 2013 — Indira Gandhi's birth anniversary. Starting with seven branches, it grew to 103.
Written by Harikrishnan Nair | New Delhi | Updated: April 1, 2017 3:15:46 am
AT THE Nehru Place branch of the Bharatiya Mahila Bank (BMB) Friday, a 40-year-old man arrived with a worried look. He had heard that the three-year-old public sector bank would be merged with the State Bank of India on April 1. The announcement was made earlier this month. Pointing to messages he had received from the bank announcing the merger, the man asked if he would have to open a new account and apply for a new ATM card. "You will be treated as an SBI customer," said a clerk, adding that the ATM card and other facilities of BMB would be deactivated from April 2. "An SBI card has already been dispatched to your address." Nine floors above, at the bank's registered office, employees were busy dealing with end-of-fiscal business. Apart from a few people talking about the "last day", there were no discussions. In the lobby, the first real sign of change was seen when the office clerk, Rakesh, arrived with a knife to remove the BMB banner. "This is an emotional moment for us," S N Swati, executive director, BMB, said. "Building a bank from scratch took time and effort. We had beaten many expectations. One report had pegged us in loss. But here we are, making a decent profit for a bank of three years," she said.
The bank was established by the UPA government on November 19, 2013 — Indira Gandhi's birth anniversary. Starting with seven branches, it grew to 103. It was intended to provide financial services, mostly to women and women self-help groups. However, the uniqueness of the bank was slowly eroded as other competitors launched all-women branches and customised products for women. The Centre decided to merge it with SBI to consolidate public sector banks. Around 500 employees and 4 lakh customers will now be absorbed by the SBI.
1 RBI to open separate enforcement department on April 3: Santosh Kumar Gangwar
2 Kotak Mahindra board clears fundraising
3 Problem of bad loans: Banks need Rs 6.50 lakh crore to tackle stressed assets, says Kotak


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Autism Spectrum Disorder – Post-Doctoral Fellowship
Fellowship - Experienced
The Autism Assessment, Research, Treatment, & Services (AARTS) Center in the Department of Psychiatry and Behavioral Sciences at Rush University Medical Center in Chicago is recruiting for a two-year, full-time Postdoctoral Psychologist, Research Track for our growing, interdisciplinary program serving individuals with autism spectrum disorder (ASD) across the lifespan. Our research program seeks to integrate the latest findings in biological, behavioral, and implementation sciences to build effective, accessible treatments for individuals with autism spectrum disorders (ASD) – of all ages.
The research-oriented fellowship seeks doctoral-level clinical psychologists with substantial expertise in the diagnosis and treatment of ASD and a scholarship track-record in neurodevelopment disorders. The fellowship will involve participation in on-going ASD clinical-research studies, support to build an independent research program, and participation in clinical, training, and didactic activities in the Department of Psychiatry and Behavioral Sciences. Competitive applicants will have established reliability on gold-standard diagnostic assessments (e.g., Autism Diagnostic Observation Schedule, 2nd Edition and the Autism Diagnostic Interview, Revised). Supervision towards clinical licensure will also be available. Salary will be competitive with NIH stipends with full benefits. Additional funds for pilot research activities may also be available to competitive applicants.
Rush University Medical Center, located in downtown Chicago, is a national leader in academic medicine. The Department has been expanding its vibrant research, clinical and educational programs. It is recognized as the only Department of Psychiatry in Chicago ranked as a "high-performer" in US News and World Report. The AARTS Center at Rush is a supportive program with unparalleled expertise in diagnosing and treating individuals with ASD. Through our clinical, outreach and research cores, the AARTS Center is dedicated to meeting the unmet needs of individuals with ASD and their families across the lifespan.
This postdoctoral fellowship consists of the following components: clinical research, independent research, and didactics. Each is detailed below
Clinical-Research (75%)
Conduct diagnostic and study-specific evaluations for research studies
Serve as study clinician on intervention trials
Support study protocol development and review for new trials
Assistant in grant development, preparation and submission
Participate in collaborative poster presentations, community presentations, and other scholarship activities
Assist in supervision of research coordinators
Independent research (15%)
Develop mentorship plan to support independent research program in collaboration with Rush Research Mentoring Program (RRMP)
Submit grant applications supporting independent research
Scholarship activities supporting independent research program – conference submissions, manuscripts
Didactics (10%)
Support community-based outreach events
Attend weekly clinic rounds; case conference; departmental grand rounds
Teach or co-teach psychiatry residents and fellow courses
The position will run from 7/1/2020 to 6/31/2022 (some flexibility on start date)
Salary consistent with NIH standards
22 days professional/personal leave
LEAP funds, pay for 90% of continuing education expenses up to $1,000.00
Low cost health & dental insurance
Malpractice insurance paid by University
Rush University System for Health (RUSH) is an academic health system whose mission is to improve the health of the individuals and the diverse communities it serves through the integration of outstanding patient care, education, research and community partnerships. RUSH comprises Rush University Medical Center, Rush University, Rush Copley Medical Center and Rush Oak Park Hospital, as well as numerous outpatient care facilities. Rush University, with more than 2,500 students, is a health sciences university that comprises Rush Medical College, the College of Nursing, the College of Health Sciences and the Graduate College. Numerous national organizations have recognized Rush for the quality of care we provide, including U.S. News & World Report, which ranked Rush University Medical Center among the top 50 hospitals in five specialties in its 2019-2020 Best Hospitals issue. Rush University Medical Center has received Vizient's Quality Leadership Award, ranking first among 99 academic medical centers across the country. In 2019, Rush was designated a leader in LGBTQ health care equality by the Human Rights Campaign for the 10th consecutive year, and was one of only five hospitals from across the nation to be recognized by the American Hospital Association as an honoree for its annual Equity of Care Award.
Interested candidates, please send your CV and cover letter to:
Allie Wainer, PhD
Research Director, AARTS Center
Allison_Wainer@rush.edu
Rush is an equal opportunity / Affirmative Action employer
Internal Number: 1329
About Rush University Medical Center
Rush University System for Health (RUSH) is an academic health system whose mission is to improve the health of the individuals and the diverse communities it serves through the integration of outstanding patient care, education, research and community partnerships. RUSH comprises Rush University Medical Center, Rush University, Rush Copley Medical Center and Rush Oak Park Hospital, as well as numerous outpatient care facilities. Rush University, with more than 2,500 students, is a health sciences university that comprises Rush Medical College, the College of Nursing, the College of Health Sciences and the Graduate College. Numerous national organizations have recognized Rush for the quality of care we provide, including U.S. News & World Report, which ranked Rush University Medical Center among the top 50 hospitals in five specialties in its 2019-2020 Best Hospitals issue. Rush University Medical Center has received Vizient's Quality Leadership Award, ranking first among 99 academic medical centers across the country. In 2019, Rush was designated a leader in LGBTQ health care equality by the Human Rights Campaign for the 10th consecutive year, and was one of only five hospitals fr...om across the nation to be recognized by the American Hospital Association as an honoree for its annual Equity of Care Award.