text:
\section{Introduction
We start by introducing some notation relating to the order of certain elements modulo $n$. For every positive integer $n$ and every prime number $p$, we denote by $v_p(n)$ the \textit{$p$-adic valuation} of $n$, i.e. the greatest exponent $e\geq0$ for which $p^e$ divides $n$. The prime factorization of $n$ may then be written as
$$
n=\prod_{p\in\mathbb{P}}p^{v_p(n)},
$$
where $\mathbb{P}$ denotes the set of all prime numbers. We denote by $\rad(n)$ \textit{the radical of $n$}, i.e. the largest square-free divisor of $n$, namely
$$
\rad(n) = \prod_{\stackrel{p\in\mathbb{P}}{p|n}}p.
$$
\par For every positive integer $n$ and every integer $a$ coprime to $n$, we denote by $\mathcal{O}_n(a)$ \textit{the multiplicative order of $a$ modulo $n$}, i.e. the smallest positive integer $e$ such that $a^e\equiv1\pmod{n}$, namely
$$
\mathcal{O}_n(a) = \min\left\{e\in\mathbb{N}^*\ \middle|\ a^e\equiv1\pmod{n}\right\}
$$
and we denote by $\mathcal{R}_n(a)$ the \textit{multiplicative remainder} of $a$ modulo $n$, i.e. the multiple of $n$ defined by
$$
\mathcal{R}_n(a) = a^{\mathcal{O}_n(a)}-1.
$$
The multiplicative order of $a$ modulo $n$ also corresponds with the order of the element $\pi_n(a)$, where $\pi_n : \mathbb{Z} \relbar\joinrel\twoheadrightarrow \Z/n\Z$ is the canonical surjective morphism, in the multiplicative group ${\left(\Z/n\Z\right)}^*$, the group of units of $\Z/n\Z$. Note that $\mathcal{O}_n(a)$ divides $\varphi(n)$, $\varphi$ being the Euler's totient function.
\par For every positive integer $n$, we define and denote by $\alpha_n$ the arithmetic function
$$
\alpha_n :
\begin{array}[t]{ccl}
\mathbb{Z} & \longrightarrow & \mathbb{N}\\
a & \longmapsto & \left\{
\begin{array}{cl}
\mathcal{O}_n\left(a^n\right) & \text{for\ all}\ \gcd(a,n)=1,\\
0 & \text{otherwise},
\end{array}\right.
\end{array}
$$
where $\gcd(a,n)$ denotes the greatest common divisor of $a$ and $n$, with the convention that $\gcd(0,n)=n$. Observe that, for every $a$ coprime to $n$, the integer $\alpha_n(a)$ divides $\varphi(n)/\gcd(\varphi(n),n)$. This follows from the previous remark on $\mathcal{O}_n(a)$ and the equality $\alpha_n(a)=\mathcal{O}_n(a^n)=\mathcal{O}_n(a)/\gcd(\mathcal{O}_n(a),n)$.
\par For every positive integer $n$ and every integer $a$ coprime to $n$, we denote by $\mathcal{PO}_n(a)$ \textit{the projective multiplicative order of $a$ modulo $n$}, i.e. the smallest positive integer $e$ such that $a^e\equiv\pm1\pmod{n}$, namely
$$
\mathcal{PO}_n(a) = \min\left\{e\in\mathbb{N}^*\ \middle|\ a^e\equiv\pm1\pmod{n}\right\}.
$$
The projective multiplicative order of $a$ modulo $n$ also corresponds with the order of the element $\overline{\pi_n(a)}$ in the multiplicative quotient group ${\left(\Z/n\Z\right)}^*/\{-1,1\}$.
\par For every positive integer $n$, we define and denote by $\beta_n$ the arithmetic function
$$
\beta_n :
\begin{array}[t]{ccl}
\mathbb{Z} & \longrightarrow & \mathbb{N}\\
a & \longmapsto & \left\{
\begin{array}{cl}
\mathcal{PO}_n\left(a^n\right) & \text{for\ all}\ \gcd(a,n)=1,\\
0 & \text{otherwise}.
\end{array}\right.
\end{array}
$$
Observe that we have the alternative $\alpha_n=\beta_n$ or $\alpha_n=2\beta_n$.
\par In this paper, we study in detail these two arithmetic functions. In particular, we prove that, for every positive integers $n_1$ and $n_2$ such that
$$
\left\{
\begin{array}{lcl}
\rad(n_1)|n_2|n_1 & \text{if} & v_2(n_1)\leq1,\\
2\rad(n_1)|n_2|n_1& \text{if} & v_2(n_1)\geq2,
\end{array}
\right.
$$
the integer $\alpha_{n_1}(a)$ (respectively $\beta_{n_1}(a)$) divides $\alpha_{n_2}(a)$ (resp. $\beta_{n_2}(a)$), for every integer $a$. More precisely, we determine the exact relationship between the functions $\alpha_{n_1}$ and $\alpha_{n_2}$ and between $\beta_{n_1}$ and $\beta_{n_2}$. We prove that we have
$$
\alpha_{n_1}(a) = \frac{\alpha_{n_2}(a)}{\gcd\left(\alpha_{n_2}(a),\frac{\gcd(n_1,\mathcal{R}_{n_2}(a))}{n_2}\right)}\ \text{for\ all}\ \gcd(a,n_1)=1
$$
in Theorem~\ref{thm21} of Section 2 and that we have
$$
\beta_{n_1}(a) = \frac{\beta_{n_2}(a)}{\gcd\left(\beta_{n_2}(a),\frac{\gcd(n_1,\mathcal{R}_{n_2}(a))}{n_2}\right)}\ \text{for\ all}\ \gcd(a,n_1)=1
$$
in Theorem~\ref{thm31} of Section 3. Thus, for every integer $a$ coprime to $n$, the determination of $\alpha_n(a)$ is reduced to the computation of $\alpha_{\rad(n)}(a)$ and $\mathcal{R}_{\rad(n)}(a)$ if $v_2(n)\leq1$ and of $\alpha_{2\rad(n)}(a)$ and $\mathcal{R}_{2\rad(n)}(a)$ if $v_2(n)\geq2$. These theorems on the functions $\alpha_n$ and $\beta_n$ are derived from Theorem~\ref{thm2} of Section~2, which states that
$$
\mathcal{O}_{n_1}(a) = \mathcal{O}_{n_2}(a)\frac{n_1}{\gcd(n_1,\mathcal{R}_{n_2}(a))},
$$
for all integers $a$ coprime to $n_1$ and $n_2$. This result generalizes the following theorem of Nathanson which, in the above notation, states that for every odd prime number $p$ and for every positive integer $k$, we have the equality
$$
\mathcal{O}_{p^k}(a)=\mathcal{O}_p(a)\frac{p^k}{\gcd(p^k,\mathcal{R}_p(a))}
$$
for all integers $a$ not divisible by $p$.
\begin{nathanson}
Let $p$ be an odd prime, and let $a\neq\pm1$ be an integer not divisible by $p$. Let $d$ be the order of $a$ modulo $p$. Let $k_0$ be the largest integer such that $a^d\equiv1\pmod{p^{k_0}}$. Then the order of $a$ modulo $p^k$ is $d$ for $k=1,\ldots,k_0$ and $dp^{k-k_0}$ for $k\geq k_0$.
\end{nathanson}

For every finite sequence $S=\left(a_1,\ldots,a_m\right)$ of length $m\geq1$ in $\Z/n\Z$, we denote by $\Delta S$ the \textit{Steinhaus triangle} of $S$, that is the finite multiset of cardinality $\binom{m+1}{2}$ in $\Z/n\Z$ defined by
$$
\Delta S = \left\{ \sum_{k=0}^{i}{\binom{i}{k}a_{j+k}}\ \middle|\ 0\leq i\leq m-1\ ,\ 1\leq j\leq m-i \right\}.
$$
A finite sequence $S$ in $\Z/n\Z$ is said to be \textit{balanced} if each element of $\Z/n\Z$ occurs in its Steinhaus triangle $\Delta S$ with the same multiplicity. For instance, the sequence $\left(2,2,3,3\right)$ of length $4$ is balanced in $\mathbb{Z}/5\mathbb{Z}$. Indeed, as depicted in Figure~\ref{fig2}, its Steinhaus triangle is composed by each element of $\mathbb{Z}/5\mathbb{Z}$ occuring twice.

\begin{figure}[!h]
\centering
\begin{pspicture}(3,1.9485)
\pspolygon(0.375,1.9485)(2.625,1.9485)(1.5,0)
\rput(1.5,0.433){$0$}
\rput(1.25,0.866){$4$}
\rput(1.75,0.866){$1$}
\rput(1,1.299){$4$}
\rput(1.5,1.299){$0$}
\rput(2,1.299){$1$}
\rput(0.75,1.732){$2$}
\rput(1.25,1.732){$2$}
\rput(1.75,1.732){$3$}
\rput(2.25,1.732){$3$}
\end{pspicture}
\caption{\label{fig2}The Steinhaus triangle of a balanced sequence in $\mathbb{Z}/5\mathbb{Z}$}
\end{figure}

Note that, for a sequence $S$ of length $m\geq1$ in $\Z/n\Z$, a necessary condition on $m$ for $S$ to be balanced is that the integer $n$ divides the binomial coefficient $\binom{m+1}{2}$. In 1976, John C. Molluzzo \cite{Molluzzo1978} posed the problem to determine whether this necessary condition on $m$ is also sufficient to guarantee the existence of a balanced sequence. In \cite{Chappelon}, it was proved that, for each odd number $n$, \textit{there exists a balanced sequence of length $m$ for every $m\equiv0$ or $-1\pmod{\alpha_n(2)n}$ and for every $m\equiv0$ or $-1\pmod{\beta_n(2)n}$}. This was achieved by analyzing the Steinhaus triangles generated by arithmetic progressions. In particular, since $\beta_{3^k}(2)=1$ for all $k\geq1$, the above result implies a complete and positive solution of Molluzzo's Problem in $\Z/n\Z$ for all $n=3^k$.
\section{The arithmetic function $\alpha_n$

The table and the graphic depicted in Figure~\ref{fig3} give us the first values of $\alpha_n(a)$ for every positive integer $n$, $1\leq n\leq 20$, and for every integer $a$, $-20\leq a\leq 20$.

\begin{figure}[!p]
\centering
\begin{tabular}{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
$n\backslash a$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$ & $12$ & $13$ & $14$ & $15$ & $16$ & $17$ & $18$ & $19$ & $20$\\
\hline
\hline
$1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$\\
\hline
$2$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$\\
\hline
$3$ & $1$ & $2$ & $0$ & $1$ & $2$ & $0$ & $1$ & $2$ & $0$ & $1$ & $2$ & $0$ & $1$ & $2$ & $0$ & $1$ & $2$ & $0$ & $1$ & $2$\\
\hline
$4$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$\\
\hline
$5$ & $1$ & $4$ & $4$ & $2$ & $0$ & $1$ & $4$ & $4$ & $2$ & $0$ & $1$ & $4$ & $4$ & $2$ & $0$ & $1$ & $4$ & $4$ & $2$ & $0$\\
\hline
$6$ & $1$ & $0$ & $0$ & $0$ & $1$ & $0$ & $1$ & $0$ & $0$ & $0$ & $1$ & $0$ & $1$ & $0$ & $0$ & $0$ & $1$ & $0$ & $1$ & $0$\\
\hline
$7$ & $1$ & $3$ & $6$ & $3$ & $6$ & $2$ & $0$ & $1$ & $3$ & $6$ & $3$ & $6$ & $2$ & $0$ & $1$ & $3$ & $6$ & $3$ & $6$ & $2$\\
\hline
$8$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$\\
\hline
$9$ & $1$ & $2$ & $0$ & $1$ & $2$ & $0$ & $1$ & $2$ & $0$ & $1$ & $2$ & $0$ & $1$ & $2$ & $0$ & $1$ & $2$ & $0$ & $1$ & $2$\\
\hline
$10$ & $1$ & $0$ & $2$ & $0$ & $0$ & $0$ & $2$ & $0$ & $1$ & $0$ & $1$ & $0$ & $2$ & $0$ & $0$ & $0$ & $2$ & $0$ & $1$ & $0$\\
\hline
$11$ & $1$ & $10$ & $5$ & $5$ & $5$ & $10$ & $10$ & $10$ & $5$ & $2$ & $0$ & $1$ & $10$ & $5$ & $5$ & $5$ & $10$ & $10$ & $10$ & $5$\\
\hline
$12$ & $1$ & $0$ & $0$ & $0$ & $1$ & $0$ & $1$ & $0$ & $0$ & $0$ & $1$ & $0$ & $1$ & $0$ & $0$ & $0$ & $1$ & $0$ & $1$ & $0$\\
\hline
$13$ & $1$ & $12$ & $3$ & $6$ & $4$ & $12$ & $12$ & $4$ & $3$ & $6$ & $12$ & $2$ & $0$ & $1$ & $12$ & $3$ & $6$ & $4$ & $12$ & $12$\\
\hline
$14$ & $1$ & $0$ & $3$ & $0$ & $3$ & $0$ & $0$ & $0$ & $3$ & $0$ & $3$ & $0$ & $1$ & $0$ & $1$ & $0$ & $3$ & $0$ & $3$ & $0$\\
\hline
$15$ & $1$ & $4$ & $0$ & $2$ & $0$ & $0$ & $4$ & $4$ & $0$ & $0$ & $2$ & $0$ & $4$ & $2$ & $0$ & $1$ & $4$ & $0$ & $2$ & $0$\\
\hline
$16$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$\\
\hline
$17$ & $1$ & $8$ & $16$ & $4$ & $16$ & $16$ & $16$ & $8$ & $8$ & $16$ & $16$ & $16$ & $4$ & $16$ & $8$ & $2$ & $0$ & $1$ & $8$ & $16$\\
\hline
$18$ & $1$ & $0$ & $0$ & $0$ & $1$ & $0$ & $1$ & $0$ & $0$ & $0$ & $1$ & $0$ & $1$ & $0$ & $0$ & $0$ & $1$ & $0$ & $1$ & $0$\\
\hline
$19$ & $1$ & $18$ & $18$ & $9$ & $9$ & $9$ & $3$ & $6$ & $9$ & $18$ & $3$ & $6$ & $18$ & $18$ & $18$ & $9$ & $9$ & $2$ & $0$ & $1$\\
\hline
$20$ & $1$ & $0$ & $1$ & $0$ & $0$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $1$ & $0$ & $0$ & $0$ & $1$ & $0$ & $1$ & $0$\\
\hline
\end{tabular}\\[4ex]
\includegraphics[height=8cm]{alpha}
\caption{The first values of $\alpha_n(a)$}
\label{fig3}
\end{figure}

The positive integer $\alpha_n(a)$ seems to be difficult to determine. Indeed, there is no general formula known to compute the multiplicative order of an integer modulo $n$ but, however, we get the following helpful propositions.

\begin{lem}\label{lem11}
Let $n_1$ and $n_2$ be two positive integers such that $\rad(n_1)=\rad(n_2)$. Then, an integer $a$ is coprime to $n_1$ if, and only if, it is also coprime to $n_2$.
\end{lem}

\begin{proof}
This follows from the definition of the greatest common divisor of two integers and from the definition of the radical of an integer.
\end{proof}

\begin{prop}\label{prop2}
Let $n_1$ and $n_2$ be two positive integers such that $\rad(n_1)|n_2|n_1$. Then, for every integer $a$, the integer $\alpha_{n_1}(a)$ divides $\alpha_{n_2}(a)$.
\end{prop}

\begin{proof}
If $a$ is not coprime to $n_1$ and $n_2$, then, by definition of the functions $\alpha_{n_1}$ and $\alpha_{n_2}$ and by Lemma~\ref{lem11}, we have
$$
\alpha_{n_1}(a)=\alpha_{n_2}(a)=0.
$$
Suppose now that $a$ is coprime to $n_1$ and $n_2$. Let $p$ be a prime factor of $n_1$ such that $v_p(n_1)\geq2$. We shall show that $\alpha_{n_1}(a)$ divides $\alpha_{n_1/p}(a)$. By definition of $\alpha_{n_1/p}(a)$, there exists an integer $u$ such that
$$
a^{\alpha_{n_1/p}(a)\frac{n_1}{p}} = 1+u\frac{n_1}{p}.
$$
Therefore, by the binomial theorem, we have
$$
a^{\alpha_{n_1/p}(a)n_1} = {\left(a^{\alpha_{n_1/p}(a)\frac{n_1}{p}}\right)}^{p} = {\left(1+u\frac{n_1}{p}\right)}^{p} = 1+ un_1 + \sum_{k=2}^{p}{\binom{p}{k}u^k\left(\frac{n_1}{p}\right)^k}.
$$
Since $v_p(n_1)\geq2$, it follows that ${\left(n_1/p\right)}^k$ is divisible by $n_1$ for every integer $k\geq2$ and so
$$
a^{\alpha_{n_1/p}(a)n_1} \equiv 1 \pmod{n_1}.
$$
Hence $\alpha_{n_1}(a)$ divides $\alpha_{n_1/p}(a)$. This completes the proof.
\end{proof}

An exact relationship between $\alpha_{n_1}(a)$ and $\alpha_{n_2}(a)$, for every integer $a$ coprime to $n_1$ and $n_2$, is determined at the end of this section. We first settle the easy prime power case.

\begin{prop}\label{prop3}
Let $p$ be a prime number and $a$ be an integer. Then we have
$$
\alpha_{p^k}(a) = \mathcal{O}_p(a)
$$
for every positive integer $k$.
\end{prop}

\begin{proof}
Let $k$ be a positive integer. If $a$ is not coprime to $p$, then  we have $\alpha_{p^k}(a)=\alpha_{p}(a)=0$. Suppose now that $a$ is coprime to $p$. By Proposition~\ref{prop2}, the integer $\alpha_{p^k}(a)$ divides $\alpha_{p}(a)$. It remains to prove that $\alpha_{p}(a)$ divides $\alpha_{p^k}(a)$. The congruence
$$
a^{\alpha_{p^k}(a)p^k} \equiv 1 \pmod{p^k}
$$
implies that
$$
a^{\alpha_{p^k}(a)p^k} \equiv 1 \pmod{p},
$$
and hence, by Fermat's little theorem, it follows that
$$
a^{\alpha_{p^k}(a)p} \equiv a^{\alpha_{p^k}(a)p^k} \equiv 1 \pmod{p}.
$$
Therefore $\alpha_{p}(a)$ divides $\alpha_{p^k}(a)$. Finally, we have
$$
\alpha_{p^k}(a) = \alpha_{p}(a) = \mathcal{O}_{p}(a^p) = \mathcal{O}_{p}(a).
$$
This completes the proof.
\end{proof}

\begin{rem}
If $p=2$, then, for every positive integer $k$, we obtain
$$
\alpha_{2^k}(a) = \mathcal{O}_{2}(a) = \left\{
\begin{array}{l}
0\ \text{for}\ a\ \text{even},\\
1\ \text{for}\ a\ \text{odd}.
\end{array}
\right.
$$
\end{rem}

\begin{prop}
Let $n_1$ and $n_2$ be two coprime numbers and $a$ be an integer. Then $\alpha_{n_1n_2}(a)$ divides $\lcm(\alpha_{n_1}(a),\alpha_{n_2}(a))$, the least common multiple of $\alpha_{n_1}(a)$ and $\alpha_{n_2}(a)$.
\end{prop}

\begin{proof}
If $\gcd(a,n_1n_2)\neq1$, then $\gcd(a,n_1)\neq1$ or $\gcd(a,n_2)\neq1$ and so
$$
\alpha_{n_1n_2}(a) = \lcm(\alpha_{n_1}(a),\alpha_{n_2}(a)) = 0.
$$
Suppose now that $\gcd(a,n_1n_2)=1$ and hence that the integers $a$, $n_1$ and $n_2$ are coprime pairwise. Let $i\in\{1,2\}$. The congruences
$$
a^{\alpha_{n_i}(a)n_i}\equiv1\pmod{n_i}
$$
imply that
$$
a^{\lcm(\alpha_{n_1}(a),\alpha_{n_2}(a))n_1n_2}\equiv1\pmod{n_i}.
$$
Therefore $\alpha_{n_1n_2}(a)$ divides $\lcm(\alpha_{n_1}(a),\alpha_{n_2}(a))$ by the Chinese remainder theorem.
\end{proof}

Let $n_1$ and $n_2$ be two positive integers such that
$$
\left\{
\begin{array}{lcl}
\rad(n_1)|n_2|n_1 & \text{if} & v_2(n_1)\leq1,\\
2\rad(n_1)|n_2|n_1& \text{if} & v_2(n_1)\geq2.
\end{array}
\right.
$$
By definition, we know that $\alpha_{n_1}(a)=\alpha_{n_2}(a)=0$ for every integer $a$ not coprime to $n_1$ and $n_2$. We end this section by determining the exact relationship between $\alpha_{n_1}(a)$ and $\alpha_{n_2}(a)$ for every integer $a$ coprime to $n_1$ and $n_2$.

\begin{thm}\label{thm21}
Let $n_1$ and $n_2$ be two positive integers such that
$$
\left\{
\begin{array}{lcl}
\rad(n_1)|n_2|n_1 & \text{if} & v_2(n_1)\leq1,\\
2\rad(n_1)|n_2|n_1& \text{if} & v_2(n_1)\geq2.
\end{array}
\right.
$$
Then, for every integer $a$ coprime to $n_1$ and $n_2$, we have
$$
\alpha_{n_1}(a) = \frac{\alpha_{n_2}(a)}{\gcd\left(\alpha_{n_2}(a),\frac{\gcd(n_1,\mathcal{R}_{n_2}(a))}{n_2}\right)}
$$
\end{thm}

This result is a corollary of the following theorem.

\begin{thm}\label{thm2}
Let $n_1$ and $n_2$ be two positive integers such that
$$
\left\{
\begin{array}{lcl}
\rad(n_1)|n_2|n_1 & \text{if} & v_2(n_1)\leq1,\\
2\rad(n_1)|n_2|n_1& \text{if} & v_2(n_1)\geq2.
\end{array}
\right.
$$
Then, for every integer $a$ coprime to $n_1$ and $n_2$, we have
$$
\mathcal{O}_{n_1}(a) = \mathcal{O}_{n_2}(a)\frac{n_1}{\gcd(n_1,\mathcal{R}_{n_2}(a))}.
$$
\end{thm}

The proof of this theorem is based on the following lemma.

\begin{lem}\label{lem1}
Let $n$ be a positive integer and let $a$ be an integer coprime to $n$. Let $m$ be an integer such that $\rad(m)|\rad{n}$. Then, there exists an integer $u_m$, coprime to $n$ if $m$ is odd, or coprime to $n/2$ if $m$ is even, such that
$$
a^{\mathcal{O}_n(a)m} = 1 + u_m\mathcal{R}_n(a)m.
$$
\end{lem}

\begin{proof}
We distinguish different cases upon the parity of $m$. First, we prove the odd case by induction on $m$. If $m=1$, then, by definition of the integer $\mathcal{R}_n(a)$, we have
$$
a^{\mathcal{O}_n(a)} = 1 + \mathcal{R}_n(a).
$$
Therefore the assertion is true for $m=1$. Now, let $p$ be a prime factor of $m$ and suppose that the assertion is true for the odd number $m/p$, i.e. there exists an integer $u_{m/p}$, coprime to $n$, such that
$$
a^{\mathcal{O}_{n}(a)\frac{m}{p}} = 1 + u_{m/p}\mathcal{R}_n(a)\frac{m}{p}.
$$
Then, we obtain
$$
a^{\mathcal{O}_n(a)m}
\begin{array}[t]{l}
= \displaystyle\left(a^{\mathcal{O}_n(a)\frac{m}{p}}\right)^p = \left( 1 + u_{m/p}\mathcal{R}_n(a)\frac{m}{p} \right)^p\\[2ex]
= \displaystyle1 + u_{m/p}\mathcal{R}_n(a)m + \sum_{k=2}^{p-1}{\binom{p}{k}\left(u_{m/p}\mathcal{R}_n(a)\frac{m}{p}\right)^k} + \left(u_{m/p}\mathcal{R}_n(a)\frac{m}{p}\right)^p\\[2ex]
= \displaystyle 1 + \left( u_{m/p} + \sum_{k=2}^{p-1}{\frac{\binom{p}{k}}{p}{u_{m/p}}^{k}{\mathcal{R}_n(a)}^{k-1}{\left(\frac{m}{p}\right)}^{k-1}} + {u_{m/p}}^p\frac{{\mathcal{R}_n(a)}^{p-1}}{p}{\left(\frac{m}{p}\right)}^{p-1} \right)\mathcal{R}_n(a)m\\[2ex]
= \displaystyle 1 + u_m\mathcal{R}_n(a)m.
\end{array}
$$
Since $n$ divides $\mathcal{R}_n(a)$ which divides
$$
u_m-u_{m/p} = \sum_{k=2}^{p-1}{\frac{\binom{p}{k}}{p}{u_{m/p}}^{k}{\mathcal{R}_n(a)}^{k-1}{\left(\frac{m}{p}\right)}^{k-1}} + {u_{m/p}}^p\frac{{\mathcal{R}_n(a)}^{p-1}}{p}{\left(\frac{m}{p}\right)}^{p-1},
$$
it follows that $\gcd(u_m,n)=\gcd(u_{m/p},n)=1$. This completes the proof for the odd case. Suppose now that $n$ and $m$ are even. We proceed by induction on $v_2(m)$. Suppose that there exists an integer $u_{m/2}$, coprime to $n/2$, such that
$$
a^{\mathcal{O}_n(a)\frac{m}{2}} = 1 + u_{m/2}\mathcal{R}_n(a)\frac{m}{2}.
$$
We know that this statement is true for $v_2(m)=1$. Indeed, if $v_2(m)=1$, then $m/2$ is odd, this implying that there exists such an integer $u_{m/2}$, coprime to $n$ thus to $n/2$. Then, we obtain
$$
a^{\mathcal{O}_n(a)m}
\begin{array}[t]{l}
= \displaystyle \left(a^{\mathcal{O}_n(a)\frac{m}{2}}\right)^2 = \left( 1 + u_{m/2}\mathcal{R}_n(a)\frac{m}{2} \right)^2 = \displaystyle 1 + u_{m/2}\mathcal{R}_n(a)m + \left(u_{m/2}\mathcal{R}_n(a)\frac{m}{2}\right)^2\\[2ex]
= \displaystyle 1 + \left( u_{m/2} + {u_{m/2}}^2\frac{\mathcal{R}_n(a)}{2}\frac{m}{2} \right)\mathcal{R}_n(a)m = 1 + u_m\mathcal{R}_n(a)m.
\end{array}
$$
Since $n/2$ divides $\mathcal{R}_n(a)/2$ which divides $u_m-u_{m/2}$, it follows that $\gcd(u_m,n/2)=\gcd(u_{m/2},n/2)=1$. This completes the proof.
\end{proof}

We are now ready to prove Theorem~\ref{thm2}.

\begin{proof}[Proof of Theorem~\ref{thm2}]
The proof is by induction on the integer $n_1/n_2$. If $n_1=n_2$, then we have
$$
\frac{n_1}{\gcd(n_1,\mathcal{R}_{n_2}(a))} = \frac{n_1}{\gcd(n_1,\mathcal{R}_{n_1}(a))} = \frac{n_1}{n_1} = 1,
$$
since $\mathcal{R}_{n_1}(a)$ is divisible by $n_1$, and thus the statement is true. Let $p$ be a prime factor of $n_1$ and $n_2$ such that $n_2$ divides $n_1/p$ and suppose that
$$
\mathcal{O}_{n_1/p}(a) = \mathcal{O}_{n_2}(a)\frac{n_1/p}{\gcd(n_1/p,\mathcal{R}_{n_2}(a))}.
$$
First, the congruence
$$
a^{\mathcal{O}_{n_1}(a)} \equiv 1 \pmod{n_1}
$$
implies that
$$
a^{\mathcal{O}_{n_1}(a)} \equiv 1 \pmod{\frac{n_1}{p}}
$$
and so $\mathcal{O}_{n_1/p}(a)$ divides $\mathcal{O}_{n_1}(a)$. Now, we distinguish different cases upon $v_p(n_1)$.\\[2ex]
\textbf{1st case:} $v_p(n_1)\leq v_p\left(\mathcal{R}_{n_2}(a)\right)$.\\
Since $n_2$ divides $n_1/p$, it follows that $\mathcal{O}_{n_2}(a)$ divides $\mathcal{O}_{n_1/p}(a)$. Therefore $\mathcal{R}_{n_1/p}(a)$ is divisible by $\mathcal{R}_{n_2}(a)$ because we have
$$
\mathcal{R}_{n_1/p}(a)
\begin{array}[t]{l}
= \displaystyle a^{\mathcal{O}_{n_1/p}(a)}-1 = a^{\mathcal{O}_{n_2}(a)\frac{\mathcal{O}_{n_1/p}(a)}{\mathcal{O}_{n_2}(a)}}-1 = \left( a^{\mathcal{O}_{n_2}(a)}-1 \right) \sum_{k=0}^{\mathcal{O}_{n_2}(a)-1}{a^{\frac{k\mathcal{O}_{n_1/p}(a)}{\mathcal{O}_{n_2}(a)}}}\\
= \displaystyle\mathcal{R}_{n_2}(a)\sum_{k=0}^{\mathcal{O}_{n_2}(a)-1}{a^{\frac{k\mathcal{O}_{n_1/p}(a)}{\mathcal{O}_{n_2}(a)}}}.
\end{array}
$$
This leads to
$$
v_{p}(n_1)\leq v_p\left(\mathcal{R}_{n_2}(a)\right)\leq v_p\left(\mathcal{R}_{n_1/p}(a)\right).
$$
Therefore $\mathcal{R}_{n_1/p}(a)$ is divisible by $n_1$ and hence we have
$$
a^{\mathcal{O}_{n_1/p}(a)} = 1 + \mathcal{R}_{n_1/p}(a) \equiv 1 \pmod{n_1}.
$$
This implies that $\mathcal{O}_{n_1}(a)=\mathcal{O}_{n_1/p}(a)$. Moreover, the hypothesis $v_p(n_1)\leq v_p\left(\mathcal{R}_{n_2}(a)\right)$ implies that $\gcd(n_1/p,\mathcal{R}_{n_2}(a))=\gcd(n_1,\mathcal{R}_{n_2}(a))/p$. Finally, we obtain
$$
\mathcal{O}_{n_1}(a) = \mathcal{O}_{n_1/p}(a) = \mathcal{O}_{n_2}(a)\frac{n_1/p}{\gcd(n_1/p,\mathcal{R}_{n_2}(a))} = \mathcal{O}_{n_2}(a)\frac{n_1}{\gcd(n_1,\mathcal{R}_{n_2}(a))}.
$$\ \\
\textbf{2nd case:} $v_p(n_1)>v_p\left(\mathcal{R}_{n_2}(a)\right)$.\\
If $v_2(n_1)\leq1$, then $(n_1/p)/\gcd(n_1/p,\mathcal{R}_{n_2}(a))$ is odd. Otherwise, if $v_2(n_1)\geq2$, then $v_{2}(n_2)\geq2$ and every integer coprime to $n_2/2$ is also coprime to $n_2$. In both cases, $v_2(n_1)\leq 1$ or $v_2(n_1)\geq2$, we know, by Lemma~\ref{lem1}, that there exists an integer $u$, coprime to $n_2$, such that
$$
a^{\mathcal{O}_{n_1/p}(a)} = a^{\mathcal{O}_{n_2}(a)\frac{n_1/p}{\gcd(n_1/p,\mathcal{R}_{n_2}(a))}} = 1 + u\mathcal{R}_{n_2}(a)\frac{n_1/p}{\gcd(n_1/p,\mathcal{R}_{n_2}(a))} = 1 + u\frac{\mathcal{R}_{n_2}(a)}{\gcd(n_1/p,\mathcal{R}_{n_2}(a))}\frac{n_1}{p}.
$$
As $v_p\left(\mathcal{R}_{n_2}(a)\right)\leq v_p\left(n_1/p\right)$, it follows that $\mathcal{R}_{n_2}(a)/\gcd(n_1/p,\mathcal{R}_{n_2}(a))$ is coprime to $p$, and hence $\mathcal{O}_{n_1/p}(a)$ is a proper divisor of $\mathcal{O}_{n_1}(a)$ since
$$
a^{\mathcal{O}_{n_1/p}(a)} \not\equiv 1 \pmod{n_1}.
$$
Moreover, by Lemma~\ref{lem1} again, there exists an integer $u_p$ such that
$$
a^{\mathcal{O}_{n_1/p}(a)p} = 1 + u_p\mathcal{R}_{n_1/p}(a)p \equiv 1 \pmod{n_1}.
$$
This leads to
$$
\mathcal{O}_{n_1}(a) = \mathcal{O}_{n_1/p}(a)p = \mathcal{O}_{n_2}(a)\frac{n_1}{\gcd(n_1/p,\mathcal{R}_{n_2}(a))} = \mathcal{O}_{n_2}(a)\frac{n_1}{\gcd(n_1,\mathcal{R}_{n_2}(a))}.
$$
This completes the proof of Theorem~\ref{thm2}.
\end{proof}

We may view Theorem~\ref{thm2} as a generalization of Theorem~$3.6$ of \cite{Nathanson2000}, where $n_2=p$ is an odd prime number and $n_1=p^k$ for some positive integer $k$. Note that the conclusion of Theorem~\ref{thm2} fails in general in the case where $v_2(n_1)\geq2$ and $n_2=\rad(n_1)$. For instance, for $n_1=24=3\cdot2^3$, $n_2=6=3\cdot2$ and $a=7$, we obtain that $\mathcal{O}_{n_1}(a)=2$ while $\mathcal{O}_{n_2}(a)n_1/\gcd(n_1,\mathcal{R}_{n_2}(a))=24/\gcd(24,6)=4$.

We now turn to the proof of the main result of this paper.

\begin{proof}[Proof of Theorem~\ref{thm21}]
From Theorem~\ref{thm2}, we obtain
$$
\alpha_{n_1}(a) = \mathcal{O}_{n_1}(a^{n_1}) = \frac{\mathcal{O}_{n_1}(a)}{\gcd(\mathcal{O}_{n_1}(a),n_1)} = \frac{\mathcal{O}_{n_2}(a)\frac{n_1}{\gcd(n_1,\mathcal{R}_{n_2}(a))}}{\gcd\left(\mathcal{O}_{n_2}(a)\frac{n_1}{\gcd(n_1,\mathcal{R}_{n_2}(a))},n_1\right)} = \frac{\mathcal{O}_{n_2}(a)}{\gcd(\mathcal{O}_{n_2}(a),n_1,\mathcal{R}_{n_2}(a))}.
$$
Thus,
$$
\frac{\alpha_{n_2}(a)}{\alpha_{n_1}(a)}
\begin{array}[t]{l}
= \displaystyle\frac{\frac{\mathcal{O}_{n_2}(a)}{\gcd(\mathcal{O}_{n_2}(a),n_2)}}{\frac{\mathcal{O}_{n_2}(a)}{\gcd(\mathcal{O}_{n_2}(a),n_1,\mathcal{R}_{n_2}(a))}} = \frac{\gcd(\mathcal{O}_{n_2}(a),n_1,\mathcal{R}_{n_2}(a))}{\gcd(\mathcal{O}_{n_2}(a),n_2)}\\[4ex]
= \displaystyle\gcd\left(\frac{\mathcal{O}_{n_2}(a)}{\gcd(\mathcal{O}_{n_2}(a),n_2)},\frac{n_2}{\gcd(\mathcal{O}_{n_2}(a),n_2)}\frac{\gcd(n_1,\mathcal{R}_{n_2}(a))}{n_2}\right).
\end{array}
$$
Finally, since we have
$$
\gcd\left(\frac{\mathcal{O}_{n_2}(a)}{\gcd(\mathcal{O}_{n_2}(a),n_2)},\frac{n_2}{\gcd(\mathcal{O}_{n_2}(a),n_2)}\right) = \frac{\gcd(\mathcal{O}_{n_2}(a),n_2)}{\gcd(\mathcal{O}_{n_2}(a),n_2)} = 1,
$$
it follows that
$$
\frac{\alpha_{n_2}(a)}{\alpha_{n_1}(a)} = \gcd\left(\frac{\mathcal{O}_{n_2}(a)}{\gcd(\mathcal{O}_{n_2}(a),n_2)},\frac{\gcd(n_1,\mathcal{R}_{n_2}(a))}{n_2}\right) = \gcd\left(\alpha_{n_2}(a),\frac{\gcd(n_1,\mathcal{R}_{n_2}(a))}{n_2}\right).
$$
\end{proof}

Thus, the determination of $\alpha_n$ is reduced to the case where $n$ is square-free.

\begin{cor}
Let $n$ be a positive integer such that $v_2(n)\leq 1$. Then, for every integer $a$, coprime to $n$, we have
$$
\alpha_n(a) = \frac{\alpha_{\rad(n)}(a)}{\gcd\left(\alpha_{\rad(n)}(a),\frac{\gcd(n,\mathcal{R}_{\rad(n)}(a))}{\rad(n)}\right)}.
$$
\end{cor}

\begin{cor}
Let $n$ be a positive integer such that $v_2(n)\geq 2$. Then, for every integer $a$, coprime to $n$, we have
$$
\alpha_n(a) = \frac{\alpha_{2\rad(n)}(a)}{\gcd\left(\alpha_{2\rad(n)}(a),\frac{\gcd(n,\mathcal{R}_{2\rad(n)}(a))}{2\rad(n)}\right)}.
$$
\end{cor}

\section{The arithmetic function $\beta_n$

First, we can observe that, by definition of the functions $\alpha_n$ and $\beta_n$, we have
$$
\alpha_n(a) = \beta_n(a) = 0
$$
for every integer $a$ not coprime to $n$ and
$$
\frac{\alpha_n(a)}{\beta_n(a)} \in \left\{1,2\right\}
$$
for every integer $a$ coprime to $n$. There is no general formula known to compute $\alpha_n(a)/\beta_n(a)$ but, however, we get the following proposition.

\begin{prop}\label{prop4}
Let $n_1$ and $n_2$ be two positive integers such that $\rad(n_1)=\rad(n_2)$. Let $a$ be an integer coprime to $n_1$ and $n_2$. If $v_{2}(n_1)\leq 1$, then we have
$$
\frac{\alpha_{n_1}(a)}{\beta_{n_1}(a)} = \frac{\alpha_{n_2}(a)}{\beta_{n_2}(a)}.
$$
If $v_2(n_1)\geq2$, then we have
$$
\alpha_{n_1}(a) = \beta_{n_1}(a).
$$
\end{prop}

\begin{proof}
Let $n_1$ be a positive integer such that $v_2(n_1)\leq 1$ and $a$ be an integer coprime to $n_1$. Let $p$ be an odd prime factor of $n_1$ such that $v_{p}(n_1)\geq2$. We will prove that
$$
\frac{\alpha_{n_1}(a)}{\beta_{n_1}(a)} = \frac{\alpha_{n_1/p}(a)}{\beta_{n_1/p}(a)}.
$$
If $\alpha_{n_1}(a) = 2\beta_{n_1}(a)$, then
$$
{a}^{\beta_{n_1}(a)n_1} \equiv -1 \pmod{n_1}
$$
and thus
$$
{a}^{\beta_{n_1}(a)p\frac{n_1}{p}} \equiv -1 \pmod{\frac{n_1}{p}}.
$$
This implies that $\alpha_{n_1/p}(a) = 2\beta_{n_1/p}(a)$. Conversely, if $\alpha_{n_1/p}(a) = 2\beta_{n_1/p}(a)$, then we have
$$
a^{\beta_{n_1/p}(a)\frac{n_1}{p}} \equiv -1 \pmod{\frac{n_1}{p}}.
$$
Since $v_p(n_1)\geq2$, it follows that
$$
a^{\beta_{n_1/p}(a)\frac{n_1}{p}} \equiv -1 \pmod{p}
$$
and thus
$$
{a}^{\beta_{n_1/p}(a)n_1} + 1 = 1 - {\left( - {a}^{\beta_{n_1/p}(a)\frac{n_1}{p}} \right)}^p = \left( 1 + {a}^{\beta_{n_1/p}(a)\frac{n_1}{p}} \right) \sum_{k=0}^{p-1}{{\left( - {a}^{\beta_{n_1/p}(a)\frac{n_1}{p}} \right)}^k} \equiv 0 \pmod{n_1}.
$$
This implies that $\alpha_{n_1}(a) = 2\beta_{n_1}(a)$. Therefore, by induction on $n_1/n_2$, we obtain that
$$
\frac{\alpha_{n_1}(a)}{\beta_{n_1}(a)} = \frac{\alpha_{n_2}(a)}{\beta_{n_2}(a)},
$$
for every integer $n_2$ such that $\rad(n_1)=\rad(n_2)$. Now, let $n_1$ be a positive integer such that $v_2(n_1)\geq2$ and $a$ be a non-zero integer. Suppose that we have $\alpha_{n_1}(a)=2\beta_{n_1}(a)$. Since
$$
{a}^{\beta_{n_1}(a)n_1} \equiv -1 \pmod{n_1}
$$
it follows that
$$
{\left({a}^{\beta_{n_1}(a)\frac{n_1}{4}}\right)}^{4} \equiv -1 \pmod{4}
$$
in contradiction with
$$
{\left({a}^{\beta_{n_1}(a)\frac{n_1}{4}}\right)}^{4} \equiv 1 \pmod{4}.
$$
\end{proof}

If $n$ is a prime power, then $\beta_n=\beta_{\rad(n)}$, in analogy with Proposition~\ref{prop3} for $\alpha_n$.

\begin{prop}
Let $p$ be a prime number and $a$ be an integer. Then we have
$$
\beta_{p^k}(a) = \beta_p(a)
$$
for every positive integer $k$.
\end{prop}

\begin{proof}
This result is trivial for every integer $a$ not coprime to $n$. Suppose now that $a$ is coprime to $n$. For $p=2$, then, by Proposition~\ref{prop4}, we have
$$
\beta_{2^k}(a) = \alpha_{2^k}(a) = 1
$$
for every positive integer $k$. For an odd prime number $p\geq3$, Proposition~\ref{prop4} and Proposition~\ref{prop3} lead to
$$
\beta_{p^{k}}(a) = \frac{\alpha_{p^k}(a)}{\alpha_p(a)}\beta_p(a) = \beta_p(a)
$$
for every positive integer $k$. This completes the proof.
\end{proof}

Let $n_1$ and $n_2$ be two positive integers such that
$$
\left\{
\begin{array}{lcl}
\rad(n_1)|n_2|n_1 & \text{if} & v_2(n_1)\leq1,\\
2\rad(n_1)|n_2|n_1& \text{if} & v_2(n_1)\geq2.
\end{array}
\right.
$$
It immediately follows that $\beta_{n_1}(a)=\beta_{n_2}(a)=0$ for every integer $a$ not coprime to $n_1$ and $n_2$. Finally, we determine the relationship between $\beta_{n_1}(a)$ and $\beta_{n_2}(a)$ for every integer $a$ coprime to $n_1$ and $n_2$.

\begin{thm}\label{thm31}
Let $n_1$ and $n_2$ be two positive integers such that
$$
\left\{
\begin{array}{lcl}
\rad(n_1)|n_2|n_1 & \text{if} & v_2(n_1)\leq1,\\
2\rad(n_1)|n_2|n_1& \text{if} & v_2(n_1)\geq2.
\end{array}
\right.
$$
Let $a$ be an integer coprime to $n_1$ and $n_2$. Then, we have
$$
\beta_{n_1}(a) = \frac{\beta_{n_2}(a)}{\gcd\left(\beta_{n_2}(a),\frac{\gcd(n_1,\mathcal{R}_{n_2}(a))}{n_2}\right)}.
$$
\end{thm}

\begin{proof}
If $v_2(n_1)\leq 1$, then Theorem~\ref{thm21} and Proposition~\ref{prop4} lead to
$$
\frac{\beta_{n_2}(a)}{\beta_{n_1}(a)} = \frac{\alpha_{n_2}(a)}{\alpha_{n_1}(a)} = \gcd\left(\alpha_{n_2}(a),\frac{\gcd(n_1,\mathcal{R}_{n_2}(a))}{n_2}\right).
$$
Since $v_2(n_2)=v_2(n_1)\leq1$, it follows that $\gcd(n_1,\mathcal{R}_{n_2}(a))/n_2$ is odd and hence, we have
$$
\frac{\beta_{n_2}(a)}{\beta_{n_1}(a)} = \gcd\left(\alpha_{n_2}(a),\frac{\gcd(n_1,\mathcal{R}_{n_2}(a))}{n_2}\right) = \gcd\left(\beta_{n_2}(a),\frac{\gcd(n_1,\mathcal{R}_{n_2}(a))}{n_2}\right).
$$
If $v_{2}(n_1)\geq2$, then $\beta_{n_1}(a)=\alpha_{n_1}(a)$ and $\beta_{n_2}(a)=\alpha_{n_2}(a)$ by Proposition~\ref{prop4} and the result follows from Theorem~\ref{thm21}.
\end{proof}

Thus, as for $\alpha_n$, the determination of $\beta_n$ is reduced to the case where $n$ is square-free.

\begin{cor}
Let $n$ be a positive integer such that $v_2(n)\leq 1$. Then, for every integer $a$, coprime to $n$, we have
$$
\beta_n(a) = \frac{\beta_{\rad(n)}(a)}{\gcd\left(\beta_{\rad(n)}(a),\frac{\gcd(n,\mathcal{R}_{\rad(n)}(a))}{\rad(n)}\right)}.
$$
\end{cor}

\begin{cor}
Let $n$ be a positive integer such that $v_2(n)\geq 2$. Then, for every integer $a$, coprime to $n$, we have
$$
\beta_n(a) = \frac{\beta_{2\rad(n)}(a)}{\gcd\left(\beta_{2\rad(n)}(a),\frac{\gcd(n,\mathcal{R}_{2\rad(n)}(a))}{2\rad(n)}\right)}.
$$
\end{cor}

\section*{Acknowledgments

The author would like to thank Shalom Eliahou for his help in preparing this paper.

\nocite{*}
\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


Renewable energy subsidies will gross up to £46 billion by 2020," declared Prof Ian Fells.
"It's a totally staggering sum and effectively means the poor are supporting the rich.
"The subsidies allow upper middle class Guardian-readers with solar panels to sell electricity, while it's all being paid for by the poor who cannot afford to fit their homes with solar equipment.
Prof Fells, emeritus professor of energy conversion at Newcastle University, was one of the hundred-or-so delegates at The Power Debate: A UK generation game at The Sage, Gateshead, last week.
Organised by Newcastle law firm Watson Burton it featured a five-strong panel of regional and national energy industry experts.
One of these – Andrew Mill, chief executive at the Blyth-based National Renewable Energy Centre (Narec) – took issue with Prof Fells, who is a prominent supporter of nuclear power.
Mill said: "Let's not forget nuclear subsidies are huge, but unlike renewable subsidies they are hidden, we cannot be sure how much nuclear power actually costs.
He then elaborated on how a subsidy regime should operate in the UK.
Narec is one of the UK's leading technology players in the developing offshore energy sector and Mill made a few salient points about the state of the UK energy market.
"Despite Ofgem highlighting the declining capacity margins in the UK electricity system, the Government has the ability to provide subsidies to gas-plant operators to bring their mothballed capacity back on line.
Despite this positive message Mill highlighted the risks in the current UK electricity market and signalled out the lack of certainty, brought about by Government prevarication evident in the ongoing discussions around the Energy Bill.
In the offshore wind industry, to which Narec is particularly entwined, Mill supported the Government's desire to cut the cost by over a quarter to £100 per MWh.
Mill highlighted how the shift to renewable energy meant the UK would be more reliant on intermittent forms of power generation – sun and wind – in the coming years.
This loss of base load power was a point of great concern for two of the panel.
Last year coal accounted for 40% of UK power generation but EU directives and the punitive Carbon Floor Price are likely to mean almost all of this capacity being shut down over the next five years.
Martin Needham, head of sales at commercial gas supplier CNG, said: "The Government has made the wrong decisions.
"No other country is facing the problems we are facing. In Poland where most of its electricity comes from coal, they just said 'no'.
"Wind is unreliable, offshore wind is the most expensive electricity source there is. We have to have a balanced mix which includes coal and gas.
These points were taken up by James Poyner, a director of construction business Miller Argent.
He said: "The UK coal industry is fighting for its survival.
"There is no clarity from the Government on UK energy policy. Coal can act as a bridge to a low-carbon future. And coal-powered plants fitted with carbon capture and storage are a viable and low carbon long-term answer.
Poyner highlighted how 1,000 new coal plants are being built in more than 50 countries, with ten new ones being built in Europe's renewable energy standard bearer – Germany.
"The Government seems hell-bent on closing our existing coal capacity before new alternatives are built, we are playing Russian roulette with the power system," he added.
Colin Henry, business development manager for Siemens Smart Grid Division in the UK had a different slant on the UK market.
Supporting initiatives to increase the role of renewables in the system, he highlighted how 'uncertainty can be good', and that while sustainability can also mean volatility this can drive innovation.
He thinks the answers to many of the problems facing the UK lie in developing smart grid and local energy systems, complete with 'smart buildings' and demand reduction initiatives.
Henry also felt that opposition leader Ed Miliband's proposal to freeze energy prices should be welcomed.
When it was thrown open to a vote from the floor only one member of the audience supported Henry's position.
Prof Ian Fells questioned whether the Labour Party would be able to fulfil this promise.
Watson Burton partner Richard Palmer, head of construction and insurance, chaired the debate. Post-event he told Journal Energy: "I was delighted that we had such a lively debate with wide ranging views on some issues and general consensus on others.
He summarised the morning's ruminations as follows: "The panel and audience agreed there should not be an over-reliance on one particular type of fuel, but rather a portfolio of energy sources.
"The general consensus was that we could not simply switch off the fossil fuels – coal, oil and gas – but rather they would need to be utilised in order to bridge the gap between what we currently have in the way of resources, and what we hope to have, and are developing, in the form of renewables.
"On the suggested prize freeze, there were mixed views – some thought that a shake-up of the industry was a good thing, bearing in mind the dominance of the Big Six, whereas the vast majority felt that in an industry where investment was needed, energy companies would only do this if guaranteed a return, and so the suggestion of a price freeze was 'naive' and 'a gimmick'.
"From a regional point of view, there needs to be a balancing of local and national needs from planning, economic and environmental perspectives.
"There is a need for the relevant regional bodies, including local authorities and the Local Enterprise Partnerships to engage in the energy debate.
The current furore over energy prices, with its simplistic 'we are being ripped off' overtone, hides a multiplicity of complex ingredients in a mix which has its roots buried in the political dogma of privatisation some 25 years ago.
Electricity prices in the 21st Century are now higher than ever before. One of the many reasons for this are the EU directives to close the low-cost coal-fired power stations – many which have not inconsiderable residual life.
Earlier this month, Ernst & Young reported that there was about 27,000 MW of consented power projects on the stocks in the UK which had planning permission. This is more than one third of the current capacity.
Not a spade has yet touched the ground on these projects. Political risk associated with Government intentions on Electricity Market Reform (EMR) and, more recently, threatened intervention by Government to freeze prices is to blame.
The costs of upgrading the National Grid add still further to the costs of electricity. It is worth remembering that the original plans for the 275 and 400kV Supergrids that were built from the 1950s through to the 1970s, were predicated on 'coal by wire'.
That is, they were designed so that the power stations could be built on the coalfields and the electricity carried to the main load centres in the South East and the Midlands by overhead lines, avoiding the need for coal trains or colliers to transport coal to power stations near the major conurbations.
Most of the large coal stations have now gone and the fitness for present-day purposes of those original power networks might be questioned.
This is particularly so given the fact that large gas-fired stations have tended to be built near gas terminals – which generally speaking are not co-located with coal fields – and the further need to connect offshore wind farms at coastal locations often far-distant from the Grid.
New overhead lines and substations have to be built to serve these developments and someone – the electricity consumer – has to pay for them.
The last links in the chain which supplies us with electricity are the local distribution systems which take power from the regional grid supply points and transport it to the domestic, commercial and industrial consumers.
A major revolution is taking place in this element of the electricity infrastructure to effectively, convert a passive, one-way system into an active system that can cope with variability of supply and demand, such as small generators embedded in the network, in a much more intelligent and efficient way.
The era of the smart grid is soon to dawn! All this new technology is expensive and someone – the electricity consumer – has to pay for it.
Current concerns over electricity prices in the domestic sector often include criticism of the structure of the industry, particularly the apparent lack of competition between suppliers.
There are at least two reasons for this dis-functionality of the domestic market: the first is the difficulty many consumers experience in changing a supplier. Unless a consumer is savvy and log their electricity and gas consumptions regularly, and are computer literate, difficulty will be experienced in obtaining competitive quotations.
The second reason for the dis-functionality in the market is the close tie between the generation divisions of the Big Six utilities and their supply businesses which actually sell the product at the retail level.
Open competition at the generating level and the breaking of the link between wholesale production and retail selling would open up the market to more supply companies and eliminate much of this dis-functionality.


Apple will accomplish this accomplishment independent of its "Make in India" technique with the normal level of three million shipments it does in a year, it included.
Apple is apparently kickstarting creation on its 2018 iPhone in India trio sooner than earlier year to maintain a strategic distance from the disaster which occured for the iPhone X. The handset had become reported in September 2017, however just hit accessibility in November.
Apple is set to be the eleventh brand in cell phones in India to have in excess of 10 million clients in the nation, as per another write about Friday.
India will soon have in excess of 10 million individuals utilizing Apple iPhones, making it the eleventh cell phone brand to cross that check, a report by CyberMedia Exploration said on Friday.
It's a wonderful achievement for Apple "in a value delicate market like India where 56% of the cell phones sold are still sub Rs 10,000", CyberMedia Exploration examiner Faisal Kawoosa said in a note.
Generally speaking however, Apple's number could not hope to compare with those of new cell phone showcase pioneer, Xiaomi, which had 28 million clients starting at 2017 end, and No. 2 Samsung, which has more than 120 million cell phone clients.
What about 'Make in India'?
Kawoosa said that Apple will accomplish the 10 million number this year regardless of its 'Make in India' technique, with the normal level of three million shipments it does in a year.
As of end 2017, there were 8.9 million iPhone clients in the nation and with an expansion of assessed three million, it would outperform the 10 million check even subsequent to marking down the switch inside the brand, which is around 35%, according to CyberMedia Exploration's appraisals.


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Officials release video of Memphis police beating Tyre Nichols, who later died from his injuries
LiveScience.com
'Monolith' Object on Mars? You Could Call It That
Natalie Wolchover
Amateur stargazers have discovered an intriguing object jutting out from the surface of Mars. The seemingly perfectly rectangular, upright structure, found in NASA images of the Red Planet, bears a striking resemblance to the monoliths planted on Earth and the moon by aliens in the classic sci-fi film "2001: A Space Odyssey."
The object in question was first spotted several years ago after being photographed by the HiRISE camera onboard the Mars Reconnaissance Orbiter, a NASA space probe; every so often, it garners renewed interest on the Internet. But is it unnatural — a beacon erected by aliens for mysterious reasons, and even more mysteriously paralleled in the imaginations of Stanley Kubrick and Arthur C. Clarke, creators of "2001"? Or is this rock the work of nature? [Photo ]
According to Jonathon Hill, a research technician and mission planner at the Mars Space Flight Facility at Arizona State University, who processes many of the images taken during NASA's Mars missions, the object in question is no more than a roughly rectangular boulder.
The HiRISE camera that photographed it has a resolution of approximately 1 foot (30 centimeters) per pixel — impressive considering the 180-mile (300-kilometer) altitude from which it photographs the Martian surface, but not quite sharp enough to capture the cragginess of a mid-size boulder. "When your resolution is too low to fully resolve an object, it tends to look rectangular because the pixels in the image are squares. Any curve will look like a series of straight lines if you reduce your resolution enough," Hill told Life's Little Mysteries.
The location of the boulder at the bottom of a cliff near many other boulders suggests it broke off the cliff and tumbled to its current spot sometime in the distant past, Hill said. Such a perilous location is itself an argument against deliberate placement by aliens: "If I was going to build a monolith somewhere, that's the last place I would put it!" he said. "The debris falling from the cliff would cover it up pretty quickly, on geologic timescales." [Vote: Do You Believe in Life on Mars?]
Hill added that the height of the boulder is being exaggerated in the photo by a low sun angle. Photographed when the sun was near the horizon, the boulder casts an especially long shadow.
The ufologists aren't necessarily wrong in calling it a monolith — the word simply translates from Latin as "one stone." But this monolith isn't the masonry of Martians.
Follow Natalie Wolchover on Twitter @nattyover. Follow Life's Little Mysteries on Twitter @llmysteries and join us on Facebook.
Seeing Things On Mars: A History of Martian Illusions
7 Things Most Often Mistaken for UFOs
If We Discover Aliens, What's Our Protocol for Making Contact?
The U.N. chief warned on the day to remember victims of the Holocaust that "many parts of the internet are becoming toxic waste dumps for hate and vicious lies," and urgently appealed for guardrails against hate speech. Secretary-General Antonio Guterres said Friday that anti-Semitism is everywhere, and it's increasing in intensity.
Bilal al-Sudani was a key figure in growing the Islamic State's presence in Africa, the US says.
Amazon will start charging Prime members up to $10 in delivery fees on grocery orders under $150
Delivery charges range from $3.95 to $9.95, depending on the size of the order, and go into effect February 28, Amazon announced Friday.
With the Angels off the block, will Shohei Ohtani be on the move? | The Mets Pod
On an all-new episode of The Mets Pod, Connor Rogers and Joe DeMayo discuss the news that the Angels are no longer for sale, and how that may affect the future of Shohei Ohtani. Will the Angels potentially trade their superstar, and is Ohtani destined for free agency after the 2023 season? Connor Rogers and Joe DeMayo bring you The Mets Pod, a podcast dedicated to all things New York Mets! SNY's signature podcast is a weekly show that delivers everything a Mets fan wants to hear about the team from Queens, including news, analysis, exclusive interviews, special guests, and more! About The Mets Pod: Subscribe below to The Mets Pod to hear Connor Rogers and Joe DeMayo talk New York Mets all year long! Apple: on.sny.tv/Fjog278 Spotify: on.sny.tv/mox84ql
Five former Memphis police officers charged with the murder of Tyre Nichols
Five Memphis police officers were fired and arrested with murder following an internal investigation into the death of Tyre Nichols.
On most weekends, Tyre Nichols would head to the city park, train his camera on the sky and wait for the sun to set. "My vision is to bring my viewers deep into what I am seeing through my eye and out through my lens," Nichols wrote. Nichols, a 29-year-old father, was on his way home from taking pictures of the sky on Jan. 7, when police pulled him over.

redpajama_set_name: RedPajamaCommonCrawl


An Enchantment of Ravens by Margaret Rogerson | Art, Autumn & Faeries!
An Enchantment of Ravens was my most anticipated debut of 2017. I even did a Waiting on Wednesday post about it! The incredibly talented Charlie Bowater tweeted about this book and I fell instantly in love with it because it's about Faeries! I listened to the audiobook last December, but it really wasn't for me. Unfortunately, the narration didn't feel right and iI was really bothered by i. Luckily, I got a beautiful hardback copy for Christmas! I really do love this book so much more now I don't have to listen to "bad" narration! Here's my spoiler free review.
Books about Faeries are my absolute favourite. I've read quite a lot of them, and although all these stories are different, there are also quite a few similarities. I was very curious to see what An Enchantment of Ravens was going to be like. Would it be all sunshine and rainbows? Or would it be super dark and twisted? Well, I'm beyond excited to say it was definitely that last one. It was dark, gruesome and even a little bit disgusting at times and I absolutely LOVED it! There were a lot of new elements in the book that I hadn't seen before (like the Fair Folk not being able to do Craft) and of course we also had the "familiar" elements (like how they use glamours on everything all the time so they seem beautiful while in fact their disgusting creatures). Oh boy, I loved it so, so, SO MUCH.
For me, the characters were a bit "ruined" because of the audiobook narration, which I listened to first. When I started reading, I had to force myself not to imagine the character's voices, because I didn't like them at all. Unfortunately, I couldn't quite put it out of my mind and it slipped in every once in a while. I just couldn't forget and because of that, I liked some of the characters (like Rook, Gadfly and Lark) much less. I did love Rook much better than before, but there was still something about him… I did love Isobel, though! I felt like she was such down to earth, real character who loves art and makes odd comparisons when talking about the Fair Folk but it worked.
Something I hadn't realised when listening to the audiobook, is the insta-love between Isobel and Rook. An Enchantment of Ravens starts a bit slow, but the love was there – literally out of nowhere – from the start. There was literally no reason for them to be in love other than the fact that Isobel painted his portrait and it started to bother me a little bit towards the end. There isn't much dept to their relationship. But to be honest, though, I really didn't mind. This was really the kind romance where you could just shut your brain off and enjoy the ride. I love how the bickered with each other and complicated they sometimes made it for each other. It was also quite hilarious sometimes! The humour was very welcome, as it was an unexpected part of the book.
As for the Autumn vibes… I didn't really feel them. I mean, didn't we all feel like this was the perfect Autumn book? I mean, look at that cover! And an Autumn prince? Yes, please! Unfortunately, it wasn't very Autumn-ish. Thinking about it, I missed some essential world building. There was enough, but I'm still curious. There's something about the Fearie Courts and Whimsy that's apparently different from the World Beyond. But how/why is there a difference? Is Whimsy a part of that World Beyond? I don't think we got an explanation and I kind of really wish to know! I'd also love to see even more of the Courts, but again, we didn't really get that, either.
I liked the plot twists at the end, but it also went a little fast. Like, it's as if we've fast forwarded to the end because suddenly things went so fast and I gotta admit that I could not keep up when I was listening to the audiobook. It was easier to follow along when just reading the book but… It just missed something. Or maybe some things should've been different? The villain, for example, really only came into the story towards the end, which was a bit confusing. We'd read of him before, but we didn't get enough info to truly say "Omg, that's a badass villain!" because he really wasn't. The other sort of "villain" was much more epic and sinister. I loved it! I was so confused but it was honestly something I wanted to be confused about because it's a Fair One and it made sense!
I do desperately wanted to rate An Enchantment of Ravens five stars because I really do love it, but I just don't think it would be honest. Yes, I love this book, but it's not worth five stars. It's wonderful and whimsical and dark and twisted and it's definitely my kind of book, I just think it could've been a little bit better.
Isn't An Enchantment of Raven's cover beautiful? Have you read the book yet?


In an interview on a Lubbock radio show, the governor ramped up the pressure for lawmakers to approve the conservative and controversial items he wants passed during the special session -- from a bathroom bill to property tax reform to new limits on cities to annex and enact local ordinances on tree cutting, development and other things.
Pressure, perhaps. But his crack may just make some lawmakers angry, especially those in the Texas House, after they just completed a fractious 140-day session that left the Senate and the House far apart on many of those same issues.
Meanwhile, talk to Texans and many have little clue what the political hubub in Austin is all about -- even that a special legislative session is scheduled to start or what many of the issues are.
Some -- including many Republican voters -- say they couldn't care less about why Abbott and Lt. Gov. Dan Patrick think the issues are so important they need attention this summer. Some say they would prefer that the Legislature stay home.
Quite a number aren't even familiar with the issues, even after five months of headlines about the legislative fighting over them.


Naître et grandir wins a Boomerang award!
Learning to live with others starts early!
Special Report on Fathers: What Role Do They Play at Present?
Over 3,500 young Quebecers come together for the third Robotics Festival!


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Tom Ackerley's biography: what is known about Margot Robbie's husband?
Wednesday, December 15, 2021 at 2:19 PM by Cyprine Apindi
Tom Ackerley is a talented English film director and producer. He is best known as the husband of actress Margot Elise Robbie. In 2014, he co-founded LuckyChap Entertainment, a film and television show production company, with his wife.
Tom Ackerley attends the premiere of the movie Terminal at ArcLight Cinemas in Hollywood, California. Photo: @ David Livingston/Getty Images)
Tom Ackerley developed a passion for film production quite early in his life. He studied the same and started his career in 2011. Read on to know more about his childhood, educational background, career, marriage, and much more.
Full name: Thomas Ackerley
Nickname: Tom
Date of birth: 1st January 1990
Age: 31 years (as of December 2021)
Place of birth: Guildford, Surrey, England, United Kingdom
Current residence: Los Angeles, California, United States of America
Nationality: British
Height in feet: 6
Height in metres: 1.83
Weight in kilogrammes: 75
Eye colour: Blue
Hair colour: Dark brown
Brothers: 2
Spouse: Margot Robbie
Alma mater: Godalming College
Profession: Film director and producer
Leo Robinton's biography: what is known about Emma Watson's partner?
Tom Ackerley's bio
The film producer was born in Guildford, Surrey, England, United Kingdom. His nationality is British, and his ethnicity is White.
Although he is yet to reveal the names of his parents, it is known that his father was a real estate agent. On the other hand, his mother owns and runs a property management firm.
The young producer has two brothers. One of them is named Ben. He was raised in the United Kingdom.
Producer Tom Ackerley attends the Kodak Motion Picture Awards Season Celebration in Los Angeles, California. Photo: @Phillip Faraone/Getty Images for Kodak
Tom Ackerley's age
The producer is 31 years old as of December 2021. He was born on 1st January 1990, and his Zodiac sign is Capricorn.
Margot Robbie's husband went to school in the United Kingdom. He is a graduate of Godalming College, where he pursued a programme in film studies.
Timothy Hawking's biography: what is known about Stephen Hawking's son?
Tom Ackerley's movies and TV shows
The young producer started his career in 2011. After graduating from college, he worked as a floor runner in four episodes of the BBC drama, The Hour. He later worked as a crowd runner and set production assistant before bagging an assistant director's job.
Since starting his career, he has earned many credits as a producer, assistant director, or unit director. His movies and television shows are listed below.
I, Tonya (2017)
Terminal (2018)
Dreamland (2019)
Promising Young Woman (2020)
The Humming of the Beast (2021)
Dollface (2022)
Boston Str*ngler (filming)
Tank Girl (announced)
Iron Mike (announced)
Barbie (announced)
Assistant/ unit director credits
Playhouse Presents (2012)
The Last Days on Mars (2013)
Final Prayer (2013)
The Two Faces of January (2014)
Da Vinci's Demons (2014)
Pride (2014)
Everly (2014)
Suite Française (2014)
American Odyssey (2015)
Yussef Is Complicated (2015)
MI-5 (2015)
The Brothers Grimsby (2016)
Helvellyn Edge (2022)
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Who was Tom Ackerley in Harry Potter? In 2001, 2002, and 2004, he was a student in different Harry Potter video games. These were Harry Potter and the Philosopher's Stone, Harry Potter and the Chamber of Secrets, and Harry Potter and the Prisoner of Azkaban (uncredited).
In 2014, he and his then-girlfriend, Margot Robbie, co-founded LuckyChap Entertainment, a film production company. Some of the productions done by the company are Dreamland, I, Tonya, and Terminal. The couple has run the production house to date.
Tom Ackerley's net worth
The producer has an estimated net worth of $4 million. His primary source of income is his work as an assistant director and producer.
Actor Margot Robbie and producer Tom Ackerley attend The 75th Annual Golden Globe Awards at The Beverly Hilton Hotel. Photo: @Stefanie Keenan/Getty Images for FIJI Water
How long have Margot Robbie and Tom Ackerley been together?
Margot Robbie and Tom Ackerley have been together for about seven years now. How did Margot Robbie Meet Tom Ackerley? The two met for the first time in 2013 on the set of Suite Francaise. Margot was one of the main cast members while he was an assistant director for the war romantic drama movie.
Thomas Elms' biography: age, birthday, height, partner, movies
The couple did not start dating until 2014. Their relationship thrived, leading to a proposal and an engagement. The lovebirds got married in December 2016.
Margot and Tom Ackerley's wedding was a low-key event held in New South Wales, Australia. The ceremony was hosted at a private property in Coorabell, Byron Bay, and the bride wore a cream, lace, off-the-shoulder dress.
Tom Ackerley's height and weight
Margot Robbie's husband is 6 feet or 183 centimetres tall. He weighs about 166 pounds or 75 kilogrammes. Also, Tom has blue eyes and dark brown hair.
Tom Ackerley is a talented film producer and director. While he has been in the film and television production industry for a while, he got public attention after dating and eventually marrying actress Margot Robbie.
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One Million Iraqis Killed?
ROBERT NAIMAN
National coordinator and senior policy analyst at Just Foreign Policy, Naiman said: "Just Foreign Policy is publishing updated estimates of Iraqi deaths due to the U.S. invasion and occupation starting in 2003. And the way that we constructed this estimate is to extrapolate from the Lancet study that was published last fall.
"The Lancet study was based on a survey completed in July 2006. So, although at the time it was the best, indeed the only scientific estimate of Iraqi deaths, it is now a year old. … We wanted to provide an update of that so we have extrapolated from the numbers that the Lancet study reported, using the trend provided by the deaths reported by Iraq Body Count, which counts deaths reported in Western media. And our online estimate now stands at 999,987 — so it's about to cross the one million mark."
LES ROBERTS
Available for a very limited number of interviews, Roberts is co-author of the study "Mortality after the 2003 Invasion of Iraq: A Cross-Sectional Cluster Sample Survey," published in October 2006 by the leading medical journal The Lancet. The study estimated 655,000 excess deaths following the 2003 U.S. invasion of Iraq.
When The Lancet published the estimate, President George W. Bush said: "600,000 or whatever they guessed at is just, it's not credible."
Roberts has produced estimates using the same methods in other conflicts, for example in Congo, which have been quoted by U.S. government officials.
Roberts said today: "It's a little disappointing to see the belligerents in this particular conflict just rejecting these findings outright, and in doing so they have actually been expressing indifference in a time when they should be expressing compassion.
"There's a lot of complimentary evidence out there that our study has to be in the right ball park. For example … a recent report from Oxfam of hundreds of thousands of war widows in Iraq — or the U.N.'s claim some months ago that there are hundreds of thousands of war orphans in Iraq."
For more information, contact the Institute for Public Accuracy at (202) 347-0020; or David Zupan at (541) 484-9167.


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Acronym vs. Initialism vs. Abbreviation With Examples
By Matt Salter, B.A. , Staff Writer
Computer keypad of initialisms
Vladimir Godnik / Getty Images
Used under license
Acronyms, initialisms, abbreviations... say what? When it comes to the wonderful world of abbreviations, they are everywhere. Abbreviations are just words or phrases that are shortened. But, what is the difference between an acronym and initialism? It comes down to how you pronounce an abbreviation that is going to make it an initialism, an acronym or both.
Initialisms vs. Acronyms vs. Abbreviations
You've now walked into the fun world of abbreviations. In short:
An abbreviation is when a phrase or word is shortened to a word or letters.
It could be abbreviated into one sound like "Rom." for Romeo and Juliet. Or, it could be shortened to its first letters like NRA for the National Rifle Association. When it comes to abbreviations, you will commonly see two different types: initialisms and acronyms. So, what's the difference between an acronym and initialism?
Initialisms are when you abbreviate a word to its initials. Then, you say those initials as individual letters, like how "National Security Administration" becomes N-S-A. The United States becomes U-S.
Acronyms are abbreviations that also use initials, but those initials are pronounced as a word rather than saying the individual initials. For example, "National Aeronautical and Space Administration" becomes NASA and is pronounced "nah-sah."
To truly understand the difference between an initialism and acronym, you can look at examples of each in business, medicine and politics/government. You can then see how some abbreviations can be both initialisms and acronyms.
Initialism Examples
Initialisms are a common component of both official and conversational English. Initialisms are featured in politics, corporate branding, scientific and medical terminology, and casual conversation alike. Review a few common examples.
Initialisms in Business
Businesses frequently use initialisms as part of their branding, usually as a way of improving out-of-date, extra long, or not-quite-global phrasing.
AT&T: American Telephone and Telegraph demonstrates the modernizing effect of using an initialism. Telegraphs aren't in common use anymore, but AT&T keeps its legacy as the very first telecom company (founded by Alexander Graham Bell!) while modernizing its branding with an initialism.
BMW: Bayerische Motoren Werke (translation: Bavarian Motor Works) is a classic example of using initialism to globalize a brand.
KFC: Founded by Colonel Harland Sanders in Corbin, Kentucky, KFC (Kentucky Fried Chicken) has become a global brand. Its initialism can be found on stores from Alaska to Singapore.
IBM: The original tech giant's full name is International Business Machines, but virtually all of their branding relies on this initialism.
Medicine Initialisms
Doctors and nurses are busy. This means that they don't have time to say or write all the complicated words that are found in the field of medicine. Therefore, initialisms can come in handy. Check out a few initialisms used in the medical field.
ALS: Amyotrophic lateral sclerosis is just one of many diseases more widely referred to by its initialism. The unwieldy full name is necessary for doctors to specify the particular nature of the illness, but an abbreviation serves for quicker communication.
EMT: Emergency medical technicians are often the first responders in medical emergencies. It only makes sense for them to shorten their professional title using an initialism.
ENT: Not to be confused with EMT above, an Ear, Nose and Throat specialist treats illness and injury in those parts of the body.
PTSD: Initialisms are particularly common when referring to mental illnesses like post-traumatic stress disorder. It simplifies phrasing while keeping things specific.
Politics/Government
Initialisms in government are so common that they've coined their own joke: "alphabet soup." The use of initialisms in politics around the world reflects a shared need to save time and space without losing specificity.
FBI: The Federal Bureau of Investigation, America's police force investigating violations of federal law, is universally known simply as the FBI.
MP: A common initialism throughout the United Kingdom and Commonwealth countries, MP is shorthand for Member of Parliament. Ironically, the head of a government made up of MPs is usually the PM, or Prime Minister.
NHS: The United Kingdom has plenty of alphabet soup of its own. The NHS, or National Health Service, is the universal health coverage enjoyed by British natives.
RCMP: The Royal Canadian Mounted Police, or "Mounties," is Canada's federal police force, also providing law enforcement on the provincial and local level in various parts of the country.
Acronym Examples
Now, it's time for acronyms. While not as common in some areas, you can find acronyms all over the board or world for that matter. Check out how acronyms compare to initialisms.
Big businesses might use acronyms to make them stand out or just to shorten a long name. A few famous examples include:
Fiat: This is actually an acronym the motor company uses. Fiat stands for Fabbrica Italiana Automobili Torino (Italian automobile factory of Turin).
MAC: You might not have realized it, but MAC actually stands for Makeup Art Cosmetics. Your new lipstick brand is an acronym.
D.A.R.E.: Many people know this one in the U.S., but D.A.R.E. stands for Drug Abuse Resistance Education.
Geico: It might come as a shock, but the creators of Geico Insurance weren't obsessed with geckos they were targeting government employees. Geico actually stands for Government Employees Insurance Company.
Medicine Acronyms
Medicine is full of complicated words and phrases that doctors just don't have time for. While you'll find initialisms a plenty, acronyms are right in the thick of it too.
HIPAA: If you've ever been to the doctor in the United States, then you've likely signed the HIPAA regulations. HIPAA stands for Health Insurance Portability and Accountability Act.
AIDS: Acquired Immune Deficiency Syndrome is an illness that is better known as AIDS.
CPAP: If you have breathing issues at night, you might need to use a CPAP machine. CPAP actually stands for Continuous Positive Airway Pressure but it is better known as "see-paap."
SAD: SAD stands for Seasonal Affective Disorder. The changing of the seasons typically from fall to winter causes depression.
Acronyms for Politics/Government
The government is not without its acronyms. You can find them all over the place from the different departments to laws. Check out a few that you might already know.
NATO: You might have heard NATO mentioned in the news and wondered what it was. NATO stands for North Atlantic Treaty Organization. It's an alliance formed between several countries.
NAFTA: Going right along with NATO, you have NAFTA (pronounced "naaf-tah"). This is the North American Free Trade Agreement.
OPEC: OPEC typically comes into play in government when they are talking about oil. That is because OPEC (pronounced "oh-peck") is the Organization of the Petroleum Exporting Countries.
AMMO - This isn't just a name for bullets for a gun. AMMO is the acronym for the Aircraft Munitions Maintenance Organization.
Combining Initialisms and Acronyms
It is the nature of active, living languages that as soon as a rule is set down, exceptions arise. Initialisms and acronyms are no different. English speakers split the difference between initialisms and acronyms simply by pronouncing certain abbreviations per their individual letters, and other times as single words.
FAQ: As gamers and other digital natives know well, FAQ stands for "Frequently Asked Questions," a type of online guide that answers questions commonly asked by users of a given digital resource. We have one of our own! As to whether it is pronounced letter by letter (i.e., "F-A-Q") or as the single word "fack," both are considered acceptable.
JPG: Here's another one for digital natives. Whether this common image extension is sounded out by letter (i.e., "J-P-G") or pronounced as "jay-peg" is, just like FAQ, very much a matter of opinion.
LSAT: It would be tempting to assume that the LSAT, the admission test for American law school, comes from the same abbreviation as SAT. It doesn't. It simply stands for Law School Admission Test. However, it can be pronounced letter by letter or as "elle-sat."
MCAT: Another initialism/acronym from the world of standardized testing, the Medical College Admission Test is taken by students who hope to become doctors. Both "M-C-A-T" and "em-cat" are valid pronunciations.
UCLA: The difference between acronyms and initialisms can also be a deliberate choice on the part of the speaker. In the world of college sports, for instance, it's commonplace for a team's fans (or detractors) to joke about an institution known by an initialism by pronouncing its name as a word instead. For instance, they might say "uhk-lah" instead of "you-see-elle-ay" for the University of California, Los Angeles.
Examples of Abbreviations
You've looked at abbreviations that can be acronyms, initialisms and both, but what about those abbreviations that are neither? There are a few of them out there. When you say these abbreviations, you'll say the full word but you know exactly what they mean in writing. Check out the list to see some common abbreviations that you might use everyday.
Govt.: When writing, you might use "govt." as a quick way to discuss the government. It makes great short-hand and everyone knows what you are talking about.
Corp.: Short for corporation, this is one that you might see on the internet. Corporations will also shorten their name with "Corp." on letterhead and business cards.
Abbr.: If you are going to abbreviate a word, then you might give someone a warning by using "abbr." for abbreviation.
Add.: When you're shortening an address for someone, you might use "add." to abbreviate it. This is especially common on forms. For example, "Add. - 1234 Broadway."
St.: Another common abbreviation that is neither an acronym nor an initialism is street, which you would abbreviate as "st." You might also see "blvd." for boulevard, "rd." for road, and "hwy" for highway.
Letters in a Line
The distinction between acronyms and initialisms - acronyms are pronounced as a single word, initialisms are pronounced as a series of individual letters - is, like many English language rules, very simple and slightly subjective. For some more entertaining abbreviations, check out our list of funny acronyms. Happy learning!


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Q: QueryBuilder with One To Many relationship and several conditions using Doctrine I will try to explain my issue as well as I can. I have a Message related to another Entity called Reply (one message can have zero or n replies) and I want to take into account the following cases:

*

*If UserA have one message created to UserB (UserA is the creator)
but the UserB didn't replied the message I don't want get message

*Same as above but the UserB replied to the message. I want get the
message with its replies if replies exists

*If the UserB sent a message to UserA, I want to get the message with
its replies if replies exists

My Message Entity (I will just put the OneToMany relantionship):
<?php
namespace AppBundle\Entity;

use Doctrine\ORM\Mapping as ORM;
use Symfony\Component\Validator\Constraints as Assert;
use Symfony\Bridge\Doctrine\Validator\Constraints\UniqueEntity;

/**
* @ORM\Table(name="messages")
* @ORM\Entity
* @ORM\Entity(repositoryClass="AppBundle\Repository\MessageRepository")
*/
class Message {

/**
* @ORM\OneToMany(targetEntity="ReplyMessage", mappedBy="message")
*/
private $replies;

/**
 * Add replies
 *
 * @param \AppBundle\Entity\ReplyMessage $replies
 * @return Message
 */
public function addReply(\AppBundle\Entity\ReplyMessage $replies)
{
    $this->replies[] = $replies;

    return $this;
}

/**
 * Remove replies
 *
 * @param \AppBundle\Entity\ReplyMessage $replies
 */
public function removeReply(\AppBundle\Entity\ReplyMessage $replies)
{
    $this->replies->removeElement($replies);
}

/**
 * Get replies
 *
 * @return \Doctrine\Common\Collections\Collection
 */
public function getReplies()
{
    return $this->replies;
}

And the code that I am using to achieve my objective is the following:
$query = $this->createQueryBuilder('message')
    ->where('message.creator = :username or message.receiver = :username')
    ->leftJoin('message.replies', 'replies')
    ->andWhere('replies.user = :username')
    ->setParameter('username', $username)
    ->getQuery();

$sent = $query->getResult();

return $sent;

I am not an expert building queries with Doctrine and I don't know how can control all my cases right now, if someone could help I'd be very grateful

A: That's the beauty of QueryBuilders. You can create statements like this:
$query = $this->createQueryBuilder('message')
    ->where('message.creator = :username or message.receiver = :username')
    ->leftJoin('message.replies', 'replies')
    ->andWhere('replies.user = :username')
    ->setParameter('username', $username);

if (some_condition) {
    $query->andWhere('some_property');
}

if (some_other_condition) {
    $query->andWhere('some_property');
}

return $query->getQuery()->getResults();

A: Don't use ->getResult(); in you AppBundle\Repository\MessageRepository repository, becose sometimes you need use him meny times, and for example: BreadcrumbsBundle need query (not result, еhis is affected by performance)

redpajama_set_name: RedPajamaStackExchange


text:
Geojit News Releases
Stay updated with every development at Geojit. From products and services to corporate events and media coverage, get a host of Geojit-related information here. You can get access to press releases or browse through the photo gallery. If you are a media person, contact our public relations team.
Geojit announces 3rd Quarter results of FY 2022-2023
Geojit's Smartfolios launches exclusive equity baskets based on MSCI indexes
Geojit announces financial results of the quarter and year ended 31 March 2022
Geojit launches IPO application facility through WhatsApp
Geojit announces 1st Quarter results of FY 2022-2023
Geojit announces 3rd Quarter results of FY 2021_2022
Geojit appoints Jones George as Executive Director
Geojit launches first of its kind digital platform for loan against shares
Geojit appoints new independent directors to its Board
Geojit launches India's first online service for NRI investors to open demat, trading accounts
Geojit Image Library
Geojit launched the first of its kind digital platform for Loan Against Shares. It was launched by Ms Padmaja Chunduru, Managing Director of National Securities Depository Ltd (NSDL) in the presence of Mr C J George, Founder and Managing Director of Geojit, Mr A Balakrishnan, Executive Director of Geojit and Mr Prashant Vagal, Executive Vice President of NSDL at a function held in Kochi.
2019 - Inauguration of STEPS-a financial planning division of Geojit by C.J. George (MD, Geojit). Also seen are Satish Menon (ED, Geojit), Sanjeev Rajan (CFO, Geojit) and other staff members.
2019 – Launch of Dakshin, a South-India based multi-cap PMS. Seen are A. Balakrishnan (ED, Geojit) A.P. Kurian (Chairman, Geojit), C.J. George (MD, Geojit), and Satish Menon (ED, Geojit).
2017 - Launch of Geojit's Online Financial Planning Tool by Milind Barve (MD of HDFC Asset Management Company) along with A.P. Kurian (Chairman, Geojit), R. Bupathy (Independent Director of Geojit) and Satish Menon (ED, Geojit)
2017 - The company is renamed as Geojit Financial Services Ltd. and a new logo is unveiled. Seen are A.P. Kurian (Chairman, Geojit), C.J. George (MD, Geojit), A. Balakrishnan (the then MD, Geojit Technologies) and Satish Menon (ED, Geojit).
2018 – Launch of mutual fund app FundsGenie by Aruna Sundararajan, chairman of Telecom Commission and secretary of department of telecommunication. Also seen are C.J. George (MD, Geojit), A. Balakrishnan (the then MD, Geojit Technologies) and Satish Menon (ED, Geojit).
Media Contact Details
Elizabeth V
Head Corporate Communications
Tel : 0484 – 2901112
Email :communicate@geojit.com
elizabeth_v@geojit.com
Mr. C. J. George
Managing Director and Promoter
The Company was founded by Mr. C. J.George in 1987. He has over 31 years of professional..
The Company was founded by Mr. C. J.George in 1987. He has over 31 years of professional experience in the securities market. He has presented numerous papers related to the industry in seminars at national and international fora. He frequently contributes articles to financial publications and serves as a guest faculty at reputed management institutes. Mr George, has a Masters Degree in Commerce and is a Certified Financial Planner.
Managing Director & Promoter
The Company was founded by Mr. C. J.George in 1987. He has over 31 years of professional experience in the securities market. He has presented numerous papers related to the industry in seminars at national and international fora. He frequently contributes articles to financial publications and serves as a guest faculty at reputed management institutes. Mr George, has a Masters Degree in Commerce and is a Certified Financial.


Find hotel near Blackwood Spiritualist Ch ..
The Blackwood Spiritualist Church in England is a very sacred institution widely known for their belief in spiritualism. They are known to promote the ideas and concepts of spiritualism and religion among people. Their ambition is to unite all of them who believe in Spiritualism. Many other churches and holy institutions support them and have expressed their interest to join them. This Church is a great monument to visit as well. With the building being built beautifully, you can also feel the sense of purity and spiritual calmness in the atmosphere once you enter the church.
Book perfect hotels in Blackwood like Castle Hotel, Holiday Inn Cardiff North M4 Jct 32 and Castle View Apartment - J32 M4.
There are plenty of ready-to-book Blackwood vacation packages available that will cover the major attractions, accommodation concerns including the prime one like how to reach Blackwood.
Love this? Explore the entire list of places to visit in Blackwood before you plan your trip.
Fancy a good night's sleep after a tiring day? Check out where to stay in Blackwood and book an accommodation of your choice.


text:
Arno ter Burg
Composities
Pianist, composer & arranger
Arno ter Burg (1973) started organ lessons at the age of 8 with Richard Kroes. When he was 9, he studied piano with Guy de Werd at the School of Music in Eindhoven (CKE). Arno studied piano and chamber music with Alexandre Hrisanide at the Brabants Conservatory in Tilburg and followed lessons composition and orchestra conducting. He gained his diploma in 2000. He also studied piano with Lena Basova and lessons at het fortepiano with Paul Komen. Arno has taken part in several masterclasses: piano and composition with Menachem Wiesenberg, chamber music with Sherban Lupu and a masterclass with operadirector Peter Sellars.
Arno is now pianist/organist and director of music at the Holy Brigid Church in Geldrop.
He teaches piano and composition at the School for Music in Eindhoven (CKE).
Arno is an accompanist for singers. He studied song-interpretation with Bernard Kruysen and Erica Pauwels (Belgium). In 1999 he was laureate in Gent (Belgium), as accompanist, at a singing competition.
Also as a composer Arno is very busy. He composed a variety of pieces (for orchestra, choir, chamber music, solo instruments and songs for voice and piano).
In 2013 he was commissioned to compose a piece, 'Te Deum', for choir, soloists and orchestra. In 2020 he composed two solo-violin pieces for Carla Leurs and Lisa Jacobs.
Arno is arranger for the Netherlands Wind Ensemble (NBE).
On april the 16th 2017, Arno was knighted in the order of Oranje-Nassau.
Arno ter Burg © 2021


Tigers News · Feb 4 Girls Swimmers Heading to State!
Emily Weiss, Munthira Raju, Kaitlynn Santino, Ellie Alexander, Grace Huisman, Catherine Grasso, Tyra Boyd, Neely Agnew, McKenna Alexander, and Anisse Stoops.
Emily Weiss was the sectional champion in the 200 yard IM with a time of 2:05.36 and the 100 Breaststroke with a time of 1:00.97.


Love this place, it's sad that that other restaurant with a similar name made people think this place was anything less than 100% professional and tasty! New name, same great food!
Pros: Great food at a great price. Their $1 Taco special is the best.
I used to work near this restaurant about 5 years ago. Now I stop by whenever I am heading up the veteran's expressway. It is clean and inexpensive as well as good. I reminds me of my favorite mex restaurant in Atlanta, El Toro.


text:
Bell Schedule & School Hours
School Dance Information
Archive for month: July, 2018
You are here: Home / Frequently Asked Questions / 2018 / July
VHSL Physical Form Available for Download [PDF]
July 25, 2018 /in Headline List /by WJCC Public Schools
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2018-19 School Supply List Available Online [PDF]
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July 18, 2018 /in Athletics, Headline List /by WJCC Public Schools
Football – Head Coach, Dr. Mike Brown – There will be an informational football meeting Wednesday, August 1 at 5:30 p.m. in the Berkeley Middle School auditorium. The first practice is August 6 from 5:30-7 p.m. on Cooley Field. If you have questions, contact Coach Brown at dr.michael.brown76@gmail.com
Volleyball – Head Coach, Jayme Phillips – Open Gym will be held Monday August 20-22, 5:30-7:30 p.m. Tryouts are August 23 and 24, 5:30-7:30 p.m. Team Practice Begins August 28 at 4:00 p.m. Please contact Coach Phillips at Jayme.Phillips@wjccschools.org or via phone at 443-669-3247.
Cheer – Head Coach, Kristine Robinson – Tryouts will be August 23 and August 24 from 4-5:30 p.m. in the Blair Middle School gym. Parents can contact Coach Robinson at Kristine.Robinson@wjccschools.org if you have questions.
Field Hockey – Head Coach, Maggie Lombardo – Tryouts will begin on Monday, August 13 at 6:00 p.m. in the Blair Middle School Gym.
*Athletes MUST have a physical dated AFTER June 1, 2018 in order to participate. Athletes should bring a water bottle and wear appropriate athletic clothing and tennis shoes to practice. Football and Field Hockey players should always bring cleats AND a pair of tennis shoes.
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7/16/2018 – Construction Update
July 16, 2018 /in James Blair Construction Updates /by WJCC Public Schools
Work Performed this Week:
Continued interior CMU pointing and patching
Continued window framing and glazing in the research commons and main entrance
Continued installing exterior composite metal panels along the radius wall
Continued hanging and finishing gypsum board in administration and auditeria
Painted the auditeria
Dropped ceiling tile in Unit AB2
Continued rubber floor tile in Unit AB3 and began in Unit AB2
Continued polishing terrazzo flooring in the auditeria
Continued hanging Marlite wood panels in Unit A
Set operable glass partition walls in Units B3, B2, and B1
Continued food service electrical connections
Installed drinking fountains and restroom fixtures in Unit B3
Continued installing ductwork at the stage
Continued controls to the air handling units and programming the mechanical system
Installed pulling data wiring in Units C and D
Placed topsoil on the ball field
Note: Areas D & C are the single-story section of the school. That includes the gym, cafeteria, kitchen, music, art, and fitness rooms. Areas A & B are the 3-story classroom sections.
Window framing and composite metal panels along the radius wall and main entrance
Storefront windows and Marlite wall panels at the stairs
Rubber tile flooring in Unit B3
Operable glass partition wall at Classroom B109
Interior finishes at the Gathering Stairs noting painted surfaces and Marlite panels
Gypsum wall board hung along the radius wall
Gypsum board wall between Commons A131 and Auditeria C101
https://wjccschools.org/jbms/wp-content/uploads/sites/39/2018/07/Operable-glass-partition-wall-at-Classroom-B109-1.jpg 375 710 WJCC Public Schools https://wjccschools.org/jbms/wp-content/uploads/sites/39/2018/07/Website-Logo-01.png WJCC Public Schools2018-07-16 11:08:112018-07-16 11:08:117/16/2018 – Construction Update
GALLERY: James Blair Middle School Tour
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Cheerleading Tryout Information
The countdown is on: Only 64 days before James Blair Middle School opens its doors to students and staff! In the meantime, there are a whole lot of preparations going on. For example, cheer tryouts will be held on 8/23 and 8/24 from 4-5:30 p.m. Anyone wishing to try out must have a physical dated AFTER June 1, 2018. Parents can contact cheer coach Kristine Robinson at Kristine.Robinson@wjccschools.org if they have questions.
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July 2, 2018 /in James Blair Construction Updates /by WJCC Public Schools
Continued window framing and glazing in the research commons
Continued installing exterior composite metal panels
Applied final coat of paint in Unit AB3
Continued rubber floor tile in Unit AB3
Continued placing and polishing terrazzo flooring
Continued hanging Marlite wood panels in the auditeria and began in Unit A
Continued electrical work in food services
Continued installing ductwork and mechanical pipe in Units C and D
Continued mechanical pipe and controls to the air handling units
Installed equipment racks in the IT closets
Continued fine grading the ball field
Installed the Cooley Field scoreboard
Terrazzo flooring
Rubber tile flooring in Foreign Language A303 and A304 Photo
Research Commons second story window framing
Plumbing fixtures and hard ceiling installed
New scoreboard
Final coat of paint
Early delivery of school supplies and equipment including 40 bicycles
https://wjccschools.org/jbms/wp-content/uploads/sites/39/2018/07/Terrazzo-flooring-1.jpg 375 710 WJCC Public Schools https://wjccschools.org/jbms/wp-content/uploads/sites/39/2018/07/Website-Logo-01.png WJCC Public Schools2018-07-02 10:56:052018-07-02 10:56:056/25/2018 – Construction Update


text:
Q: Coercive or not? I had this problem in the exam.
Let $X = [x_1,...,x_d]^T$, $a\space \epsilon$ $\mathbb{R}^d$ and $C$$\epsilon$$\mathbb{R}$.
Argue for or against. $f(X) = a^TX + C||X||^2$ is coercive only for $C > 0$?
My answer: True.
Case $1$: If $C = 0$, then $f(x) = a^TX$. Linear functions are not coercive.
Case $2$ : If $C < 0$, then the function can go to $-\infty$.
Is this reasoning correct?

A: If $C\ne 0$, then
$$
f(x)=C\left\|x-\tfrac{1}{2C}a\right\|^2-\frac{\|a\|^2}{2C}.
$$
Hence for $C>0$:
$$
f(x)=C\left\|x-\tfrac{1}{2C}a\right\|^2-\frac{\|a\|^2}{2C}\ge-\frac{\|a\|^2}{2C}
$$
while for $C<0$, $f(x)\to-\infty$, as $\|x\|\to\infty$.
Hence, $f$ coercive iff $C>0$.