text:
Seal Beach, CA (Rossmoor)
About Seal Beach, CA (Rossmoor)
Rossmoor median real estate price is $1,060,421, which is more expensive than 85.5% of the neighborhoods in California and 97.0% of the neighborhoods in the U.S.
The average rental price in Rossmoor is currently $5,064, based on NeighborhoodScout's exclusive analysis. The average rental cost in this neighborhood is higher than 99.0% of the neighborhoods in California.
Rossmoor is an urban neighborhood (based on population density) located in Seal Beach, California.
Rossmoor real estate is primarily made up of medium sized (three or four bedroom) to large (four, five or more bedroom) single-family homes. Most of the residential real estate is owner occupied. Many of the residences in the Rossmoor neighborhood are older, well-established, built between 1940 and 1969. A number of residences were also built between 1970 and 1999.
In Rossmoor, the current vacancy rate is 1.4%, which is a lower rate of vacancies than 94.2% of all neighborhoods in the U.S. This means that the housing supply in Rossmoor is very tight compared to the demand for property here.
Many things matter about a neighborhood, but the first thing most people notice is the way a neighborhood looks and its particular character. For example, one might notice whether the buildings all date from a certain time period or whether shop signs are in multiple languages. This particular neighborhood in Seal Beach, the Rossmoor neighborhood, has some outstanding things about the way it looks and its way of life that are worth highlighting.
Most neighborhoods have a mixture of ages of homes in them, from new to old, but this neighborhood stands out due to its concentration of residential real estate built in one time frame: from 1940 through 1969, generally considered older, well-established homes. This was a busy time in America for home construction. After the end of World War II, as GIs came home, bought newly built homes on the edges of cities with the help of the GI Bill, and began their families. This housing era generally coincides with the 'Baby Boom' generation (1945 - 1964), and many baby boomers grew up in homes built in this era. But what is so interesting about the Rossmoor neighborhood, is that an incredible 91.0% of the homes here were built in this era. So when you walk its streets or drive through, this neighborhood has a look and feel that harkens to that era in American life, a very important slice of Americana.
In addition, the Rossmoor neighborhood has earned the amazing distinction of having one of the highest rates of detached, single-family homes of any neighborhood in the U.S. With 100.0% of the residential real estate here made up of free-standing single-family homes, there is a greater proportion of single-family homes here than in 99.3% of all neighborhoods in America.
Furthermore, owner-occupied real estate dominates the Rossmoor neighborhood. In fact, according to NeighborhoodScout research, the percentage of residential real estate occupied by its owner is higher here than in 98.2% of neighborhoods in America.
According to NeighborhoodScout's exclusive analysis, Rossmoor is among the best neighborhoods for families in California. In fact, this neighborhood is more family-friendly than 99.4% of neighborhoods in the entire state of California. Its combination of top public schools, low crime rates, and owner-occupied single family homes gives this area the look and feel of a "Leave It to Beaver" episode. Many other families also live here, making it easy to socialize and develop a strong sense of community. In addition, the high number of college-educated parents influences the academic success of the local schools. Overall, you will find all of the amenities a family needs to thrive in the Rossmoor neighborhood. In addition to being an excellent choice for families with school-aged children, this neighborhood is also a very good choice for urban sophisticates and highly educated executives.
In addition, priests and therapists would like to think they know the secrets to a truly successful marriage, but according to NeighborhoodScout's research, the folks of the Rossmoor neighborhood may actually hold the key. 65.5% of its residents are married, which is a higher percentage than is found in 95.5% of the neighborhoods in America.
Did you know that the Rossmoor neighborhood has more Austrian and Iranian ancestry people living in it than nearly any neighborhood in America? It's true! In fact, 1.8% of this neighborhood's residents have Austrian ancestry and 0.9% have Iranian ancestry.
Rossmoor is also pretty special linguistically. Significantly, 0.9% of its residents five years old and above primarily speak Persian at home. While this may seem like a small percentage, it is higher than 97.3% of the neighborhoods in America.
There are two complementary measures for understanding the income of a neighborhood's residents: the average and the extremes. While a neighborhood may be relatively wealthy overall, it is equally important to understand the rate of people - particularly children - who are living at or below the federal poverty line, which is extremely low income. Some neighborhoods with a lower average income may actually have a lower childhood poverty rate than another with a higher average income, and this helps us understand the conditions and character of a neighborhood.
The neighbors in the Rossmoor neighborhood in Seal Beach are wealthy, making it among the 15% highest income neighborhoods in America. NeighborhoodScout's exclusive analysis reveals that this neighborhood has a higher income than 94.7% of the neighborhoods in America. In addition, 4.9% of the children seventeen and under living in this neighborhood are living below the federal poverty line, which is a lower rate of childhood poverty than is found in 72.8% of America's neighborhoods.
A neighborhood is far different if it is dominated by enlisted military personnel rather than people who earn their living by farming. It is also different if most of the neighbors are clerical support or managers. What is wonderful is the sheer diversity of neighborhoods, allowing you to find the type that fits your lifestyle and aspirations.
In the Rossmoor neighborhood, 55.7% of the working population is employed in executive, management, and professional occupations. The second most important occupational group in this neighborhood is sales and service jobs, from major sales accounts, to working in fast food restaurants, with 24.0% of the residents employed. Other residents here are employed in clerical, assistant, and tech support occupations (10.4%), and 9.9% in manufacturing and laborer occupations.
The most common language spoken in the Rossmoor neighborhood is English, spoken by 89.3% of households. Some people also speak Spanish (2.9%).
In the Rossmoor neighborhood in Seal Beach, CA, residents most commonly identify their ethnicity or ancestry as Irish (13.8%). There are also a number of people of English ancestry (13.5%), and residents who report Mexican roots (12.5%), and some of the residents are also of German ancestry (12.0%), along with some Asian ancestry residents (8.9%), among others.
Even if your neighborhood is walkable, you may still have to drive to your place of work. Some neighborhoods are located where many can get to work in just a few minutes, while others are located such that most residents have a long and arduous commute. The greatest number of commuters in Rossmoor neighborhood spend between 15 and 30 minutes commuting one-way to work (41.5% of working residents), which is shorter than the time spent commuting to work for most Americans.
AGE OF Seal Beach, CA (Rossmoor) HOMES
TYPE OF Seal Beach, CA (Rossmoor) HOMES
SIZE OF Seal Beach, CA (Rossmoor) HOMES
Employment Industries in Rossmoor
in Rossmoor
in Seal Beach
Seal Beach VIOLENT CRIMES
Seal Beach Property CRIMES
Rates the quality of all K-12 public schools that your children would be exposed to if you lived in this neighborhood.
Schools In This Neighborhood
Depending on where you live in this neighborhood, your children may attend these schools or other schools outside the neighborhood. Always check with your local school department to determine which schools your children may attend based on your specific address and your child's grade-level.
District Quality Compared to California
Better than of CA school districts.
Schools In This District
Quality Rating Compared to CA*
Quality Rating Compared to Nation*
Francis Hopkinson Elementary School
12582 Kensington Rd.
KG-05
J. H. Mcgaugh Elementary School
1698 Bolsa Ave.
Jack L. Weaver Elementary School
11872 Wembley Rd.
Los Alamitos Elementary School
10862 Bloomfield St.
Los Alamitos High School
3591 Cerritos Ave.
Oak Middle School
Richard Henry Lee Elementary School
11481 Foster Rd.
Rossmoor Elementary School
3272 Shakespeare Dr.
Sharon Christa Mcauliffe Middle School
Los Angeles-Long Beach-Anaheim, CA Metro Area regional investment potential
Housing Affordability Trends: Los Angeles-Long Beach-Anaheim, CA Metro Area
Popular real estate near Rossmoor
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84% Match ‐ Seal Beach, CA (Lampson Ave / Seal Beach Blvd)
81% Match ‐ Huntington Beach, CA (Brookhurst St / Atlanta Ave)
80% Match ‐ Santa Ana, CA (Hewes Ave / Foothill Blvd)
80% Match ‐ Torrance, CA (Torrance Blvd / Palos Verdes Blvd)
79% Match ‐ Yorba Linda, CA (Esperanza Rd / Yorba Linda Blvd)


Come on, who doesn't love waffles? This place is bound to covert those anti-waffle Joes and Janes.
I noticed there were fewer items on the regular menu than my first visit last summer, yet I recognized a few mainstays like the Maple Bacon, Milk and Honey, and Chunky Monkey.
My eyes darted to the Caramel Apple Waffle ($6). It has apples in it, so that should automatically make it a pretty healthy option, right? OK fine, it's sauteed spiced apples, but that's still awesome. Let's see...peanuts and peanut butter are nice for unsaturated fats and protein! Gotta fill this stomach somehow!
Oh, the whole caramel sauce deal and the pastry cream portion of the description? Um...well, yes, that could rock me to sleep, but it's all good! We want flavor! Sweet baby Jesus flavor!
And a sweet baby Jesus of a presentation. Will I go to hell if I eat this? I can go to one of those confession booths later, if need be.
And for eating the waffle, God nuked thy gluttonous panda with severe food coma.
It might be the most decadent dish I've had in a while. The abundance of peanut butter and Spanish peanuts attacked my stomach with multiple textures, attempting to score a premature surrender. The apples were this warm, heavenly gift countering the heavy peanut aspects. I wanted to get on my knees and start praising the Lord. The waffle had a firm, crisp outside and a soft interior. My belly wanted to wave the white flag near the end, but my eyes and heart refused. In layman's terms? It's a meal of a waffle and it's holy delicious.
Next time, I might consider a less filling waffle. The peanut butter and Spanish peanuts became an anchor as the minutes rolled by. I might have won the battle against the Caramel Apple Waffle, but completely lost the war. I was immobile and listless back home, forgoing any dinner. On the bright side, Charlene saved me money! Thanks!
You will also see frequent specials at this cart. Charlene's even done specials with Southern influences, such as a waffle with sausage gravy and a waffle topped with pulled pork. Yes, it's like I torture myself as I write this review.
How about this for a waffle special?!
Biscuit waffle and sausage gravy? Yes!
Head to Gaufre Gourmet in downtown Portland and get one of Charlene's enlightening waffles!


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Crazy Crow Trading Post offers this list of general crafts, craft supply, blade and outdoor living Internet Resources to help you in your search for information about these and related topics. Inclusion in this list does not represent an endorsement by Crazy Crow, although we do try to be selective – and reserve the right to do so.
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Here's a story from today's Washington Post about Michael Hiltzik, a Pulitzer Prize winning columnist and blogger for the LA Times. Howard Kurtz writes that Hiltzik's blog was suspended when editors discovered that he had been posting comments to other blogs using fake names.
Hiltzik has apparently been sniping at conservative bloggers like Hugh Hewitt anonymously, and the LAT says that's a violation of its policy for reporters and editors to "identify themselves when dealing with the public." They're investigating the matter.
"This is amusing, because most of the comments posted on his websiteare anonymous or pseudonymous. . . . Anonymity for commenters is afeature of his blog, as it is of mine. It's a feature that he canwithdraw from his public any time he wishes. He has chosen to do thatin one case only, and we might properly ask why. The answer is thathe's ticked off that someone would disagree with him."
Well, maybe. But my gut reaction to this is that he shouldn't have been commenting anonymously. Anonymous commenters get to side-step responsibility for their remarks, and generally tend to dillute the conversation. People feel much freer to unleash their spiteful sides anonymously, and that's not really productive. Look no further than this thread about that Explore New England blogger plagiarist for evidence. I say responsible bloggers and journalists need to set high standards for themselves if they want to be taken seriously as sources of reliable information.


text:
Home / Locations / United States / Texas / Sugar Land
Local Self Storage Units in Sugar Land, TX
Find Self Storage Near Sugar Land
About Sugar Land
Sugar Land is a thriving city just 20 miles from downtown Houston, home to a rapidly-growing population that has been drawn by a booming job market, plentiful amenities and upscale residential communities. The city offers a quieter, more suburban lifestyle within easy reach of the busy city of Houston, as well as plenty of local shopping, dining and entertainment.
The History of Sugar Land
Sugar Land's unusual name is derived from its history as a sugarcane plantation and major hub of the American sugar refining industry. Colonel E. H. Cunningham bought the large plantation following the previous owners' deaths, shortly after the Civil War. He constructed a sugar refining plant and began to establish a community surrounding it in 1879. In the early 1900s, the plantation changed hands once again, and the sugar refinery was renamed Imperial Sugar Co.
Until the late 1950s, Sugar Land was primarily a company town, with residences constructed for the purpose of housing sugar refinery workers. Imperial Sugar Co. worked to expand the town in the mid-20th century, establishing a new, waterfront subdivision. The next few decades saw the construction of several master-planned communities, including the town's first country club neighborhood. Throughout the 1990s, Sugar Land grew very rapidly. It welcomed many highly-educated residents who worked in energy-industry jobs in nearby Houston, and also experienced major commercial and retail growth.
In recent years, annexations and ongoing growth have continued to expand Sugar Land's population and boundaries.
Living in Sugar Land
Sugar Land has been honored with a number of national and state awards, many of which recognize the quality of life it provides its residents. It has been named one of the best places to live in Texas, as well as one of the most scenic. Local events like KidsFish and the Eggstravaganza have also earned attention from several state organizations.
As a very family-oriented community, Sugar Land offers a wide range of amenities that include youth sports leagues, community parks, recreational facilities and well-rated schools. Many of the city's master-planned communities provide residential benefits like golf courses, swimming pools, country clubs, community centers and neighborhood events.
You'll also find plenty of shopping and dining options throughout the city. Sugar Land Town Square is a popular local destination, home to many restaurants and cafes, shops and a large public plaza. The First Colony Mall is another Sugar Land highlight, with more than 130 stores and many well-known retailers.
Sugar Land Schools
Students in Sugar Land attend Fort Bend Independent School District and Lamar Consolidated Independent School District. The schools in the city are very highly-rated and provide a number of special academic programs for students. You also have the option of enrolling your child in one of the many private and charter schools in Sugar Land, including parochial, secular and college-prep campuses.
Resources for Moving to Sugar Land
If you're preparing for a move to Sugar Land, the following information will be useful:
Utilities: Sugar Land receives its water and sewer services from the city's utility department. Like other Texas cities, Sugar Land doesn't have a designated energy provider. Instead, you have the freedom to choose your own company.
Garbage and Recycling: Trash and recycling are both collected on a regular schedule that's determined by your home's specific location.
Transportation: Public transportation options in Sugar Land are very limited, but you can utilize a commuter service line between Sugar Land and Houston.
Sugar Land Housing
Homes in Sugar Land tend to be priced higher than the national average, and are also on the high end of the Texas state averages. The city's housing market is very large, however, so you have a wide range of options. Spacious family homes on generously-sized lots are the norm throughout Sugar Land, but there are also more budget-friendly options, including town houses and older homes.

redpajama_set_name: RedPajamaCommonCrawl


text:
\section{Introduction}
The study of additive structures in certain sets of integers has a long history.
In particular, the description of arithmetic progressions has attracted the attention of many researchers
and is still an actively studied topic with a
vast literature. Without trying to be exhaustive and going into details we only
mention a few interesting directions.

Bremner \cite{bremner} found infinitely many elliptic curves such that among the
first coordinates of rational points one can find an eight-term arithmetic
progression. Campbell \cite{campbell}, modifying ideas of Bremner, gave the
first example of a progression of length twelve which was later improved to
fourteen by MacLeod \cite{macleod}. For more details on the various
generalizations to other families of curves and recent progress we refer to the
paper of Ciss and Moody \cite{cissmoody} and the references given there.

The study of arithmetic progressions in solution sets of Pellian equations is another interesting direction.
Dujella, Peth\H{o} and Tadi\'c \cite{dujellapethotadic} showed that for every four term arithmetic progression $(y_1,y_2,y_3,y_4)\neq (0,1,2,3)$ there exist infinitely many pairs $(d,m)$ such that the equation $x^2-dy^2=m$ admits solutions with $y=y_i$ for $i=1,2,3,4$.
On the other hand, Peth\H{o} and Ziegler \cite{pethoziegler} proved that for five-term
arithmetic progressions only a finite number of such equations are possible. For
further progress we refer to the papers of Aguirre, Dujella and Peral
\cite{aguirredujellaperal} and Gonz\'alez-Jim\'enez \cite{gonzalezjimenez}.

In this paper, we connect to the field by considering arithmetic progressions in
Lucas sequences. Let $A$ and $B$ be non-zero integers such that the roots
$\alpha$ and $\beta$ of the polynomial
\begin{equation}
\label{charpoly}
x^2-Ax-B
\end{equation}
satisfy that $\alpha/\beta$ is not a root of unity. Define the sequences
$u=(u_n)_{n=0}^\infty$ and $v=(v_n)_{n=0}^\infty$ via the binary recurrence
relations
\begin{equation}
\label{recurrence}
u_{n+2}=Au_{n+1}+Bu_n\qquad \mathrm{and} \qquad v_{n+2}=Av_{n+1}+Bv_n \qquad (n\geq 0)
\end{equation}
with initial values $u_0=0,\ u_1=1$ and $v_0=2,\ v_1=A$. We call $u$ and $v$ the
Lucas sequences of first and second kind corresponding to the pair $(A,B)$,
respectively.

\begin{remark}
The classical definition of $u$ and $v$ requires $\gcd(A,B)=1$ which is needed for
certain $p$-adic properties to hold. Hence our definition is more general.
\end{remark}

\begin{remark}
The assumption on the roots of the companion polynomial \eqref{charpoly}
coincides with the non-degeneracy property. The study of all linear recurrences
can effectively be reduced to such sequences, see
\cite{everestpoortenshparlinskiward}.
\end{remark}

The investigation of arithmetic progressions in linear recurrences is not a completely new topic.
Pint\'er and Ziegler \cite{pinterziegler} gave a criterion
for linear recurrences of general order to contain infinitely many three term
arithmetic progressions. Note that the study of three term arithmetic
progressions is the first non-trivial problem and is the most general.
Further, they proved that the
sequence of Fibonacci and Jacobsthal numbers are the only increasing Lucas
sequences having infinitely many three term progressions.

In this paper, we connect to the results of Pint\'er and Ziegler. On one hand,
we give an upper bound for the number of three term arithmetic progressions provided
that there are only finitely many. On the other hand, we explicitly list all those sequences which
contain infinitely many arithmetic progressions. Finally, by restricting
ourselves to sequences with companion polynomials having a dominant root, we
find all sequences which contain a three term arithmetic progression and give a
complete list of the occurrences.

Our main result is as follows:

\begin{theorem}
\label{theorem1}
Let $u=(u_n)_{n=0}^\infty$ and $v=(v_n)_{n=0}^\infty$ be the Lucas sequences of
first and second kind, respectively, corresponding to the pair $(A,B)$. Then $u$
and $v$ admit at most $6.45\cdot 10^{2340}$ triples $(k,l,m)$ such that $u_k<u_l<u_m$ is
a three term arithmetic progression,
except when $u$ corresponds to $(A,B)=(\pm 1,1),(\pm 1,2),(-1,-2)$ and $v$
corresponds to $(A,B)=(\pm1, 1),(-1,\pm 2)$, in which cases they admit
infinitely many.
\end{theorem}

\begin{remark}
We expect that the upper bound in Theorem \ref{theorem1} is very far from being sharp.
It would be an interesting problem to study the average number of three term arithmetic progressions in Lucas sequences of the first and second kind.
\end{remark}

By imposing stronger restrictions on the sequences we can state
much more. Namely, if the characteristic polynomial \eqref{charpoly} has a
dominant root, then it is possible to explicitly list all three term arithmetic
progressions in the corresponding sequence.

\begin{theorem}
\label{theorem2}
Let $u=(u_n)_{n=0}^\infty$ and $v=(v_n)_{n=0}^\infty$ be the Lucas sequences of
first and second kind, respectively, corresponding to the pair $(A,B)$.
Assume that $A^2+4B>0$. Then $u$ and $v$ admit no three term
arithmetic progression, except the cases listed in Tables \ref{table1} and
\ref{table2}, respectively.
\end{theorem}

\begin{table}[ht]
\centering
\begin{tabular}{|c|c|}
\hline
$(A,B)$ & $(k,l,m)$\\
\hline
$(1,1)$ & $(0,1,3),\ (2,3,4),\ (t,t+2,t+3),\ t\geq 0$\\
\hline
$(-1,1)$ & $(1,0,2),\ (t,t+1,t+3),\ t\geq 0$\\
\hline
$(1,2)$ & $(1,2t+1,2t+2),\ (2,2t+1,2t+2),\ t\geq 1$\\
\hline
$(-1,2)$ & $(t+2,t,t+1),\ t\geq 0$\\
\hline
$(2,B),\ B\geq 1$ & $(0,1,2)$\\
\hline
$(1,B),\ B\geq 3$ & $(1,3,4),\ (2,3,4)$\\
\hline
$(-1,B),\ B\geq 3$ & $(1,0,2)$\\
\hline
\end{tabular}
\caption{Triples $(k,l,m)$ s.t. $(u_k,u_l,u_m)$ is a three term AP.}
\label{table1}
\end{table}

\begin{table}[ht]
\centering
\begin{tabular}{|c|c|}
\hline
$(A,B)$ & $(k,l,m)$\\
\hline
$(1,1)$ & $(1,0,2),\ (t,t+2,t+3),\ t\geq 0$\\
\hline
$(-1,1)$ & $(t,t+1,t+3),\ t\geq 0$\\
\hline
$(-1,2)$ & $(t,t-1,t+1),\ t\geq 1$\\
\hline
$(-2,1)$ & $(1,0,2)$\\
\hline
$(1,3)$ & $(1,4,5)$\\
\hline
$(-3,-1)$ & $(1,0,2)$\\
\hline
\end{tabular}
\caption{Triples $(k,l,m)$ s.t. $(v_k,v_l,v_m)$ is a three term AP.}
\label{table2}
\end{table}

As an immediate consequence of Theorem \ref{theorem2} we have the following:

\begin{corollary}
\label{cor}
Let $u=(u_n)_{n=0}^\infty$ (resp. $v=(v_n)_{n=0}^\infty$) be the Lucas sequence
of first (resp. second) kind corresponding to the pair $(A,B)$. Assume that
$A^2+4B>0$. If $u$ (resp. $v$) contains more than two (resp. one)
three term arithmetic progressions, then $u$ (resp. $v$) contains infinitely
many three term arithmetic progressions.
\end{corollary}

The ineffective methods we use in the proof of Theorem \ref{theorem1} do not
give such sharp bounds as in Corollary \ref{cor}. However, it would be
interesting to find optimal bounds also in the general case, i.e., if we drop the
condition $A^2+4B>0$ in Corollary \ref{cor}.

In the next section, we prove Theorem \ref{theorem2}. The main idea is to show
that under a testable condition, the growth of the sequence would contradict the
existence of any arithmetic progression. Further, this condition leaves only
finitely many explicitly given possibilities for $(A,B)$ which we treat one by
one, usually in an
elementary way. The proof of Theorem \ref{theorem1} in Section \ref{Sec:imag} is
based on the theory of $S$-unit equations.

\section{The dominant root case}\label{Sec:real}

This section is devoted to the case, where $A^2+4B>0$, that is, the roots of the
companion polynomial are real. First, we remark that in any linear recurrence,
the terms can be represented as polynomial-exponential sums depending on the
roots of the companion polynomial. In our situation, this leads to the
well-known formulas
\[
u_n=\dfrac{\alpha^n-\beta^n}{\alpha-\beta} \qquad \mathrm{and} \qquad
v_n=\alpha^n+\beta^n \qquad (n\geq 0).
\]
We will regularly use this definition of Lucas sequences
instead of the recurrence relation \eqref{recurrence}.

The following lemma concerns equations satisfied by three term arithmetic
progressions.

\begin{lemma}
\label{lemmatech}
Let $u=(u_n)_{n=0}^\infty$ be any sequence. Suppose that for some $n_0$ we
have $|u_n/u_{n'}|>3$ for every $n>n_0$ and every $n'<n$. If $m>n_0$, then
neither of the
equations
\begin{equation}
\label{equation:klm}
u_k-2u_l+u_m=0, \qquad u_l-2u_k+u_m=0,\qquad u_k-2u_m+u_l=0
\end{equation}
has a solution $(k,l,m)$ with $k<l<m$. In particular, $u$ does not admit any
three term arithmetic progression with terms corresponding to the indices $k,l$
and $m$ with $\max\{k,l,m\}>n_0$.
\end{lemma}

\begin{proof}
Assuming that $n_0<m$ gives us either $3|u_k|<3|u_l|<|u_m|$ or
$3|u_l|<3|u_k|<|u_m|$. In the former case, we have
\[
|u_k-2u_l+u_m|\geq |u_m|-|u_k-2u_l|\geq |u_m|-(|u_k|+2|u_l|)>3|u_l|-3|u_l|=0.
\]
Thus the first equation of \eqref{equation:klm} has no solution with $k<l<m$.
In an analogous way, one can get the same deduction for $3|u_l|<3|u_k|<|u_m|$
and it is easy to see that the second and third equations of
\eqref{equation:klm} are unsolvable as well.
\end{proof}

We are in need of a good criterion for the applicability of Lemma
\ref{lemmatech}.

\begin{lemma}
\label{lem:growth_Lucas}
Let $u=(u_n)_{n=0}^\infty$ be the Lucas sequence of first kind corresponding to
the pair
$(A,B)$ with $A^2+4B>0$. Assume that $n>n'$. Then $|u_n/u_{n'}|>3$ if $n$ is odd
or $n\geq 8$, unless
\begin{itemize}
 \item $B<0$ and $|A|\leq 6$ or
 \item $|A|=1$ and $0<B\leq 9$ or
 \item $|A|=2$ and $0<B\leq 3$.
\end{itemize}
\end{lemma}

\begin{proof}
Under our conditions the polynomial $X^2-AX-B$ has a
dominant root, say $\alpha$. Note that if $|A|>6$ or $|B|>9$, then we have
$|\alpha|>3$ and may write
\[
\left|\dfrac{u_n}{u_{n-1}}\right|=\left|\dfrac{\dfrac{\alpha^n-\beta^n}{
\alpha-\beta}}{\dfrac{\alpha^{n-1}-\beta^{n-1}}{\alpha-\beta}}
\right|=\left|\dfrac{\alpha^n}{\alpha^{n-1}}\dfrac{1-\left(\frac{\beta}{\alpha}
\right)^{n}}{1-\left(\frac{\beta}{\alpha}\right)^{n-1}}
\right|=|\alpha|\left|\dfrac{1-\theta^n}{1-\theta^{n-1}}\right|>3\left|\dfrac{
1-\theta^n}{1-\theta^{n-1}}\right|
\]
with $\theta=\beta/\alpha$. It suffices to prove that
\[
\left|\dfrac{
1-\theta^n}{1-\theta^{n-1}}\right|>1.
\]
Observe that there are trivial cases. These include $\theta>0$ and also
$\theta<0$ when $n$ is odd. What remains to consider is the case when $\theta<0$
and $n$ is even.

Next, we show that the quotients $|u_{2n+2}/u_{2n+1}|$ increase with
$n$, that is, they satisfy the inequality
\[
\left|\dfrac{u_{2n+2}}{u_{2n+1}}\right|>\left|\dfrac{u_{2n}}{u_{2n-1}}\right|.
\]
Equivalently, we may write
\[
|\alpha|\left|\dfrac{1-\theta^{2n+2}}{1-\theta^{2n+1}}
\right|>|\alpha|\left|\dfrac{1-\theta^{2n}}{1-\theta^{2n-1}}\right|.
\]
Since $0<|\theta|<1$, it reduces to
\[
\dfrac{1-\theta^{2n+2}}{1-\theta^{2n+1}}>\dfrac{1-\theta^{2n}}{1-\theta^{2n-1}}
\]
giving us
\[
1-\theta^{2n-1}-\theta^{2n+2}+\theta^{4n+1}>1-\theta^{2n+1}-\theta^{2n}+\theta^{
4n+1}.
\]
Collecting the terms on the left hand side and dividing both sides of the
inequality by $-\theta^{2n-1}>0$ we obtain
\[
\theta^3-\theta^2-\theta+1=(\theta-1)^2(\theta+1)>0,
\]
which is obviously true for any $\theta$ with $0<|\theta|<1$. Since
$|u_2/u_1|=|A|\geq 1$, we have, in particular, that $|u_{n}/u_{n-1}|>1$ for all
$n\geq 0$. This proves the lemma for all odd $n$ and all $n>2n_0$ with
$|u_{2n_0}/u_{2n_0-1}|>3$. Thus we need to show that $|u_{8}/u_{7}|>3$ under our
hypotheses.

We distinguish between the cases where $B>0$ and $B<0$. Let us start with $B<0$.
We have
\[
|\theta|=\left|\frac{\beta}{\alpha}\right|=\frac{|A|-\sqrt{A^2+4B}}{|A|+\sqrt{
A^2+4B}}\leq \frac{|A|-1}{|A|+1}=1-\frac{2}{|A|+1}.
\]
Note that in this case we also have that $|\alpha|>\frac{|A|+1}2$. Write $x=\frac{|A|+1}2$.
We show that
\[
\left|\dfrac{u_{8}}{u_{7}}\right|\geq x \frac{1-\left(1-\dfrac
1x\right)^{8}}{1+\left(1-\dfrac 1x\right)^{7}}>3.
\]
Solving this inequality for $x$ shows that the second inequality holds unless
$x<3.55$, i.e., $|A|<6.1$.

When $B>0$, we have
\[
|\theta|=\left|\frac{\beta}{\alpha}\right|=\frac{\sqrt{A^2+4B}-|A|}{\sqrt{A^2+4B
}+|A|}=1-\frac{2|A|}{\sqrt{A^2+4B}+|A|}
\]
and with $|A|\geq 3$ we obtain
\[
|\theta|\geq 1+\frac{6}{\sqrt{9+4B}+3}
\]
and
\[
|\alpha|>\dfrac{\sqrt{9+4B}+3}2=:y.
\]
We need to show that
\[
\left|\dfrac{u_{8}}{u_{7}}\right|\geq y \frac{1-\left(1-\dfrac
3y\right)^{8}}{1+\left(1-\dfrac 3y\right)^{7}}>3.
\]
The inequality holds unless $y\leq 3$. However, $y\leq 3$ is impossible, since
it would imply $B\leq 0$ which we excluded.

Similar arguments for the cases $|A|=2$ and $|A|=1$ yield that
$\left|\dfrac{u_{8}}{u_{7}}\right|>3$ provided that $B>9$ and $B>3$,
respectively.
\end{proof}

A similar argument for Lucas sequences of second kind reveals the following.

\begin{lemma}
\label{lem:growth_comp}
 Let $v=(v_n)_{n=0}^\infty$ be the Lucas sequence of second kind corresponding to the pair
$(A,B)$, with $A^2+4B>0$. Assume that $n>n'$. Then $|u_n/u_{n'}|>3$ if $n$ is
even or $n\geq 7$, unless
\begin{itemize}
 \item $B<0$ and $|A|\leq 7$ or
 \item $|A|=1$ and $0<B\leq 14$ or
 \item $|A|=2$ and $0<B\leq 3$.
\end{itemize}
\end{lemma}

\begin{proof}
 Since the proof of this Lemma is similar to that of Lemma
\ref{lem:growth_Lucas}, we omit most of the details. Note that when $|A|=1$ we
have $|v_1/v_0|=1/2<1$ and we can only deduce $|v_n/v_{n-1}|\geq 1$ provided
that $n\geq 2$. Thus we shall ensure $|v_7|\geq 6$. Direct computation yields
that indeed
\[
|v_7|=7B^3+14B^2+7B+1
\]
when $|A|=1$. Since $|A|=1$ implies that $B\geq 1$, we get $|v_7|>6$.
\end{proof}

Now we apply Lemma \ref{lemmatech} in combination with Lemma
\ref{lem:growth_Lucas} and find that if $u$ admits a three term
arithmetic progression, then one of the equations in \eqref{equation:klm} must
have a solution with even $m\leq 6$. Writing the terms of $u$ as polynomials in
$A$ and $B$ we have to deal with finitely many equations. We only present the main
ideas through one example for each different type of possible equations, we treat the rest in a similar manner.

\textbf{Case $m=2$.} Since $u_0=0,\ u_1=1$ and $u_2=A$, we only get trivial
equations like $A-2=0$.

\textbf{Case $m=4$.} The corresponding equations are linear in $B$. For
example, the triple $(k,l,m)=(1,2,4)$ substituted into the second equation of
\eqref{equation:klm} gives us
\[
A^3+2AB+A-2=0
\]
implying
\[
B=\dfrac{-A^3-A+2}{2A}.
\]
Hence $A\mid 2$, i.e., $A=\pm1,\pm2$. However,
we see that one of the conditions $AB\neq 0$ and $A^2+4B>0$ fails in all cases. One can handle the other equations similarly.

\textbf{Case $m=6$.} The corresponding equations are quadratic in $B$. Our
general strategy is to push the discriminant of each equation between two squares and solve the
parametric equation for each square between these lower and upper bounds. For
example, the triple $(k,l,m)=(0,3,6)$ and the third equation of
\eqref{equation:klm} gives
\[
-6AB^2+(-8A^3+1)B-2A^5+A^2=0.
\]
The discriminant is $D=16A^6+8A^3+1=(4A^3+1)^2$. However, solving this for $B$
we get the pair $(A,B)=(A,-A^2)$ and $A^2+4B>0$ fails to hold.

Let us consider another example. For instance, take the triple
$(k,l,m)=(0,2,6)$. Then we obtain $D=A^6+6A^2-3A$. But
$$A^6<A^6+6A^2-3A<(|A|^3+1)^2$$ provided that $|A|>3$ and we deduce that $D$
is never a square if $|A|>3$.
For those with $|A|\leq 3$ and $A\neq 0$ we obtain that $D$ is a square if
and only if $A=1$. In this case, $B=0$ which contradicts our assumption.

The same approach for each triple gives the complete list of three term arithmetic
progressions.

We are left with the finite number of sequences satisfying
\begin{itemize}
 \item $B<0$ and $|A|\leq 6$ or
 \item $|A|=1$ and $0<B\leq 9$ or
 \item $|A|=2$ and $0<B\leq 3$.
\end{itemize}
In these finitely many cases, we can use a growth argument to find
all three term arithmetic progressions.

Let us demonstrate this by an example, say $A=2$ and $B=1$ and $(u_n)_{n=0}^\infty$ being a Lucas sequence of first kind. One may handle all the other finitely many equations by a similar reasoning. First note
that in this specific case we have that $\alpha=1+\sqrt{2}$ and
$\beta=1-\sqrt{2}$. Assume that $u_k<u_l<u_m$ is a three term arithmetic
progression and consider the first equation of \eqref{equation:klm} which is
equivalent to
\[
\alpha^k+\alpha^m-2\alpha^l=\beta^k+\beta^m-2\beta^m.
\]
Let us assume for the moment that $m>l+1>k+1$. Then
\[
|\alpha|^m-2|\alpha|^{m-2}-|\alpha|^{m-3}
<|\alpha^k+\alpha^m-2\alpha^l|=|\beta^k+\beta^m-2\beta^m|<4|\beta|^k.
\]
Plugging in the concrete values of $\alpha$ and $\beta$ we immediately get
\[
(1+\sqrt{2})^{m-3}\left((1+\sqrt{2})^3-2(1+\sqrt{2})-1\right)=(4+3\sqrt
2)(1+\sqrt{2})^{m-3}<4
\]
which yields a contradiction unless $m<3$. Now it is easy to find all three term
arithmetic progressions in this case. Note that in case of $l>m$ or $k>m$ we may draw similar conclusions.

Hence we can assume that $m=l+1>k+1$. Then
\[
|\alpha|^m-2|\alpha|^{m-1}<|\alpha|^{k}+4|\beta|^k
\]
and inserting the concrete values of $\alpha$ and $\beta$ we get
\[
(1+\sqrt 2)^{m-2}<(1+\sqrt 2)^{k}+4.
\]
Assuming $k\leq m-3$ yields
\[
\sqrt{2}(1+\sqrt 2)^{m-3}<4,
\]
whence $m\leq 4$ and we easily find all three term arithmetic progressions
listed in Table \ref{table1} for this case. Finally, we have to consider the
case when $m=l+1=k+2$. Here we get the inequality
\[
\left|\alpha^m-2\alpha^{m-1}-\alpha^{m-2}\right|<4|\beta|^k
\]
which yields
\[
2(1+\sqrt 2)^{m-2}<4,
\]
that is, $m\leq 3$ and we find no additional three term arithmetic progressions.
Similar arguments for different orderings of $u_k,u_l,u_m$ reveal no further
solutions. Thus the case $A=2$ and $B=1$ is completely solved.

\section{The case $A^2+4B<0$}\label{Sec:imag}

Unfortunately, we found no effective method to resolve this case completely. The
reason has its roots in the use of the theory of $S$-unit equations. Indeed, the
indices $k,l,m$ corresponding to a non-trivial arithmetic progression
$u_k<u_l<u_m$ contained in a Lucas sequence $(u_n)_{n= 0}^\infty$ of first kind or a
Lucas sequence $(v_n)_{n= 0}^\infty$ of second kind satisfies the Diophantine equation
\begin{equation}\label{eq:AP}
\alpha^k+\alpha^m-2\alpha^l=\pm \left(\beta^k+\beta^m-2\beta^l\right),
\end{equation}
where the $\pm$ sign depends on whether we consider Lucas sequences of first or
second kind. In view of Theorem \ref{theorem1}, it is
crucial to discuss the number of solutions to $S$-unit equation
\eqref{eq:AP}.
In the real case (cf. Section \ref{Sec:real}), we resolved this $S$-unit
equation by
using the fact that $|\alpha|>|\beta|$ and that $\left|\dfrac{u_n}{u_{n-1}}\right|\simeq |\alpha|$ as $n\rightarrow \infty$.
If $A^2+4B<0$, then this is no longer true and we have to apply
the deep theory of $S$-unit equations.

In order to determine an upper bound for the number of solutions we prove a
series
of Lemmas which culminate in a proof for Theorem \ref{theorem1}.

Before we start with the first lemma, let us note that both $\alpha$ and $\beta$
are algebraic integers. Moreover, due to Theorem \ref{theorem2}, which was proved in
the previous section, we may assume that $\alpha$ and $\beta$ are conjugate
imaginary quadratic integers. Hence we can suppose that $\alpha$ and
$\beta$ are not units, otherwise $\alpha/\beta$ would be a root of unity. Thus
we may suppose that $|B|\geq 2$. Write $K=\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$ and denote by
$\sigma$ the unique, nontrivial $\mathbb{Q}$-automorphism of $K$, i.e.,
$\mathrm{Gal}(K/\mathbb{Q})=\{\mathrm{id},\sigma\}$. Since $K$ is an imaginary quadratic
field, we also have that $\sigma(\epsilon)=\epsilon^{-1}$, where $\epsilon$ is
any unit in $K$.

For fixed $\alpha$ and $\beta$ we denote by $C_1$ and $C_2$ the number of
non-trivial three term arithmetic progressions of $(u_n)_{n=0}^{\infty}$ and
$(v_n)_{n=0}^{\infty}$, respectively and write $C=\max\{C_1,C_2\}$. Actually,
our upper bounds for $C_1$ and $C_2$ coincide in each instance and hence we use
only the quantity $C$.

Let $\Gamma=\langle\alpha,\beta\rangle\leq \bar\mathbb{Q}^*$ be the
multiplicative group generated by $\alpha$ and $\beta$. We denote by
$A=A(a_0,\dots,a_r)$ the number of non-degenerate, projective solutions
$\lbrack x_0:\dots:x_r \rbrack\in\mathbb{P}^r(\Gamma)$ of the weighted, homogeneous $S$-unit
equation
\begin{equation}\label{eq:S-unit}
 a_0u_0+\dots+a_ru_r=0.
\end{equation}
with $a_0,\dots,a_r\in \mathbb{C}$, in unknowns $u_0,\dots,u_r\in\Gamma$. By a non-degenerate solution we mean a solution to equation \eqref{eq:S-unit} such that no subsum on the left hand side
of equation \eqref{eq:S-unit} vanishes. Upper bounds $A(a_0,\dots,a_r)\leq
A(r,s)$, where $s$ denotes the rank of $\Gamma$, have been found most prominently
by Evertse, Schlickewei and Schmidt \cite{Evertse:2002} and Amoroso and Viada
\cite{Amoroso:2009} in the most general case. In particular, we use the bounds
due to Schlickewei and Schmidt \cite{Schlickewei:2000} which in our case yield better
results although they depend on $d=\lbrack K:\mathbb{Q} \rbrack$, as well.

As already noted above, three term arithmetic progressions in the Lucas
sequences $(u_n)_{n=0}^\infty$ and $(v_n)_{n=0}^\infty$ yield solutions to
\eqref{eq:AP}. Thus we study the weighted $S$-unit equation
\begin{equation}\label{eq:S-unit2}
x_1+x_2-2x_3=\pm(y_1+y_2-2y_3),
\end{equation}
where $x_1,x_2,x_3,y_1,y_2,y_3\in \Gamma=\langle\alpha,\beta\rangle$ and, in
particular, where $x_1=\alpha^k,x_2=\alpha^m,x_3=\alpha^l,
y_1=\beta^k,y_2=\beta^m,y_3=\beta^l$.

Before we start investigating $S$-unit equations of type \eqref{eq:S-unit2} we state a theorem due to Beukers~\cite[Theorem 2]{Beukers:1980}
and draw some simple conclusions from it.

\begin{lemma}[Beukers~\cite{Beukers:1980}]
Let the non-degenerate recurrence sequence of rational integers $(u_n)_{n=0}^\infty$ be given by $u_0\geq 0$, $\gcd(u_0, u_1)=1$,
$u_n = Au_{n-1} - Bu_{n-2}$, with $A, B\in\mathbb{Z}$ and $A\geq 0$. Assume that $A^2 - 4B < 0$. If $u_m =\pm u_0$ has
more than three solutions $m$, then one of the following cases holds:
\begin{align*}
A=1, B=2,& u_0=u_1=1& \text{when\ \ }& \ m=0,1,2,4,12;\\
A=1, B=2,& u_0=1, u_1=-1& \text{when\ \ }& \ m=0,1,3,11;\\
A=3, B=4,& u_0=u_1=1& \text{when\ \ }& \ m=0,1,2,6;\\
A=2, B=3,& u_0=u_1=1& \text{when\ \ }& \ m=0,1,2,5.
\end{align*}
\end{lemma}

From this Lemma we can easily conclude the following statement.

\begin{lemma}\label{lem:multiplicity}
Let $(u_n)_{n=0}^\infty$ be a Lucas sequence of first or second kind. Then for a fixed integer $\lambda$ the equation
$\lambda=u_m$ has at most $3$ solutions $m$.
\end{lemma}

\begin{proof}
Let $m_0$ be the smallest solution to $\lambda=u_m$. Consider instead of the original sequence the sequence $a_k=u_{m_0+k}/g(-1)^{k\delta}$,
where $g=\mathrm{sign}(u_{m_0})\gcd(u_{m_0},u_{m_0+1})$ and $\delta=\frac{1-\mathrm{sign}(A)}2$ (see also the remarks following Theorem 2 in \cite{Beukers:1980}).
Therefore we conclude that an equation of the form  $\lambda=|u_m|$ has more than three solutions if for the smallest solution $m_0$, also $m_0+1$ is a solution and $(A,B)=(\pm 1,-2),(\pm 3,-4)$ or $(\pm 2,-3)$.

However, the equation $|u_{m_0}|=|u_{m_0+1}|$ is equivalent to the equation
$$\frac{\alpha \pm 1}{\beta \pm 1}=\left(\frac \beta \alpha\right)^{m_0}$$
in case that $(u_n)_{n=0}^\infty$ is a Lucas sequence of first kind and
$$\frac{\alpha \pm 1}{\beta \pm 1}=- \left(\frac \beta \alpha\right)^{m_0}$$
in case that $(u_n)_{n=0}^\infty$ is a Lucas sequence of second kind.
For all pairs of $(\alpha,\beta)$ we solve these equations and find for instance, that $m_0\leq 2$ (if a solution exists at all). By computing the
values of all possible further solutions we find that indeed no equation of the form $\lambda=u_m$ has more than three solutions.
\end{proof}

\begin{remark}
Beukers \cite[Corollary on page 267]{Beukers:1980} already showed this result in case of Lucas sequences of first kind. In fact, he proved more and
listed all the cases when there can be three solutions. To the authors knowledge the case of Lucas sequences of second kind
has not been treated since then.
\end{remark}

Let us start by investigating the number of non-degenerate solutions to $S$-unit equation \eqref{eq:AP}:

\begin{lemma}\label{lem:nondeg}
There are at most $A(5,2)$ three term arithmetic progressions contained in a
Lucas sequence $(u_n)_{n=0}^\infty$ of first kind that yield non-degenerate solutions to \eqref{eq:AP}.

The same statement also holds for Lucas sequence $(v_n)_{n=0}^\infty$ of second kind.
\end{lemma}

\begin{proof}
Consider the weighted $S$-unit equation
\begin{equation}\label{eq:S-unit2-special}
x_1+x_2-2x_3=y_1+y_2-2y_3
\end{equation}
with unknowns $x_1,x_2,x_3,y_1,y_2,y_3\in\Gamma$. This has at most
$A(5,2)$ non-degenerate, projective solutions. Thus there exists a set
$\mathcal C$, with $|\mathcal C|\leq A(5,2)$, of sextuples
$(1,c_2,c_3,c_4,c_5,c_6)$ such that for any solution we have
$x_2/x_1=c_2$, $x_3/x_1=c_3$, $y_1/x_1=c_4$, $y_2/x_1=c_5$ and
$y_3/x_1=c_6$. Assume now that a solution comes from an arithmetic
progression $u_k<u_l<u_m$ in a Lucas sequence $(u_n)_{n=0}^\infty$
of first kind. Then we have
\begin{equation}\label{eq:only-one}
\frac{\alpha^k}{\beta^k}=\frac{1}{c_4}, \quad
\frac{\alpha^m}{\beta^m}=\frac{c_2}{c_5},\quad
\frac{\alpha^l}{\beta^l}=\frac{c_3}{c_6}
\ \ \ \text{or}\ \ \ \frac{\beta^k}{\alpha^k}=\frac{1}{c_4}, \quad
\frac{\beta^m}{\alpha^m}=\frac{c_2}{c_5},\quad
\frac{\beta^l}{\alpha^l}=\frac{c_3}{c_6}
\end{equation}
for some $(1,c_2,c_3,c_4,c_5,c_6)\in{\mathcal C}$. Since by assumption
$\alpha/\beta$ is not a root of unity, for every sextuple
$(1,c_2,c_3,c_4,c_5,c_6)\in\mathcal C$, there exists at most one triple
$(k,l,m)$ such that any set of identities of \eqref{eq:only-one} is
satisfied, i.e., $u_k<u_l<u_m$ is an arithmetic progression. Hence there
exist at most $A(5,2)$ three term arithmetic progressions in a Lucas
sequence of first kind $(u_n)_{n=0}^\infty$, yielding a
non-degenerate solution to \eqref{eq:AP}.

In the case of a Lucas sequence of second kind $(v_n)_{n=0}^\infty$,
we consider instead of \eqref{eq:S-unit2-special} the Diophantine
equation
\[
x_1+x_2-2x_3=-(y_1+y_2-2y_3)
\]
and derive the same conclusion as before.
\end{proof}

Before we start dealing with vanishing subsums in equation \eqref{eq:AP} we note
that $\alpha$ and $\beta$ are always multiplicatively independent. Otherwise, we
would have $\alpha^t=\beta^s$ for some integers $t,s$. Taking the conjugates we
get $\alpha^s=\beta^t$ and hence $(\alpha/\beta)^{t+s}=1$ which we did not allow
in the definition.

To deal with vanishing two-term subsums we use the following lemma.

\begin{lemma}\label{lem:mult-dep-2}
For fixed non-zero complex numbers $a,b$ and $c$ either there exists at most one solution
$x,y\in\mathbb{Z}$ to $a^x=cb^y$ or $a$ and $b$ are
multiplicatively dependent.
\end{lemma}

\begin{proof}
Assume that two distinct solutions $x,y$ and $x',y'$ exist. Then we have
\[
\frac{a^x}{b^y}=c=\frac{a^{x'}}{b^{y'}},
\]
hence
\[
a^{x-x'}=b^{y-y'}.
\]
Thus $a$ and $b$ are multiplicatively dependent since by
assumption at least one exponent does not vanish.
\end{proof}

In view of Lemma \ref{lem:nondeg}, we are left to deal with vanishing subsums,
which may occur in equation~\eqref{eq:AP}. Of course, no one-term vanishing
subsum exists. First, consider two-term vanishing subsums.

\begin{lemma}\label{lem:two-vanish}
There are at most $3A(3,2)+30$  three term arithmetic progressions in a Lucas
sequence $(u_n)_{n=0}^\infty$ of first kind or a Lucas sequence
$(v_n)_{n=0}^\infty$ of second kind that yield a solution to \eqref{eq:AP} such
that a two-term subsum vanishes.
\end{lemma}

\begin{proof}
First, we consider the Lucas sequences $(u_n)_{n=0}^\infty$ of first kind. In this case, there
exist exactly $15$ possible vanishing two-term subsums in equation
\eqref{eq:AP}. We can divide these $15$ subsums into five classes, namely
\begin{description}
  \item[Case I] $x_1=-x_2$, $y_1=-y_2$,
  \item[Case II] $x_1=y_2$, $x_2=y_1$,
  \item[Case III] $x_1=2x_3$, $x_2=2x_3$, $y_1=2y_3$, $y_2=2y_3$,
  \item[Case IV] $x_1=-2y_3$, $x_2=-2y_3$, $y_1=-2x_3$, $y_2=-2x_3$.
  \item[Case V] $x_1=y_1$, $x_2=y_2$, $x_3=y_3$,
\end{description}

For each one we pick an equation and discuss it in detail. Since one can treat the other
equations in the same class by exactly the same arguments, we do
not give details for them.

\noindent\textbf{Case I:} The equation $x_1=-x_2$ implies that
$\alpha^k=-\alpha^m$ and we get $\alpha^{k-m}=-1$. Thus $\alpha$ and hence
$\beta$ are roots of unity, which is excluded.

\noindent\textbf{Case II:} The equation $x_1=y_2$ implies that
$\alpha^k=\beta^m$, i.e., $\alpha$ and $\beta$ are multiplicatively dependent,
which cannot be the case.

\noindent\textbf{Case III:} The equation $x_1=2x_3$ implies that
$\alpha^k=2\alpha^l$, i.e., $\alpha^{k-l}=2$. Thus $\alpha=\pm \sqrt 2$ or
$\alpha=2$, both a contradiction to our assumption that $\alpha$ is imaginary
quadratic.

\noindent\textbf{Case IV:} If $x_1=-2y_3$ we get the equation
$\alpha^k=-2\beta^l$ and an by an application of Lemma \ref{lem:mult-dep-2}
there exists at most one pair $(k,l)$ satisfying $\alpha^k=-2\beta^l$ or
$\alpha$ and $\beta$ are multiplicatively dependent. The latter is not
possible. However, if the pair $(k,l)$ is fixed then $u_l$ and $u_k$ and therefore
also $u_m$ is fixed. Due to Lemma \ref{lem:multiplicity} there are at most three possiblities for $m$,
i.e., there are at most three ways to extend the pair $(k,l)$ to a triple $(k,l,m)$ such that
$u_k<u_l<u_m$ is an arithmetic progression.

\noindent\textbf{Case V:} If $x_1=y_1$, then we have $\alpha^k=\beta^k$ and
therefore $(\alpha/\beta)^k=1$. Hence $\alpha/\beta$ is a
root of unity unless $k=0$.

So far we have proved that one of the following statements holds:
\begin{itemize}
\item we obtain at most $12$ additional solutions coming from Case IV,
\item $klm=0$.
\end{itemize}

Concerning the last case we assume that $k=0$ first. We get the $S$-unit
equation
\begin{equation}\label{eq:AP-k=0}
x_2-2x_3=y_2-2y_3.
\end{equation}
By the same arguments as in the proof of Lemma \ref{lem:nondeg} we obtain that
there are at most $A(3,2)$ three term arithmetic progressions which come from
non-degenerate solutions of \eqref{eq:AP-k=0}. Further, we have two vanishing
two-term subsums falling into Case III, two falling into Case IV and two
into Case V. Case III yields no additional solutions. Case IV yields at most $3$
additional solution, hence in total at most six additional solutions. Finally,
note that in Case V we can now exclude that $l$ or $m$ vanishes, otherwise we
would obtain $k=l$ or $k=m$ and we get no additional solution. Thus $k=0$ yields
at most $A(3,2)+6$ additional solutions.

By the same arguments it is possible to treat the cases of $l=0$ and $m=0$. We
omit the details. However, the case that $klm=0$ yields at most $3A(3,2)+18$
additional solutions. Therefore we have at most $3A(3,2)+30$ solutions.
\end{proof}

Let us note that the proof of Lemma \ref{lem:two-vanish}, in particular its last
part, shows the following.

\begin{lemma}\label{lem:kmn=0}
There are at most $A(3,2)+6$ three term arithmetic progressions $u_k<u_l<u_m$ in
a Lucas sequence of first or second kind such that $k=0$. The same statements hold
if we replace $k=0$ by $m=0$ or $l=0$, respectively.
\end{lemma}

Since a vanishing four- or five-term subsum implies a two- or one-term vanishing
subsum, respectively, we are left to consider vanishing three term subsums.

\begin{lemma}\label{lem:three-vanish}
At least one of the following statements holds:
\begin{itemize}
\item There are at most $18A(2,2)+9$ three term arithmetic progressions in a
Lucas sequence of first or second kind that yield a solution to \eqref{eq:AP}
such that a three term subsum vanishes,
\item or $\alpha$ and $\beta$ are quadratic irrational numbers and if
$u_k<u_l<u_m$ is an arithmetic progression, then the companion polynomial of the
Lucas sequence $(u_n)_{n=0}^\infty$ of first kind divides $X^k-2X^l+X^m$,
\item or $\alpha$ and $\beta$ are quadratic irrational numbers and if
$v_k<v_l<v_m$ is an arithmetic progression, then the companion polynomial of the
Lucas sequence $(v_n)_{n=0}^\infty$ of second kind divides $X^k-2X^l+X^m$.
\end{itemize}
\end{lemma}

\begin{proof}
Let us consider the case of Lucas sequences of first kind. Lucas sequences of second kind can
be treated by the same means.

There are $20$ possible, different vanishing three term subsums of
\eqref{eq:AP}. These appear in pairs and we can differentiate three different
types:
 \begin{align*}
  I  &:& x_1+x_2-2x_3&=0,& 0&=y_1+y_2-2y_3,\\
  II &:& x_1+x_2&=y_1,& -2x_3&=y_2-2y_3,\\
  II &:& x_1+x_2&=y_2,& -2x_3&=y_1-2y_3,\\
  III&:& x_1+x_2&=-2y_3,& -2x_3&=y_1+y_2,\\
  II &:& x_1-2x_3&=y_1,& x_2&=y_2-2y_3,\\
  III&:& x_1-2x_3&=y_2,& x_2&=y_1-2y_3,\\
  II &:& x_1-2x_3&=-2y_3,& x_2&=y_1+y_2,\\
  II &:& x_1&=y_1+y_2,& x_2-2x_3&=-2y_3,\\
  II &:& x_1&=y_1-2y_3,& x_2-2x_3&=y_2,\\
  III&:& x_1&=y_2-2y_3,& x_2-2x_3&=y_1.
 \end{align*}

Let us take the equations of type II first. Pick the first pair of equations in
the list above (the other cases run completely analogously). Consider the left
hand side equation
\begin{equation}\label{eq:CaseII}
x_1+x_2=y_1.
\end{equation}
By the theory of $S$-unit equations there are at most $A(2,2)$ pairs $(c_1,c_2)$
such that $x_1/y_1=(\alpha/\beta)^k=c_1$ and $x_2/y_1=\alpha^m/\beta^k=c_2$.
These two relations determine the pair $(k,m)$ uniquely. Hence there are at most
$A(2,2)$ pairs $(k,m)$ that fulfill the equation. With the pair $(k,m)$ fixed, the value of $u_l$ is
fixed as well and by Lemma \ref{lem:multiplicity} there are at most three ways to extend the pair
$(k,l)$ to a triple $(k,l,m)$ such that $u_k<u_l<u_m$ is an arithmetic progression.
We deduce that there exist at most $3A(2,2)$ triples $(k,l,m)$ satisfying a pair of equations of type II.

Consider now equations of type I. Rewriting the two equations we get
$\alpha^k+\alpha^m-2\alpha^l=0=\beta^k+\beta^m-2\beta^l$. Thus in this case, the
minimal
polynomial of $\alpha$ and $\beta$, which is also the companion polynomial of
the sequence, divides $X^k-2X^l+X^m$.

Finally, we turn to equations of type III and consider the first equation of
this type in detail. We can handle the others similarly. Assume for the moment
that there exists a prime ideal $\mathfrak p$ with $\mathfrak p\nmid (2)$ and
$v_{\mathfrak p}(\alpha)>0$. Let $\mathfrak p_2$ be a prime ideal lying above
$(2)$ and suppose that $m>k$. We consider the equation $x_1+x_2=-2y_3$ from a
$\mathfrak p$-adic point of view. Write $v_{\mathfrak p}(\alpha)=\xi_{\mathfrak p}$ and
$v_{\mathfrak p}(\beta)=\eta_{\mathfrak p}$. Since the two smallest $\mathfrak
p$-adic valuations of the summands have to coincide, we have $v_{\mathfrak
p}(x_2)=k\xi_{\mathfrak p}=l\eta_{\mathfrak p}=v_{\mathfrak p}(2y_3)$.
Considering $\mathfrak p_2$-adic valuations we obtain $k\xi_{\mathfrak
p_2}=l\eta_{\mathfrak p_2}+\delta$, where $\delta=1,2$ depending on whether
$(2)$ is ramified or not. We get the following system of linear equations
\begin{align*}
 k\xi_{\mathfrak p_2}-l\eta_{\mathfrak p_2}&=\delta\\
 k\xi_{\mathfrak p}-l\eta_{\mathfrak p}&=0
\end{align*}
which has either no solution or at most one solution. Note that by assumption $\xi_{\mathfrak p}>0$.
Thus in this case, equations of type III yield at most one pair $(k,l)$ that extends to a triple $(k,l,m)$
such that $u_k<u_l<u_m$ is an arithmetic progressions. Thus by Lemma \ref{lem:multiplicity} we have at most
three additional three term arithmetic
progression.

If $v_\mathfrak{p}(\alpha)=0$, but
$v_\mathfrak{p}(\beta)>0$, then we take the equation $-2x_3=y_1+y_2$ instead and
draw the same conclusions.

In view of the paragraph above, we may assume that the only prime ideals
dividing $(\alpha)$ and $(\beta)$ are lying above $(2)$. We distinguish now
whether $(2)$ splits, ramifies or is inert above $K$.

We start with the case when $(2)$ is inert, i.e., $(2)$ is a prime in $K$. Then
$(\alpha) = (2)^x$
and $(\beta) = (2)^y$ with some non-negative integers $x,y$. Since
$\alpha$ and
$\beta$ are conjugate, we get $x=y$. Hence $\alpha/\beta$ is a unit.
However,
recalling that $\alpha,\beta$ are from a quadratic imaginary field, this
yields that
$\alpha/\beta$ is a root of unity, which is a contradiction.

One may prove the case where $(2)$ ramifies by similar means as the
inert case. In this
case,we have $(2)=\mathfrak{p}^2$ with some prime ideal $\mathfrak{p}$.
Note that $\mathfrak p$ need not be principal although $\mathfrak p^2$ is principal.
Thus $(\alpha)=\mathfrak{p}^x$
and
$(\beta) = \mathfrak{p}^y$  with some non-negative integers $x,y$. Taking norm, we
get $x=y$,
and similarly as above we obtain that $\alpha/\beta$ is a root of unity,
which is impossible.

Finally, let us consider the case when $(2)$ splits, that is, we can write
$(2)=\mathfrak p_1\mathfrak p_2$. Assume that $(\alpha)=\mathfrak
p_1^{\xi}\mathfrak p_2^{\eta}$, then $(\beta)=\mathfrak p_1^{\eta}\mathfrak
p_2^{\xi}$. Considering the equation $x_1+x_2=-2y_3$ from a $\mathfrak p_1$-adic
and a $\mathfrak p_2$-adic viewpoint, respectively we get the following system
of linear equations under the assumption that $m>k$:
\begin{equation}\label{eq:valuations}
\begin{split}
v_{\mathfrak p_1}(x_2)=k\xi &= l\eta +1= v_{\mathfrak p_1}(2y_3),\\
v_{\mathfrak p_2}(x_2)=k\eta &= l\xi +1= v_{\mathfrak p_2}(2y_3).
\end{split}
\end{equation}
Note that we obtained this system due to the fact that the smallest
$\mathfrak{p}$-adic valuations must coincide. Solving linear system
\eqref{eq:valuations} yields $k(\xi^2-\eta^2)=(\xi-\eta)$. Therefore either
$\xi=\eta$ or $k(\xi+\eta)=1$. The first case yields $\alpha=2^\xi\epsilon$ and
$\beta=2^\xi\epsilon^{-1}$ for some unit $\epsilon$ which yields that
$\alpha/\beta=\epsilon^2$ is a root of unity, a contradiction. When
$k(\xi+\eta)=1$ we have $k=1$ and $(\xi,\eta)=(0,1),(1,0)$. In any case, we
obtain from the linear system \eqref{eq:valuations} that $l=-1$, a contradiction
to our assumption that $k,l,m$ are nonnegative integers.
\end{proof}

So we are left with the problem which quadratic polynomials divide
$X^k-2X^l+X^m$. In particular, we may assume that the companion polynomial is
irreducible, otherwise $\alpha$ and $\beta$ would be rational integers which
case is covered by Theorem \ref{theorem2}. We need to find all quadratic factors
of polynomials of the type:
\begin{itemize}
\item $X^a-2X^b+1$ or
\item $X^a+X^b-2$ or
\item $2X^a-X^b-1$.
\end{itemize}

Since all roots of the first polynomial are units, and all the roots of the
third polynomial are either non-integral algebraic numbers or units, it is enough to
study only polynomials of the form $X^a+X^b-2$.

We state the following result due to Pint\'{e}r and Ziegler \cite[Lemma
3]{pinterziegler}.

\begin{lemma}\label{lem:Pinter}
 Let $a>b>0\in\mathbb{Z}$ and let $f(X)=X^a+X^b-2$ be a
polynomial. Then $f(X)=h(X)g(X)$ factors into the
monic polynomials $h(X)$ and $g(X)$ over $\mathbb{Z}$, where the only roots of
$h(X)$ are roots of unity and $g(X)$ is irreducible. Moreover
$h(X)|X^{\gcd(a,b)}\pm 1$.
\end{lemma}

We are interested in finding the quadratic factors of $X^{a}+X^{b}-2$. By Lemma
\ref{lem:Pinter} we only have to consider the cases $(a,b)=(3,2),(3,1),(2,1)$ or
$(4,2)$. Thus the companion polynomial is either $X^2+2X+2$ or $X^2+X+2$ or
$X^2+2$.

Since $X^2+X+2$ is the only polynomial with negative discriminant which is also a companion polynomial of a
non-degenerate binary recurrence, we obtain that either $(A,B)=(-1,-2)$ or there
are at most $A(5,2)$ three term arithmetic progressions coming from
non-degenerate solutions to \eqref{eq:AP} (cf. Lemma \ref{lem:nondeg}),
$3A(3,2)+30$ three term arithmetic progressions coming form two-term vanishing
subsums of \eqref{eq:AP} (cf. Lemma \ref{lem:two-vanish}) and $18A(2,2)+9$
three term arithmetic progressions coming form three term vanishing subsums of
\eqref{eq:AP} (cf. Lemma \ref{lem:three-vanish}). We summarize these results.

\begin{proposition}
\label{prop:exact}
Let $u=(u_n)_{n=0}^\infty$ and $v=(v_n)_{n=0}^\infty$ be Lucas sequences of
first and second kind corresponding to the pair $(A,B)$ with $A^2+4B<0$. Then
$u$ and $v$ admit at most
\[
C=A(5,2)+3A(3,2)+18A(2,2)+39
\]
three term arithmetic progressions unless $u$ or $v$ correspond to
$(A,B)=(-1,-2)$ respectively. In case of $(A,B)=(-1,-2)$ the recurrences $u$
and $v$ admit infinitely many three term arithmetic progressions.
\end{proposition}

\begin{remark}
Since $X^2+X+2|X^3+X-2$, we get that $u_k<u_l<u_m$ and $v_k<v_l<v_m$ with
$(k,l,m)=(t+1,t,t+3)$ yield infinite families of three term arithmetic
progressions in $u$ and $v$, respectively.
\end{remark}

Theorem \ref{theorem1} is now an easy Corollary of Proposition
\ref{prop:exact}. Indeed, due to a result of Schlickewei and Schmidt
\cite{Schlickewei:2000}, we know that $A(k,s)\leq 2^{35B^3}d^{6B^2}$, where
$B=\max\{k+1,s\}$ and $d=\lbrack K:\mathbb{Q} \rbrack$, hence
\[
C\leq A(5,2)+3A(3,2)+18A(2,2)+39 <6.45\cdot 10^{2340}.
\]

\section{Acknowledgment}

The authors are grateful to the referee for her/his useful comments and suggestions.

\def$'${$'$}

redpajama_set_name: RedPajamaArXiv


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text:
Top 12 Zac Efron Movies You Should Have Seen By 2022
by Candice Burns.
Candice Burns - Candice Burns has worked in the real estate industry for a long time. She understands the importance of a home for long-term happiness and has dedicated her career to placing people in the home of their dreams. Candice's passions for helping others during difficult times and a strong interest in high-end, luxury homes drove her to explore real estate. She's been in the real estate business for ten years and has helped over 3,500 people find homes during that period. She earned a bachelor's degree in business administration from the University of Los Angeles. She's worked with some of Los Angeles' most prestigious real estate firms.
17 Again
Extremely Wicked, Disgracefully Evil, And Vile
High School Musical 2
Me And Orson Welles
What Is Zac Efron's Best Movie?
How Old Was Zac Efron In 17 Again?
Efron was born in San Luis Obispo, California, and subsequently relocated to Arroyo Grande. His father, David Efron, works at Diablo Canyon Power Plant as an electrical engineer, and his mother, Starla Baskett, works as an administrative assistant there as well.
Efron has a brother, Dylan, and grew up in a middle-class household, as he has put it. His surname is Hebrew in origin. Although Efron was reared agnostically and did not follow religion as a kid, his paternal grandfather is Jewish, and he has identified himself as Jewish.
Efron has previously said that if he received a "B" instead of an "A" in school, he would "freak out," as well as that he was a class clown. Following that, Efron worked in The Great American Melodrama and Vaudeville theaters and started taking singing training.
Gypsy, Peter Pan, or The Boy Who Wouldn't Grow Up, Little Shop of Horrors, and The Music Man were among the productions in which he appeared. Robyn Metchik, his acting instructor, introduced him to an agency in Los Angeles (the mother of actors Aaron Michael Metchik and Asher Metchik).
The Creative Artists Agency subsequently signed Efron. In 2006, Efron graduated from Arroyo Grande High School and was accepted to the University of Southern California, but he chose not to go.
COPYRIGHT_WI: Published on https://washingtonindependent.com/w/zac-efron-movies/ by Candice Burns on 2022-10-17T15:06:48.884Z
He also appeared in 2000 and 2001 at Pacific Conservatory of the Performing Arts, a theatrical group based at Allan Hancock Institution, a community college in Santa Maria, California.
Baywatch (2017) - Official Trailer - Paramount Pictures
I had high expectations coming into this, and that is all my fault. However, as a child, I loved the program, and I appreciate The Rock's character, as well as seeing shirtless men play on the beach.
However, I believe it is more regressive than the original program since it seems as though it should know better but doesn't. Zac's role, rookie lifeguard Matt Brody, is a major element of the toxic masculinity in the film. Baywatch is a 2017 American action comedy film based on the Michael Berk, Douglas Schwartz, and Gregory J. Bonann television series.
Seth Gordon directed the film, which stars Dwayne Johnson and Zac Efron and was written by Mark Swift and Damian Shannon (from a story by Jay Scherick, David Ronn, Thomas Lennon, and Robert Ben Garant). It also features Jon Bass, Alexandra Daddario, Kelly Rohrbach, Priyanka Chopra, and David Hasselhoff in supporting roles.
The tale follows lifeguard Mitch Buchannon and his crew as they attempt to rescue their beach by taking down a drug kingpin. On February 22, 2016, principal photography started in Florida and Savannah, Georgia.
The film was released in the United States on May 25, 2017, by Paramount Pictures and earned $177 million worldwide. It got mixed reviews from reviewers, with some praising Johnson's performance despite the film's coarse comedy, plot, and character development.
At the 38th Golden Raspberry Awards, Baywatch earned five nominations, including Worst Picture and Worst Actor for Efron.
Neighbors - Trailer
Who wouldn't want Zac to live next door to them? In Neighbors, Dad Mac (Seth Rogen) gets a taste of what it's like when the home next door turns into a frat house, with Teddy (Zac Efron) and the rest of the fraternity moving in.
When things become too loud, they agree that Mac would phone Teddy instead of the police, but when Mac breaks his promise, he and his wife, Kelly, find themselves on the receiving end of some pranks and hazing.
Neighbors 2: Sorority Rising picks up two years later with the couple and Teddy as he attempts to assist a sorority seeking to move into his former fraternity's home.
Dirty Grandpa Official Trailer #1 (2016) Zac Efron, Robert De Niro Comedy Movie HD
Jason Kelly works for his father as a lawyer. Jason's grandma passes away, and his Army veteran grandpa, Lt. Colonel Dick Kelly, requests that Jason drive him to Boca Raton, Florida, since his driver's license has been revoked.
Jason's domineering fiancée, Meredith, is being married in a week, but he chooses to take his grandpa regardless.
On their way there, they run across Shadia, Jason's former photography student, and her pals Lenore and Bradley. Dick and Lenore are drawn to one other right away.
Dick introduces himself to the girls as a professor and Jason as a photographer. They all go their own ways, but Dick persuades Jason to meet the females at Daytona Beach, Florida since Dick is interested in having sex with Lenore.
Dick flirts with two ladies at a golf course where the two meet. They meet up with the ladies and their pals, Cody and Brah, at Daytona Beach, where they compete in beer chugging.
That night, a drunk Jason party and is duped into consuming crack cocaine by a drug dealer called Pam while wearing nothing but a Hornet fanny pack. He borrows a motorbike and wakes up on the beach the following day.
During an uncomfortable FaceTime conversation with Meredith, a toddler grabs Jason's fanny pack and pulls it off; the boy's father accuses Jason of child molesting and contacts the cops, who quickly arrest Jason. Dick bails him out, and the two go to a nursing home to see Dick's old Army buddy Stinky.
They meet up with the ladies again and compete in a flexing contest with Cody and Brah; when they lose, Dick modifies a T-shirt cannon to shoot a beer can at the winners, putting them in the hospital.
Jason and Dick steal their hotel room while Cody and Brah are in the hospital. Jason and Dick head to a nightclub with the girls after Dick confesses that he was a Green Beret.
After several guys harass Bradley because of his homosexuality, Dick gets into a fight with them. The following day, Jason and Shadia attend a demonstration, where she informs him that she will be departing shortly to live on a ship for a year.
The Greatest Showman - The other side [1080P] Subtitled
The Greatest Showman, starring Efron and Hugh Jackman as PT Barnum, brings PT Barnum's legendary circus to life on the big screen. Barnum's business associate, Phillip Carlyle, is played by Zac, who contributes to the man's circus idea while falling in love with a pink-haired trapeze performer.
He performed two musical duets with each of the aforementioned characters, named "The Other Side" and "Rewrite the Stars," which brought him back to his adolescent singing origins. Not only did he sing well, but he also looked dashing in a top hat.
17 Again (2009) Official Trailer - Zac Efron, Matthew Perry Movie HD
Do you think Matthew Perry is getting younger? In this 2009 film, Zac Efron portrays the younger version of their common character, Mike. Mike, 37, feels as if he's losing everything: his marriage is gone, he's growing apart from his children, and he just left his work.
Mike inadvertently becomes 17 again, as the title implies, and now he must attempt to restore the damage in his life as his 17-year-old self. When your estranged wife is dating someone else and your daughter believes you're into her, it's not easy.
Extremely Wicked, Shockingly Evil and Vile Trailer #2 (2019) | Movieclips Trailers
While Zac Efron may be a bit of a heartbreaker in real life, his character Ted Bundy did exactly that in Extremely Wicked, Shockingly Evil, and Vile, a 2019 Netflix thriller.
Bundy was an American serial murderer in the 1970s who abducted, tortured, raped, and killed over 30 women before being apprehended and sentenced to death. Zac not only had Bundy's dreamy and beautiful appearance down, but he also delivered a flawless performance.
THE BEACH BUM [Official Red Band Trailer] - In Theaters March 29, 2019
Moondog, a lethargic stoner and writer portrayed by Matthew McConaughey, discovers that his part of his late wife's inheritance has been frozen and is under his daughter's hands in this 2019 R-rated film.
Growing up, taking life more seriously, and completing his present work-in-progress novel are all requirements. Flicker, played by Zac Efron, is a guy who, together with Moondog, escapes out of a rehabilitation facility.
Tweens got their heads in the game with Zac on Disney Channel in the mid-2000s, thanks to High School Musical. The adolescent idol portrayed a great basketball player who fell in love with theater and a girl, and the films depict his struggle to balance his three loves.
The series was a big hit with Disney fans and served as a solid foundation for Efron's successful career.
Liberal Arts - Zac Efron's Scenes
When 35-year-old Jesse (Josh Radnor) returns to the town where he attended college and meets his former teacher's daughter, Zibby, he is homesick about that (Elizabeth Olsen).
Zac Efron portrays Nat, a college student who is well-intentioned yet a touch strange. After they run into her at a dorm party, he arranges a coffee date for Jesse and Zibby. Zac appears on film for less than 10 minutes, yet the movie would not exist if it weren't for his matchmaking role!
Me and Orson Welles Trailer
Zac portrayed yet another character who was a rising sensation on stage shortly after the previous High School Musical picture was released. After meeting Orson Welles, 17-year-old Richard is unexpectedly given a part on Broadway in the production of Caesar. Richard quickly discovers that theater is not simple, particularly when Welles is in charge of the production.
Zac's character, like HSM, falls in love with a girl who is also in the play, and he must learn to control his desire and jealousy.
The Disaster Artist | Tommy | Official Trailer 2 HD | A24
Zac Efron also had a minor role in The Disaster Artist. This real tale begins with Tommy Wiseau (James Franco) and Greg Sestero (Dave Franco) becoming friends in this cinematic version of the same-named book.
They go to Los Angeles in order to pursue a career as actors, but instead get engaged in the production of their own project, which leads to the real-life film The Room. Dan Janjigian, who portrays Chris-R in The Room, is Efron's character.
Hairspray (2007) Official Trailer #1 - John Travolta Movie HD
Zac appeared in Hairspray, a film set in 1962 Baltimore, in the midst of the High School Musical movies. The 2007 picture was a remake of a 1988 film that was also staged on Broadway.
Link Larkin, a regular on the fictitious TV show The Corny Collins Show, is played by Zac Efron.
Tracy, Efron's on-screen love interest, is played by Nikki Blonsky, a young girl who joins the show and catches the globe by storm with her plus-size figure and anti-racism stance.
Neighbors (2014)
The Greatest Showman (2017)
Neighbors 2: Sorority Rising (2016)
The film follows 37-year-old Mike (Matthew Perry) who becomes his 17-year-old self (Zac Efron) after a chance accident.
1783fb327d7ad6c753ab5a7001dd954b683a7cf38ce2dc2fbbeae676845c6431b89f381338e0bd026cc5b964ec4144d84fb915fc000f31e0c41c21e80b84d710


text:
November>
Italian Women's Membership
Valeria Kleiner (r) Of Germany Heads For The Ball With Roberta Filippozzi Of Italy
Other Statistics On The Topicsports In Italy
Labor, Delivery, And Postpartum Health
Furthermore, the neonatal end result may additionally have an impact on women's emotional start expertise. Some danger factors for infant health (for example, gestational age, a low PH index, and so forth.) might fear or frighten the girl and could change her emotional expertise. Finally, our sample is somewhat small, and we wouldn't have the exact worth of the response fee. Although the research was carefully explained to all the women who met the inclusion standards, the response rate ought to prompt us to reflect on components that will have interfered with recruitment. Certainly, the predictors that emerged in our research clarify solely a limited a part of the worldwide delivery expertise (23% of total variance) and fail to explain the variables that have an effect on women's ideas and feelings. The childbirth experience is a multidimensional assemble, which requires the consideration of many variables. For example, a number of studies show how attentive assistance from midwives has a pivotal function in promoting constructive start expertise , serving to the women really feel safe and cared for .
Because of this, casting administrators choose tall, slender women with medium to giant breasts and slender hips—my guess is that they're toying with innocence versus eroticism. In Italy, there's a huge dissonance between real our bodies and the "perfect," to the purpose that many Italian women keep away from activities and even garments because of their bodies. By "bravery", she didn't just imply sporting a short, type-fitting dress with fabric slightly draped on the sides, which gave it the sultry appearance of one thing out of a late-Forties comic strip.
In this way, the help from hospital employees could in the end assist moms-to-be really feel more self-assured and ready to deal with the labor course of . In addition, the obstetric characteristics of our pattern are consultant of the Tuscan inhabitants, a area in central Italy, but not of the general Italian population. Tuscany, in reality, has a CS fee of 21.7%, which is decrease than the nationwide price, which is round 35% .
In other words, it appears that it is the ultimate stage of the whole labor course of that weighs closely on the quality of the woman's emotional experience. Some psychological research on memory processes could provide a contribution to explaining this end result. For instance, some studies on recollections of disagreeable medical procedures recommend that sufferers' reminiscences of painful medical procedures mirror the intensity of the ultimate part of the expertise more than the entire expertise in itself. In an analogous means, it's possible that the ache experience in the course of the second stage of labor has a deeper effect on women's memories and emotions than the primary stage, even when the latter lasts longer.
In Italian society, especially in previous a long time, for instance, the woman's capability to tolerate delivery pain was usually embraced as a part of a "full childbirth expertise" and an indication of a lady's self-efficacy. Some women might, due to this fact, really feel efficacious in knowing that they haven't "surrendered" to the reduction of pain, which ends up in a more positive delivery . What adversely affects the start experience in our research is the length of the expulsive phase of start. If the first stage of labor doesn't considerably affect women's emotional and cognitive appraisal of childbirth, the longer the second stage, the more serious the girl's subjective experience.
Other research have analyzed the complicated relations between childbirth satisfaction, labor pain, and analgesia, confirming that satisfaction throughout labor isn't necessarily directly correlated with pain or ache aid . It is hypothesized, for example, that some women require analgesia because of a chronic and tough labor, so they might appraise their birth as tense or less passable than women who give delivery naturally. Pain reduction, subsequently, is one, but not the most italian mail order brides significant, issue that contributes to the evaluation of the birthing expertise. In the identical way, women who have vaginal natural births could perceive extra personal efficacy in the delivery course of and extra capability to give start even when dealing with difficulties. Van der Gucht and Kiara highlight how some parturient women view the ache of giving delivery as a challenging life factor, capable of enhance their coping ability.
We ought to make clear that health care in Italy is managed at a regional degree with strongly unbiased policies, and childbirth statistics differ tremendously from one region to a different. For this reason, it might be helpful to increase the analysis to different Italian areas so that the outcomes could be made generalizable to the complete Italian population. Moreover, as already described, this research didn't think about the impression of intra-partum variables, for example, the impact that relational and organizational components might have on a woman's subjective delivery experience .
We can even hypothesize that one-to-one assistance provided by the midwife to the lady during the first stage of labor – which in nulliparous women could last a number of hours – contributes to reducing the impact of pain on the woman's birth satisfaction. Indeed, literature underscores that relational variables, corresponding to practical and emotional support from a partner or a midwife during labor and delivery, can affect the childbirth experience as well .
Despite the restrictions described above, our examine confirms the function played by obstetric variables in shaping the lady's subjective experience of childbirth, although in a partially sudden direction. In explicit, it isn't the kind of delivery – neither a CS nor an operative start – that impacts the subjective start expertise, but the period of the expulsion phase and the use of analgesia. The woman's experience is extra positive when the expulsion section lasts less time and when analgesics usually are not used. Our outcomes additionally demonstrate that psychological components, linked to expectations about childbirth and, more particularly, to the extent of worry with which women strategy the event, are also essential. Future insights into this facet should higher clarify the impact of worry.
She, sadly, meant having the audacity to wear that kind of garment as a lady who was not stick-thin.
The use of epidural analgesia was also implicated in women's perception of their start expertise, however in a counterintuitive direction. A variety of research have found that much less pain skilled during labor and childbirth is related to extra satisfaction . For the ladies in our study, use of pharmacological ache aid fostered birth experiences that have been much less constructive as in comparison with these during which the lady had a pure childbirth without pain aid.
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