text:
After mental health break, a turnaround at Detroit Mercy
Carlos Osorio/AP
Detroit Mercy head coach Mike Davis talks with guard Antoine Davis (0) during the first half of an NCAA college basketball game against Michigan State, Friday, Dec. 4, 2020, in East Lansing, Mich. (AP Photo/Carlos Osorio)
By: Noah Trister, AP Sports Writer
DETROIT (AP) — Detroit Mercy's men's basketball team began the season with three games in five days. A week later, a matchup with Kentucky was canceled. Then on the last weekend of December, there was some uncertainty about whether a couple games against rival Oakland would be played.
They were, but Detroit Mercy coach Mike Davis wasn't happy about it.
"I said enough is enough," Davis recalled. "We're taking some time off."
After the two games against Oakland, the Detroit Mercy program paused activities, citing not positive COVID-19 tests or contact tracing issues but the mental health of its athletes. The Titans didn't play another game until Jan. 15. Since then, they've found their stride, winning 10 of their final 12 regular-season games going into Thursday night's game against Robert Morris in the Horizon League Tournament.
In this season unlike any other, players throughout the country have had to adjust to coronavirus protocols, scheduling uncertainty and just general anxiety about what's to come. Many have been largely isolated on campuses that are far emptier than normal.
Even for those who avoid catching the virus, the mental strain can be a real issue.
"We don't have the normal balance of our lives that we usually do, while being concerned about not getting COVID and trying to be as healthy as you can be and control what you can control," Iowa star Luka Garza said. "It's definitely been a different year than any other for all of us."
An NCAA study conducted in the fall on the well-being of athletes shed some light on the challenges they face during the pandemic.
"Student-athletes reported fewer sleep difficulties and lower levels of loneliness, loss, anger and sadness in fall 2020 than at the outset of the pandemic," the study said. "However, similar to spring, elevated rates of mental exhaustion, anxiety, hopelessness and feelings of being depressed were reported."
Davis said he thought playing some games would be good, but that can create stress of its own — something as simple as staying at a hotel takes a different toll than it did a year ago.
Then there were the two games against Oakland at the end of December. Davis was concerned because of a game the Golden Grizzlies had played shortly before that.
"We found out that the referee that was in that game had the virus, and then we had to play Oakland six days later," Davis said. "Our guys did not want to play, I didn't want to coach, and you could just see it in them."
Oakland coach Greg Kampe said at the time that he didn't want to play the games either, but they were held. Detroit Mercy lost both, falling to 1-7 on the season. Then the Titans took their break.
It wasn't a completely unusual move. Earlier in December, Duke canceled its remaining nonconference games to help players spend time with family over the holidays.
At Detroit Mercy, Davis felt the virus was wearing on his team in a variety of ways.
"You have the threat of someone getting it every day, anybody. Most people who get it don't know where they got it from," he said. "I think that our fans have been really patient with us, and our administration, our AD is just unbelievable."
The pause was initially supposed to be a week and wasn't due to any positive tests in the program. Davis said there was eventually a positive test, and the Titans didn't play Milwaukee the weekend of Jan. 9 as planned. They returned the following week against Green Bay.
Since coming back, the Titans have had only conference games, and they were all on Fridays and Saturdays. Those back-to-backs weren't necessarily easy, but they allowed the team to establish a sense of routine.
Davis said the Titans would play their games, then take Sunday off. Then they'd practice Monday and Tuesday before another day off Wednesday.
Marquell Fraser, a guard for Detroit Mercy, said the lengthy break did have some drawbacks — any time away from the court can make it tough to build team chemistry. But he said it has been helpful to have a consistent schedule from week to week — something that wasn't the case during the nonconference season.
"Going into every week, we all know what to expect," Fraser said. "We all know that we're going to be playing Friday, Saturday, Sunday off, and after that, Monday we're back to work. I think it is good just knowing your set schedule for the rest of the week."
Of course, now Detroit Mercy begins the conference tournament with its first Thursday game of the entire season. But the Titans have a lot of reasons to feel optimistic. They not only made it through the regular season, they now look like a threat in the Horizon.
And Davis has learned a few things from this experience that might be useful even in a non-pandemic year.
"You realize that you can give guys off days. Sometimes, coaches, we're so nervous about not being in the gym and not practicing," he said. "But two days on and one day off, we're playing better than we ever played."
(Copyright 2022 The Associated Press. All rights reserved. This material may not be published, broadcast, rewritten or redistributed.)


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Chelsea's winners and losers from win vs Man City which.
If more than one player slap at the same time, the lowest hand or the hand directly on the card wins the pile. Players sometimes slap the wrong card, meaning any card other than a jack. If this happens they give one card to the player that placed the card they mistakenly slapped. Players who run out of cards can slap back in the game. However.
Rummy is a group of matching-card games notable for similar gameplay based on matching cards of the same rank or sequence and same suit. The basic goal in any form of rummy is to build melds which consist of sets, three or four of a kind of the same rank; or runs, three or more cards in sequence, of the same suit.If a player discards a card, making a run in the discard pile, it may not be.
Lionel Messi escapes red card for second game in row after.
Stores opening from Monday across Liverpool and rules that will be in places Catherine Murphy. 1 day ago 'Unlawful': 14-day quarantine rules for UK arrivals come into force. Single people.Liverpool Rugby Referees. Pitchero. Join Website or existing users Login. Liverpool Rugby Referees. Home; News; Events; Leagues. 2019-20; Discipline. Red Card and Abuse Forms and Reports; On-Field Discipline; Discipline Process (inc key contacts) Referee Abuse; Club actions to help prevent referee abuse; RFU Discipline Reporting Writing Guide; Contacts. Exec Comittee Members; Important.How To Play Indian Rummy (13 Card Game Rules) Objective is to make the best combinations of cards in the form of Sets or Sequences. To win, one has to declare first and then meld the cards in hand correctly. Game is played with two decks of cards consisting of 52 cards and 2 face Jokers each. Joker. All cards with the same rank as the card below the closed deck are considered jokers. A joker.
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A yellow card is issued as a caution, giving a player a second chance to remain on the pitch for the rest of the game, but being shown a red card results in immediate ejection from the field of play.
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redpajama_set_name: RedPajamaCommonCrawl


text:
Rashid Khan on his way to 100 T20 wickets in 2018https://indianexpress.com/article/sports/cricket/rashid-khan-on-his-way-to-100-t20-wickets-in-2018-5502423/
Rashid Khan on his way to 100 T20 wickets in 2018
With three wickets for Adelaide Strikers in their 2018/19 Big Bash League opener, Rashid Khan has taken his tally for the year to 92.
By Sports Desk |	Published: December 20, 2018 5:08:35 pm
World Cup 2019: Impatient Rashid Khan faces toughest test
World Cup 2019: Afghanistan aim to stay afloat against India in Southampton
People forget 10 good days and conveniently remember one bad outing: Rashid Khan
Rashid Khan has already broken the record for most T20 wickets in a year. (PTI Photo)
Afghanistan's star spinner Rashid Khan is on his way to becoming the first player to take 100 T20 wickets in a year. He took three wickets playing for Adelaide Strikers in the opening match of the 2019-19 Big Bash League, thus taking his tally to 92 for the calendar year.
Strikers play three more matches this year which means Rashid is on track to reaching the milestone. He has already broken the record for most T20 wickets in a year. The previous record holder was West Indies' Dwayne Bravo with 87 in 2016. Rashid reappears in the list in third place with 80 wickets in 56 matches that he played in 2017.
Rashid finished with figures of 3/19 at the end of Adelaide's opening match against Brisbane Heat. "He's really smart," Carey said of Rashid after the match. "He's obviously got the skills, we all know that, but I think he's got a lot between the ears. He knows how he's going to get batters out – he analyses them pretty quickly out in the middle, he analyses the wicket, and he knows what's going to be effective.
"He's obviously got the talent and the freak skills, but the stuff he does behind the scenes … he gets in the gym after hours, he goes for extra runs, he bowls more balls than most guys I've seen.
"So the stuff he does on the field is amazing, but the stuff we see him doing off the field is a real credit to him."
The 20-year-old Rashid Khan is the no.1 ranked bowler in T20s on the ICC rankings. He is also Afghanistan's frontline spinner in the ODI format with 118 wickets to his name. In June, he was part of the first Afghanistan Test squad in their one-off match against India in Dharamsala.
Follow the Cricket World Cup 2019 live updates and real-time analysis on IndianExpress.com. Check the ICC Cricket World 2019 Schedule, Teams and Points Table.
1 Appointment of India Women's team coach not on behalf of BCCI: Treasurer Anirudh Chaudhry
2 PSL 2019: Multan Consortium bags franchise rights for sixth team
3 Bangladesh vs West Indies Highlights: Bangladesh win 2nd T20I by 36 runs


text:
Bears Sports NFL
QB Andy Dalton doesn't need this preseason game, and that's good for the Bears
Unlike Mitch Trubisky and Nick Foles, Dalton really is good to go. That allows coach Matt Nagy to concentrate on getting rookie Justin Fields the snaps he needs.
By Jason Lieser
SHARE QB Andy Dalton doesn't need this preseason game, and that's good for the Bears
Andy Dalton has 218 touchdown passes and 126 interceptions in his 10-year career.
Nam Y. Huh/AP
Unlike the quarterbacks the Bears had the last few seasons, Andy Dalton doesn't need this preseason game Saturday against the Dolphins. He's not a star by any means, but he's also not Mitch Trubisky or Nick Foles.
With 8,151 snaps and 142 starts in his career — more than Trubisky and Foles combined — and such a thorough mastery of the offense that he has committed only one mental error since the start of training camp, Dalton continues to make coach Matt Nagy's job easier.
Dalton will exit early, possibly after the first possession, and Nagy can tend to the ultra-important, franchise-shifting development of rookie Justin Fields.
Bears plan to play rookie QB Justin Fields extensively in preseason game vs. Dolphins
''If he wasn't at the point that we wanted, we would play him more in the preseason,'' Nagy said. ''He's exceeded that. He could play tomorrow in a game, and I would feel great about it.''
Nagy's plan is to let Dalton feel out how many snaps he wants — in the single digits, certainly — and that's how it should be with an experienced quarterback. It would be concerning if the Bears were fretting about Dalton's readiness at this point.
Nagy convinced himself that Trubisky was equally prepared heading into the ill-fated 2019 season, and it quickly backfired on him. He knew going into 2020 that both his quarterbacks needed those snaps, but the pandemic prompted the NFL to cancel the preseason.
With Dalton, however, there's not nearly as much teaching and adjusting. Trubisky still was learning how to run the offense and decode defenses. Foles had minimal history with Nagy, and it became clear as the season progressed that they were on very different wavelengths.
Nagy erupted at Foles over a delay-of-game penalty at the 7-yard line against the Panthers and criticized him for another one against the Saints when he was too preoccupied with the plays on his wristband to get the Bears to the line of scrimmage quickly enough.
ESPN analyst Brian Griese also shared a conversation in which Foles said Nagy didn't grasp that some of his play-calls were doomed.
Those little snags are highly unlikely with Dalton. He already seems fluent in the offense and in step with Nagy.
''I feel really comfortable with everything that we've got in [the playbook], but you get out there to get into a little bit of a rhythm,'' Dalton said when asked whether there was anything he needed to get out of this preseason game. ''We'll see how long I'm actually out there.''
The translation on that last part is that he barely will play. Nagy will have to be careful about how much he uses Fields, too, because the Bears might be missing three injured starters on the offensive line.
Dalton's most valuable work might be helping out on the sideline when Fields is playing. He has embraced the role of mentoring him this season, even as Fields strives to replace him, and has been serving as somewhat of a part-time assistant coach. That will continue Saturday.
In the meantime, Dalton will get one more full session of game-like competition in a joint practice with the Dolphins on Thursday.
Everything about the offense seems to be sharper since Dalton arrived, which is probably a big factor in Nagy feeling ''as calm as I've ever been in my life'' as he heads into a season in which his job appears to be at stake. Their relationship has been much closer to a partnership than the teacher-student dynamic he had with Trubisky and the choppy ride he had with Foles.
''Our guys, they're playing fast, playing confident,'' Nagy said. ''It's not like in Year 1, when they were thinking about the play. They know the play, and now they can take it to the next level.''
Next Up In Chicago Bears
Bears make staff changes, including return of defensive assistant Jon Hoke
Tom Brady retires again. Ever get the feeling you're being played for a fool?
Bears want running to be part of QB Justin Fields' game, but not the majority of it
Luke Getsy, assistants give Bears inside track on Senior Bowl draft prospects
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North Lawndale beats Marshall in West Side city tournament showdown
North Lawndale, which opened in 1998 and started varsity basketball in 2000, has never finished a season with a losing record. So there is plenty of pride to play for over the next few weeks.
By Michael O'Brien


text:
Severe Psoriasis May Make Diabetes Increasingly Likely: MedlinePlus Health News
Severe Psoriasis May Make Diabetes Increasingly Likely
By Robert Preidt
MONDAY, Nov. 20, 2017 (HealthDay News) -- People with the skin disease psoriasis are at increased risk for type 2 diabetes, and the more severe the psoriasis, the greater their risk, a new study finds.
Researchers examined data on nearly 85,000 adults in the United Kingdom, including 8,100 who had psoriasis. Compared with people who did not have psoriasis, the risk for diabetes was 21 percent higher among those with psoriasis on 2 percent or less of their body. It was 64 percent higher among those with psoriasis on more than 10 percent of their body.
For every 10 percent increase in body area with psoriasis, the risk for diabetes rose another 20 percent. For instance, people with psoriasis on 20 percent of their body had a nearly 84 percent increased risk for diabetes. And those with psoriasis on 30 percent of their body had a 104 percent increased risk, the study authors said.
When applying their findings to the number of people worldwide who have psoriasis, the researchers estimated that psoriasis is linked to 125,650 new cases of type 2 diabetes each year.
The study was published online recently in the Journal of the American Academy of Dermatology.
Psoriasis, which affects about 7.5 million Americans, is a disease of the immune system in which inflammation causes skin cells to multiply faster than normal.
"The type of inflammation seen in psoriasis is known to promote insulin resistance, and psoriasis and diabetes share similar genetic mutations suggesting a biological basis for the connection between the two conditions we found in our study," said senior author Dr. Joel Gelfand. He is a professor of dermatology and epidemiology at the University of Pennsylvania.
"We know psoriasis is linked to higher rates of diabetes, but this is the first study to specifically examine how the severity of the disease affects a patient's risk," he explained in a university news release.
"These findings are independent of traditional risk factors for diabetes and still show a strong connection between the increasing severity of psoriasis and the increasing risk of developing diabetes, which makes a strong argument for a causal relationship between the two," according to Gelfand.
However, the study does not prove a cause-and-effect link, just an association.
People with psoriasis should be checked routinely for how much of their body is affected by the disease, Gelfand noted. In addition, these patients should be targeted for diabetes prevention -- especially those with psoriasis on 10 percent or more of their body.
SOURCE: University of Pennsylvania, news release, Nov. 14, 2017


text:
\section{Introduction and statement of results}\label{sec:intro}

A partition of a natural number $n$
is a finite weakly decreasing sequence of positive integers
that sums to~$n$.
For $k\in \mathbb{N}$, $k>1$,
we consider the generating function $p_k(n)$
that enumerates $k$-regular partitions of $n$,
i.e., it counts partitions of $n$
where no part
is divisible by~$k$.
These generating functions arise
in many different contexts, in particular in connection
with the representation theory of the symmetric groups, Hecke algebras,
and related groups and algebras;
for a long time, this has been studied
both in combinatorics and number theory.

For the classical (unrestricted) partition function $p(n)$,
explicit formulae are known due to
the work of Hardy, Ramanujan and Rademacher, and more recent work
of Bruinier and Ono~\cite{BruinierOno}.
Based on a result due to Lehmer, the following inequality was
shown in a recent article by the second author and Ono~\cite{BesOno}:
\medskip

\centerline{
For any integers $a,b$ such that  $a,b >1$ and $a+b>9$, we have
$p(a)p(b)>p(a+b)$.}
\medskip

Also the cases of equality were determined
in \cite{BesOno}. The inequality above
was then used to study an ``extended partition function",
given by defining for a partition $\mu=(\mu_1,\mu_2,\ldots)$:
$$
p(\mu)=\prod_{j\ge 1} p(\mu_j).
$$
With $P(n)$ denoting the set of all partitions of~$n$,
the maximum
$$\text {\rm maxp}(n)=\max(p(\mu) \mid \mu \in P(n))$$
was determined explicitly in \cite{BesOno};
we recall this below in Theorem~\ref{thm:BO-max}.

Our aim is to prove a corresponding result for an
extension of the generating function $p_k(n)$ to
a function on the set $P_k(n)$ of
all $k$-regular partitions of $n$,
defined for $\mu=(\mu_1,\mu_2,\ldots)\in P_k(n)$ by:
$$
p_k(\mu)=\prod_{j\ge 1} p_k(\mu_j).
$$
We then determine on which partitions the maximum
$$\text {\rm maxp}_k(n)=\max(p_k(\mu) \mid \mu \in P_k(n))$$
is attained, and we use this to give an explicit formula for the
maximum.

By Theorem~\ref{thm:BO-max},  for $k>6$
nothing new happens, as all the partitions providing the maximal values
$\text {\rm maxp}(n)$ are already $k$-regular; hence we may restrict our considerations to
the cases where $2\le k\le 6$.
For this case,
we first show  in Theorem~\ref{thm:reg-inequality}
that $p_k(n)$ satisfies a similar inequality
as the one given for $p(n)$ above,
where again we specify the corresponding bounds explicitly.

For the maximum problem,
we find that the behavior is quite similar to the one
observed in~\cite{BesOno}, though we lose uniqueness for small~$k$;
see Theorem~\ref{thm:max-regular} for the detailed results.

\section{An analytic result on the generating function for $k$-regular partitions}

The main result of this section is
the following analytic inequality for the generating function~$p_k(n)$.
As mentioned above,
Theorem \ref{thm:reg-inequality} is the analogue of a result for the
ordinary partition function $p(n)$ in recent work by the second author and Ono~\cite{BesOno}.

\begin{theorem}\label{thm:reg-inequality}
For $k\in \mathbb{N}$, $2\leq k\leq 6$, we define parameters $n_k,m_k$ by the following table:
\begin{center}

\begin{tabular}{ | c | c|  c | c | c | c |}
\hline
$k$             & 2 & 3 &4  & 5 & 6\\
\hline
$n_k$       	&3 & 2 & 2 & 2  &2 \\

$m_k$           & 22& 17 & 9 & 9 & 9 \\
\hline

\end{tabular}
\end{center}
Then for any $a, b\in \mathbb{N}$ with $a,b\geq n_k$ and $a+b\geq m_k$
we have
   $$
       p_k(a) p_k(b) > p_k(a+b).
   $$
Furthermore, all the pairs $(a,b)$ with $2 \le a \le b$ for which this inequality
fails are given in the table below.
$$
\begin{array}{|l|l|l|}
\hline
k & (a,b) \text{ with } p_k(a) p_k(b) = p_k(a+b) & (a,b) \text{ with } p_k(a) p_k(b) < p_k(a+b)\\
\hline
2&  (3, 3), (3, 5), (3, 6), (3, 7), (3, 8), (4, 15),
& (2, *), (3, 4), (4, 4), (4, 5),(4, 6), (4, 7),
\\
  &   (4, 16), (4, 17),(5, 6), (5, 7), (5, 8)
  &  (4, 8), (4, 9), (4, 10),(4, 11), (4, 12), \\
  &
  & (4, 13), (4, 14),  (5, 5)   \\
\hline
3&  (2, 2), (2, 3), (3, 3), (3, 4), (3, 5), (3, 6),
& (3, 11), (3, 13)\\
  & (3, 7), (3, 8), (3, 9),(3, 10)  &  \\
\hline
4&  (2, 2), (2, 3),  (2, 5), (3, 3)  & (2, 4), (3, 5)  \\
\hline
5&   (2,3), (2,4)  & (2,2), (2,5), (3, 3), (3, 5)\\
\hline
6&   (2, 4), (2, 5), (2, 6) & (2, 2), (2, 3),  (3, 3)\\
\hline
\end{array}
$$
\end{theorem}

\iffalse{
Thanks to Theorem \ref{thm:reg-inequality} we are able to classify all the pairs $(a,b)$ satisfying $2 \le a \le b$ for which the inequality in
Theorem~\ref{thm:reg-inequality}
fails. These pairs are in the table below.
\ \\
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
$k$& $(a,b)$ with $p_k(a) p_k(b) = p_k(a+b)$ & $(a,b)$ with $p_k(a) p_k(b) < p_k(a+b)$\\
\hline
2&  (3, 3), (3, 5), (3, 6), (3, 7), (3, 8), (4, 15),
& (2, *), (3, 4), (4, 4), (4, 5),(4, 6), (4, 7),
\\
  &   (4, 16), (4, 17),(5, 6), (5, 7), (5, 8)
  &  (4, 8), (4, 9), (4, 10),(4, 11), (4, 12), \\
  &
  & (4, 13), (4, 14),  (5, 5)   \\
\hline
3&  (2, 2), (2, 3), (3, 3), (3, 4), (3, 5), (3, 6),
& (3, 11), (3, 13)\\
  & (3, 7), (3, 8), (3, 9),(3, 10)  &  \\
\hline
4&  (2, 2), (2, 3),  (2, 5), (3, 3)  & (2, 4), (3, 5)  \\
\hline
5&   (2,3), (2,4)  & (2,2), (2,5), (3, 3), (3, 5)\\
\hline
6&   (2, 4), (2, 5), (2, 6) & (2, 2), (2, 3),  (3, 3)\\
\hline
\end{tabular}
\end{center}
\ \\
}\fi

\bigskip

The main tool for deriving Theorem~\ref{thm:reg-inequality} is an analogue of a classical result of D.~H.\ Lehmer~\cite{Lehmer}.
To prove Theorem~\ref{thm:reg-inequality} we need precise approximations for $p_k(n)$ which have explicitly bounded error terms.
We will use work of Hagis \cite{Hagis} to obtain sufficient approximations in Theorem \ref{thm:Lehmer}.

\subsection{Preliminaries}
Hagis \cite{Hagis} proved an explicit formula for $p_k(n)$ that is analogous to Rade\-macher's formula for $p(n)$.
Before describing his theorem, we introduce several necessary quantities, most importantly the Kloosterman-type
sums $A(m,t,n,s,D)$ and the expressions $L(m,t,n,s,D)$.

Let $D$ divide $t+1$, let $J=J(t,D) := \frac{(t/D) - D}{24D}$, and let $a=a(t) := \frac{t}{24}$.
Let $I_1$ be the order one modified Bessel function of the first kind, and let $L(m,t,n,s,D)$ be given by
\begin{equation}\label{eq:Lsum}
L(m,t,n,s,D) := D^{\frac{3}{2}} m^{-1} \left(\frac{J-s}{n+a} \right)^{\frac{1}{2}} I_1 \left (4 \pi D m^{-1} \left( \frac{(J-s)(n+a)}{(t+1)} \right)^{\frac{1}{2}} \right).
\end{equation}

Several definitions are needed to define the modified Kloosterman sums $A(t,m,n,s,D)$.
First $g=g(m)$ is defined to be $\gcd(3,m)$ when $m$ is odd, and $8\gcd(3,m)$ when $m$ is even.
We define $M=M(m,D):= \frac{m}{D}$.
Additionally, we define $f=f(m)  := \frac{24}{g}$,
and define $r=r(m)$ to be any integer such that $fr \equiv 1 \pmod{gm}$.
Further, $G$ is defined to be analogous to $g$,
in that $G=G(m,D) := \gcd(3,M)$ when $M$ is odd and $G := 8\gcd(3,M)$ when $M$ is even.
Then we also let $B=B(m,D) := \frac{g}{G}$, and we define $A$ to be any integer such
that $AB \equiv 1 \pmod{GM}$. We also let $T=T (t,D):= \frac{t+1}{D}$, and choose $T'=T'(t,D)$
to satisfy $T T' \equiv 1 \pmod{GM}$. More importantly:
\begin{equation*}
U=U(t,m,D) := 1 - AB(t+1), V=V(t,m,D):=ABT'D - 1.
\end{equation*}

Hagis defines special roots of unity, $w(h,t,m,D)$, which satisfy the following:
\begin{equation*}
w (h,t,m,D) = C(h,t,m,D) \exp (2 \pi i(rUh + rV h')/gm).
\end{equation*}
The $C(h,t,m,D)$ satisfy $|C(h,t,m,D)| = 1$, and are independent of $h$ if $m$ is odd, or if $m$ is even and we restrict to $h \equiv d \pmod{8}$ for some odd $d$. In what follows we will not explicitly use the definitions of $C(h,t,m,D)$.

Then we define $A(m,t,n,s,D)$ to be the Kloosterman sum with multiplier system given by
\begin{equation}
A(m,t,n,s,D) = \sum_{\substack{h \pmod{m}, \\ \gcd(h,m)=1} } w(h,t,m,D) \exp(- 2 \pi i( nh  - DT' sh')/ m),
\end{equation}
where $h h' \equiv 1 \pmod{gm}$.

Let $p'(s)$ be the number of partitions of $s$ into an even number of distinct parts
minus the number of partitions of $s$ into an odd number of distinct parts;
by Euler's pentagonal number theorem, $p'(s)$ is $\pm 1$ if $s$ is a pentagonal number, and 0 otherwise.
Recall Glaisher's partition identity saying that
the number $p_k(n)$ of $k$-regular partitions of $n$
is equal to the number of partitions of $n$ where no part
has a multiplicity $\geq k$.
Using the previous notation, Hagis proved the following for the numbers $p_k(n)$
in \cite[Theorem 3]{Hagis}.
\begin{theorem}\label{thm:Hagis}
For all $k \ge 2$, the number of $k$-regular partitions of $n\in \mathbb{N}$ is given by
\begin{equation}
p_k(n) = \frac{2 \pi}{k} \sum_{\substack{D | k \\  D < k^{\frac{1}{2}}}}
\; \sum_{\substack{m\\ \gcd(k,m) = D}}^{\infty} \mbox{ }\sum_{s < J(k,D)} p'(s) A(m, k-1,n,s,D) L(m,k-1,n,s,D).
\end{equation}
\end{theorem}
For $2 \le k \le 6$, in the summations above, we only have $s = 0$ and $D \le 2$.
Thus, the formulae needed for Theorem \ref{thm:reg-inequality}
consist of one or two of the inner sums in Theorem~\ref{thm:Hagis}.

\subsection{Estimates in the theorem of Hagis}
In this section, we obtain an asymptotic for $p_k(n)$ with an explicitly bounded error term.

Let $\alpha_k$ be defined as follows:

\begin{equation}\label{eq:alpha}
\alpha_k := \left\{
  \begin{array}{lr}
   1.8  & \mbox{ if } k=2 \\
   9.84 & \mbox{ if } k=3 \\
   1.8 \cdot 3^{\frac{1}{2}}  & \mbox{ if } k=4 \\
   14.37 & \mbox{ if } k=5 \\
   1.23 \cdot 5^{\frac{1}{2}}  & \mbox{ if } k=6
  \end{array}
\right .
\end{equation}
We also let $\alpha_6 ':= 19.68$.
\begin{theorem}\label{thm:Lehmer}
For $n\in \mathbb{N}$, let $\mu=\mu(n,k) := \frac{\pi ((k-1)^2 + 24n(k-1))^{\frac{1}{2}}}{6 k^{\frac{1}{2}}}$.
\begin{enumerate}
 \item
 For $2\leq k\leq 5$ we have:
\begin{equation*}
p_k (n) = \frac{2 \pi}{k} \left( \frac{k-1}{k-1 + 24n} \right)^{\frac{1}{2}} I_1(\mu) + E_k(n)
\end{equation*}
where
\begin{equation*}
|E_k(n)| <  \frac{\alpha_k \pi}{k} \left( \frac{k-1}{(k-1)+ 24 n} \right)^{\frac{1}{2}} \frac{1}{\mu} e^{\mu} (1 + 5 \mu^2 e^{- \mu}).
\end{equation*}

 \item
 For $k=6$ we have:
\begin{equation*}
p_6(n) = \frac{\pi}{3} \left(\frac{5}{24n + 5} \right)^{\frac{1}{2}} I_1 (\mu) + E_6(n)
\end{equation*}
where
\begin{equation*}
\begin{split}
| E_6(n) | &< \frac{\pi}{3} \left(\frac{5}{24n + 5} \right)^{\frac{1}{2}} \frac{\alpha_6}{2} \frac{e^{\mu}}{\mu} \left( 1 + \delta (n) \right)
+ \frac{\pi}{3} \left(\frac{1}{24n + 5}\right)^{\frac{1}{2}} I_1 \left( \frac{ \mu}{10^{\frac{1}{2}}} \right) .
\end{split}
\end{equation*}
\end{enumerate}

where $\delta(n)  := 5 \mu^2 e^{-\mu} + \frac{2^{\frac{4}{3}} \alpha_6'}{\alpha_6} e^{\mu \left( \frac{1}{\sqrt{10}} -1 \right)} \left( 1 + e^{- \frac{\mu}{\sqrt{10}}} \frac{\mu^2}{2} \right) $.
\end{theorem}

\begin{remark}
Theorem \ref{thm:Lehmer} is analogous to \cite[(4.14)]{Lehmer} in the case of $p(n)$.
\end{remark}
To prove this theorem, we need some preparations.
The first is a bound on the divisor counting function $d(n)$.
\begin{lemma}\label{thm:divisorbound}
Let $d(n)$ denote the number of positive divisors of a positive integer $n$.
\begin{enumerate}
\item
For all $n$, $d(n) \le 3.57 n^{\frac{1}{3}}$.

\item
If $n$ is odd, then $d(n) \le 1.8 n^{\frac{1}{3}}$.

\item
If $\gcd(n,3) =1$, then $d(n) \le 2.46 n^{\frac{1}{3}}$.

\item
If $\gcd(n,5) = 1$, then $d(n) \le 3.05 n^{\frac{1}{3}}$.

\item
If $\gcd(n,6)=1$, then $d(n) \le 1.23 n^{\frac{1}{3}}$.
\end{enumerate}
\end{lemma}
\begin{remark}
Actually, it is known that $d(n) = O(n^{\epsilon})$ for any $\epsilon >0$ (see ~\cite{Wigert}). However,
to prove our main theorem it is necessary that we have exact constants for the bounds. We chose these exponents and constants carefully to ease the calculations in the proof of Theorem \ref{thm:reg-inequality}.
\end{remark}
\begin{proof}
Let $n = \prod_{i =1}^M  {p_i}^{a_i}$, where each $p_i$ is prime. Then $d(n) = \prod_{i = 1}^M (1 + a_i)$. We follow the classical method in \cite{Hardy} of bounding $\prod_{i = 1}^M \frac{a_i+ 1}{p_i^{\frac{a_i}{3}}}$. For $p_i \ge 11$, we have $\frac{a_i + 1}{p_i^{\frac{ a_i}{3}}} \le 1$ for $a_i \ge 1$. For the remaining $p_i$, the quantity $\frac{a_i + 1}{p_i^{\frac{ a_i}{3}}}$ is maximized when $a_i$ is equal to $ 3,2,1$ and $1$ for $p_i$ equal to $2,3,5$ and $7$, respectively. The lemma follows by maximizing $\prod_{i = 1}^M \frac{a_i+ 1}{p_i^{\frac{a_i}{3}}}$ over $n$ which respect each of the given divisibility constraints.
\end{proof}
The next lemma is a bound on $A(m,k-1,n,0,D)$, which is related to the classical Kloosterman sums defined below;
it is a slight modification  of \cite[Theorem 12]{Lehmer}.
\begin{definition}
Let $a,b,m\in \mathbb{N}$. The Kloosterman sum $S(a,b,m)$ is defined by
\begin{equation*}
S(a,b,m) : = \sum_{\substack{1 \le h \le m-1 \\ \gcd(h,m) = 1}} e^{2 \pi i (a h + bh')/m},
\end{equation*}
where $h'$ is the multiplicative inverse of $h$ modulo $m$.
\end{definition}

 Weil proved the following bound (see \cite[Theorem 4.5]{Iwaniec}):
\begin{theorem}\label{thm:weilbound}
Let $a,b,m\in \mathbb{N}$.
\begin{equation*}
|S(a,b,m)| \le d(m) m^{\frac{1}{2}} \gcd(a,b,m)^{\frac{1}{2}}.
\end{equation*}
\end{theorem}

We will use this bound in the following lemma.

\begin{lemma}\label{thm:Asums}
\begin{enumerate}
\item
For $2 \le k \le 6$, and for all $n,m \ge 1$ with $\gcd(k,m) = 1$, we have
\begin{equation*}
|A(m,k-1,n,0,1)| < \alpha_k m^{\frac{5}{6}}.
\end{equation*}

\item
For all $n,m \ge 1$ with $\gcd(6,m) = 2$, we have
\begin{equation*}
|A (m,5,n,0,2) | < \alpha_6' m^{\frac{5}{6}}.
\end{equation*}
\end{enumerate}
\end{lemma}
\begin{proof} We will follow Hagis' argument in \cite[Theorem 2]{Hagis}.
Our strategy is to rewrite $A(m,k-1,n,0,D)$ as a sum of ordinary Kloosterman sums and apply
Theorem~\ref{thm:weilbound}.

In order to bound the ordinary Kloosterman sums, we will need to be able to bound certain greatest common divisors.
We use the notation introduced at the beginning of the section,
and we begin by stating a series of bounds for
$\gcd(Ur - gn, rV, gm)$ and $\gcd(Ur - gn, rV + \frac{wgm}{8}, gm)$ which depend on $k$ and $D$.
These are straightforward to verify from their definitions.

For $D = 1$, $ 2 \le k \le 6$ we have $\gcd(r, gm) = 1 $ and $\gcd(k,gm) = 1$, thus \\
$\gcd(rV, gm) = \gcd(kV,gm)$. Then since $kV = k(T' - 1)  \equiv 1-k \pmod{gm}$, we have
\[ \gcd(rV, gm)  = (1-k, gm) \le k-1. \]

Let $k= 3,5$, let $D = 1$, and let $m$ be even. Note that $\gcd(r,g) = 1$ and $U= k-1$, which implies
$\gcd(Ur - gn , g) = \gcd(k-1, g)$. Also for $1 \le w \le 8$, we have
\[\gcd(rV + \frac{wgm}{8} , m) = \gcd(V, m) = \gcd( 1-k,m). \]
Therefore $\gcd(Ur -gn, rV + \frac{wgm}{8} , gm)$ divides $(k-1)^2$, so it must be $1,2,4,8$, or 16.
However, the highest power of $2$ that $Ur - gn$ can be divisible by is $k-1$, because $g$ is divisible by $8$, and $r$ is odd, and $Ur - gn = r (1-k) - gm$. Thus we have:
\begin{equation*}
\gcd(Ur - gn, Vr + \frac{wgm}{8}, gm) \le k-1.
\end{equation*}

For the last bound, we let $k=6$ and $D=2$. Then we have $g= 8$, $T = 3$, $M = \frac{m}{2}$, and $\gcd(6,m) = 2$.
So $\gcd(rV + \frac{wgm}{8}, m) = \gcd(V, m) = \gcd(2 ABT' - 1,m)$. Now we have \\ $2 AB \equiv 2 \pmod{m}$ and $6T' \equiv 2 \pmod{m}$, thus \[ \gcd(V,m) = \gcd(2T' - 1,m) = \gcd(3(2 T' - 1),m )  = 1. \] Therefore we have $\gcd( rU - gn, rV + \frac{wgm}{8},gm)\le g = 8.$

To use these bounds, we rewrite $A(m,k-1,n,0,D)$ as a sum over a reduced residue class modulo $gm$:
\begin{equation*}
A(m,k-1,n,0,D) =
\frac{1}{g} \sum_{\substack{h \mod{m} \\ \gcd(h,m) = 1}} C(h,k-1,m,D) \exp{(2 \pi i ((Ur-gn)h + rVh')/gm))}.
\end{equation*}
For odd $m$, $C(h,k-1,m,D)$ does not depend on $h$. Therefore we have
\begin{equation*}
A(m,k-1,n,0,1) = C(1,k-1,m,1)  \frac{1}{g} \sum_{\substack{h \mod{m} \\ \gcd(h,m) = 1}} \exp(2 \pi i  ((rU-gn)h + rVh')/gm) .
\end{equation*}
The sum on the right is an ordinary Kloosterman sum, so by Theorem \ref{thm:weilbound} we have, for all odd~$m$:
\begin{equation*}
|A(m,k-1,n,0,1)|  = |S(Ur - gn, rV, gm)| \le \frac{1}{g} d(gm) \gcd(Ur - gn, rV, gm)^{\frac{1}{2}} (gm)^{\frac{1}{2}}.
\end{equation*}
Then by Lemma \ref{thm:divisorbound} and the bounds at the beginning of the proof, it follows that for all $m$ such that $2 \nmid m$ and $\gcd(k,m) = 1$, we have:
\begin{equation*}
|A(m,k-1,n,0,1)| \le (k-1)^{\frac{1}{2}} \cdot 1.8 \cdot m^{\frac{5}{6}}.
\end{equation*}
This proves the lemma for $k=2,4$, and for $k=3,5$ in the case of $m$ being odd.
Similarly, for $k=6$, Lemma \ref{thm:divisorbound}, we have:
\begin{equation*}
|A(m,5,n,0,1)| \le (k-1)^{\frac{1}{2}} \cdot 1.23  \cdot m^{\frac{5}{6}}.
\end{equation*}
For $k=6$, $D=1$, the proof is complete.

If $m$ is even, we write
\begin{align*}
A(m,k-1,n, 0, D) &= A_1(m,k-1,n,0,D) + A_3(m,k-1,n,0,D) \\
&\qquad + A_5(m,k-1,n,0,D) + A_7(m,k-1,n,0,D),
\end{align*}
where
\begin{equation*}
A_d (m,k-1,n,0,D) = \frac{1}{g} \sum_{\substack{h \mod{gm}, \\ h \equiv d \mod{8}, \\ \gcd(h,m) = 1}} C(h,k-1,m,D) \exp{(2 \pi i  ((rU-gn)h +rVh')/gm)}.
\end{equation*}
Over each $d$, the coefficient $C(h,k-1,m,D)$ does not depend on $h$, so
\begin{equation*}
A_d (m,k-1,n,0,D) = C(d,k-1,m,D) \frac{1}{g} \sum_{\substack{h \pmod{gm},  \\ h \equiv d \pmod{8},\\ \gcd(h,m) = 1 }} \exp{(2 \pi i  ((rU-gn)h +rVh')/gm)}.
\end{equation*}
By the formula on page 266 of \cite{Salie}, for $d d' \equiv 1 \pmod{8}$, we have:
\begin{equation*}
A_d (k-1,m,n,0,D) = \frac{1}{8g} C(d,k-1,m,D) \sum_{w = 1}^8 e^{2\pi i \frac{d' w}{8}}  S(Ur-gn, Vr + \frac{w gm}{8}; gm).
\end{equation*}
By Theorem \ref{thm:weilbound},
\begin{equation*}
A_d (m,k-1,n,0,D) = \frac{1}{8g} C(d,k-1,m,D) \sum_{w = 1}^8 e^{- \frac{2 \pi i}{8} d'w} \gcd(Ur-gn, Vr + \frac{w gm}{8},gm)^{\frac{1}{2}} d(gm) (gm)^{\frac{1}{2}}.
\end{equation*}
For $k=3$, by the bounds at the beginning of the proof we have:
\begin{equation*}
A_d(m,2,n,0,1) \le 8 \cdot \frac{1}{8g} \cdot 2^{\frac{1}{2}}\cdot 2.46 (gm)^{\frac{1}{3}} \cdot (gm)^{\frac{1}{2}} \le 2.46 m^{\frac{5}{6}}.
\end{equation*}
Similarly for $k=5$, if $3 | m$, by our previous bounds we have:
\begin{equation*}
A_d (m,4,n,0,1) \le  8 \cdot \frac{1}{8 \cdot 24} \cdot 4^{\frac{1}{2}} \cdot 3.05 \cdot (24m)^{\frac{1}{3}} \cdot (24m)^{\frac{1}{2}} \le 3.592 m^{\frac{5}{6}}.
\end{equation*}
If $3 \nmid m$, then we have:
\begin{equation*}
|A_d (m,4,n,0,1)| \le  8 \cdot \frac{1}{8\cdot 8} \cdot 4^{\frac{1}{2}} \cdot 2.46 \cdot (8m)^{\frac{1}{3}} \cdot (8m)^{\frac{1}{2}} \le 3.48 m^{\frac{5}{6}} .
\end{equation*}
We note that $|A(m,k-1,n,0,D)| \le 4 |A_d (m,k-1,n,0,D)|$. Comparing these bounds to the bounds in the odd $m$ case, we conclude that for $k = 3,5$, the desired bound holds whenever $\gcd(m,k) = 1$.

For $\gcd(6,m) = 2$, we have:
\begin{equation*}
|A(m,5,n,0,2) | \le 4 |A_d (m,5,n,0,2)| \le 4 \cdot (8\cdot 8^{\frac{1}{2}} \cdot \frac{1}{8g}  \cdot 2.46 (gm)^{\frac{1}{3}} \cdot (gm)^{\frac{1}{2}}) \le 19.6 m^{\frac{5}{6}}.
\end{equation*}
This completes the proof.
\end{proof}

\bigskip

Now we come to the {\bf proof of Theorem \ref{thm:Lehmer}}.
For $2 \le k \le 5$, Theorem \ref{thm:Hagis} says
\begin{equation}\label{eq:hagis1}
p_k(n) = \frac{2 \pi}{k} \sum_{\substack{m=1\\ \gcd(k,m) = 1}}^{\infty} m^{-1} \left(\frac{k-1}{(k-1)+ 24 n} \right)^{\frac{1}{2}} A(m,k-1,n,0,1) I_1 \left( \frac{\mu}{m} \right),
\end{equation}
and for $k=6$, Theorem \ref{thm:Hagis} says
\begin{equation}\label{eq:hagis2}
\begin{split}
p_6(n) = &\frac{\pi}{3} \frac{5^{1/2}}{(5 + 24n)^{\frac{1}{2}}} \sum_{m= 1}^{\infty} \frac{1}{m} A(m,5,n,0,1) I_1 \left(\frac{\mu}{m} \right) \\
&+ \frac{\pi}{3} \frac{1}{(5 + 24n)^{\frac{1}{2}}} \sum_{(3,a) = 1}^{\infty} \frac{1}{a} A(2a,5,n,0,2) I_1 \left( \frac{ \mu}{10^{\frac{1}{2}} a} \right).
\end{split}
\end{equation}

Let $\alpha = \frac{1}{6}$. Our proof works by bounding the sums in~(\ref{eq:hagis1}) and~(\ref{eq:hagis2}).
We have, for any $\nu \neq 0$,
\begin{align*}
|\sum_{m= N+1}^{\infty} m^{-1} A(m,k-1,n,0,1) I_1 \left(\frac{\nu}{m} \right)|  &\le \sum_{m= N+1}^{\infty} \alpha_k m^{- \alpha} \sum_{j=0}^{\infty} \frac{ (\frac{\nu}{2m})^{2j+1}}{j! (j+1)!} \\
&< \alpha_k \int_N^{\infty} x^{-\alpha} \sum_{j=0}^{\infty} \frac{ (\frac{\nu}{2x})^{2j+1}}{j! (j+1)!} \mathbf{D}{x}.
\end{align*}
We substitute $t = \frac{\nu}{2x}$.
\begin{align*}
|\sum_{m= N+1}^{\infty}m^{-1} A(m,k-1,n,0,1) I_1 \left(\frac{\nu}{m} \right)| &\le \alpha_k  \int_0^{\frac{\nu}{2N}} (\frac{\nu}{2t})^{-\alpha} \sum_{j=0}^{\infty} \frac{t^{2j+1}}{j! (j+1)!} \frac{\nu}{2t^2} \mathbf{D}{t} \\
&= \alpha_k  (\frac{\nu}{2})^{1 - \alpha} \int_0^{\frac{\nu}{2N}} \sum_{j=0}^{\infty} \frac{(t^{2j -1 +\alpha})}{j! (j+1)!} \mathbf{D}{t} \\
&\le \alpha_k \ (\frac{\nu}{2})^{1 - \alpha}  \sum_{j=0}^{\infty} \frac{(\frac{\nu}{2N})^{2j+\alpha}}{j! (j+1)! (2j + \alpha)} \\
&\le \alpha_k  (\frac{\nu}{2})^{1- \alpha}  \left(\frac{ (\frac{\nu}{N})^{\alpha}}{2 \alpha} + \sum_{j=2}^{\infty} \frac{((\frac{\nu }{N})^{2j-2 + \alpha})}{(2j)!}  \right)2^{1 - \alpha}\\
&\le \alpha_k N^{2 + \alpha} \frac{1}{\nu} \left( \cosh(\nu/N) -1 + \frac{5}{2} \left(\frac{\nu}{N} \right)^2 \right). \\
\end{align*}

To bound $\sum_{a = N+1}^{\infty} (2a)^{-1} A(2a,5,n,0,2) I(\frac{\nu}{2a})$, we replace $\alpha_6$ with $\alpha_6'$ in the previous argument.
To complete the proof, we let $N = 1$, and apply the above inequality to the sums in Theorem~\ref{thm:Hagis}, where $\nu=\mu$ for $2 \le k \le 5$,
and for $k = 6$, $\nu$ is set to be $\mu$ and $\frac{\mu}{\sqrt{10}}$ in the first and second sum, respectively.
\qed

\subsection{Proof of Theorem \ref{thm:reg-inequality}}\label{sec:proof-reg-inequality}
Our proof is analogous to the proof of  \cite[Theorem 2.1]{BesOno}.

By well known properties of Bessel functions, such as the bounds in
(9.8.4) of \cite{Abram}, for $x \ge 37.5$ the modified Bessel function $I_1(x)$
is bounded by
\begin{equation*}
N \le x^{\frac{1}{2}} e^{-x} I_1(x) \le M
\end{equation*}
where $N = 0.394$, $M = 0.399$.

First, let $2 \le k \le 5$, and let $\beta := \frac{\alpha_k}{2}$. Then by Theorem \ref{thm:Lehmer}, for $n \ge 450$ we have:
\begin{align*}
\frac{2 \pi}{k} \left( \frac{k-1}{k-1 + 24n} \right)^{\frac{1}{2}} & \left( N - \frac{\beta}{\sqrt{\mu}} \left( 1 + 5 \mu^2 e^{-\mu} \right) \right) \frac{e^{\mu}}{\sqrt{\mu}} < p_k(n) \\
&< \frac{2 \pi}{k}  \left(\frac{k-1}{k-1 + 24n}\right)^{\frac{1}{2}} \frac{e^{\mu}}{\sqrt{\mu}} \left(M + \frac{\beta}{\sqrt{\mu}}\left( 1 + 5 \mu^2 e^{-\mu} \right) \right).
\end{align*}
We assume $a \le b$ and write $b = \lambda a$ for some $\lambda \ge 1$. Then it is sufficient to show
\begin{equation*}\label{lambda}
e^{\mu(a) + \mu(\lambda a) - \mu(\lambda a + a)} > S_{a,k}(\lambda) (k-1 + 24a)^{\frac{3}{4}},
\end{equation*}
where
\begin{equation*}
S_{a,k}(\lambda) :=C_k \frac{ \left(M+ \frac{\beta}{\sqrt{\mu(\lambda a +  a)} }\left( 1 + 5 \mu(\lambda  a+ a)^2 e^{-\mu (\lambda a + a)} \right)  \right) }{\left( N - \frac{\beta}{\sqrt{\mu( \lambda a )}}\left( 1 + 5 \mu(\lambda a)^2 e^{-\mu(\lambda a)} \right) \right) \left(  N - \frac{\beta}{\sqrt{\mu(a)}}  \left( 1 + 5 \mu(a)^2 e^{-\mu (a)} \right)\right) },
\end{equation*}
for $C_k := \frac{k^{\frac{3}{4}}}{2 (\pi(k-1))^{\frac{1}{2}}}$.
For a fixed $a$, the left-hand side of the inequality is increasing for all $\lambda \ge 1$, and the right-hand side is decreasing. Thus, for any given $a$, to prove Theorem \ref{thm:reg-inequality} for $b \ge a$, it suffices to verify the inequality for $\lambda =1$. Taking the natural logarithm of each side, it is straightforward to verify that the inequality holds for $a \ge 1000$ for $k=2,4$, and holds for $a \ge 5 \cdot 10^4$ for $k=3,5$. Then for each of the remaining $a$, we wish to find $\lambda_{a,k}$ such that for $\lambda \ge \lambda_{a,k}$:
\begin{align*}
p_k(a) & \frac{2 \pi}{k} \left( \frac{k-1}{k-1 + 24\lambda a} \right)^{\frac{1}{2}} \left( N - \frac{\beta}{\sqrt{\mu (\lambda a)}} \left( 1 + 5 \mu(\lambda a)^2 e^{-\mu (\lambda a)} \right) \right) \frac{e^{\mu (\lambda a)}}{\sqrt{\mu( \lambda a)}}  > \\
& \frac{2 \pi}{k}  \left(\frac{k-1}{k-1 + 24(\lambda a + a)}\right)^{\frac{1}{2}} \frac{e^{\mu (\lambda a + a)}}{\sqrt{\mu (\lambda a+ a)}} \left(M + \frac{\beta}{\sqrt{\mu (\lambda a + a)}}\left( 1 + 5 \mu( \lambda a + a)^2 e^{-\mu (\lambda a + a)} \right) \right) .
\end{align*}
For $a \ge 20$, $k = 2,4$, $\lambda_{a,k} = \frac{1000}{a}$ suffices. For $a \le 20$, $k = 3,5$, $\lambda_{a,k} = \frac{50000}{a}$ suffices. For smaller $a$, the needed $a \lambda_{a,k} $ values can be as large as $4 \cdot 10^5$, except when $k=5$ and $a=2$, where the larger bound in Theorem \ref{thm:Asums} for $k=5$ causes the needed $\lambda_{a,k}$ values to be much larger. All other cases are reduced to checking a large but finite number of pairs $(a,b)$, where $a \le 5\cdot 10^4$ and $b \le \lambda_{a,k} a$. We carried out these calculations using Sage mathematical software \cite{Sage}. To ease our calculation,
we proved the inequality $ p_5(2) \cdot p_5(b) > p_5(b + 2)$  for $b \ge 75$ with a combinatorial argument (see the end of the section), and used Sage \cite{Sage} to check the remaining pairs.

Now we handle the $k=6$ case. This case is very similar to the cases for
$2 \le k \le 5$, but because of the second summation in (\ref{eq:hagis2}), we have additional, non-dominant terms in our expressions. Using Theorem \ref{thm:Lehmer} and factoring out the leading term, we obtain
\begin{align*}
\frac{\pi}{3} \frac{\sqrt{5}}{\sqrt{24n+5}} \frac{e^{\mu}}{\sqrt{\mu_6}} &\left( N \left(1 -  \eta(n) \right) - \frac{\beta}{\sqrt{\mu}} \left( 1 + \delta(n) \right) \right)
< p_6(n) \\
&< \frac{\pi}{3} \frac{\sqrt{5}}{\sqrt{24n+5}} \frac{e^{\mu}}{\sqrt{\mu_6}} \left( M \left(1 + \eta(n) \right) + \frac{\beta}{\sqrt{\mu}} \left( 1 + \delta(n) \right) \right),
\end{align*}
where $\eta(n) := \left( \frac{2}{5}\right)^{\frac{1}{4}} e^{ \mu \left( 10^{- \frac{1}{2}} - 1 \right) }$.
The desired inequality is implied by
\begin{equation*}
e^{\mu(a) + \mu(\lambda a) - \mu(\lambda a + a)} > S_{a,k}(\lambda) (k-1 + 24a)^{\frac{3}{4}},
\end{equation*}
where
\begin{equation*}
S_{a}(\lambda) =  C_6 \frac{ \left( M( 1 + \eta(\lambda a + a) ) +\frac{\beta}{ \sqrt{\mu((\lambda +1)a)}} \left(1  +  \delta(\lambda a + a) \right)  \right)}{\left(N ( 1 - \eta(a))  - \frac{\beta}{\sqrt{\mu(\lambda a)}}\left(1 +\delta (\lambda a) \right) \right) \left( N( 1 - \eta(a)) - \frac{\beta}{2 \sqrt{\mu_6(a)}}\left(1 +\delta(a) \right) \right) },
\end{equation*}
and $C_6 =  \frac{3 }{ \pi \sqrt{5}} (\frac{6^{\frac{3}{2}}}{\sqrt{5}\pi})^{\frac{1}{2}}$. As before, it suffices to verify that this is true for $\lambda  = 1$, which is straightforward for $a \ge 3500$.
Then for each $a \le 3500$, we wish to find $\lambda_{a,6}$ such that for  all $\lambda \ge \lambda_{a,6}$,
\begin{align*}
p_6(a) &\frac{\pi}{3} \frac{\sqrt{5}}{\sqrt{24(\lambda a)+5}} \frac{e^{\mu(\lambda a)}}{\sqrt{\mu(\lambda a)}} \left(  N( 1 - \eta(\lambda a)) - \frac{\beta}{\sqrt{\mu (\lambda a) }}\left(1  + \delta(\lambda a) \right) \right) > \\
&\frac{\pi}{3} \frac{\sqrt{5}}{\sqrt{24(\lambda a +a)+5}} \frac{e^{\mu(\lambda a + a)}}{\sqrt{\mu ((\lambda + 1)a)}}  \left( M(1 + \eta(\lambda a + 1))+ \frac{\beta}{\sqrt{\mu(\lambda a + a)}} \left( 1 + \delta(\lambda a + a) \right) \right) .
\end{align*}
 It is straightforward to verify that the inequality holds for $\lambda \ge \frac{3500}{a}$ for all $a \ge 4$. For $a = 2,3,4$, the inequality holds for $\lambda \ge \frac{50000}{a}$. This reduces the $k=6$ case to a finite number of pairs $(a,b)$ to check, which we computed with Sage \cite{Sage}.

Finally, we prove that for $b \ge 75$, we have $p_5(b + 2) < 2 \, p_5(b)$.
To do this, we separate the 5-regular partitions of $b+2$ into two disjoint sets.
Let $S_1$ be the set of 5-regular partitions of $b+2$ which contain 1 as a part with multiplicity at least two.
Let $S_2$ contain all the other 5-regular partitions of $b + 2$.
Let $S$ be the set of 5-regular partitions of $b$.
We map $S_1$ and $S_2$ each injectively into $S$.
To map $S_1$ injectively into $S$, for each partition in $S_1$, simply remove two parts~1.

Next, we define an injective map from $S_2$ into $S$.
Let $\gamma=(\gamma_1,\gamma_2,\ldots,\gamma_\ell)$ be a partition in~$S_2$.
If $\gamma_\ell \ge 2$ and $\gamma_1 \ge 7$, then $\gamma$ is mapped to $(\gamma_2, \ldots,\gamma_\ell,1^{\gamma_1-2})$
(here, we use exponential notation for multiplicities).
If $\gamma_\ell \ge 2$ and $\gamma_1 < 7$, then if 2 has multiplicity at least 5 in $\gamma$,
replace five parts~2 with eight parts~1.
Otherwise, if $\gamma$ has five parts 3, we replace them with thirteen parts~1.
If $\gamma$ has five parts 4, then we replace them with eighteen parts~1.
Otherwise, $\gamma$  must have at least five parts 6, which we replace with 28 parts~1.
Finally, assume $\gamma_\ell = 1$.
If $\gamma_{\ell-1} \equiv 1 \pmod{5}$, then we map $\gamma$ to
$(\gamma_1, \ldots,\gamma_{\ell-2},\gamma_{\ell-1}-4,1^3)$.
Otherwise,  $\gamma$ is mapped to $(\gamma_1, \ldots,\gamma_{\ell-2},\gamma_{\ell-1}-1)$.
Note that the mapping from $S_2$ to $S$
is not onto by considering any 5-regular partition of $b$ which contains exactly two ones.
Thus we obtain the inequality $p_5(b + 2) < 2 p_5(b)$ for $b \ge 75$.

This completes the proof of the inequality stated in Theorem~\ref{thm:reg-inequality}.

The exceptional pairs given in the table are then easily obtained by direct computations.
\qed

\bigskip

\section{The maximum property}\label{sec:max}

We first recall \cite[Theorem 1.1]{BesOno}.
\begin{theorem}\label{thm:BO-max}
Let $n\in \mathbb{N}$. For $n\geq 4$ and $n\neq 7$,
the maximal value $\text {\rm maxp}(n)$
of the partition function on $P(n)$
is attained exactly at the partitions (in exponential notation)
$$
\begin{array}{ll}
(4^{\frac n4 }) & \text{when } n \equiv 0 \pmod 4\\
(5,4^{\frac{n-5}4}) & \text{when } n \equiv 1 \pmod 4\\
(6,4^{\frac{n-6}4}) & \text{when } n \equiv 2 \pmod 4\\
(6,5,4^{\frac{n-11}4}) & \text{when } n \equiv 3 \pmod{4}
\end{array}
$$
For $n=7$, the maximal value is $\text {\rm maxp}(7)=15$,
attained at the two partitions $(7)$ and $(4,3)$.

In particular, if $n\geq 8$, then
$$
\text {\rm maxp}(n)=\begin{cases}
5^{\frac{n}{4}}\ \ \ \ \ \ \ \
 & \text {\rm if $n\equiv 0\pmod 4$},\\
7\cdot 5^{\frac{n-5}{4}}\ \ \ \ \ \ \ \ &\text {\rm if $n\equiv 1\pmod 4$},\\
11\cdot 5^{\frac{n-6}{4}}\ \ \ \ \ \ \ \ &\text {\rm if $n\equiv 2\pmod 4$},\\
11\cdot7\cdot 5^{\frac{n-11}{4}}\ \ \ \ \ \ \ \ &\text {\rm if $n\equiv 3\pmod 4$}.
\end{cases}
$$
\end{theorem}

Since the partitions where the maximum of $p(n)$ is attained
on $P(n)$ are $k$-regular for any $k>6$, in the following
it suffices to consider the cases $k\in \{2,3,4,5,6\}$.

\begin{theorem}\label{thm:max-regular}
Let $k\in \mathbb{N}$, $k>1$. Let $n\in \mathbb{N}$.

\begin{enumerate}
\item[(i)] $k=2$.
For $n\geq 9$ and $n\neq 11$,
the maximal value $\text {\rm maxp}_2(n)$
of $p_2(n)$ on $P_2(n)$
is attained exactly at the partitions
$$
\begin{array}{ll}
(9^a,3^b) & \text{when } n \equiv 0 \pmod 3 \\
(9^a,7,3^b) & \text{when } n \equiv 1 \pmod 3 \\
(9^a,7,7,3^b) & \text{when } n \equiv 2 \pmod 3 \\
\end{array}
$$
where $a,b\in \mathbb{N}_0$ may be chosen arbitrarily
as long as we have partitions of~$n$.
\\
In particular, we have
$$
\text {\rm maxp}_2(n)= \left\{
\begin{array}{ll}
2^{\frac n3} & \text{when } n \equiv 0 \pmod 3\\
5\cdot 2^{\frac {n-7}3} & \text{when } n \equiv 1 \pmod 3\\
5^2\cdot 2^{\frac {n-14}3} & \text{when } n \equiv 2 \pmod 3\\
\end{array}
\right.
$$
\item[(ii)] $k=3$.
For $n\geq 2$ and $n\neq 3$,
the maximal value $\text {\rm maxp}_3(n)$
of $p_3(n)$ on $P_3(n)$
is attained exactly at the partitions
$$
\begin{array}{ll}
(4^a,2^b) & \text{when } n \equiv 0 \pmod 2\\
(5,4^a,2^b) & \text{when } n \equiv 1 \pmod 2\\
\end{array}
$$
where $a,b\in \mathbb{N}_0$ may be chosen arbitrarily
as long as we have partitions of~$n$.
\\
In particular, we have
$$
\text {\rm maxp}_3(n)= \left\{
\begin{array}{ll}
2^{\frac n2} & \text{when } n \equiv 0 \pmod 2\\
5\cdot 2^{\frac {n-5}2} & \text{when } n \equiv 1 \pmod 2\\
\end{array}
\right.
$$
\item[(iii)] $k=4$.
For $n\geq 2$,
the maximal value $\text {\rm maxp}_4(n)$
of $p_4(n)$ on $P_4(n)$
is attained exactly at the partitions
$$
\begin{array}{ll}
(6^a,3^b) & \text{when } n \equiv 0 \pmod 3\\
(6^a,3^b,2,2),(7,6^a,3^b),(6^a,5,3^b,2),
(6^a,5,5,3^b) & \text{when } n \equiv 1 \pmod 3\\
(6^a,3^b,2) , (6^a,5,3^b)& \text{when } n \equiv 2 \pmod 3\\
\end{array}
$$
where $a,b\in \mathbb{N}_0$ may be chosen arbitrarily
as long as we have partitions of~$n$,
and with the understanding that
partitions with given parts $2,5,7$ of positive multiplicity
do not occur when $n$ is too small.
\\
In particular, we have
$$
\text {\rm maxp}_4(n)= \left\{
\begin{array}{ll}
3^{\frac n3} & \text{when } n \equiv 0 \pmod 3\\
4\cdot 3^{\frac {n-4}3} & \text{when } n \equiv 1 \pmod 3\\
2\cdot 3^{\frac {n-2}3} & \text{when } n \equiv 2 \pmod 3\\
\end{array}
\right.
$$
\item[(iv)] $k=5$.
For $n\geq 2$,
the maximal value $\text {\rm maxp}_5(n)$
of $p_5(n)$ on $P_5(n)$
is attained exactly at the partitions
$$
\begin{array}{ll}
(4^{\frac n4 }) & \text{when } n \equiv 0 \pmod 4\\
(4^{\frac{n-5}4},3,2),(6,4^{\frac{n-9}4},3) & \text{when } n \equiv 1 \pmod 4\\
(4^{\frac{n-2}4},2),(6,4^{\frac{n-6}4})  & \text{when } n \equiv 2 \pmod 4\\
(4^{\frac{n-3}4},3)  & \text{when } n \equiv 3 \pmod 4\\
\end{array}
$$
with the understanding that
partitions with given parts $2,3,6$ of positive multiplicity
do not occur when $n$ is too small.
\\
In particular, we have
$$
\text {\rm maxp}_5(n)= \left\{
\begin{array}{ll}
5^{\frac n4} & \text{when } n \equiv 0 \pmod 4\\
6\cdot 5^{\frac {n-5}4} & \text{when } n \equiv 1 \pmod 4\\
2\cdot 5^{\frac {n-2}4} & \text{when } n \equiv 2 \pmod 4\\
3\cdot 5^{\frac {n-3}4} & \text{when } n \equiv 3 \pmod 4\\
\end{array}
\right.
$$
\item[(v)] $k=6$.
For $n\geq 2$,
the maximal value $\text {\rm maxp}_6(n)$
of $p_6(n)$ on $P_6(n)$
is attained exactly at the partitions
$$
\begin{array}{ll}
(4^{\frac n4 }) & \text{when } n \equiv 0 \pmod 4\\
(5,4^{\frac{n-5}4}) & \text{when } n \equiv 1 \pmod 4\\
(4^{\frac{n-2}4},2)  & \text{when } n \equiv 2 \pmod 4\\
(4^{\frac{n-3}4},3)  & \text{when } n \equiv 3 \pmod 4\\
\end{array}
$$
\\
In particular, we have
$$
\text {\rm maxp}_6(n)= \left\{
\begin{array}{ll}
5^{\frac n4} & \text{when } n \equiv 0 \pmod 4\\
7\cdot 5^{\frac {n-5}4} & \text{when } n \equiv 1 \pmod 4\\
2\cdot 5^{\frac {n-2}4} & \text{when } n \equiv 2 \pmod 4\\
3\cdot 5^{\frac {n-3}4} & \text{when } n \equiv 3 \pmod 4\\
\end{array}
\right.
$$
\end{enumerate}
\end{theorem}

\begin{proof
(i) We will need the partitions where $\text {\rm maxp}_2(n)$ is attained
for $n\leq 22$;
these are given in Table~\ref{table:4} (computed by Maple).
We see that the assertion holds as stated up to $n=22$.
\medskip

\begin{table}[h]
\caption{\label{table:4} Maximum value partitions $\mu$ for $k=2$}
\begin{tabular}{|r|cc|cc|cc| }
\hline \rule[-3mm]{0mm}{8mm}
$n$ && $p_2(n)$ && $\text {\rm maxp}_2(n)$ && $\mu$ \\
\hlin
$1$  && 1 && 1 && (1) \\
$2$  && 1 && 1 && (1,1) \\
$3$  && 2 && 2 && (3) \\
$4$  && 2 && 2 && (3,1) \\
$5$ && 3 && 3 && (5) \\
$6$&&  4 && 4 && $(3^2)$ \\
$7$&& 5 && 5 && (7)\\
$8$&& 6 && 6 && (5,3) \\
$9$&&  8 && 8 && $(9), (3^3)$ \\
$10$&& 10  && 10 && (7,3) \\
$11$&& 12 && 12 && $(11), (5,3^2)$ \\
$12$&& 15 &&  16 && $(9,3),(3^4)$ \\
$13$&& 18 &&  20 && $(7,3^2)$ \\
$14$&& 22 && 25 && $(7^2)$ \\
$15$&& 27 && 32 && $(9,3^2), (3^5)$ \\
$16$&& 32 && 40 && $(9,7), (7,3^3)$\\
$17$&& 38 && 50 && $(7^2,3)$\\
$18$&& 46 && 64 && $(9^2), (9,3^3),(3^6)$\\
$19$&& 54 && 80 && $(9,7,3),(7,3^4)$ \\
$20$&& 64 && 100&& $(7^2,3^2)$ \\
$21$&& 76 && 128&& $(9^2,3), (9,3^4), (3^7)$\\
$22$&& 89 && 160&& $(9,7,3^2), (7,3^5)$\\
\hline
\end{tabular}
\end{table}
\medskip

Now take $n>22$.
Let $\mu \in P_2(n)$ be such that $p_2(\mu)$ is maximal;
let $m_j$ be the multiplicity of a part $j$ in $\mu$.
Suppose $\mu$ has a part $j=2h+1\geq 19$;
let $\{h,h+1\}=\{2l,h'\}$.
Then by Theorem~\ref{thm:reg-inequality} and Table~\ref{table:4},
replacing $j$ by the parts $h',2l-3,3$ in $\mu$
would produce a partition $\nu\in P_2(n)$
such that $p_2(\nu)>p_2(\mu)$.
Thus  $\mu$ has no parts $j\geq 19$.
By Table~\ref{table:4},  a part $j\in \{13,15,17\}$
could be replaced in $\mu$ by a partition in $P_2(j)$
giving a partition $\nu\in P_2(n)$ of larger
$p_2$-value.
Thus $\mu$ only has odd parts $j\leq 11$.

Any two parts $(11^2)$, $(11,9)$, $(11,7)$, $(11,5)$, $(11,3)$, $(11,1)$
can be replaced by a $2$-regular partition to obtain a higher $p_2$-value,
see Table~\ref{table:4};
thus $m_{11}=0$.
Also $(7^3)$, $(7,5)$, $(7,1)$ can be replaced to obtain a higher
$p_2$-value. Thus $m_7\le 2$, and the part 7 can only occur when
$\mu$ is of the form $(9^a,7,3^b)$ or $(9^a,7^2,3^b)$;
in the first case  $n\equiv 1 \mod 3$, in the second case we have $n\equiv 2 \mod 3$.
Also $(5^2)$ can be replaced by $(7,3)$ to obtain a higher $p_2$-value,
so $m_5\le 1$; then replacing $(9,5)$ or $(5,3^3)$ by $(7^2)$,
and $(5,1)$ by $(3^2)$ shows that $\mu$ has no part 5. Hence if $\mu$ has no part 7, then $\mu$ is of the form $(9^a,3^b)$,
and $n\equiv 0 \mod 3$.
As $p_2((9))=p_2((3^3))$, the part 9 and the parts $3,3,3$ can always be used interchangeably.
Now for $n\ge 19$ and any congruence $n\equiv c \mod 3$,
$c\in \{0,1,2\}$,
we have found just one type of 2-regular partition
maximizing the $p_2$-value, namely $(9^a,7^c,3^b)$,
with $a,b\in \mathbb{N}_0$ such that $(3a+b)\cdot 3+c\cdot 7=n$,
where $p_2((9^a,7^c,3^b))=2^{3a+b}5^c=\text {\rm maxp}_2(n)$.
This proves the claim for $k=2$.
\medskip

(ii) By Table~\ref{table:5} the claim holds for $n\le 16$.
So we assume now that $n>16$.
\begin{table}[h]
\caption{\label{table:5} Maximum value partitions $\mu$ for $k=3$}
\begin{tabular}{|r|cc|cc|cc| }
\hline \rule[-3mm]{0mm}{8mm}
$n$ && $p_3(n)$ && $\text {\rm maxp}_3(n)$ && $\mu$ \\
\hlin
$1$  && 1 && 1 && (1) \\
$2$  && 2 && 2 && (2) \\
$3$  && 2 && 2 && (2,1) \\
$4$  && 4 && 4 && $(4),(2^2)$ \\
$5$ && 5 && 5 && (5) \\
$6$&&  7 && 8 && $(4,2),(2^3)$ \\
$7$&& 9 && 10 && (5,2)\\
$8$&&  13 && 16 && $(4^2),(4,2^2),(2^4)$ \\
$9$&&  16 && 20 && $(5,4), (5,2^2)$ \\
$10$&& 22  && 32 && $(4^2,2),(4,2^3),(2^5)$ \\
$11$&& 27 && 40 && $(5,4,2),(5,2^3)$ \\
$12$&& 36 &&  64 && $(4^3),(4^2,2^3),(4,2^4),(2^6)$ \\
$13$&& 44 &&  80 && $(5,4^2),(5,4,2^2),(5,2^4)$ \\
$14$&& 57 && 128 && $(4^3,2),(4^2,2^3),(4,2^5),(2^7)$ \\
$15$&& 70 && 160&& $(5,4^2,2),(5,4,2^3),(5,2^5)$ \\
$16$&& 89 && 256&& $(4^4),(4^3,2^2),(4^2,2^4),(4,2^6),(2^8)$ \\
\hline
\end{tabular}
\end{table}

Let $\mu \in P_3(n)$
be such that $p_3(\mu)$ is maximal.
Suppose $\mu$ has a part $j\geq 17$.
Replace $j$ by $\nu_j=(j-2,2)$ if $j\equiv 1 \mod 3$, and by $\nu_j=(j-4,4)$
if $j\equiv 2 \mod 3$.
By Theorem~\ref{thm:reg-inequality} we have
$p_3(j) < p_3(\nu_j)$.
Thus $\mu$ only has parts $\le 16$.
By Table~\ref{table:5}, any of these can be replaced by a partition
of the form $(5^a,4^b,2^c,1^d)$ to increase the $p_3$-value,
and we note that the parts 4 and $2,2$ can be used interchangeably.
Hence only parts $1,2,4,5$ can appear in~$\mu$.
By Table~\ref{table:5},  the following replacements would increase
the $p_3$-value:
$(5^2) \to (2^5)$, $(5,1)\to (2^3)$, $(4,1),(2^2,1)\to (5)$,
$(1^2)\to (2)$.
This implies that $\mu$ can only have one of the forms
$(4^a,2^b)$ or $(5,4^a,2^b)$, where in the first case $n\equiv 0 \mod 2$, in the second case
$n \equiv 1 \mod 2$.
Hence $\text {\rm maxp}_3(n)=2^{\frac n2}$ when $n$ is even, and
$\text {\rm maxp}_3(n)=5\cdot 2^{\frac {n-5}2}$ when $n$ is odd.

\medskip

(iii)
By Table~\ref{table:6} the claim holds for $n\le 15$,
so now take $n\ge 16$.
Let
$\mu \in P_4(n)$ be such that $p_4(\mu)$ is maximal.
Note that by Table~\ref{table:6}, we may always exchange a part
$6$ against the parts $3,3$ without
changing  the $p_4$-value.
Suppose $\mu$ has a part $j\geq 9$.
Replace $j$ by $\nu_j=(j-2,2)$, when $j\not\equiv 2 \mod 4$, or by $\nu_j=(j-3,3)$
when $j\equiv 2 \mod 4$.
By Theorem~\ref{thm:reg-inequality}, $p_4(j)<p_4(\nu_j)$; hence $\mu$ only has parts $\le 7$.
Replacing $(7^2)$ by $(6^2,2)$, $(7,5)$ by $(6^2)$, $(7,2)$ by $(6,3)$,
$(7,1)$ by $(6,2)$ shows that $\mu$ can have a part $7$ only when it is
of the form $(7,6^a,3^b)$, and then $n\equiv 1 \mod 3$.
By Table~\ref{table:6}, in these partitions we may exchange $7$ with $(5,2)$ or $(3,2^2)$,  and $(7,3)$ with $(5^2)$ without changing the $p_4$-value.

Now assume that $\mu$ has no part~7.
Replacing $(5^3)$ by $(6^2,3)$, $(5^2,2)$ by $(6^2)$, $(5,1)$ by $6$,
shows that $\mu$ can have a part $5$ only when  $n\equiv 1 \mod 3$
and it is of the form $(6^a,5,3^b,2)$ or $(6^a,5^2,3^b)$ already discussed above,
or  $n\equiv 2 \mod 3$ and it is of the form $(6^a,5,3^b)$.
Note that $5$ can be exchanged with $(3,2)$ without changing the $p_4$-value.

Finally, when $\mu$ has no parts 5 and 7, the replacements of
$(6,1)$ by $7$, $(2^3)$ by~6,
$(3,1)$ by $(2^2)$,
$(2,1)$ by~$3$, $(1^2)$ by~$2$
show that $\mu$ can have no part~1 and $m_2\le 2$.
Then $\mu$ has one of the forms $(6^a,3^b)$, $(6^a,3^b,2)$ or $(6^a,3^b,2^2)$,
when $n$ is congruent to $0,2,1 \mod 3$, respectively.

Together with the remarks above, we then have
$\text {\rm maxp}_4(n)=3^{\frac n3}$ when $n\equiv 0 \mod 3$,
$\text {\rm maxp}_4(n)=4\cdot 3^{\frac {n-4}3}$ when  $n\equiv 1 \mod 3$,
and
$\text {\rm maxp}_4(n)=2\cdot 3^{\frac {n-2}3}$ when  $n\equiv 2 \mod 3$,
attained at the partitions as stated in the claim for $k=4$.

\begin{table}[h]
\caption{\label{table:6} Maximum value partitions $\mu$ for $k=4$}
\begin{tabular}{|r|cc|cc|cc| }
\hline \rule[-3mm]{0mm}{8mm}
$n$ && $p_4(n)$ && $\text {\rm maxp}_4(n)$ && $\mu$ \\
\hlin
$1$  && 1 && 1 && (1) \\
$2$  && 2 && 2 && (2) \\
$3$  && 3 && 3 && (3) \\
$4$  && 4 && 4 && (2,2) \\
$5$ && 6 && 6 && (5), (3,2) \\
$6$&&  9 && 9 && $(6),(3^2)$ \\
$7$&& 12 && 12 && $(7),(5,2),(3,2^2)$\\
$8$&&  16 && 18 && $(6,2),(5,3),(3^2,2)$ \\
$9$&&  22 && 27 && $(6,3), (3^3)$ \\
$10$&& 29  && 36 && $(7,3),(6,2^2),(5^2),(5,3,2)(3^2,2^2)$ \\
$11$&& 38 && 54 && $(6,5),(6,3,2), (5,3^2), (3^3,2)$\\
$12$ && 50 && 81 && $(6^2), (6,3^2), (3^4)$\\
$13$ && 64 && 108 && $(7,6), (7,3^2),(6,5,2), (6,3,2^2),(5^2,3),
(5,3^2,2),(3^3,2^2)$\\
$14$ && 82 && 162 && $(6^2,2),(5^2,3),(6,3^2,2),(5,3^3),(3^4,2)$\\
$15$ && 105 && 243 && $(6^2,3),(6,3^3),(3^5)$\\
\hline
\end{tabular}
\end{table}
\medskip

(iv)
Table~\ref{table:7} shows that the assertion is true for $n\le 12$.
Take $n\ge 13$, and
let $\mu \in P_5(n)$ be such that $p_5(\mu)$ is maximal.
Note that by Table~\ref{table:7} we may always exchange a part
$6$ against the parts $4,2$ without
changing  the $p_5$-value.
Suppose $\mu$ has a part $j\geq 7$.
Replace $j$ by $\nu_j=(j-3,3)$, when $j\not\equiv 3 \mod 5$, or by $\nu_j=(j-4,4)$
when $j\equiv 3 \mod 5$.
By Table~\ref{table:7} and Theorem~\ref{thm:reg-inequality}
$p_5(j)<p_5(\nu_j)$; hence $\mu$ only has parts $\le 6$.

Replacing $(6^2)$ by $(4^3)$, $(6,2)$ by $(4^2)$, $(6,1)$ by $(4,3)$,
$(3^2)$ by $6$, $(3,1)$ or $(2^2)$ by 4, $(2,1)$ by 3 and $(1^2)$ by 2
increases the $p_5$-value.
Hence $\mu$ can only have the forms stated in (iv), and the
assertion about the $\text {\rm maxp}_5$-value also follows.

\begin{table}[h]
\caption{\label{table:7} Maximum value partitions $\mu$ for $k=5$}
\begin{tabular}{|r|cc|cc|cc| }
\hline \rule[-3mm]{0mm}{8mm}
$n$ && $p_5(n)$ && $\text {\rm maxp}_5(n)$ && $\mu$ \\
\hlin
$1$  && 1 && 1 && (1) \\
$2$  && 2 && 2 && (2) \\
$3$  && 3 && 3 && (3) \\
$4$  && 5 && 5 && (4) \\
$5$ && 6 && 6 && (3,2) \\
$6$&&  10 && 10 && (6), (4,2) \\
$7$&& 13 && 15 && (4,3)\\
$8$&&  19 && 25 && $(4^2)$ \\
$9$&&  25 && 30 && (6,3), (4,3,2) \\
$10$&& 34  && 50 && $(6,4), (4^2,2)$ \\
$11$&& 44  && 75 && $(4^2,3)$ \\
$12$&& 60  &&  125 && $(4^3)$ \\
\hline
\end{tabular}
\end{table}
\medskip

(v)
Table~\ref{table:8} shows that the assertion is true for $n\le 10$.
Let $n\ge 11$, and
let $\mu \in P_6(n)$ be such that $p_6(\mu)$ is maximal.
Suppose $\mu$ has a part $j\geq 7$.
Replace $j$ by $\nu_j=(j-3,3)$, when $j\equiv 4 \mod 6$, or by $\nu_j=(j-4,4)$
when $j\not\equiv 4 \mod 6$.
By Table~\ref{table:8} and Theorem~\ref{thm:reg-inequality}
$p_6(j)<p_6(\nu_j)$; hence $\mu$ only has parts $\le 5$.
Replacing $(5^2)$ by $(4^2,2)$, $(5,1)$ by $(4,2)$, $(5,2)$ by $(4,3)$,
$(5,3)$ by $(4^2)$, $(3^2)$ by $(4,2)$, $(3,2)$ by 5,
$(3,1)$ or $(2^2)$ by 4, $(2,1)$ by 3 and $(1^2)$ by 2
increases the $p_6$-value.
Hence $\mu$ can only have the forms stated in (v), and the
assertion about the $\text {\rm maxp}_6$-value also follows in this final case.

\begin{table}[h]
\caption{\label{table:8} Maximum value partitions $\mu$ for $k=6$}
\begin{tabular}{|r|cc|cc|cc| }
\hline \rule[-3mm]{0mm}{8mm}
$n$ && $p_6(n)$ && $\text {\rm maxp}_6(n)$ && $\mu$ \\
\hlin
$1$  && 1 && 1 && (1) \\
$2$  && 2 && 2 && (2) \\
$3$  && 3 && 3 && (3) \\
$4$  && 5 && 5 && (4) \\
$5$ && 7 && 7 && (5) \\
$6$&&  10 && 10 && (4,2) \\
$7$&& 14 && 15 && (4,3)\\
$8$&&  20 && 25 && $(4^2)$ \\
$9$&&  27 && 35 && (5,4) \\
$10$&& 37  && 50 && $(4^2,2)$ \\
\hline
\end{tabular}
\end{table}
\end{proof}

\section{Concluding remarks}

We note that recently also other multiplicative properties of the partition function have been studied
and one might ask whether those also hold for the generating function for $k$-regular partitions.
Originating in a conjecture by William Chen, DeSalvo and Pak in \cite{DeSalvoPak} have proved
log-concavity for the partition function for all $n>25$;
do we have an analogue of this?

Indeed, there is computational evidence for a version of Chen's conjecture
for $k$-regular partitions, i.e.,
when $n>n_0$ (with $n_0$ being relatively small) then for all
$m\in \{2,3, \ldots, n-1\}$:
$$
p_k(n)^2 > p_k(n-m)p_k(n+m)\:.
$$

The inequality
$p_k(1)p_k(n) = p_k(n) < p_k(n+1)$
has an easy combinatorial proof by an injection $P_k(n) \to P_k(n+1)$.
One may ask whether there is also a combinatorial argument for
proving the inequality in Theorem~\ref{thm:reg-inequality}.

As mentioned before,
the number $p_k(n)$ is equal to the number of partitions where no part
has a multiplicity $\geq k$.
But when we extend the corresponding (same)
generating function $p_k(n)$ to the set of partitions with all multiplicities being $<k$
in analogy to the extension to the set $P_k(n)$,
the behavior is quite different.
In particular, the maximal values on the
two different partition sets to a given $n\in\mathbb{N}$ are in general different,
and for the second extension, the sets of partitions giving the maximal value
are more complicated.

Hagis' formulae play a crucial role in this paper;
as pointed out by the referee,
results of this type have been obtained recently
in a much wider context.
Indeed, Bringmann and Ono \cite{BringmannOno}
give exact formulae for the coefficients
of all weight 0 modular functions
and also all harmonic Maass forms of non-positive weight.
This work might be employed to study other
partition-related functions
defined similarly as our maxp-functions.

\bigskip

{\bf Acknowledgement.}
The authors thank Michael Griffin for assisting with the calculations at the end of Section~\ref{sec:proof-reg-inequality}.

redpajama_set_name: RedPajamaArXiv


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Next Story → What are the different central heating services available?


We expanded our representative network in Sydney, Australia. So far this is the furthest, a staggering 16,000 km away from Andivi headquarters in Slovenia.
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text:
Can Mike Kelley's Memory Ware Flats Give the Artist Market Power to Match His Reputation?
December 21, 2016 by Marion Maneker
Memory Ware Flat #3
Normally it makes more sense to write about an exhibition as it opens. But it may be more in keeping with the effect of of Hauser + Wirth's Autumn show in New York of Mike Kelley's Memory Ware Flat works that we take stock of the exhibition as it closes.
The reason is fairly simple. This is the most comprehensive and scholarly show of the Memory Ware Flat series that has ever been mounted. And Hauser + Wirth worked with the Mike Kelley Foundation to both organize the show and present it in the early days as John Welchman, the foundation's director, and Mary Clare Stevens, who worked closely with the artist, were both available to discuss the body of work and the show at its opening.
The significance of the Memory Ware Flat show goes far beyond its scholarly thoroughness. The Memory Ware Flat series is the most commercial body of work from Kelley. It's also the body of work with the most commercial potential. In this brief essay, we're going to look at the market history of the Memory Ware Flats and the way in which Hauser's recent presentation is setting up the works for what may be another market run.
Mike Kelley's reputation as an artist has only grown in the years after his early death in 2012. Since 2015, Hauser + Wirth has represented the estate by mounting a massive Kelley installation in London and a show of Kelley's equally ambitious Kandors in New York. (Another show of Kelley's Kandors was also held this Fall in New York at Adam Lindemann's Venus Over Manhattan.) His wide range of work, often in large formats or unconventional materials is difficult for the market to absorb.
The Memory Ware Flats deal with many of the important themes in Kelley's work but also present themselves as domestically-sized, wall-friendly works of art that are both intellectually complex and visually arresting. According to Hauser, there are approximately 100 memory ware works. Some nearly two-thirds of those are Memory Ware Flats. Over the past 13 years, since the first of these works appeared at auction, public and private sales of the 60 or so works have increased in value and frequency.
In the last two years, prices for Memory Ware Flats have jumped in value to sell for prices in the $2-3m range. Through a series of four auction sales in late 2105 and early 2016, prices and so-called hammer ratios spiked briefly and then cooled. This is the second time the Memory Ware Flats have appeared on the auction market and broadcast a new level of pricing. Each of the previous runs was preceded by an influential gallery show.
From Dysfuncational Memory to Abstract Expressionism
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text:
Anything but consensus on electoral reform
Go Back to the Homepage / 42nd Parliament (Archive) • News / Anything but consensus on elec …
Sep 15, 2016 Administrator42nd Parliament (Archive), News
The Ottawa-based Institute on Governance and the Environics Institute in Toronto conducted a comprehensive survey on electoral reform. The below were results from the over 2000 responses from respondents over 18 (MP Brassard's survey results in brackets – where applicable):
24 per cent thought the system needed a major overhaul (Brassard survey: 30%)
29 per cent didn't think any changes were needed (Brassard survey: 42%)
38 per cent said they thought only minor changes were needed. (Brassard survey: 28%)
45 per cent opposed making it mandatory to vote in federal elections. (Brassard survey: 40%)
36 per cent Mixed Member Proportional
34 per cent Single Member Plurality – i.e. current system
21 per cent Pure Proportional Representation
17 per cent Ranked or Preferential Ballot
(Brassard survey: FPTP – 42%, minor changes only 28%, Proportional – 25%, Ranked – 4%, Other – 1%)
To see MP Brassard's full results of his survey click here
Ottawa's Institute on Governance and the Environics Institute in Toronto provided a brief description of the 4 examples of voting models for their survey as follows:
Single Member Plurality (current system): Canadians vote for a single candidate running in their electoral district. The candidate who wins the most votes in the electoral district is elected to Parliament.
Pure Proportional Representation: Canadians would vote for a political party and the number of seats each party gets in Parliament is based on the number of votes it receives nationally.
Mixed Member Proportional: Canadians would have two votes. First, they vote for a single candidate running in their electoral district (like in the current system), and second, they cast a separate vote for a party. The number of seats each party gets in Parliament is proportionate to the number of votes each party receives from both types of ballots.
Ranked or Preferential Ballot: Canadians would rank all of the candidates running in their electoral district from most preferred to least preferred. If a candidate wins 50 per cent or more of the first choice ballots they are declared the winner. If no candidate wins at least 50 per cent, the candidate with the least first choice votes is eliminated from the race. If a voter's preferred candidate is eliminated, their vote is automatically transferred to their second choice on the list. This repeats until one candidate gets more than 50 per cent of the votes.


The human body has protective layers against pathogens such as e.g. mucosal layers in the stomach and the gastrointestinal tract, in the airway system and lungs, in the urogenital and female reproductive tract. While this is helpful against infections, the mucosal layer also presents an obstacle for the passage of pharmaceuticals. This leads to a strong need for a more efficient uptake of these drugs as most hydrophobic drugs are rapidly cleared from the delivery site at the mucus barrier.
There is no doubt that a lot of diseases could be treated more effectively if the penetration of pharmaceuticals through the human mucus layers could be improved. Therefore, a number of novel drug delivery technologies, based also on nanoparticles, have been developed. However, conventional nanoparticles still are likely to be eliminated or cannot penetrate without irreversibly damaging the mucus barrier.
In a project funded by Baden-Württemberg Stiftung gGmbH, scientists at the University of Stuttgart and the Max Planck Institute for Intelligent Systems have developed novel conjugate complexes that can be used for enhanced transport/diffusion performance through mucus. The complexes comprise a biological entity; a cargo moiety and an effector moiety that is capable of influencing a human protective layer e.g. mucus and also facilitate the movement of this carrier. The biological entity is a viral nanoparticle (VNP), virus like particle (VLP), bacteriophage or a proteinaceous assembly. In addition, further moieties providing functions such as sensing functions, catalytic activities, and tracking and/or marking functions can be included.
The conjugate complex (so called 'phagetool') is self-assembling, i.e. it forms spontaneously when all components are brought together in a suitable solution. The 'phagetool' diffuses and/or moves autonomously or can be moved in a directed manner by an external force e.g. an externally applied field.
Here we present an innovative and fully biocompatible delivery system that can, at the same time, transport a useful cargo and influence its environment to increase diffusion/transport.
Technologie-Lizenz-Büro GmbH is responsible for the exploitation of this technology and assists companies in obtaining licenses.


text:
Q: Skip some array in PHP I need some help. In my database have fields
ID, Domain, Key, Status
Below is what i have currently in my code
$connection = mysql_connect('localhost', 'username', 'password'); //The Blank string is the password
mysql_select_db('database');

$query = "SELECT * FROM tblhosting Where domain = 'LiCf8124d0c5724f32d0eaf14fc17fae6f44'";

$result = mysql_query($query);

$row = mysql_fetch_array($result);

echo json_encode($row);

mysql_close();

The question is how can I display only field domain and status?
Thanks in advance

A: Use this
$query = "SELECT Domain, Status FROM tblhosting Where domain = 'LiCf8124d0c5724f32d0eaf14fc17fae6f44'";

A: This is quite a simple one. After the SELECT simply add the fields you need comma separated.
So for example
SELECT domain, status FROM tblhosting Where domain = 'LiCf8124d0c5724f32d0eaf14fc17fae6f44'

A: I think you are looking for this. If your status and domain field name is 'Status' and 'Domain' :
echo $row['Domain'];
echo $row['Status'];

A: Just select only the fields you want:
$query = "SELECT Domain,Status FROM tblhosting Where domain = 'LiCf8124d0c5724f32d0eaf14fc17fae6f44'";

redpajama_set_name: RedPajamaStackExchange


A record field of over 54,000 took part in the 2007 Vancouver Sun Run on Sunday, making a 10 km lap of the downtown area starting at the Hotel Vancouver and finishing at B.C. Place.
Among the field were 81 athletes from Whistler, 16 from Pemberton, 86 from Squamish and 42 from Garibaldi Highlands. All results are posted online at www.sunrun.com .
The top Sea to Sky athlete was Whistler's John Blok with a time of 37 minutes and 49 seconds. He finished 13 th in the men's 50 to 54 age category, and 196 th overall. That's also in the top 0.3 per cent of all runners.
Blok credited his promotion into the elite category based on his time from previous years. In the front pack there is more room to run from the start, as well as an opportunity to warm-up. For his part, he did a lap of Lost Lake in Stanley Park, adding 5 km to his run total.
Second up for Whistler was Paul Suter in a time of 38:28, placing 246 th overall and 17 th in the men's 45 to 49 category.
Daniel Havens was third, the only other local athlete to finish under 40 minutes with a time of 38:40. He was 16 th in the men's 50 to 54 category, and 266 th overall.


text:
Gadacz, René R.. "Wampum". The Canadian Encyclopedia, 05 November 2020, Historica Canada. www.thecanadianencyclopedia.ca/en/article/wampum. Accessed 31 January 2023.
Gadacz, R. (2020). Wampum. In The Canadian Encyclopedia. Retrieved from https://www.thecanadianencyclopedia.ca/en/article/wampum
Gadacz, René R.. "Wampum." The Canadian Encyclopedia. Historica Canada. Article published February 07, 2006; Last Edited November 05, 2020.
The Canadian Encyclopedia, s.v. "Wampum," by René R. Gadacz, Accessed January 31, 2023, https://www.thecanadianencyclopedia.ca/en/article/wampum
Wampum
Article by René R. Gadacz
Published Online February 7, 2006
Last Edited November 5, 2020
Wampum are tubular purple and white beads made from shells. Wampum are used primarily by Indigenous peoples of the Eastern Woodlands for ornamental, ceremonial, diplomatic and commercial purposes. Belts made of wampum were used to mark agreements between peoples. (See also Covenant Chain and Treaties with Indigenous Peoples in Canada.)
Wampum was used by Indigenous peoples and to record treaties and settle disputes (courtesy NMC/CMC/575-620).
Nicholas Isawanhanhi, Huron chief, shown wearing the regimental coat commonly awarded to Indigenous trading captains during the fur trade. Painting by Chatfield (courtesy Library and Archives Canada/C-38948).
What is Wampum?
Wampum — which comes from a Narragansett (Algonquian language family) word meaning a string of white shell beads — are tubular beads manufactured from Atlantic coast seashells. While a variety of shells may be used, most often the white shells are formed from the whelk shell and the purple from the quahog clam shell.
Belts made of wampum were used to mark agreements between peoples. Wampum belts are of particular significance with regards to treaties and covenants made between Indigenous peoples and European colonial powers.
Function and Uses of Wampum
Indigenous peoples living along the coast collected the shells, produced the beads, and traded them in-land, for example to the Haudenosaunee, for furs, corns, beans and squash.
The beads had considerable value in eastern and maritime Canada for ornament, ceremony, the fur trade and diplomacy, particularly in the 17th and 18th centuries.
Wampum was threaded on string or woven into bracelets, necklaces, collars, sashes and later into belts that served as physical representations of political agreements. Particular patterns symbolized events, alliances or kinship relations between different peoples, and wampum could be used to confirm relationships, propose marriage, atone for murder or ransom captives. Beads and belts also validated treaties and were used to recall oral tradition. Many communities had wampum keepers, who protected the belts and interpreted the history contained therein.
Wampum Belts, Treaties and Alliances
Before Confederation some groups of Indigenous people, particularly those in the Eastern Woodlands, indicated their agreement to certain treaties by presenting long wampum belts to Crown officials. Some examples include the Hiawatha Belt (the belt of the Haudenosaunee), the Covenant Chain Wampum of 1764, and the Two Row Wampum Belt (Kaswentha).
Two Row Wampum Belt (Kaswentha)
The Two Row Wampum Belt (Kaswentha) of the Haudenosaunee is a well-known example of a wampum belt. It still symbolizes an agreement of mutual respect and peace between the Haudenosaunee and European newcomers (initially the Dutch) to North America. The principles were embodied in the belt by virtue of its design: two rows of purple wampum beads on a background of white beads represent a canoe and a European ship. The parallel paths represent the rules governing the behaviour of the Indigenous and European peoples. The Kaswentha stipulates that neither group will force their laws, traditions, customs or language on each other, but will coexist peacefully as each group follows their own path.
The Kaswentha continues to represent the Covenant Chain alliance between the Six Nations and their European partners, extending from the 17th century to modern times. As a foundational philosophical principle of respect and friendship, the Belt may be able to function as a framework for improved relations between Indigenous peoples and various levels of government.
Contemporary Significance
Wampum belts and other wampum bead artifacts continue to hold significance for Indigenous peoples and any treaties or agreements they might represent.
Numerous wampum artifacts exist in private collections and in local, provincial and national museums both in Canada and the United States. Wampum in public collections have become the subject of repatriation negotiations between community members and institutional officials who seek to uphold the rights of a given community to their cultural artifacts while balancing traditional care and preservation techniques.
Indigenous Perspectives Education Guide
Indigenous Peoples Collection
Indigenous Treaties in Canada Collection
Indigenous Peoples in Canada
Penelope Myrtle Kelsey, Reading the Wampum: Essays on Haudenosaunee Visual Code and Epistemological Discovery (2014)
Marc Shell, Wampum and the Origins of American Money (2013)
Lynn Gehl, The Truth that Wampum Tells: My Debwewin on the Algonquin Land Claims Process (2014)
Onondaga Nation Learn more about wampum
Open Library Read more about wampum belts, their uses and significance
Museum of Ontario Archeology Learn more about wampum
Indigenous Treaties in Canada
Treaties with Indigenous Peoples in Canada
Covenant Chain
Indigenous Oral Histories and Primary Sources
Repatriation of Artifacts


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text:
One of the most interesting aspects of the film project was collaborating with so many people - directors, filmmakers, and writers - over a five-year period. I learned that there are two components to this.
Yo-Yo Ma's quotes
Many of the Central Asians know Russian, and Ted Levin speaks it fluently. I speak Chinese, but Mongolian is completely different, so we had to have translators.
One is that you have to take time, lots of time, to let an idea grow from within. The second is that when you sign on to something, there will be issues of trust, deep trust, the way the members of a string quartet have to trust one another.
I'm not likely to forget where I've been and what I've done and learned. I think it's just as important to play new instruments as to play new pieces. The old ones are getting scarcer and the new ones more and more wonderful.
The tango is really a combination of many cultures, though it eventually became the national music of Argentina.
Good things happen when you meet strangers.
I've been traveling all over the world for 25 years, performing, talking to people, studying their cultures and musical instruments, and I always come away with more questions in my head than can be answered.
I love grocery shopping when I'm home. That's what makes me feel totally normal. I love both the idea of home as in being with my family and friends, and also the idea of exploration. I think those two are probably my great interests.
After reaching 50, I began to wonder what the root of life is.
Things can fall apart, or threaten to, for many reasons, and then there's got to be a leap of faith. Ultimately, when you're at the edge, you have to go forward or backward; if you go forward, you have to jump together.
Music has always been transnational; people pick up whatever interests them, and certainly a lot of classical music has absorbed influences from all over the world.
The role of the musician is to go from concept to full execution. Put another way, it's to go from understanding the content of something to really learning how to communicate it and make sure it's well-received and lives in somebody else.
We may be coming to a new golden age of instrument making.
It took me way beyond what I knew, into places of which I was totally scared, but as I became less frightened, I welcomed new ways of thinking and approaching something. It made me an infinitely richer person, and I think a better musician.
But an innovation, to grow organically from within, has to be based on an intact tradition, so our idea is to bring together musicians who represent all these traditions, in workshops, festivals, and concerts, to see how we can connect with each other in music.
I learn something not because I have to, but because I really want to. That's the same view I have for performing. I'm performing because I really want to, not because I have to bring bread back home.
When you learn something from people, or from a culture, you accept it as a gift, and it is your lifelong commitment to preserve it and build on it.
My best film composing experience was with Elia Kazan.
Alex North
I was lucky enough to occasionally break out of that racist situation that prevails in the Hollywood film production community. But it was racist then and it will always be that way. It will never be otherwise.
J. J. Johnson
'Batman' took 10 months to film, and by the time I stopped working on it, it took a long time before my English accent came out again. I was actually having to try for it.
Luck is everything... My good luck in life was to be a really frightened person. I'm fortunate to be a coward, to have a low threshold of fear, because a hero couldn't make a good suspense film.
I still feel that a movie has to attempt to say something - even if it fails miserably. But I've sort of given up on believing that I'm going to change the world with every film I choose to act in.
Sarah Polley
It's the difficulty we had with Mr. Bean, actually, when it went from TV to film. You certainly discover that you need to explain more about a character.
To make a documentary is one thing, to make a feature film is quite another.
I mean when the play was on in New York I was starting to get film offers coming through, and since the film's come out I get offered more than I used to, but it happens incrementally.
Patrick Marber
I think it would be very boring dramatically to have a film where everybody was a lawyer or doctor and had no faults. To me, the most important thing is to be truthful.
I like doing them and they're ridiculous and the actors can improvise a lot, and they don't have to be really realistic characters that hit a very specific tone as in a feature film. They're really fun, I want to make more of them definitely.
The best thing I can think of would be to create a union between something as beautiful and powerful and wonderful as Hollywood films and a criticism of the status quo. That's my dream, to make such a German film.
Rainer W. Fassbinder
If he didn't fall in love he would have never come back near the end of the film. Because, what man is going to dishonor himself so that he comes back in front of the man that took a woman away from him... and warns her to save her life?
Robert Altman's Nashville is my all-time favorite film because it covers all the bases - it's original, moving, and has something to say, but also funny and incredibly entertaining.
Douglas Wood
I never see a novel as a film while I'm writing it. Mostly because novels and films are so different, and I'm such an internal novelist.
I have a movie coming out called Spun, which will be at the Toronto Film Fest.
Well ironically my last three roles have all been a mother. One was a Canadian film where the baby was taken away because she is a drug addict, in Irish Jam I play a mother to a four year old. I think in the future I'll be able to handle the role with a lot more depth.
And when you see it the first time you put the film together, the roughest cut, is when you want to go home and open up your veins and get in a warm tub and just go away. And then it gradually, maybe, works its way back, somewhere toward that spot you were at before.
A good film script should be able to do completely without dialogue.
My advice to young film-makers is this: don't follow trends, start them!
The whole aspect of cinema and film festivals should be a moment to come together and celebrate art and humanity. It would be a shame if there was such a divide.
It's really rare for film directors to be that interested in things other than themselves.
When I said I was going to audition for a film, I got a hearty laugh from all my family.
I don't think any actor feels comfortable watching themselves in movies. You must be very narcissistic. The problem with your own opinion of yourself is that contrary to the normal spectators, when you watch a film you are in, you only watch yourself.
We're making a commercial Hindi film catering to all types of audiences.
They all matter to me, whether I'm working on a Sam Jackson film for a week or I'm the star of my own TV series - I take it all very seriously, and I have a healthy respect for the work in general, despite the role.
Anthony Michael Hall
There are no black film composers doing the likes of Star Wars, doing the likes of E.T., doing the likes of Jurassic Park. There are none, nor will there ever be one. That ain't about to happen!
I've explored the worship side, the pop side, and the film scoring side of me.
After the war, there was no industry. We lost the war. We had our whole city destroyed. No money. No studio. No film. No camera. No equipment. We would shoot in the street. We had no actors. Nothing. But we wanted to do movies. And we did the best movies in the world.
Dino De Laurentiis
That would be getting up at 5 am... I don't understand why film's shoot such brutal hours. I think it'd be worth it to not be so strictly cost-effective and have an 8 hour day. The film's would benefit in the end.
Terry Zwigoff
Because when the film was first mooted, the Beatles didn't like the idea at all. In fact they wouldn't have any part in it. And when Brian had committed them, it was part of a deal he did with United Artists, I think.