text:
Swing out Sister new cd
Re: Swing out Sister new cd
Postby pljms » Fri Mar 30, 2018 2:35 pm
When this thread was started in 2004 the 'new' album under discussion was 'Where Our Love Grows', and here we are in 2018 with Swing Out Sister's latest album set for general release in June. 'Almost Persuaded' has already had limited availability since the end of 2017 as it was self released by SOS as a PledgeMusic project. The album features ten new songs by Andy Connell and Corinne Drewery, two of which are reprised in different versions to make twelve tracks in all. Here from the album is 'Happier Than Sunshine', a blissful track about bliss, appropriately enough.
https://m.youtube.com/watch?v=FtwxrlJ-M-0
Postby Martin Johnson » Mon Apr 02, 2018 1:08 am
Scenes of Swing Out Sister in the studio last summer recording one of the new album's many beautiful and haunting tracks, 'All in a Heartbeat':
https://m.youtube.com/watch?v=wHbQ5yWcJps
Postby Sara D » Wed May 02, 2018 8:12 am
Getting into this album big time. Love the vocal arrangement, flute and percussion on 'Which Wrong Is Right':
https://m.youtube.com/watch?v=hlZoV8qu22M
Blair N. Cummings
Postby Blair N. Cummings » Fri May 04, 2018 1:29 pm
Okay, enough!
I understand why the Nikki Jean track prompted a retaliation (unmelodic , I know, but I - like Burt - have always had a soft spot for city kids). Nevertheless, SOS isn`t all THAT great.
Here are a couple of madeleines from the `60s for the old folks:
https://www.youtube.com/watch?v=BF4pN19mXws
https://www.youtube.com/watch?v=YtedKPGsf7M
The first was sung by Lani Hall, a founding group member who later married label-owner Herb Alpert and continues to record and tour with him.
The second was performed by Karen Philip who left the group to pursue an acting career after recording an unreleased solo album. I have been unable to learn what became of her after her character in the first season of the TV version of MASH was written out of the series.
Postby pljms » Wed May 23, 2018 2:14 am
Here's a snippet from 2015 of what was to become 'Don't Give the Game Away' from the new album set for general release next month. While I don't think the album as a whole is up to the standard of 'Where Our Love Grows', two or three of the tracks must rank among the best things they've ever recorded. More on those if and when they become available on YT.
https://m.youtube.com/watch?v=4SpHFLlimkI
Postby Martin Johnson » Tue Jun 12, 2018 5:02 pm
SOS did a London club gig a few weeks ago to preview the new album and here they are performing one of my favourites from it, the Michel Legrand influenced 'Everybody's Here':
https://m.youtube.com/watch?v=8mUtnyskJOs
Postby Martin Johnson » Mon Jun 18, 2018 12:09 am
From the same press launch gig last month here's their performance of a shortened 'Almost Persuaded', the utterly sublime title track of the new album. Not even the amateurish nature of the recording can obscure or diminish this song's sumptuous chord sequence:
https://m.youtube.com/watch?v=ug6x2PzbYTM
Postby pljms » Tue Jun 19, 2018 1:02 am
Martin Johnson wrote: From the same press launch gig last month here's their performance of a shortened 'Almost Persuaded', the utterly sublime title track of the new album. Not even the amateurish nature of the recording can obscure or diminish this song's sumptuous chord sequence:
Here is the album version of Almost Persuaded. I'm just delighted that after over thirty years in the business they're continuing to fly the flag for beautiful and sophisticated pop music and recording tracks of this quality.
https://m.youtube.com/watch?v=RHgAwhH6JRA
nowmedical
Postby nowmedical » Tue Jun 19, 2018 2:52 am
All we need now is some concert dates....
Postby nowmedical » Fri Jun 22, 2018 4:50 pm
Last edited by nowmedical on Sun Jun 24, 2018 9:35 am, edited 2 times in total.
Postby Martin Johnson » Sat Jun 23, 2018 2:24 am
I've been having a personal love affair with the track 'Everybody's Here' since I first heard the new album and Andy Connell's orchestration for the intro is so excuisite it gets me every time I hear it:
https://m.youtube.com/watch?v=ZvjsSvphMak
Postby pljms » Mon Jul 23, 2018 3:55 am
A London date has just been confirmed by SOS and it's at Islington Assembly Hall on Tuesday November 13th, which was where I saw them last five or six years ago.
Here's 'I Wish I Knew' from the new album.
https://m.youtube.com/watch?v=dEAvvIbMB1A
Postby Martin Johnson » Sat Oct 13, 2018 6:09 am
A couple of my former work colleagues in Japan saw Swing Out Sister perform in Tokyo on Thursday evening and told me they were exceptional, which augers well for their London show next month. Here they are with 'Chuo Freeway', a song that Corinne Drewery wrote with Yumi Arai. The video features Corinne on a nocturnal drive but not on the famous Japanese expressway but in central London:
https://m.youtube.com/watch?v=UkaFLnRyEA0
Postby pljms » Tue Nov 13, 2018 6:32 pm
Just on my way home from SOS's gig at Islington Assembly Hall in North London and I have to say the show was marred for me by the exaggerated volume of the bass guitar which not only made for a very muddy sound overall but which for a large part of the evening managed to drown out both the keyboards and the lead guitar, obscuring their songs' often beautiful chord changes. I much prefer them to play live with a string double bass and a percussionist but without a drummer. I also missed the appearance of their regular trumpet/flugelhorn player. However, having said all that, both Corinne Drewery and her co-vocalist Gina Foster were in splendid voice and it was a busy night with a packed and very enthusiastic house. It was no coincidence that the song highlights for me were those where both the bassist and the drummer eased up a bit, namely 'Almost Persuaded' and 'Everybody's Here'. Of the upbeat songs, 'Am I The Same Girl' and 'Not Gonna Change' came off the best or were the least marred by the dodgy sound balance.
Postby nowmedical » Wed Nov 14, 2018 2:46 am
I entirely agree. I was in the balcony, and by the time the sound reached there it was often a thumping mess. Bit like a school disco from the 1970s.


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text:
Q: Signals not received in correct thread using PyQt5 So having followed my best understanding of how to correctly use threads in Qt, I've written a small toy example in which a QObject is moved to a running thread and a signal is called on that object when the window is clicked.
The problem I'm having is the slot is never called in the correct thread - it is always called in the main thread.
This is implemented with the following code:
from PyQt5 import QtWidgets, QtCore

class WidgetWithInput(QtWidgets.QWidget):

    widgetClicked = QtCore.pyqtSignal()

    def mousePressEvent(self, event):
        self.widgetClicked.emit()

class MyObject(QtCore.QObject):

    aSignal = QtCore.pyqtSignal()

    def __init__(self, *args, **kwargs):
        super(MyObject, self).__init__(*args, **kwargs)

        self.aSignal.connect(self.onASignal)

    def onASignal(self):
        print('MyObject signal thread: %s'
               % str(int(QtCore.QThread.currentThreadId())))

class SimpleDisplay(QtWidgets.QMainWindow):
    ''' Class presenting the display subsytem
    '''

    def __init__(self, my_object, *args, **kwargs):

        super(SimpleDisplay, self).__init__(*args, **kwargs)

        self._my_object = my_object

        self._widget = WidgetWithInput()

        self.setCentralWidget(self._widget)
        self._widget.widgetClicked.connect(self._do_something)

    def _do_something(self):
        self._my_object.aSignal.emit()

def main(run_until_time=None):

    import sys
    app = QtWidgets.QApplication(sys.argv)
    print ('main thread: %s' % str(int(QtCore.QThread.currentThreadId())))

    concurrent_object = MyObject()
    display = SimpleDisplay(concurrent_object)

    myobject_thread = QtCore.QThread()
    myobject_thread.start()

    concurrent_object.moveToThread(myobject_thread)

    display.show()

    exit_status = app.exec_()

    myobject_thread.quit()
    myobject_thread.wait()

    sys.exit(exit_status)

if __name__ == '__main__':
    main()

Which outputs something like:
main thread: 139939383113472
MyObject signal thread: 139939383113472

I would expect those two printed thread ids to be different.
Am I missing something here? Explicitly setting the signal to be queued doesn't change anything.

A: With thanks to @three_pinapples, his suggestion in the comments solves the problem: The slot needs to be decorated with QtCore.pyqtSlot:
class MyObject(QtCore.QObject):

    aSignal = QtCore.pyqtSignal()

    def __init__(self, *args, **kwargs):
        super(MyObject, self).__init__(*args, **kwargs)

        self.aSignal.connect(self.onASignal)

    @QtCore.pyqtSlot()
    def onASignal(self):
        print('MyObject signal thread:\t %s'
               % str(int(QtCore.QThread.currentThreadId())))

redpajama_set_name: RedPajamaStackExchange


At the IPLI Foundation, we believe that it is critically important to encourage continuous research and innovation in public policies. For this reason we dedicate our efforts to the support and facilitation of a variety of research initiatives in cooperation with our associate institutions. And also to this end, we are constantly assessing opportunites for building cooperation in projects with potential new associates.
The IPLI Foundation periodically offers grant opportunities to individuals and institutions which have innovative proposals linked to our thematic areas of interest: security and development capacity building within Europe, the Mediterranean region and Africa.


text:
Q: How to Make a UIButton fix at UIScrollView? I am having an UIButton at my UIScrollView and when i zoom my scroll view the button position is not fix...is there any way to make it fix at one location?

A: Since iOS 11, UIScrollView has an instance property called frameLayoutGuide. frameLayoutGuide has the following declaration:
var frameLayoutGuide: UILayoutGuide { get }

The layout guide based on the untransformed frame rectangle of the scroll view.

Use this layout guide when you want to create Auto Layout constraints that explicitly involve the frame rectangle of the scroll view itself, as opposed to its content rectangle.

The following Swift 5.1 / iOS 13 UIViewController implementation shows how to use frameLayoutGuide in order to center a UIButton inside a UIScrollView.
import UIKit

class ViewController: UIViewController {

    let scrollView = UIScrollView()
    let button = UIButton(type: .system)
    let imageView = UIImageView(image: UIImage(named: "LargeImage"))

    override func viewDidLoad() {
        super.viewDidLoad()

        // Set scrollView constraints
        scrollView.contentInsetAdjustmentBehavior = .never
        view.addSubview(scrollView)
        scrollView.translatesAutoresizingMaskIntoConstraints = false
        NSLayoutConstraint.activate([
            scrollView.leadingAnchor.constraint(equalTo: view.leadingAnchor),
            scrollView.trailingAnchor.constraint(equalTo: view.trailingAnchor),
            scrollView.topAnchor.constraint(equalTo: view.topAnchor),
            scrollView.bottomAnchor.constraint(equalTo: view.bottomAnchor)
        ])

        // Set imageView constraints
        scrollView.addSubview(imageView)
        imageView.translatesAutoresizingMaskIntoConstraints = false
        NSLayoutConstraint.activate([
            imageView.leadingAnchor.constraint(equalTo: scrollView.contentLayoutGuide.leadingAnchor),
            imageView.trailingAnchor.constraint(equalTo: scrollView.contentLayoutGuide.trailingAnchor),
            imageView.topAnchor.constraint(equalTo: scrollView.contentLayoutGuide.topAnchor),
            imageView.bottomAnchor.constraint(equalTo: scrollView.contentLayoutGuide.bottomAnchor)
        ])

        // Set button constraints (centered)
        button.setTitle("Button", for: .normal)
        scrollView.addSubview(button)
        button.translatesAutoresizingMaskIntoConstraints = false
        NSLayoutConstraint.activate([
            button.centerXAnchor.constraint(equalTo: scrollView.frameLayoutGuide.centerXAnchor),
            button.centerYAnchor.constraint(equalTo: scrollView.frameLayoutGuide.centerYAnchor)
        ])
    }

}

A: Place/set your button over scroll view (not inside scroll view) as shown here in this snapshot. And also set button constraints (position) with respect to super view of your scrollview.

Here is ref. snapshot of hierarchy of position of each view over each-other.

A: If you want your button to stay at a fixed size and position when you modify your scrollview, the best way would be to not have it be in the scrollview at all. Add the button to the parent view of the scrollview, then it won't be affected by any changes to the scrollview and can be overlaid on top of it.

A: You should position your button outside the scroll view and anchor it to the bottom of the page. This way the content in your scroll view will change but your button will remain statically fixed on the view.


text:
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As Moishe House continues to grow, the organization pilots innovative initiatives to meet the interests and needs of the young adults participating (and those not yet engaged). Over the years, Moishe House has provided training for these community builders through Jewish Learning Retreats and created a platform for former residents and other strong leaders to host Moishe House–style programs from their own homes (now Moishe House Without Walls or MHWOW).
Moishe House is now the global leader in peer-led Jewish young adult engagement. Every year, thousands of young Jews experience innovative, engaging, exciting Jewish programming. All programming is planned and executed by their peers, creating countless opportunities for young adults to connect with their own Jewish identities, their friends and their wider communities. By 2020, Moishe House will dramatically extend its impact on Jewish young adults by scaling our community building programs, expanding alumni efforts and successfully experimenting with innovative programs that engage new segments of the young adult population. Moishe House is providing an important pathway for young adults to take part in — and create — Jewish homes and communities.
Moishe House provides vibrant Jewish community for young adults by supporting leaders in their 20s as they create meaningful home-based Jewish experiences for themselves and their peers.
Every Jewish young adult in their 20s has the access and resources needed to build their own place to go for Jewish community, find friends and move further along their Jewish journey. By making this stage of life one of the most exciting times to be Jewish, young adults are set on a trajectory for lifelong involvement in the Jewish community.
View Our 990
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In 2016, Moishe House celebrated 10 years of empowering Jewish young adults to be the creators of their own peer-led Jewish experiences. This video (created by a pair of Moishe House alumni) commemorates the first 10 years of Moishe House and looks to the future as we continue to grow and evolve.

redpajama_set_name: RedPajamaCommonCrawl


The Coffee Co•Op Brand Identity by Jake Dugard has taken my breath away; this is some seriously cool creative talent.
The real travesty is that The Coffee Co•Op is a fictional company— this was for a grad school campaign—so you can't even enjoy the spectacular creative work that Jake has put into this.
I would even go so far as saying this has to be one of the best coffee identities I have seen, real or concept.
Before I realised this was a concept I was about to hook it over on LogoStack. No can do.
Putting all that aside; kudos to Jake, and hat-tip to The Dieline for posting it originally.


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text:
Arts & Culture » The B-List
Saved by the bells?
by Billy Manes
"I've got a proposition for you," says a mustache into his cell phone outside of lady Estefan's Bongos food emporium at Downtown Disney West Side. "I'm here with Jon Secada's agent. He wants to do New York tomorrow anytime between five and seven. I mean, if he can get on that New York station ... "
If he can get on that New York station, his flagging career will be saved? "Just Another Day" will return to relevant rotation? Wet, curly hair will once again reign supreme in the adult-contemporary world? Oh, to dream.
Well, it's neither five nor seven on this typically drippy midday, and I'm only mildly unenthused about the proposition of sitting on a theme-restaurant stool for five minutes in the floodlights of another failing career. It must be time for a Christmas album.
Oh, it is.
You'll remember that Jon Secada was once the Michael Bolton of Latin pop sap. Hits like the aforementioned "Just Another Day," "Do You Believe in Us?" and "Angel" strained ears with their overwrought earnestness long before anyone wondered whether Ricky Martin was gay. These days, Secada has donned the yoke of calculated novelty sentiment with a big-time holiday offering, The Gift, and -- surprise! -- is available for an interview.
"Jon's busy for a few more minutes," hots the agent hottie, with a polite offering of, "Do you want some coffee ... or water?" He's going nowhere.
Jon, however, is lobbying to go anywhere his semi-recognizable moniker will take him. And if he can't get to New York tomorrow between five and seven, then he'll at least do some Spanish language TV today. As señor Secada and the pancaked faces of Telemundo ambition wrap their recorded repartee, nobody around me is speaking English, which starts to recall those dreams of my demise in a Puerto Rican prison. Mira! Mira! I don't need a shower!
Jon's actually a perfectly nice guy (a prisoner only to his own impertinence) and turns to me kindly with a sparkling caramel gaze. So, why is he in Orlando, beyond the fact that his sugar momma's restaurant is here?
"Well, the record," he yuletides. "It just happened that it coincided with me being here for the 100th anniversary of Disney."
These things do just happen. Kind of like obligatory contract fulfillments.
"I've always wanted to do a holiday record," he ginger snaps. "I just got a deal with Sony Music, and that's one of the recordings they wanted me to do. We recorded it all live in a period of three days. So we did it all old-style, very old-fashioned, as a tribute to the Sinatra ... "
Perry Como? Johnny Mathis?
"Yeah. Very soothing and very timeless, I think," he thinks. "Even the two originals that I put in there, they also have that very standard, timeless feel. I'm very excited about it. We wanted to accomplish something very classic."
And very safe, presumably. Is it hard to record a Christmas record during a summer in Miami?
"It sure is," he obviates. "But the music was so beautiful. I did the album with Emilio" -- Estefan, natch -- "and 62 musicians, mostly members of the Florida Philharmonic. They did an incredible job. I mean, it's one of the most challenging things I've ever done, doing everything all live. Nowadays, without the use of all of the technologies available, it's kind of strange."
Kind of. About this time, I'm peering into Secada's caramel haze, while his entourage of cronies mills around with bent ears. Things I would like to ask, other than "Are you gay?," wander around my head with bilingual translations -- because at least Spanish sounds interesting -- but the fear of upsetting the grand corporate balance of Glo's black-bean empire and the general fear for my life that comes with this territory eventually mute everything down to a digestible pap. Um, what's your favorite Christmas song?
"I really think it's Ã?The Christmas Song'," he chestnuts. "I remember growing up and listening to the classics -- the Sinatras and the Nat King Coles -- and that's what I remember, those kind of songs and that kind of sound. It's a very peaceful, very soothing thing. It makes you feel a healing type of feeling. Especially this year. It's funny that it coincided with everything that's going on ... "
Very funny. Secada's a part of Disney's big, emotional, feel-good whitewash of America's sapped spirits, chiming in wherever he's welcome during the "100 Years of Magic" media junket. That explains his obligatory guesting on the percolating institution Live with Regis & Kelly, broadcast from Disney that morning.
Did you meet Regis? I love this line of questioning.
"Yeah, I met Regis," he nods.
I wanted to meet -- be? -- Kelly Rippa. I think she's fantastic!
"She is fantastic. She's a doll," he feys, suspicions confirmed. "They have a good chemistry."
Surely it's not your biggest career achievement. You've written for Ricky and J-Lo! Yummy Latin chemistry!
"There's a lot of versatility," he blends. "There's a lot of personality among all of these artists. I mean, we're all Latinos, but everybody's got their own thing going on. So it's very diverse, you know. Going from Shakira to J-Lo to Marc to whatever Ricky does."
What does Ricky do? And to whom? Oh, who cares anymore. He's half way to becoming Jon Secada already.
The B-List The B List


[end] Source: Human Events, p. 301. He laughed as he thought of his sexual exploits, and the fact that soon, very soon, he'd kill again. Benny moved to the big wooden door of the chapel and locked it. She was an intravenous drug user. One reason, according to a damning Wall Street Journal report, is this: For 10 years, the government has been deliberately lying to us about who is at risk of AIDS.
had a date past 1 a.
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text:
\section{Introduction and Statement of Results}\label{intro}

Let $N$ be a positive integer. For an odd integer $k$, we denote by $M^{!+\cdots +}_{k/2}(N)$ the space of weakly holomorphic modular forms of weight $k/2$ on $\Gamma_{0}(4N)$ whose $n^{\text{th}}$ Fourier coefficient vanishes unless $(-1)^{(k-1)/2}\,n$ is a square modulo $4N$. For the moment, we assume that $N$ is contained in the set
$$\mathfrak{S}=\{1,2,3,5,7,11,13,17,19,23,29,31,41,47,59,71\}.$$
Then the group $\Gamma_{0}^{*}(N)$, which is the group generated by $\Gamma_{0}(N)$ and all Atkin--Lehner involutions $W_{e}$ for $e\parallel N$, has genus $0$.
From the correspondence between Jacobi forms and half-integral weight forms (cf. \cite[Theorem 5.6]{EiZa1}),
we see that for any $D\in\mathbb{Z}_{>0}$ with $D\equiv\square\pmod{4N}$, there is a unique modular form $g_{D,N}\in M^{!+\cdots +}_{3/2}(N)$ having a Fourier expansion of the form
$$g_{D,N}(\tau)=q^{-D}+\sum_{d\geq 0}B^{(N)}(D,d)\, q^{d}\quad(q=e^{2\pi i\tau},~\tau\in\mathbb{H}).$$
Here, $\mathbb{H}$ denotes the complex upper half plane.

Let $\ell$ be a prime with $\ell\nmid 4N$. Then the Hecke operator $T_{k/2,4N}(\ell^{2})$, originally defined on the space of weakly holomorphic modular forms of weight $k/2$ on $\Gamma_{0}(4N)$, acts on $M_{k/2}^{!+\cdots+}(N)$. We define $T_{k/2,4N}(\ell^{2n})$ for $n\geq 2$ recursively by
$$T_{k/2,4N}(\ell^{2n}):=T_{k/2,4N}(\ell^{2n-2})T_{k/2,4N}(\ell^{2})-\ell^{k-2}T_{k/2,N}(\ell^{2n-4}).$$
For any positive integer $m$ with $\gcd{(m,4N)}=1$,
 define $T_{k/2,4N}(m^{2})$ multiplicatively and set
$$g_{D,N}^{(m)}:=g_{D,N}\mid T_{3/2,4N}(m^{2}).$$
We denote by $B^{(N)}_{m}(D,d)$ the $d^{\text{th}}$ Fourier coefficient of $g_{D,N}^{(m)}(\tau)$:
$$g_{D,N}^{(m)}(\tau)=\mbox{(principal part)}+\sum_{d\geq 0}B^{(N)}_{m}(D,d)\, q^{d}.$$

By the works of Zagier \cite{Zag2} and Kim \cite{Kim2}, the coefficients $B^{(N)}_{m}(D,d)$ can be interpreted as traces of CM values of certain modular functions (or traces of singular moduli).
Remarkably, the coefficients $B^{(N)}_{m}(D,d)$ show many congruence properties, and many authors studied them. In 2005, Ahlgren and Ono \cite{AhO1} showed that if $p\nmid m$ is an odd prime and $\left(\frac{-d}{p}\right)=1$, then
$$B^{(1)}_{m}(1,p^{2}d)\equiv 0\pmod{p}.$$
Edixhoven \cite{Edix1} used the $p$-adic geometry of modular curves to show that, for any $m$ and any $d$ with $\left(\frac{-d}{p}\right)=1$, we have
\begin{equation*}
B^{(1)}_{m}(1,p^{2n}d)\equiv 0\pmod{p^n}.
\end{equation*}
When $p$ is an odd prime, Jenkins \cite{Jen1} obtained a recursive formula for $B^{(1)}_{1}(D,p^{2n}d)$ in terms of $B^{(1)}_{1}(D,p^{2k}d)$ with $k<n$. As a corollary he proved that if $\left(\frac{-d}{p}\right)=\left(\frac{D}{p}\right)\neq 0$, then we have
\begin{equation*}\label{1.8}
B^{(1)}_{1}(D,p^{2n}d)=p^{n}B^{(1)}_{1}(p^{2n}D,d).
\end{equation*}
Guerzhoy \cite{Guer1} showed that if $D$ and $-d$ are fundamental discriminants with $\left(\frac{-d}{p}\right)=\left(\frac{D}{p}\right)$, then, for any $m$, we have
\begin{equation*}
B^{(1)}_{m}(D,p^{2n}d)=p^{n}B^{(1)}_{m}(p^{2n}D,d).
\end{equation*}
In 2012, Ahlgren \cite{Ahl1} proved a general theorem which implies the above results as special cases. On the other hand, Osburn \cite{Osb1} proved that if $d$ is a positive integer such that $-d$ is congruent to a square modulo $4N$ and if $p\neq N$ is an odd prime which splits in $\mathbb{Q}(\sqrt{-d})$, then
$$B^{(N)}_{1}(1,p^{2}d)\equiv 0\pmod{p}.$$
Jenkins \cite{Jen2} and Koo and Shin \cite{KS1} obtained the following generalization of Osburn's result: for a positive integer $d$ such that $-d\equiv\square\pmod{4N}$ and an odd prime $p\neq N$ which splits in $\mathbb{Q}(\sqrt{-d})$,
$$B^{(N)}_{1}(1,p^{2n}d)\equiv 0\pmod{p^{n}}$$
for all $n\geq 1$.

The purpose of this paper is to generalize all these congruences to more general modular forms. To be precise, from now on, we assume that $N\geq 1$ is odd and square-free. For an even Dirichlet character $\chi$ modulo $4N$, we denote by $M^{!}_{k/2}(4N,\chi)$ the space of weakly holomorphic modular forms of weight $k/2$ on $\Gamma_{0}(4N)$ with Nebentypus $\chi$. The subspace of holomorphic forms and that of cuspforms are denoted by $M_{k/2}(4N,\chi)$ and $S_{k/2}(4N,\chi)$ respectively.

Let $\mathcal{D}$ be a discriminant form of level $4N$ satisfying some additional conditions which will be given in Section \ref{sec-epsilon}. (For the basics on discriminant forms, see Section \ref{disc} below.) Then $\mathcal{D}$ determines an even Dirichlet character $\chi$ modulo $4N$ and a sign vector $\epsilon=(\epsilon_{p})_{p}$ over $p=2$ or $p\mid N$ with $\chi_{p}\neq 1$, where the character $\chi$ is decomposed into $p$-components: $\chi=\prod_{p}\chi_{p}$. Set $\chi'=\chi\left(\frac{4N}{\cdot}\right)$.

We define the associated modular form space $M_{k/2}^{!\epsilon}(N,\chi')$ to be the subspace of $M_{k/2}^{!}(4N,\chi')$ consisting of the forms $f\in M_{k/2}^{!}(4N,\chi')$ satisfying the so-called \emph{$\epsilon$-condition}, which will be defined in Section \ref{sec-epsilon}. We let $$M_{k/2}^{\epsilon}(N,\chi')=M_{k/2}^{!\epsilon}(N,\chi')\cap M_{k/2}(4N,\chi') \text{ and } S_{k/2}^{\epsilon}(N,\chi')=M_{k/2}^{!\epsilon}(N,\chi')\cap S_{k/2}(4N,\chi').$$

Let us give an example. Consider the following even lattice
$$L=\left\{\left(\begin{array}{cc}a & b/N \\ c & -a\end{array}\right):a,b,c\in\mathbb{Z}\right\},$$
with $Q(\alpha)=-N\det(\alpha)$ and $(\alpha,\beta)=N\mathrm{tr}(\alpha\beta)$. We denote by $L'$ the dual lattice of $L$. Then the space $M_{k/2}^{!\epsilon}(N,\chi')$ associated with the discriminant form $L'/L$ is exactly the same as the space $M_{k/2}^{!+\cdots+}(N)$. Hence the $\epsilon$-condition can be considered as a generalization of the Kohnen plus condition.

Now we further assume that $\chi_{p}\neq 1$ for each $p\mid N$, so $\chi'=1$. In \cite{Zhang2}, Zhang defined a family of forms in $M_{k/2}^{!\epsilon}(N,1)$, called \emph{reduced forms}. (For the definition, see Section \ref{sec-epsilon}.) If a reduced form $f_m$ exists for some $m \in \mathbb Z$, it must be unique and $\chi_{p}(m)\neq-\epsilon_{p}$ for each $p\mid N$. The set of reduced modular forms forms a basis for $M_{k/2}^{!\epsilon}(N,1)$. When $k=3$ and $N\in\mathfrak{S}$, the reduced form $f_{-D}$ exists for each $D>0$ which is a square modulo $4N$ (cf. Proposition \ref{prop-zhang2} below). In fact, $s(-D)f_{-D}=g_{D,N}$ for every $D$ where $s(-D)$ is a scaling constant. Thus the reduced forms are natural generalizations of the forms $g_{D,N}$.

In order to generalize the congruences mentioned above to reduced forms, we first need to check integrality of the Fourier coefficients of reduced forms. We establish the following proposition which allows us to check whether a fixed reduced form has integer Fourier coefficients.

\begin{proposition} \label{prop-1.1}
Let $k$ be an odd integer. Assume that $f=\sum_{n}a(n)q^{n}\in M_{k/2}^{!\epsilon}(N,\chi')\cap\mathbb{Q}(\!(q)\!)$ with bounded denominator, and that $a(n)\neq 0$ for some $n<0$. Furthermore, let $k'$ be the smallest positive integer which satisfies $k'\geq |\mathrm{ord}_{\infty}(f)|/4N$ and $k+12k'> 0$. If $a(n)\in\mathbb{Z}$ for $n\leq\mathrm{ord}_{\infty}(f)+\frac{k+12k'}{12}[\mathrm{SL}_{2}(\mathbb{Z}):\Gamma_{0}(4N)]$, then $a(n)\in\mathbb{Z}$ for all $n$.
\end{proposition}

Let $\mathcal{D}^{*}$ be the dual discriminant form of $\mathcal{D}$. It is known that the corresponding data to $\mathcal{D}^{*}$ is $(4N,\chi',\epsilon^{*})$ with $\epsilon_{p}^{*}=\chi_{p}(-1)\epsilon_{p}$. Denote by $M_{k/2}^{\epsilon^{*}}(N,\chi')$ the space of modular forms associated to $\mathcal{D}^{*}$. We denote by $a(m,n)$ the $n^{\mathrm{th}}$ Fourier coefficient of the reduced form $f_{m}$.
We prove the following theorem which turns the integrality problem for reduced forms into checking finitely many of them.

\begin{theorem} \label{thm-1.2}
Let $m_{\epsilon}=\max\{m : f_{m}^{*}\in M_{2-k/2}^{\epsilon^{*}}(N,\chi')~\textnormal{exists}\}$. Assume that for all $n\in\mathbb{Z}$ and $m\geq -4N-m_{\epsilon}$, we have $s(m)a(m,n)\in\mathbb{Z}$. Then $s(m)a(m,n)\in\mathbb{Z}$ for all $m,n\in\mathbb{Z}$.
\end{theorem}

Therefore, to check the integrality of reduced forms, it suffices to show the integrality of a finite number of Fourier coefficients satisfying the conditions of both Proposition \ref{prop-1.1} and Theorem \ref{thm-1.2}. We give an example to illustrate this.

\begin{example} \label{ex-int}
We consider the space {$M_{1/2}^{!+\cdots +}(7,1)$}. Then we have $m_\epsilon=-1$. Define $$E_{k}(\tau)=1-\frac{2k}{B_{k}}\sum_{n=1}^{\infty}\sigma_{k-1}(n)q^{n}\qquad (2<k\in 2\mathbb{Z})$$ to be the normalized Eisenstein series,
and denote by $[\cdot,\cdot]_{n}$ ($n\geq 1$) the $n^{\text{th}}$ Rankin--Cohen bracket (cf. \cite[pp.53-58]{BrGeHaZa1}). Set
\[
\begin{array}{llll}
\mathrm{RC}_{1}=\frac{[\theta,E_{10}(28\tau)]_{1}}{\Delta(28\tau)}, & \mathrm{RC}_{2}=\frac{[\theta,E_{8}(28\tau)]_{2}}{\Delta(28\tau)}, & \mathrm{RC}_{3}=\frac{[\theta,E_{6}(28\tau)]_{3}}{\Delta(28\tau)}, & \mathrm{RC}_{4}=\frac{[\theta,E_{4}(28\tau)]_{4}}{\Delta(28\tau)}, \\ \mathrm{RC}_{5}=\frac{[\mathrm{RC}_{1},E_{10}(28\tau)]_{1}}{\Delta(28\tau)}, & \mathrm{RC}_{6}=\frac{[\mathrm{RC}_{1},E_{8}(28\tau)]_{2}}{\Delta(28\tau)}, &
\mathrm{RC}_{7}=\frac{[\mathrm{RC}_{1},E_{6}(28\tau)]_{3}}{\Delta(28\tau)}, & \mathrm{RC}_{8}=\frac{[\mathrm{RC}_{1},E_{4}(28\tau)]_{4}}{\Delta(28\tau)}, \\ \mathrm{RC}_{9}=\frac{[\mathrm{RC}_{2},E_{10}(28\tau)]_{1}}{\Delta(28\tau)}, &
\mathrm{RC}_{10}=\frac{[\mathrm{RC}_{2},E_{8}(28\tau)]_{2}}{\Delta(28\tau)}, & \mathrm{RC}_{11}=\frac{[\mathrm{RC}_{2},E_{6}(28\tau)]_{3}}{\Delta(28\tau)}, & \mathrm{RC}_{12}=\frac{[\mathrm{RC}_{2},E_{4}(28\tau)]_{4}}{\Delta(28\tau)}, \\
\mathrm{RC}_{13}=\frac{[\mathrm{RC}_{1},E_{10}(28\tau)]_{1}}{\Delta(28\tau)}, & \mathrm{RC}_{14}=\frac{[\mathrm{RC}_{3},E_{8}(28\tau)]_{2}}{\Delta(28\tau)}, & \mathrm{RC}_{15}=\frac{[\mathrm{RC}_{3},E_{6}(28\tau)]_{3}}{\Delta(28\tau)}, & \mathrm{RC}_{16}=\frac{[\mathrm{RC}_{3},E_{4}(28\tau)]_{4}}{\Delta(28\tau)}.\\
\end{array}
\]
In addition, we set
$$f=\tfrac{1}{5600}\mathrm{RC}_{1}+\tfrac{7}{103680}\mathrm{RC}_{2}+\tfrac{1}{80640}\mathrm{RC}_{3}+\tfrac{1}{705600}\mathrm{RC}_{4}-\tfrac{41687}{1800}\theta,$$
and define
\[ \mathrm{RC}_{17}=\tfrac{[f,E_{4}(28\tau)]_{4}}{\Delta(28\tau)}.\]
By taking linear combinations of these Rankin-Cohen brackets, we find
\begin{align*}
s(0)f_{0}&=1+2q+2q^4+2q^{9}+2q^{16}+\cdots,\\
s(-3)f_{-3}&=q^{-3}-3q-2q^{4}+6q^{8}+5q^{9}-10q^{16}+\cdots,\\
s(-7)f_{-7}&=q^{-7}-10q+4q^{4}+28q^{8}-24q^{9}+60q^{16}+\cdots,\\
s(-12)f_{-12}&=q^{-12}-10q-25q^{4}-6q^{8}+46q^{9}+152q^{16}+\cdots,\\
s(-19)f_{-19}&=q^{-19}-q-50q^{4}-50q^{8}-153q^{9}+798q^{16}+\cdots,\\
s(-20)f_{-20}&=q^{-20} - 22q + 26q^{4} - 180q^{8} - 78q^{9} - 338q^{16}+\cdots,\\
s(-24)f_{-24}&=q^{-24} - 2q - 28q^{4} + 225q^{8} - 450q^{9} - 2976q^{16}+\cdots,\\
s(-27)f_{-27}&=q^{-27} + 12q + 52q^{4} - 468q^{8} + 156q^{9} - 1300q^{16}+\cdots.   \end{align*}

For example, we obtain
\begin{align*}
f_{3}&=-\tfrac{92368453}{1197504000}\, \mathrm{RC}_{1}-\tfrac{1105849}{739031040}\, \mathrm{RC}_{2}-\tfrac{7775323}{804722688000}\, \mathrm{RC}_{3}+\tfrac{31109}{68584320000}\, \mathrm{RC}_{4}\\
&\quad-\tfrac{1}{49268736000}\, \mathrm{RC}_{7}+\tfrac{1}{862202880000}\, \mathrm{RC}_{8}-\tfrac{1}{86910050304000}\, \mathrm{RC}_{12}\\
&\quad+\tfrac{1}{216309458534400}\, \mathrm{RC}_{15}+\tfrac{83841213721}{1026432000}\, \theta\\
&=q^{-3} - 3q - 2q^{4} + 6q^{8} + 5q^{9} - 10q^{16}+\cdots.
\end{align*}
By Proposition \ref{prop-1.1}, the forms $s(0)f_{0},\ldots,s(-27)f_{-27}$ have integer Fourier coefficients. It follows from Theorem \ref{thm-1.2} that every reduced form in {$M_{1/2}^{!+\cdots+}(7,1)$} has integer Fourier coefficients.
\end{example}

Now we assume that, for any reduced form
$$f_{m}=\sum_{n}a(m,n)\, q^{n}\in M_{k/2}^{!\epsilon}(N,1),$$
the form $s(m)f_{m}$ has integer Fourier coefficients. Furthermore, let $k\geq 3$ be an odd integer and set $\lambda=(k-1)/2$. Then the reduced form $f_{m}\in M_{k/2}^{!\epsilon}(N,1)$ exists for every $m\in\mathbb{Z}_{<0}$ with $\chi_{p}(m)\neq -\epsilon_{p}$ for all $p\mid N$. We write
$$F_{m}(\tau)=s(m)f_{m}(\tau)=q^{m}+\sum\limits_{\substack{d\geq 0 \\ \chi_{p}(d)\neq -\epsilon_{p}~\textnormal{for all}~p\mid N}}B^{(N)}(m,d)\, q^{d}.$$
Note that the Hecke operator $T_{k/2,4N}(\ell^{2})$ acts on the space $M_{k/2}^{!\epsilon}(N,1)$ for each prime $\ell$ with $\gcd{(\ell,4N)}=1$. For any positive integer $t$ with $\gcd{(t,4N)}=1$, define
$$F_{m}^{(t)}:=F_{m}\mid T_{k/2,4N}(t^{2}).$$
Then we obtain the coefficients $B^{(N)}_{t}(m,d)$ from the equation
$$F_{m}^{(t)}(\tau)=\mbox{(principal part)}+\sum\limits_{\substack{d\geq 0,\\ \chi_{p}(d)\neq-\epsilon_{p}~\textnormal{for all}~p\mid N}}B^{(N)}_{t}(m,d)\, q^{d}.$$

We state our main theorem which describes various relations among the coefficients $B^{(N)}_{t}(m,d)$.

\begin{theorem}\label{thm-1.4}
We have the following:
\begin{enumerate}
\item[\textnormal{(\lowerromannumeral{1})}] $B^{(N)}_{t}(m,\ell^{2n+2}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)B^{(N)}_{t}(m,\ell^{2n}d)=\ell^{(k-2)n}\left \{ B^{(N)}_{t}(\ell^{2n}m,\ell^{2}d)-B^{(N)}_{t}(\ell^{2n-2}m,d)\right \}$.
\item[\textnormal{(\lowerromannumeral{2})}] If $\ell\nmid d$, then
\begin{align*}
\ell^{(k-2)n}&B^{(N)}_{t}(\ell^{2n}m,d)=B^{(N)}_{t}(m,\ell^{2n}d)\\
&\quad+\left[\left(\frac{(-1)^{\lambda}d}{\ell}\right)-\left(\frac{(-1)^{\lambda}m}{\ell}\right)\right]\cdot\sum_{k=1}^{n}\ell^{(\lambda-1)k}\left(\frac{(-1)^{\lambda}d}{\ell}\right)^{k-1}B^{(N)}_{t}(m,\ell^{2n-2k}d).
\end{align*}
\item[\textnormal{(\lowerromannumeral{3})}] If $\ell\parallel d$, then
$$\ell^{(k-2)n}B^{(N)}_{t}(\ell^{2n}m,d)=B^{(N)}_{t}(m,\ell^{2n}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)\cdot B^{(N)}_{t}(m,\ell^{2n-2}d).$$
\end{enumerate}
\end{theorem}

As a corollary, we obtain the following congruences:

\begin{corollary}\label{cor-1.5}
Assume that $S_{k/2}^{\epsilon}(N,1)=0$.
\begin{enumerate}
\item If $\left(\frac{-d}{\ell}\right)=\left(\frac{-m}{\ell}\right)\neq 0$, or if $\ell\parallel d$ and $\ell\parallel m$, then for any positive integer $t$ with $(t,4N)=1$ and $n$, we have
$$B^{(N)}_{t}(m,\ell^{2n}d)=\ell^{(k-2)n}\, B^{(N)}_{t}(\ell^{2n}m,d)\equiv 0\pmod{\ell^{(k-2)n}}.$$
\item If $\chi_{p}(\ell d)\neq -\epsilon_{p}$ for all $p\mid N$, then for any positive integer $t$ with $(t,4N)=1$ and any $n\geq 1$, we get
$$B^{(N)}_{t}(m,\ell^{2n+1}d)\equiv\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)B^{(N)}_{t}(m,\ell^{2n-1}d)\pmod{\ell^{(k-2)n}}.$$
\end{enumerate}
\end{corollary}

As for the condition in the above corollary, we remark that if
\begin{align*}
N\in\{n\mid n~\textnormal{is an odd } & \textnormal{square-free integer with}~1\leq n< 37\}\\
&\cup\{39,41,47,51,55,59,69,71,87,95,105,119\},
\end{align*}
then $S_{3/2}^{+\cdots +}(N,1)=0$. (See \cite[Table 4]{BrEhFr1}.)

We organize this paper as follows. In Section 2, we present preliminaries on discriminant forms and modular forms of half-integral weight. Also, we recall the definitions of $\epsilon$-condition and reduced forms. In Section 3, we prove Proposition \ref{prop-1.1} and Theorem \ref{thm-1.2}, and in Section 4 we prove Theorem \ref{thm-1.4} and Corollary \ref{cor-1.5}.

\section{Preliminaries}
In this section, we review the basics on discriminant forms, modular forms of half-integral weight and present a recent result of Zhang \cite{Zhang2}.

\subsection{Discriminant forms.} \label{disc} A discriminant form is a finite abelian group {$\mathcal{D}$} with a quadratic form {$Q:\mathcal{D}\rightarrow\mathbb{Q}/\mathbb{Z}$}, such that the symmetric bilinear form defined by $(\beta,\gamma)=Q(\beta+\gamma)-Q(\beta)-Q(\gamma)$ is nondegenerate, namely, the map {$\mathcal{D}\mapsto\mathrm{Hom}(\mathcal{D},\mathbb{Q}/\mathbb{Z})$} defined by $\gamma\mapsto(\gamma,\cdot)$ is an isomorphism. We define the level of a discriminant form {$\mathcal{D}$} to be the smallest positive integer $N$ such that $NQ(\gamma)=0$ for each {$\gamma\in\mathcal{D}$}. It is known that if $L$ is an even lattice then $L'/L$ is a discriminant form, where $L'$ is the dual lattice of $L$. Conversely, any discriminant form can be obtained in this way. The signature {$\mathrm{sign}(\mathcal{D})\in\mathbb{Z}/8\mathbb{Z}$} is defined to be the signature of $L$ modulo 8 for any even lattice $L$ such that {$L'/L=\mathcal{D}$}.

Every discriminant form can be decomposed uniquely into $p$-components {$\mathcal{D}=\oplus_{p}\mathcal{D}_{p}$}. Each $p$-component can be written as a direct sum of indecomposable Jordan $q$-components with $q$ powers of $p$. Such decompositions are not unique in general. We recall the possible indecomposable Jordan $q$-components as follows.

Let $p$ be an odd prime and $q>1$ be a power of $p$. The indecomposable Jordan components with exponent $q$ are denoted by $q^{\delta_{q}}$ with $\delta_{q}=\pm 1$. These discriminant forms both have level $q$.

If $q>1$ is a power of 2, there are also precisely two indecomposable \emph{even} Jordan components of exponent $q$, denoted by $q^{\delta_{q}2}$ with $\delta_{q}=\pm 1$. Such components have level $q$. There are also \emph{odd} indecomposable Jordan components, denoted by $q_{t}^{\pm 1}$ with $\pm 1=\left(\frac{2}{t}\right)$ for each $t\in (\mathbb{Z}/8\mathbb{Z})^{\times}$. These discriminant forms have level $2q$.

We call a discriminant form {$\mathcal{D}$} \emph{transitive} if  the action of {$\mathrm{Aut}(\mathcal{D})$} is transitive on the subset of elements of norm $n$ for any $n\in\mathbb{Q}/\mathbb{Z}$. By the classification of transitive forms, the level $N=\prod_{p}N_{p}$ of a transitive form {$\mathcal{D}=\oplus_{p}\mathcal{D}_{p}$} is of the following form: $N_{p}=1$ or $p$ for an odd prime $p$ and $N_{2}=1,2,4$ or 8. In other words, $N$ is the conductor of a quadratic Dirichlet character.

Let {$\mathcal{D}$} be a transitive discriminant form of odd signature $r$ and level $N$. Then {$\mathcal{D}$} determines an even quadratic character $\chi$ modulo $N$. Explicitly, it is given as follows: Decompose $\chi=\prod_{p}\chi_{p}$ into $p$-components. If $p$ is odd,
{\begin{equation*}
\chi_{p}(d)=
\begin{cases}
1, & \textnormal{if}~p\nmid|\mathcal{D}|~\textnormal{or}~p^{2}\mid|\mathcal{D}|,\\
\left(\frac{d}{p}\right), & \textnormal{otherwise}.
\end{cases}
\end{equation*}
When $2\mid |\mathcal{D}|$,
\begin{equation*}
\chi_{2}(d)=
\begin{cases}
1, & \textnormal{if}~\left(\frac{-1}{|\mathcal{D}|}\right)=+1~\textnormal{and}~\mathcal{D}_{2}=2_{\pm 3}^{+3},~2_{\pm 1}^{+1},\\
\left(\frac{-4}{d}\right), & \textnormal{if}~\left(\frac{-1}{|\mathcal{D}|}\right)=-1~\textnormal{and}~\mathcal{D}_{2}=2_{\pm 3}^{+3},~2_{\pm 1}^{+1},\\
\left(\frac{2}{d}\right), & \textnormal{if}~\left(\frac{-1}{|\mathcal{D}|}\right)=+1~\textnormal{and}~\mathcal{D}_{2}=4_{\pm 1}^{+1},~4_{\pm 3}^{-1},\\
\left(\frac{-2}{d}\right), & \textnormal{if}~\left(\frac{-1}{|\mathcal{D}|}\right)=-1~\textnormal{and}~\mathcal{D}_{2}=4_{\pm 1}^{+1},~4_{\pm 3}^{-1}.
\end{cases}
\end{equation*}}

For more details on discriminant forms, see \cite{CS1}, \cite{Ni1}, \cite{Sch1} or \cite{Str1}.

\subsection{Metaplectic covers.} \label{meta} Throughout this paper, unless otherwise stated, {$k$ is an odd integer}. Let $\mathrm{Mp}_{2}^{+}(\mathbb{R})$ be the metaplectic cover of $\mathrm{GL}_{2}^{+}(\mathbb{R})$. The elements of $\mathrm{Mp}_{2}^{+}(\mathbb{R})$ are pairs $(A,\phi)$ where $\phi$ is a holomorphic function on $\mathbb{H}$ and
$$A=\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)\in\mathrm{GL}_{2}^{+}(\mathbb{R}),\quad\phi(\tau)=tj(A,\tau),~\textnormal{for some}~t\in\mathbb{C},~|t|=1.$$
Here $j(A,\tau)=\det(A)^{-\frac{1}{4}}(c\tau+d)^{\frac{1}{2}}$. The product of two elements $(A_{1},\phi_{1})$ and $(A_{2},\phi_{2})$ is defined by
$$(A_{1}A_{2},\phi_{1}(A_{2}\tau)\phi_{2}(\tau)).$$
To introduce the theta multiplier system, we first extend the Jacobi symbol. For an integer $c$ and an odd integer $d\neq 0$, the ``extended Jacobi symbol" $\left(\frac{c}{d}\right)$ is defined as follows.
\begin{enumerate}
\item[(1)] $\left(\frac{c}{d}\right)=0$ if $\gcd{(c,d)}>1$;
\item[(2)] $\left(\frac{0}{\pm 1}\right)=1$;
\item[(3)] If $d>0$, then $\left(\frac{c}{d}\right)$ is the usual Jacobi symbol;
\item[(4)] If $d<0$, then $\left(\frac{c}{d}\right)=\mathrm{sgn}(c)\left(\frac{c}{|d|}\right)$.
\end{enumerate}
Next we define $\varepsilon_{d}$ for odd $d$ by:
\begin{equation*}
\varepsilon_{d}=
\begin{cases}
1 & \textnormal{if}\quad d\equiv 1\pmod{4};\\
i & \textnormal{if}\quad d\equiv 3\pmod{4}.
\end{cases}
\end{equation*}
We finally define the theta multiplier system $\nu$ on $\Gamma_{0}(4)$:
$$\nu(A)=\left(\frac{c}{d}\right)\varepsilon_{d}^{-1},\quad A=\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)\in\Gamma_{0}(4).$$
Note that
$$\overline{\nu}(A)=\nu^{3}(A)=\left(\frac{-1}{d}\right)\nu(A),\quad\nu(A)\nu(A^{-1})=1,\quad A\in\Gamma_{0}(4).$$
For any $A=\left(\begin{smallmatrix}a & b \\ c & d\end{smallmatrix}\right)\in\mathrm{GL}_{2}^{+}(\mathbb{R})$, we let
$$\tilde{A}=(A,j(A,\tau))\in\mathrm{Mp}_{2}^{+}(\mathbb{R}).$$

Let $\mathrm{Mp}_{2}(\mathbb{Z})$ be the metaplectic double cover of $\mathrm{SL}_{2}(\mathbb{Z})$ inside $\mathrm{Mp}_{2}^{+}(\mathbb{R})$, consisting of pairs $(A,\phi)$ with $A=\left(\begin{smallmatrix}a & b \\ c & d\end{smallmatrix}\right)\in\mathrm{SL}_{2}(\mathbb{Z})$ and $\phi^{2}=c\tau+d$. Let $S$ and $T$ denote the standard generators of $\mathrm{SL}_{2}(\mathbb{Z})$. Then
$$\tilde{S}=\left(\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right),\sqrt{\tau}\right),\quad\tilde{T}=\left(\left(\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right),1\right)$$
generate $\mathrm{Mp}_{2}(\mathbb{Z})$.

\subsection{Modular forms.} \label{mod} Let $(A,\phi)\in\mathrm{Mp}_{2}^{+}(\mathbb{R})$ and $f:\mathbb{H}\rightarrow\mathbb{C}$ be a function. The weight {$k/2$} slash operator is defined by
{$$(f|_{k/2}(A,\phi))(\tau)=\phi^{-k}(\tau)f(A\tau),\quad A=\left(\begin{array}{cc}a & b \\ c & d\end{array}\right).$$}

Consider the Atkin--Lehner operators on the space {$M_{k/2}^{!}(4N,\chi)$}, where $N$ is odd square-free. For any odd divisor $m$ of $N$, we choose $\gamma_{m}$ in $\mathrm{SL}_{2}(\mathbb{Z})$ such that
\begin{equation*}
\gamma_{m}\equiv
\begin{cases}
S &\mod{m^{2}},\\
I &\mod{(4N/m)^{2}},
\end{cases}
\end{equation*}
and let
\begin{equation*}
\quad\gamma_{4m}:=S\gamma_{N/m}^{-1}\equiv
\begin{cases}
S &\mod{(4m)^{2}},\\
I &\mod{(4N/m)^{2}}.
\end{cases}
\end{equation*}
We shall assume for simplicity that all of the entries of $\gamma_{m}$ are positive; this can be achieved by left and/or right multiplication by matrices in $\Gamma(16N^{2})$.

For any nonzero integer $m$, let
$$\delta_{m}=\left(\begin{array}{cc}m & 0 \\ 0 & 1\end{array}\right),\quad\widetilde{\delta_{m}}=\left(\left(\begin{array}{cc}m & 0 \\ 0 & 1\end{array}\right),m^{-\frac{1}{4}}\right).$$
For any odd positive divisor $m$ of $N$, let $W(m)=\gamma_{m}^{*}\widetilde{\delta_{m}}$. Define
$$\tau_{4N}=(I,\sqrt{-i})\widetilde{\beta_{4N}}=\left(\left(\begin{array}{cc}0 & -1 \\ 4N & 0\end{array}\right),(4N)^{\frac{1}{4}}(-i\tau)^{\frac{1}{2}}\right),\quad\beta_{N}=\left(\begin{array}{cc}0 & -1 \\ 4N & 0\end{array}\right).$$
For each divisor $m$, even or odd, of $4N$, define $U(m)$ as follows:
{$$f|U(m)=m^{\frac{k}{4}-1}\sum_{j\ \textrm{mod} \ {m}}f|\widetilde{\delta_{m}}^{-1}\tilde{T}^{j}=m^{\frac{k}{4}-1}\sum_{j\ \textrm{mod} \ {m}}f|\widetilde{\delta_{m}^{-1}T^{j}}.$$}
Finally, we define $Y(p)$ for each odd prime divisor $p$ of $N$ by
{$$f|Y(p)=p^{1-\frac{k}{4}}f|U(p)W(p),$$}
and $Y(4)$ by
{$$f|Y(4)=4^{1-\frac{k}{4}}f|U(4)W(M)\tau_{4N}.$$}

\begin{proposition}[\cite{Shi1, Ueda1, Zhang2}]
Let {$f\in M_{k/2}^{!}(4N,\chi)$}.
\begin{enumerate}
\item[\textnormal{(1)}] {$f|\tau_{4N}\in M_{k/2}^{!}(4N,\chi\left(\frac{4N}{\cdot}\right))$} and $f|\tau_{4N}^{2}=f$.
\item[\textnormal{(2)}] For each $m\mid 4N$, {$f|U(m)\in M_{k/2}^{!}(4N,\chi\left(\frac{m}{\cdot}\right))$}.
\item[\textnormal{(3)}] For each $m\mid N$, {$f|W(m)\in M_{k/2}^{!}(4N,\chi\left(\frac{m}{\cdot}\right))$} and
{$$f|W(m)^{2}=\varepsilon_{m}^{-k}\chi_{m}(-1)\chi_{4N/m}(m)f.$$}
Moreover, if $m,m'\mid N$ and $\gcd{(m,m')}=1$, then $f|W(m)W(m')=\chi_{m'}(m)f|W(mm')$.
\item[\textnormal{(4)}] For any $m,m'\mid N$ with $\gcd{(m,m')}=1$,
$$f|W(m)U(m')=\chi_{m}(m')f|U(m')W(m)~\textnormal{and}~f|U(4)W(m)=f|W(m)U(4).$$
\item[\textnormal{(5)}] For any $m\mid N$, $f|\tau_{4N}U(m)W(m)=\chi_{m}(M/m)f|W(m)U(m)\tau_{4N}$.
\end{enumerate}
\end{proposition}

Now we consider the operators $Y(p)$ and $Y(4)$.

\begin{proposition}[\cite{Koh1, Ueda1, Zhang2}] \hfill
\begin{enumerate}
\item[\textnormal{(1)}] The space {$M_{k/2}^{!}(4N,\chi)$} decomposes under $Y(4)$ into eigenspaces
{$$M_{k/2}^{!}(4N,\chi)=M_{k/2}^{!}(4N,\chi)_{\mu_{2}^{+}}\oplus M_{k/2}^{!}(4N,\chi)_{\mu_{2}^{-}},$$}
where the eigenvalues are
{$$\mu_{2}^{+}=\chi_{2}(-1)^{\frac{k}{2}+1}(-1)^{\lfloor\frac{k+1}{4}\rfloor}2^{\frac{3}{2}} \quad \text{ and } \quad \mu_{2}^{-}=-2^{-1}\mu_{2}^{+}.$$}
Moreover, {$f=\sum_{n}a(n)q^{n}\in M_{k/2}^{!}(4N,\chi)_{\mu_{2}^{+}}$} if and only if
{$$a(n)=0~\text{whenever}~\chi_{2}(-1)(-1)^{\frac{k-1}{2}}n\equiv 2,3\pmod{4}.$$}
\item[\textnormal{(2)}] Assume that $p\mid N$ with $\chi_{p}=1$. Then the space {$M_{k/2}^{!}(4N,\chi)$} decomposes under $Y(p)$ into eigenspaces
$$M_{k/2}^{!}(4N,\chi)=M_{k/2}^{!}(4N,\chi)_{\mu_{p}^{+}}\oplus M_{k/2}^{!}(4N,\chi)_{\mu_{p}^{-}},$$
where the eigenvalues are $\mu_{p}^{+}=\varepsilon_{p}^{-1}p^{\frac{1}{2}}$ and $\mu_{p}^{-}=-\mu_{p}^{+}$. Moreover, {$f=\sum_{n}a(n)q^{n}\in M_{k/2}^{!}(4N,\chi)_{\mu_{p}^{+}}$} (resp. {$M_{k/2}^{!}(4N,\chi)_{\mu_{p}^{-}}$}) if and only if
$$a(n)=0~\text{whenever}~\left(\frac{n}{p}\right)=- 1   \quad ~(resp.~ \left(\frac{n}{p}\right)= 1~)~.$$
\item[\textnormal{(3)}] Assume that $p\mid N$ with $\chi_{p}=\left(\frac{\cdot}{p}\right)$. Then the space {$M_{k/2}^{!}(4N,\chi)$} decomposes under $Y(p)$ into eigenspaces
{$$M_{k/2}^{!}(4N,\chi)=M_{k/2}^{!}(4N,\chi)_{\mu_{p}^{+}}\oplus M_{k/2}^{!}(4N,\chi)_{\mu_{p}^{-}},$$}
where the eigenvalues are $\mu_{p}^{+}=-1$ and $\mu_{p}^{-}=-p$.
\item[\textnormal{(4)}] The space {$M_{k/2}^{!}(4N,\chi)$} decomposes into the direct sum of simultaneous eigenspaces for the operators $Y(4)$ and $Y(p)$.
\end{enumerate}
\end{proposition}

\subsection{$\epsilon$-condition and reduced forms}
\label{sec-epsilon} Let {$\mathcal{D}$} be a transitive discriminant form of odd signature $r$ and level $M$. Assume that {$\mathcal{D}_{2}=2_{\pm 1}^{+1}$}. Then $M=4N$ for some odd square-free $N$ and
{$$\chi_{2}(-1)=e_{4}(r-t)=\left(\frac{-1}{|\mathcal{D}|}\right),$$}
where $e_{4}(x)=e^{2\pi ix/4}$. Recall that {$\mathcal{D}$} determines an even Dirichlet character $\chi$ modulo $4N$. We define a sign vector $\epsilon=(\epsilon_{p})_{p}$ over $p=2$ or $p\mid N$ with $\chi_{p}\neq 1$ as follows:
{\begin{equation*}
\epsilon_{p}=
\begin{cases}
\chi_{p}(2N/p) & \textnormal{if}~p\mid N,~\chi_{p}\neq 1~\textnormal{and}~\mathcal{D}_{p}=p^{+1},\\
-\chi_{p}(2N/p) & \textnormal{if}~p\mid N,~\chi_{p}\neq 1~\textnormal{and}~\mathcal{D}_{p}=p^{-1},\\
\left(\frac{-1}{N}\right) & \textnormal{if}~p=2~\textnormal{and}~\mathcal{D}_{2}=2_{+1}^{+1},\\
-\left(\frac{-1}{N}\right) & \textnormal{if}~p=2~\textnormal{and}~\mathcal{D}_{2}=2_{-1}^{+1}.
\end{cases}
\end{equation*}}
Therefore, {$\mathcal{D}$} determines $(4N,\chi\left(\frac{4N}{\cdot}\right),\epsilon)$: $\chi$ an even Dirichlet character modulo $4N$ and $\epsilon$ a sign vector. We shall denote $\chi'=\chi\left(\frac{4N}{\cdot}\right)$ from now on.

Given any data $(4N,\chi',\epsilon)$ with even $\chi$ and $\epsilon_{p}=\pm 1$ for $p=2$ or $p\mid N$ with $\chi_{p}\neq 1$, we define the associated modular form space {$M_{k/2}^{!\epsilon}(N,\chi')$} to be the common eigenspace with eigenvalues $\mu_{2}^{+}$ for $Y(4)$, $\mu_{p}^{\epsilon_{p}}$ for $Y(p)$ if $\chi_{p}\neq 1$ and $\mu_{p}^{+}=-1$ for $Y(p)$ if $\chi_{p}=1$. Since
\begin{align*}
\epsilon_{2}\chi_{2}'(-1)(-1)^{\frac{k-1}{2}}&=t\chi_{2}(-1)e_{4}(k-1)\\
&=\chi_{2}(-1)e_{4}(k-1)e_{4}(1-t)\\
&=\chi_{2}(-1)e_{4}(r-t)=1,
\end{align*}
we have {$f=\sum_{n}a(n)q^{n}\in M_{k/2}^{!\epsilon}(N,\chi')$} if and only if the following two conditions are satisfied:

(1) $a(n)=0$ whenever $n\equiv 2,-\epsilon_{2}\pmod{4}$ or $\left(\frac{n}{p}\right)=-\epsilon_{p}$ for some $p\mid N$ with $\chi_{p}\neq 1$,

(2) {$f|_{k/2}Y(p)=-f$} for every $p\mid N$ with $\chi_{p}=1$.\\

Recall the even lattice $L$ introduced in Section \ref{intro}. Then the discriminant form {$\mathcal{D}=L'/L\cong\mathbb{Z}/2N\mathbb{Z}$} with {$\mathcal{D}=\prod_{p\mid 2N}\mathcal{D}_{p}$} is given by
{$$\mathcal{D}_{2}=2_{t}^{+1},\quad t=\left(\frac{-1}{N}\right),\quad \mathcal{D}_{p}=p^{\delta_{p}},\quad \delta_{p}=\left(\frac{2N/p}{p}\right)\quad\textnormal{for}\quad p\mid N.$$}
It follows that for such {$\mathcal{D}$}, we have $\epsilon_{p}=+1$ for all $p\mid N$ and $\chi'=\chi\left(\frac{4N}{\cdot}\right)=1$. From the above observation, we have {$f=\sum_{n}a(n)q^{n}\in M_{k/2}^{!\epsilon}(N,1)$ if and only if $a(n)=0$ unless $(-1)^{\frac{k-1}{2}}n$ is a square modulo $4N$}. Thus the space {$M_{k/2}^{!\epsilon}(N,1)$} is exactly the  same as the space {$M_{k/2}^{!+\ldots+}(N)$}. Eichler and Zagier denote the space {$M_{k/2}^{\epsilon}(N,1)$} by $M_{k}^{+\ldots+}(N)$ in \cite[p.69]{EiZa1}.

\medskip

Now we shall assume that $\chi_{p}\neq 1$ for each $p\mid N$, so $\chi'=1$. A form {$f\in M_{k/2}^{!\epsilon}(N,\chi')$} is called \textit{reduced} if $f=\frac{1}{s(m)}q^{m}+ \sum_{\ell \ge m+1}  a(\ell) q^\ell$ for some integer $m$ and if for each $n>m$ with  $a(n)\neq 0$, there does not exist {$g\in M_{k/2}^{!\epsilon}(N,\chi')$} such that $g=q^{n}+O(q^{n+1})$. Here, {$s(m)=\displaystyle\prod_{p\mid\gcd{(N,m)}}\left(1+\frac{p}{|\mathcal{D}_p|}\right)$}. If a reduced form exists for some $m$, it is unique and $\chi_{p}(m)\neq-\epsilon_{p}$ for each $p\mid N$; we denote it by $f_{m}$. The set of reduced modular forms is a basis for {$M_{k/2}^{!\epsilon}(N,\chi')$}.

The following proposition determines $m<0$ for which $f_{m}$ exists. To state it, we need some notation. Let {$\mathcal{D}^{*}$} be the dual discriminant form of {$\mathcal{D}$} given by the same abelian group with the quadratic form $-Q$. It is known that {$\mathcal{D}^{*}$} is also transitive and the corresponding data is $(4N,\chi',\epsilon^{*})$ with $\epsilon_{p}^{*}=\chi_{p}(-1)\epsilon_{p}$.

\begin{proposition}[\cite{Zhang2}, Proposition 6.1] \label{prop-zhang2}
Let {$B^{*}=\{m : f_{m}^{*}\in M_{2-k/2}^{\epsilon^{*}}(N,\chi')~\textnormal{exists}\}$}. Then for any $m<0$ with $\chi_{p}(m)\neq -\epsilon_{p}$ for all $p\mid N$, the reduced form {$f_{m}\in M_{k/2}^{!\epsilon}(N,\chi')$} exists if and only if $-m\notin B^{*}$.
\end{proposition}

\section{Proofs of Proposition \ref{prop-1.1} and Theorem \ref{thm-1.2}}
In this section, we prove the rationality of Fourier coefficients of reduced forms following the lines in \cite{BrBun1} and \cite{Zhang1}. For $f=\sum_{n}a(n)q^{n}$ and $\sigma\in\mathrm{Aut}(\mathbb{C})$, define $f^{\sigma}=\sum_{n}\sigma(a(n))q^{n}$.

\begin{lemma} \label{lem-1}
Let $\chi$ be a Dirichlet character modulo $N$ with values in $\mathbb{Q}$. If $f\in M_{k/2}^{!}(4N,\chi)$, so is $f^{\sigma}$.
\end{lemma}

\begin{proof}
It is known that $\mathrm{Aut}(\mathbb{C})$ acts on the space {$M_{k/2}(\Gamma_{1}(4N),\chi)$}, the space of holomorphic modular forms of weight {$k/2$} on $\Gamma_{1}(4N)$ with Nebentypus $\chi$, and {$\sigma(M_{k/2}(4N,\chi))=M_{k/2}(4N,\chi^{\sigma})$}. (See \cite{SeSt1}.) Since $\chi$ has values in $\mathbb{Q}$, $\mathrm{Aut}(\mathbb{C})$ acts on {$M_{k/2}(4N,\chi)$}.

Note that {$f\Delta^{k'}\in M_{k/2+12k'}(4N,\chi)$} for a sufficiently large positive integer $k'$. Here $\Delta$ is the unique normalized cusp form of weight $12$ for $\mathrm{SL}_{2}(\mathbb{Z})$. The above observation shows that {$(f\Delta^{k'})^{\sigma}\in M_{k/2+12k'}(4N,\chi)$}. But $\Delta$ has integral Fourier coefficients, hence {$f^{\sigma}\Delta^{k'}=(f\Delta^{k'})^{\sigma}\in M_{k/2+12k'}(4N,\chi)$} and {$f^{\sigma}\in M_{k/2}^{!}(4N,\chi)$}.
\end{proof}

\begin{proposition}
Let $k<0$ and let {$f=\sum_{n}a(n)q^{n}\in M_{k/2}^{!\epsilon}(N,\chi')$}. Suppose that $a(n)\in\mathbb{Q}$ for $n<0$. Then all the coefficients $a(n)$ are rational with bounded denominators.
\end{proposition}

\begin{proof}
Let $\sigma\in\mathrm{Aut}(\mathbb{C})$. By Lemma \ref{lem-1}, {$f^{\sigma}\in M_{k/2}^{!}(4N,\chi')$}. It is easy to check that the action of $\mathrm{Aut}(\mathbb{C})$ preserves the $\epsilon$-condition. Since $a(n)\in\mathbb{Q}$ for $n<0$, $h:=f-f^{\sigma}$ is holomorphic at $\infty$. By \cite[Corollary 5.5]{Zhang2}, {$h\in M_{k/2}^{\epsilon}(N,\chi')$}. But $k<0$, so $h=0$. It follows that $f$ has rational coefficients.

We know that {$\theta f\Delta^{k'}\in S_{(k+1)/2+12k'}(4N,\chi')\subset S_{(k+1)/2+12k'}(\Gamma_{1}(4N))$} for a sufficiently large positive integer $k'$. Shimura proved that {$S_{(k+1)/2+12k'}(\Gamma_{1}(4N))$} has a basis $\mathcal{B}$ consisting of forms whose Fourier coefficients at $\infty$ are rational integers. (See \cite[Theorem 3.52]{Shi2}.) Let {$S_{(k+1)/2+12k'}^{\mathbb{Q}}(\Gamma_{1}(4N))$} be the $\mathbb{Q}$-vector space of cusp forms in {$S_{(k+1)/2+12k'}(\Gamma_{1}(4N))$} whose Fourier coefficients at $\infty$ are rational numbers. Then $\mathcal{B}$ is a $\mathbb{Q}$-basis of {$S_{(k+1)/2+12k'}^{\mathbb{Q}}(\Gamma_{1}(4N))$} and {$f\theta\Delta^{k'}\in S_{(k+1)/2+12k'}^{\mathbb{Q}}(\Gamma_{1}(4N))$}. This implies that $f\theta\Delta^{k'}$ has coefficients with bounded denominators, and we conclude that the $a(n)$ are rational with bounded denominators.
\end{proof}

We are interested in integrality of Fourier coefficients. So we generalize Sturm's theorem to {$M_{k/2}^{!\epsilon}(N,\chi')$}. We begin with introducing the original Sturm's theorem.

\begin{theorem}[\cite{St1}] \label{thm-St}
Let $\mathcal{O}_{F}$ be the ring of integers of a number field $F$, $\mathfrak{p}$  any prime ideal, $N'$ a positive integer and $k'$  a positive integer. Assume {$f=\sum_{n}a(n)q^{n}\in M_{k'}(N',\chi)\cap\mathcal{O}_{F}[\![ q]\!]$}. If $a(n)\in\mathfrak{p}$ for $n\leq\frac{k'}{12}[\mathrm{SL}_{2}(\mathbb{Z}):\Gamma_{0}(N')]$, then $a(n)\in\mathfrak{p}$ for all $n$.
\end{theorem}

Using Theorem \ref{thm-St}, Kim, Lee and Zhang proved the following:

\begin{corollary}[\cite{KLZ1}, Corollary 3.2] \label{cor-KLZ}
Let $k'$ be a positive integer. Assume {$f=\sum_{n}a(n)q^{n}\in M_{k'}(4N,\chi)\cap\mathbb{Q}[\![ q]\!]$} with bounded denominator. If $a(n)\in\mathbb{Z}$ for $n\leq\frac{k'}{12}[\mathrm{SL}_{2}(\mathbb{Z}):\Gamma_{0}(4N)]$, then $a(n) \in\mathbb{Z}$ for all $n$.
\end{corollary}

We extend this result to half-integral weight case. Let
$$\theta(\tau)=\sum_{n\in\mathbb{Z}}q^{n^{2}}=1+2q+2q^{4}+2q^{9}+\cdots,\quad (q=e^{2\pi i\tau},~\tau\in\mathbb{H}).$$
\begin{corollary} \label{cor-let}
{Let $k>0$ be an odd integer} and assume that {$f=\sum_{n}a(n)q^{n}\in M_{k/2}(4N,\chi)\cap\mathbb{Q}[\![ q]\!]$} with bounded denominator. If $a(n)\in\mathbb{Z}$ for {$n\leq\frac{k}{12}[\mathrm{SL}_{2}(\mathbb{Z}):\Gamma_{0}(4N)]$}, then $a(n)\in\mathbb{Z}$ for all $n$.
\end{corollary}

\begin{proof}
By multiplying {$\theta^{k}$}, we have {$f\theta^{k}\in M_{k}(4N,\chi)$}. It suffices to show that all the coefficients of {$f\theta^{k}$} are integers. Since $a(n)\in\mathbb{Z}$ for $n\leq\frac{k}{12}[\mathrm{SL}_{2}(\mathbb{Z}):\Gamma_{0}(4N)]$, the same thing holds for the coefficients of {$f\theta^{k}$}. By Corollary \ref{cor-KLZ}, every coefficient of {$f\theta^{k}$} is an integer.
\end{proof}

Now we are ready to prove Proposition \ref{prop-1.1} and Theorem \ref{thm-1.2}.

\begin{proof}[Proof of Proposition \ref{prop-1.1}]
{Since $k'\geq |\mathrm{ord}_{\infty}(f)|/4N$, we see that $f(\tau)\Delta(4N\tau)^{k'}\in M_{k/2+12k'}^{\epsilon}(N,\chi)$ and that every coefficient of $f(\tau)\Delta(4N\tau)^{k'}$ less than or equal to $\frac{k+12k'}{12}[\mathrm{SL}_{2}(\mathbb{Z}):\Gamma_{0}(4N)]$ is an integer. By Corollary \ref{cor-let}, $f(\tau)\Delta(4N\tau)^{k'}$ has integer Fourier coefficients, hence so does $f$.}
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm-1.2}]
Consider any reduced form $f_{m'}$ with $m'<-4N-m_{\epsilon}$. There exist integers $-4N-m_{\epsilon}\leq m_{0}'<-m_{\epsilon}$ and $l\geq 1$ such that $m'=-4Nl+m_{0}'$. By maximality of $m_{\epsilon}$, $f_{m_{0}'}$ exists. Consider now
$$g=j(4N\tau)^{l}f_{m_{0}'}=\sum_{n}b(n)q^{n}\in M_{k/2}^{!\epsilon}(N,\chi'),$$
where $j(\tau)$ denotes the classical $j$-function. It is known that $j$ has integral Fourier coefficients. By the assumption on $f_{m_{0}'}$,  we see that $b(n)s(m_{0}')\in\mathbb{Z}$ for each $n$.

Now $s(m_{0}')g$ and $s(m')f_{m'}$ share the same lowest power term, and we must have that
$$s(m')f_{m'}=s(m_{0}')g-\sum_{m>m'}s(m_{0}')b(m)s(m)f_{m}.$$
Hence $s(m')a_{m'}(n)=s(m_{0}')b(n)-s(m)b(m)s(m_{0}')a_{m}(n)\in\mathbb{Z}$ by the assumption and induction on $m$.
\end{proof}

\section{Proofs of Theorem \ref{thm-1.4} and Corollary \ref{cor-1.5}}
From now on, we shall assume that, for any reduced form
$$f_{m}=\sum_{n}a(m,n)q^{n}\in M_{k/2}^{!\epsilon}(N,\chi),$$
the modular form $s(m)f_{m}$ has integral Fourier coefficients. We remark that such integrality for each fixed reduced form can be verified by Proposition \ref{prop-1.1}. {Also, as we showed in Example \ref{ex-int}, every reduced form in the space {$M_{1/2}^{!+\cdots +}(7,1)$} satisfies the assumption.

We begin with a lemma.

\begin{lemma} \label{lem-sm}
Let $N\geq 1$ be a square-free integer. Then we have {$M_{3/2}^{+\cdots +}(N,1)=S_{3/2}^{+\cdots +}(N,1)$}.
\end{lemma}

\begin{proof}
Let {$f=\sum_{n\geq 0}a(n)q^{n}\in M_{3/2}^{+\cdots +}(N,1)$}. By Borcherds' obstruction theorem (Theorem 3.1 of \cite{Bor1}), we get
$$s(0)a(0)b(0)=0$$
for each {$g=\sum_{n}b(n)q^{n}\in M_{1/2}^{\epsilon^{*}}(N,1)$}. Since $N$ is square-free,
{$$M_{1/2}^{\epsilon^{*}}(N,1)=\mathbb{C}\theta.$$}
Setting $g=\theta$, we obtain
$$s(0)a(0)=0.$$
Since $s(0)\neq 0$, the form $f$ vanishes at $\infty$. By \cite[Proposition 5.3]{Zhang2}, the vector-valued form $\psi(f)$ is a cusp form. Here $\psi$ is the map constructed in Chapter 5 of \cite{Zhang2}. We conclude from \cite[Corollary 5.5]{Zhang2} that {$f=\psi^{-1}(\psi(f))\in S_{3/2}^{+\cdots +}(N,1)$}.
\end{proof}

\begin{remark} \label{rmk-zero}
If $k\geq 3$ is an odd integer and $(k,\epsilon)\neq (3,+)$, then
{$$M_{2-k/2}^{\epsilon^{*}}(N,1)=0.$$}
\end{remark}

Let $N\geq 1$ be an odd square-free integer. Suppose that $\ell$ is a prime with $\ell\nmid 4N$.
Consider a modular form
$$f(\tau)=\sum_{n} a(n)q^{n}\in M_{k/2}^{!\epsilon}(N,1),$$ where the sum is over $n$ such that $\chi_{p}(n)\neq -\epsilon_{p}$ for all $p\mid N$.
Then the action of the Hecke operator $T_{k/2,4N}(\ell^{2})$ on $f$
is given by
\begin{equation}\label{1}
f(\tau)\mid T_{k/2,4N}(\ell^{2})=\sum_{n}\left(a(\ell^{2}n)+\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}n}{\ell}\right)a(n)+\ell^{2\lambda-1}a(n/\ell^{2})\right)q^{n},
\end{equation}
where $\lambda=(k-1)/2$ and the sum is over $n$ such that $\chi_{p}(n)\neq -\epsilon_{p}$ for all $p\mid N$. Here we set $a(n/\ell^{2})=0$ if $\ell^{2}\nmid n$. Note that {$f(\tau)\mid T_{k/2,4N}(\ell^{2})\in M_{k/2}^{!\epsilon}(N,1)$}. We define $T_{k/2,4N}(\ell^{2n})$ for $n\geq 2$ recursively by
$$T_{k/2,4N}(\ell^{2n}):=T_{k/2,4N}(\ell^{2n-2})T_{k/2,4N}(\ell^{2})-\ell^{k-2}T_{k/2,4N}(\ell^{2n-4}).$$

\begin{remark} For $n\geq 2$, our $T_{k/2,4N}(\ell^{2n})$ is different from the $\ell^{2n}$-th Hecke operator given in \cite{Shi1}. See \cite[p.241]{Pur1} for details.
\end{remark}

By Proposition \ref{prop-zhang2}, the reduced form {$f_{m}\in M_{k/2}^{!\epsilon}(N,1)$} exists for every $m<0$ with $\chi_{p}(m)\neq -\epsilon_{p}$ for all $p\mid N$. We write
$$F_{m}(\tau)=s(m)f_{m}(\tau)=q^{m}+\sum\limits_{\substack{d\geq 0 \\ \chi_{p}(d)\neq -\epsilon_{p}~\textnormal{for all}~p\mid N}}B(m,d)q^{d}.$$
For any positive integer $t$ with $\gcd{(t,4N)}=1$, define
$$F_{m}^{(t)}:=F_{m}\mid T(t^{2}).$$
Then we obtain the coefficients $B_{t}(m,d)$ from the equation
$$F_{m}^{(t)}(\tau)=\mbox{(principal part)}+\sum\limits_{\substack{d\geq 0,\\ \chi_{p}(d)\neq-\epsilon_{p}~\textnormal{for all}~p\mid N}}B_{t}(m,d)q^{d}.$$

\medskip

For the rest of this section, let $k\geq 3$  be an odd integer and set $\lambda=(k-1)/2$, and assume
\begin{enumerate}
\item $m \in \mathbb Z_{<0}$ such that  $\chi_{p}(m)\neq -\epsilon_{p}$ for all $p\mid N$,
\item $\ell$ is a prime with $\ell\nmid 4N$ and $\ell^{2}\nmid m$.
\end{enumerate}

\begin{proposition} \label{prop-FG}
 Assume that {$S_{k/2}^{\epsilon}(N,1)=0$}. Then, for any positive integer $t$ with $(t,4N)=1$ and any positive integer $n$, we have
$$F_{m}^{(t)}|T_{k/2,4N}(\ell^{2n})-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)F_{m}^{(t)}|T_{k/2,4N}(\ell^{2n-2})=\ell^{(k-2)n}F_{\ell^{2n}m}^{(t)}.$$
\end{proposition}

\begin{proof}

For convenience, define $G_{0}^{(t)}:=F_{m}^{(t)}$, and, for each $n\geq 1$,
\begin{equation}\label{2}
G_{n}^{(t)}:=F_{m}^{(t)}\mid T_{k/2,4N}(\ell^{2n})-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)F_{m}^{(t)}\mid T_{k/2,4N}(\ell^{2n-2}).
\end{equation}
We need to show $G_{n}^{(t)} = \ell^{(k-2)n}F_{\ell^{2n}m}^{(t)}$.
Since the Hecke operators commute, it suffices to prove the proposition in the case $t=1$, which we now assume.

We claim that
\begin{equation}\label{3}
G_{n}^{(1)}=G_{n-1}^{(1)}\mid T_{k/2,4N}(\ell^{2}) -\ell^{k-2}\cdot G_{n-2}^{(1)}\quad\textnormal{for}\quad n\geq 2.
\end{equation}
Indeed, if $n=2$, then
\begin{align*}
G_{2}^{(1)}-&G_{1}^{(1)}\mid T_{k/2,4N}(\ell^{2})+\ell^{k-2}\cdot G_{0}^{(1)}\\
&=\left(F_{m}^{(1)}\mid T_{k/2,4N}(\ell^{4})-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)F_{m}^{(1)}\mid T_{k/2,4N}(\ell^{2})\right)\\
&\quad-\left(\left(F_{m}^{(1)}\mid T_{k/2,4N}(\ell^{2})\right)\mid T_{k/2,4N}(\ell^{2})-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)F_{m}^{(1)}\mid T_{k/2,4N}(\ell^{2})\right)+\ell^{k-2}\cdot F_{m}^{(1)}\\
&=F_{m}^{(1)}\mid T_{k/2,4N}(\ell^{4})-\left(F_{m}^{(1)}\mid T_{k/2,4N}(\ell^{2})\right)\mid T_{k/2,4N}(\ell^{2})+\ell^{k-2}\cdot F_{m}^{(1)} =0 .\\
\end{align*}
For $n\geq 3$,
\begin{align*}
G_{n-1}^{(1)}\mid &T_{k/2,4N}(\ell^{2})-\ell^{k-2}\cdot G_{n-2}^{(1)}\\
&=\left(F_{m}^{(1)}\mid T_{k/2,4N}(\ell^{2n-2})\right)\mid T_{k/2,4N}(\ell^{2})-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)\cdot\left(F_{m}^{(1)}\mid T_{k/2,4N}(\ell^{2n-4})\right)\mid T_{k/2,4N}(\ell^{2})\\
&\quad-\ell^{k-2}\cdot\left(F_{m}^{(1)}\mid T_{k/2,4N}(\ell^{2n-4})-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)F_{m}^{(1)}\mid T_{k/2,4N}(\ell^{2n-6})\right)\\
&=\left(F_{m}^{(1)}\mid T_{k/2,4N}(\ell^{2n-2})T(\ell^{2})-\ell^{k-2}\cdot F_{m}^{(1)}\mid T_{k/2,4N}(\ell^{2n-4})\right)\\
&\quad-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)\cdot\left(F_{m}^{(1)}\mid T_{k/2,4N}(\ell^{2n-4})T(\ell^{2})-\ell^{k-2}\cdot F_{m}^{(1)}\mid T_{k/2,4N}(\ell^{2n-6})\right)\\
&=F_{m}^{(1)}\mid T_{k/2,4N}(\ell^{2n})-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)\cdot F_{m}^{(t)}\mid T_{k/2,4N}(\ell^{2n-2}) =G_{n}^{(1)}. \\
\end{align*}

Since $G_{0}^{(1)}=F_{m}^{(1)}=F_{m}$, the principal part of $G_{0}^{(1)}$ is $q^{m}$. By \eqref{1}, the principal part of $G_{1}^{(1)}$ is $\ell^{k-2} q^{m\ell^{2}}$. Moreover, we see from \eqref{3} that, for all $n\geq 0$, the principal part of $G_{n}^{(1)}$ is equal to $\ell^{(k-2)n}q^{m\ell^{2n}}$. Since  $F_{m\ell^{2n}}^{(1)}=F_{m\ell^{2n}}$ has principal part $q^{m\ell^{2n}}$, {$G_n^{(1)} - \ell^{(k-2)n}F_{\ell^{2n}m}$ is holomorphic at the cusp $\infty$. Arguing as in the proof of Lemma \ref{lem-sm}}, we have
{\[ G_n^{(1)} - \ell^{(k-2)n}F_{\ell^{2n}m} \in M_{k/2}^{\epsilon}(N,1).
\]}
If $(k,\epsilon)=(3,+)$, then it follows from Lemma \ref{lem-sm} that
{$$G_n^{(1)} - \ell^{(k-2)n}F_{\ell^{2n}m}  \in S_{k/2}^{\epsilon}(N,1).$$} Since {$S_{k/2}^{\epsilon}(N,1) =\{ 0 \}$} by assumption, we have $G_n^{(1)} = \ell^{(k-2)n}F_{\ell^{2n}m}$.

If $(k,\epsilon) \neq (3,+)$, then {$M_{2-k/2}^{\epsilon^{*}}(N,1)=0$} (Remark \ref{rmk-zero}). By Borcherds' obstruction theorem, there exists a reduced form $g$ such that $g=1+O(q)$. We see from the definition of reduced forms that $B(m,0)=B(\ell^{2n}m,0)=0$. Hence the constant term of $G_{n}^{(1)}-\ell^{(k-2)n}F_{\ell^{2n}m}$ is zero, and thus
{$$G_n^{(1)} - \ell^{(k-2)n}F_{\ell^{2n}m} \in S_{k/2}^{\epsilon}(N,1)=\{0\}.$$}
Therefore we have $G_n^{(1)} = \ell^{(k-2)n}F_{\ell^{2n}m}$ in this case too.

\end{proof}

Write
$$G_{n}^{(t)}=\mbox{(principal part)}+\sum\limits_{\substack{d\geq 0, \\ \chi_{p}(d)\neq -\epsilon_{p}~\textnormal{for all}~p\mid N}}C_{n}(d)q^{d}.$$
Proposition \ref{prop-FG} implies that, for all $n$ and $d$,
\begin{equation}\label{4}
C_{n}(d)=\ell^{(k-2)n}B_{t}(\ell^{2n}m,d).
\end{equation}

\begin{lemma} \label{lem-BC}
The following are true:
\begin{enumerate}
\item[\textnormal{(\lowerromannumeral{1})}] For any $d\geq 0$ with $\chi_{p}(d)\neq -\epsilon_{p}$ for all $p\mid N$, we have
$$C_{n}(\ell^{2}d)-\ell^{k-2}\cdot C_{n-1}(d)=C_{0}(\ell^{2n+2}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)C_{0}(\ell^{2n}d).$$
\item[\textnormal{(\lowerromannumeral{2})}] If $\chi_{p}\neq -\epsilon_{p}$ for all $p\mid N$ and $\ell\parallel d$, then
$$C_{n}(d)=C_{0}(\ell^{2n}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)C_{0}(\ell^{2n-2}d).$$

\item[\textnormal{(\lowerromannumeral{3})}] If $\chi_{p}\neq -\epsilon_{p}$ for all $p\mid N$ and $\ell\nmid d$, then
$$C_{n}(d)=C_{0}(\ell^{2n}d)+\left[\left(\frac{(-1)^{\lambda}d}{\ell}\right)-\left(\frac{(-1)^{\lambda}m}{\ell}\right)\right]\cdot\sum_{k=1}^{n}\ell^{(\lambda-1)k}\left(\frac{(-1)^{\lambda}d}{\ell}\right)^{k-1}C_{0}(\ell^{2n-2k}d).$$
\end{enumerate}
\end{lemma}

\begin{proof}
We first prove (\lowerromannumeral{1}). Note that
\begin{align*}
\ell^{2}d&\mbox{-th coefficient of }G_{1}^{(m)}=C_{1}(\ell^{2}d),\\
\ell^{2}d&\mbox{-th coefficient of }F_{m}^{(t)}\mid T_{k/2,4N}(\ell^{2})-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)F_{m}^{(t)}\\
&=B_{t}(m,\ell^{4}d)+\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}\ell^{2}d}{\ell}\right)B_{t}(m,\ell^{2}d)+\ell^{k-2}\cdot B_{t}(m,d)\\
&\quad-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)B_{t}(m,\ell^{2}d)\\
&=B_{t}(m,\ell^{4}d)+\ell^{k-2}\cdot B_{t}(m,d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)B_{t}(m,\ell^{2}d)\\
&=C_{0}(\ell^{4}d)+\ell^{k-2}\cdot C_{0}(d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)C_{0}(\ell^{2}d).
\end{align*}
By \eqref{2}, we have
$$C_{1}(\ell^{2}d)=C_{0}(\ell^{4}d)+\ell^{k-2}\cdot C_{0}(d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)C_{0}(\ell^{2}d).$$
Hence,
$$C_{1}(\ell^{2}d)-\ell^{k-2}\cdot C_{0}(d)=C_{0}(\ell^{4}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)C_{0}(\ell^{2}d).$$
When $n\geq 2$, we use \eqref{3} to find that
$$C_{n}(\ell^{2}d)-\ell^{k-2}\cdot C_{n-1}(d)=C_{n-1}(\ell^{4}d)-\ell^{k-2}\cdot C_{n-2}(\ell^{2}d)=\cdots=C_{1}(\ell^{2n}d)-\ell^{k-2}\cdot C_{0}(\ell^{2n-2}d).$$
From \eqref{2}, we see that
$$C_{1}(\ell^{2n}d)=C_{0}(\ell^{2n+2}d)+\ell^{k-2}\cdot C_{0}(\ell^{2n-2}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)C_{0}(\ell^{2n}d).$$
Thus we obtain
$$C_{n}(\ell^{2}d)-\ell^{k-2}\cdot C_{n-1}(d)=C_{0}(\ell^{2n+2}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)C_{0}(\ell^{2n}d).$$

We now prove (\lowerromannumeral{2}) and (\lowerromannumeral{3}). Observe that
\begin{align*}
d&\mbox{-th coefficient of }G_{1}^{(t)}=C_{1}(d),\\
d&\mbox{-th coefficient of }F_{m}^{(t)}\mid T_{k/2,4N}(\ell^{2})-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)F_{m}^{(t)}\\
&=B_{t}(m,\ell^{2}d)+\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}d}{\ell}\right)B_{t}(m,d)+\ell^{k-2}\cdot B_{t}(m,d/\ell^{2})\\
&\quad-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)B_{t}(m,d)\\
&=
\begin{cases}
B_{t}(m,\ell^{2}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)B_{t}(m,d) & \textnormal{if}\quad\ell\parallel d,\\
B_{t}(m,\ell^{2}d)+\ell^{\lambda-1}\left[\left(\frac{(-1)^{\lambda}d}{\ell}\right)-\left(\frac{(-1)^{\lambda}m}{\ell}\right)\right]B_{t}(m,d) & \textnormal{if}\quad\ell\nmid d,
\end{cases}\\
&=
\begin{cases}
C_{0}(\ell^{2}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)C_{0}(d) & \textnormal{if}\quad\ell\parallel d,\\
C_{0}(\ell^{2}d)+\ell^{\lambda-1}\left[\left(\frac{(-1)^{\lambda}d}{\ell}\right)-\left(\frac{(-1)^{\lambda}m}{\ell}\right)\right]C_{0}(d) & \textnormal{if}\quad\ell\nmid d.
\end{cases}
\end{align*}
By \eqref{2}, we have
$$C_{1}(d)=
\begin{cases}
C_{0}(\ell^{2}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)C_{0}(d) & \textnormal{if}\quad\ell\parallel d,\\
C_{0}(\ell^{2}d)+\ell^{\lambda-1}\left[\left(\frac{(-1)^{\lambda}d}{\ell}\right)-\left(\frac{(-1)^{\lambda}m}{\ell}\right)\right]C_{0}(d) & \textnormal{if}\quad\ell\nmid d.
\end{cases}
$$
On the other hand, it follows from \eqref{3} that
$$C_{n}(d)=C_{n-1}(\ell^{2}d)-\ell^{k-2}\, C_{n-2}(d)+\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}d}{\ell}\right)C_{n-1}(d)$$
for $n\geq 2$. Applying part (\lowerromannumeral{1}) to $C_{n-1}(\ell^{2}d)-\ell^{k-2}\, C_{n-2}(d)$, we obtain
$$C_{n}(d)=C_{0}(\ell^{2n}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)C_{0}(\ell^{2n-2}d)+\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}d}{\ell}\right)C_{n-1}(d).$$
If $\ell\parallel d$, then we immediately obtain part (\lowerromannumeral{2}). Now assume that $\ell\nmid d$. Then by induction we have
\begin{align*}
C_{n}(d)&=C_{0}(\ell^{2n}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)C_{0}(\ell^{2n-2}d)+\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}d}{\ell}\right)C_{n-1}(d)\\
&=C_{0}(\ell^{2n}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)C_{0}(\ell^{2n-2}d)+\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}d}{\ell}\right)C_{0}(\ell^{2n-2}d)\\
&\quad+\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}d}{\ell}\right)\cdot\left[\left(\frac{(-1)^{\lambda}d}{\ell}\right)-\left(\frac{(-1)^{\lambda}m}{\ell}\right)\right] \cdot \sum_{k=1}^{n-1}\ell^{(\lambda-1)k}\left(\frac{(-1)^{\lambda}d}{\ell}\right)^{k-1}C_{0}(\ell^{2n-2k-2}d)\\
&=C_{0}(\ell^{2n}d)+\ell^{\lambda-1}\left[\left(\frac{(-1)^{\lambda}d}{\ell}\right)-\left(\frac{(-1)^{\lambda}m}{\ell}\right)\right]\cdot C_{0}(\ell^{2n-2}d)\\
&\quad+\left[\left(\frac{(-1)^{\lambda}d}{\ell}\right)-\left(\frac{(-1)^{\lambda}m}{\ell}\right)\right]\cdot\sum_{k=2}^{n}\ell^{(\lambda-1)k}\left(\frac{(-1)^{\lambda}d}{\ell}\right)^{k-1}C_{0}(\ell^{2n-2k}d)\\
&=C_{0}(\ell^{2n}d)+\left[\left(\frac{(-1)^{\lambda}d}{\ell}\right)-\left(\frac{(-1)^{\lambda}m}{\ell}\right)\right]\cdot\sum_{k=1}^{n}\ell^{(\lambda-1)k}\left(\frac{(-1)^{\lambda}d}{\ell}\right)^{k-1}C_{0}(\ell^{2n-2k}d).
\end{align*}
This proves the identity in part (iii).
\end{proof}

We now prove Theorem \ref{thm-1.4} and Corollary \ref{cor-1.5}.

\begin{proof}[Proof of Theorem \ref{thm-1.4}] \hfill

\noindent (\lowerromannumeral{1})
By \eqref{4} and Lemma \ref{lem-BC} (\lowerromannumeral{1}), we have
\begin{align*}
& B_{t}(m,\ell^{2n+2}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)B_{t}(m,\ell^{2n}d)\\
=&C_{0}(\ell^{2n+2}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)C_{0}(\ell^{2n}d)=C_{n}(\ell^{2}d)-\ell^{k-2}C_{n-1}(d)\\
=&\ell^{(k-2)n}B_{t}(\ell^{2n}m,\ell^{2}d)-\ell^{k-2}\cdot\ell^{(k-2)(n-1)}B_{t}(\ell^{2n-2}m,d)\\
=&\ell^{(k-2)n} \left \{ B_{t}(\ell^{2n}m,\ell^{2}d)-B_{t}(\ell^{2n-2}m,d) \right \}.
\end{align*}
(\lowerromannumeral{2}) Using \eqref{4} and Lemma \ref{lem-BC} (\lowerromannumeral{3}), we obtain
\begin{align*}
&\ell^{(k-2)n}B_{t}(\ell^{2n}m,d)=C_{n}(d)\\
=&C_{0}(\ell^{2n}d)+\left[\left(\frac{(-1)^{\lambda}d}{\ell}\right)-\left(\frac{(-1)^{\lambda}m}{\ell}\right)\right]\cdot\sum_{k=1}^{n}\ell^{(\lambda-1)k}\left(\frac{(-1)^{\lambda}d}{\ell}\right)^{k-1}C_{0}(\ell^{2n-2k}d)\\
=&B_{t}(m,\ell^{2n}d)+\left[\left(\frac{(-1)^{\lambda}d}{\ell}\right)-\left(\frac{(-1)^{\lambda}m}{\ell}\right)\right]\cdot\sum_{k=1}^{n}\ell^{(\lambda-1)k}\left(\frac{(-1)^{\lambda}d}{\ell}\right)^{k-1}B_{t}(m,\ell^{2n-2k}d).
\end{align*}
(\lowerromannumeral{3}) By \eqref{4} and Lemma \ref{lem-BC} (\lowerromannumeral{2}),
\begin{align*}
\ell^{(k-2)n}B_{t}(\ell^{2n}m,d)&=C_{n}(d)=C_{0}(\ell^{2n}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)C_{0}(\ell^{2n-2}d)\\
&=B_{t}(m,\ell^{2n}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)B_{t}(m,\ell^{2n-2}d).
\end{align*}
\end{proof}

\begin{proof}[Proof of Corollary \ref{cor-1.5}]
(1) First, suppose that $\left(\frac{-d}{\ell}\right)=\left(\frac{-m}{\ell}\right)\neq 0$. Then $\ell\nmid d$. By Theorem \ref{thm-1.4} (\lowerromannumeral{2}),
$$\ell^{(k-2)n}B_{t}(\ell^{2n}m,d)=B_{t}(m,\ell^{2n}d).$$
Now assume that $\ell\parallel d$ and $\ell\parallel m$. Then by Theorem \ref{thm-1.4} (\lowerromannumeral{3}),
$$\ell^{(k-2)n}B_{t}(\ell^{2n}m,d)=B_{t}(m,\ell^{2n}d).$$

(2) If $\ell\nmid d$, then $\ell\parallel\ell d$. By Theorem \ref{thm-1.4} (\lowerromannumeral{3}),
\begin{align*}
&B_{t}(m,\ell^{2n+1}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)B_{t}(m,\ell^{2n-1}d)\\
=&B_{t}(m,\ell^{2n}(\ell d))-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)B_{t}(m,\ell^{2n-2}(\ell d))\\
=&\ell^{(k-2)n}\, B_{t}(\ell^{2n}m,\ell d)\equiv 0\pmod{\ell^{(k-2)n}}.
\end{align*}
If $\ell\mid d$, then by Theorem \ref{thm-1.4} (\lowerromannumeral{1}), we obtain
\begin{align*}
&B_{t}(m,\ell^{2n+1}d)-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)B_{t}(m,\ell^{2n-1}d)\\
=&B_{t}(m,\ell^{2n+2}(d/\ell))-\ell^{\lambda-1}\left(\frac{(-1)^{\lambda}m}{\ell}\right)B_{t}(m,\ell^{2n}(d/\ell))\\
=&\ell^{(k-2)n}\, (B_{t}(\ell^{2n}m,\ell d)-B_{t}(\ell^{2n-2}m,d/\ell))
\equiv 0\pmod{\ell^{(k-2)n}}.
\end{align*}

\end{proof}

\section*{Acknowledgement}
We would like to thank Professor Yichao Zhang for his kind and valuable comments.

redpajama_set_name: RedPajamaArXiv


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text:
Q: measure cycles spent in accessing remote cache How to measure cycles spent in accessing shared remote cache say L3. I need to get this cache access information both system-wide and for per-thread. Is there any specific tool/hardware requirements. Or can I use any formula to get an approximate value of cycles spent over a time interval

A: To get the average  latencies (when a single thread is running) to various caches present on your machine, you can use memory profiler tools such as RMMA for windows (http://cpu.rightmark.org/products/rmma.shtml) and Lmbench for linux.
You can also write your own benchmarks based on the ideas used by these tools.
See the answers posted on this StackOverflow question:
measuring latencies of memory
Or Google for how the Lmbench benchmark works.
If you want to find exact latencies for particular memory access patterns, you will need to use a simulator. This way you can trace a memory access as it flows through the memory system. However simulators will not model all the effects that are present in a modern processor or memory system.
If you want to learn how multiple threads affect the average latency to L3, I think the best bet would be to write your own benchmark.


text:
Dylan D'Agate
Plastics - Bad Idea for the Planet
Please scroll down to the social media bar at bottom of page to share this post, thanks!
Plus an interview with Bill McKibben! (scroll down)
Recently I walked into a convenience store and was faced with a garbage bin overflowing with plastic bottles. This site reminded me that plastic bottles are a ubiquitous environmental problem and that perhaps we should focus more attention on decreasing their production. The reason? According to the Natural Resources Defense Council (https://www.nrdc.org/), more than 99 percent of plastic is created from chemicals sourced from fossil fuels! And when fossil fuels (i.e., coal, oil, natural gas) are burned, they release carbon dioxide and other greenhouse gases, which trap heat in our atmosphere and cause global warming.
Plastics, Fracking, Profit
Most plastics produced in the United States are derived from fossil fuel-based petrochemicals such as natural gas. Natural gas contains ethane, a fundamental building block of plastic. Ethane is separated from natural gas and made into a liquid form that is transferred via pipeline to an "ethane cracker," where intense heat breaks apart ethane molecules to form ethylene, used to make plastics. Information services company IHS Markit Ethylene Market Outlook informs that even with delays due to the COVID-19 pandemic, most new ethane cracker projects are going ahead—especially in Asia. Profit drives overproduction of natural gas, in large part due to an increase in hydraulic fracturing, or fracking -- the process of injecting liquid at high pressure into subterranean rocks to force open existing fissures and extract oil or gas. The increase in fracking has created a decline in gas prices, an increase in profits for fossil fuel companies -- and the release of methane, a greenhouse gas many times more toxic than carbon dioxide.
What Does the Future Look Like?
Not good! According to the International Energy Agency (https://www.iea.org/), the United States, the largest exporter of ethane, currently has 40 percent of global capacity to produce ethane-based petrochemicals. IHS Markit states that if unfettered, U.S. natural gas production is projected to grow by more than 50 percent, from approximately 80 billion cubic feet per day (Bcf/d) in 2017 to 123 Bcf/d in 2040. As reported in Plastics News (click here for article), the fossil fuel and plastics industries view the change as a "once-in-a-generation opportunity" and a "coming renaissance" for North American plastics. Alarmingly, the American Chemistry Council (https://www.ACC Fact Sheets) reports that by 2023 the chemical industry is expected to spend more than $164 billion on 264 new facilities and expansion projects specifically to take advantage of shale, or sedimentary rock, gas.
Stopping the Money Pipeline
For an eye-opening take of what can be done, and is being done, to decrease the profits that fossil-fuel companies reap as they pose an existential threat to our planet, click on this link. There you can listen to solutions put forth by American environmentalist Bill McKibben, best-selling author, co-founder of 350.org, and recipient of the International Gandhi Peace Prize.
Dylan D'Agate, a Sierra Club member, is a junior at Half Hollow Hills High School East, Dix Hills, NY, and a leader in his school's environmental club.
Surviving the New Climate
Ocean Plastic – Young People Tackle the Crisis
Severe Weather: Threat to Life as We Know It
Want to Be an Earth Blogger?
Dylan and the Sierra Club LI Group are looking to get other students to post environmental information
on Earth Blogger. Interested in being heard and advocating for planet Earth? Connect with the Sierra
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1. Semble that, under the Conformity Acts, rule of evidence established by decisions of the highest court of the state apply to an action on contract between private parties in the district court. P. 275 U. S. 127.
2. A man cannot make evidence for himself by writing a letter containing the statements that he wishes to prove. He does not make the letter evidence by sending it to the party against whom he wishes to prove the facts. P. 275 U. S. 128.
3. A, having bought bonds of B through B's sales agent, wrote B that the purchase was made upon the understanding that B would repurchase at the same price at A's request, and that he desired to avail himself of that privilege. Held that B was under no duty to answer the letter, and that the letter was inadmissible to prove the salesman's authority to make the agreement asserted. P. 275 U. S. 128.
Certiorari, 273 U.S. 676, to a judgment of the circuit court of appeals affirming a judgment recovered by Peirson from the petitioner on an alleged agreement to repurchase bonds sold by the latter to the former.
"it was agreed by Mr. Mather that at any time I so desired you would take them off my hands at cost 98. I have need of some money, and will avail myself of this privilege. When shall I deliver them to you?"
The officers of the petitioner denied ever having received the letter and denied the authority of Mather to make any such agreement. It may be mentioned further, although it is not relevant to the question here, that Mather denied having made the contracts alleged. The letter was offered in evidence. It was objected to as a self-serving document, but was admitted subject to exceptions. There was no other evidence of Mather's authority, but the jury were instructed that, if the petitioner received the letter and failed to disaffirm what Peirson said Mather had done, they would be justified in finding that the petitioner acquiesced in the agreement, and that Mather had authority to do what Peirson said he did. The plaintiff got a verdict and judgment, and the judgment was affirmed by the circuit court of appeals, 16 F.2d 86. On a suggestion of conflict between this and other circuit courts of appeal and of failure to conform to the rule of evidence in Pennsylvania (a failure in no way affected by the fact that the same rule prevails in most courts of high authority), as also of a difference among the courts as to the scope of the Conformity Acts, a writ of certiorari was granted by this Court. 273 U.S. 676.
A man cannot make evidence for himself by writing a letter containing the statements that he wishes to prove. He does not make the letter evidence by sending it to the party against whom he wishes to prove the facts. He no more can impose a duty to answer a charge than he can impose a duty to pay by sending goods. Therefore, a failure to answer such adverse assertions, in the absence of further circumstances making an answer requisite or natural, has no effect as an admission. Fraley v. Bispham, 10 Pa. 320; Kann v. Bennett, 223 Pa. 36, 47; Packer v. United States, 106 F. 906; Woolsey v. Haynes, 165 F. 391; Thrush v. Fullhart, 210 F. 1, 6; Harris v. Egger, 226 F. 389, 399; Kumin v. Fine, 229 Mass. 75; Viele v. McLean, 200 N.Y. 260; Richards v. Gellatly, L.R. 7, C.P. 127, 131; Wiedemann v. Walpole,  2 Q.B. 534, 539; Thomas v. Jones,  2 K.B. 399;  1 K.B. 22.
There were no circumstances in this case to take it out of the general rule. The letter might have been admissible as a demand if a binding contract had been proved, but until evidence of Mather's authority was given, the demand was immaterial. It is true that, two days after that on which the plaintiff says that he wrote the letter that we have quoted, the petitioner lent to the plaintiff $15,000 on the security of the $20,000 bonds in question, with the usual powers of sale and the plaintiff's note. It would be the merest speculation to regard the plaintiff's story as confirmed by this loan. It may as probably have been an independent transaction, and it might be argued at least as plausibly that the plaintiff's note and assent to the severe conditions of a pledge to brokers was inconsistent with the right that he now asserts. No evidence having been given of Mather's authority to make the contract in suit, the petitioner was entitled to a verdict. The request that one should be directed should have been granted. A new trial must be awarded.


We are so excited about one of our favorite kitchen projects winning the National First Prize for the Thermador Kitchen Design Challenge in the Contemporary/Modern design category. We were also honored with the Designer's Choice Award that was chosen by our peers. The awards were announced during a gala celebration in Southern California that was attended by 32 regional finalists. "The winners of the Thermador Kitchen Design Challenge are above and beyond some of the most exceptional kitchen design professionals in the industry," said Beatriz Sandoval, Director of Brand Marketing for Thermador. "Their stunning kitchens beautifully highlight the groundbreaking design and innovation of Thermador luxury home appliances, and boldly demonstrate the customized aesthetic that is increasingly prevalent in the American luxury home." Many thanks to Thermador for recognizing us and to our wonderful clients for the opportunity to design this awesome kitchen.