An area of Melton Road has been cordoned off this morning following an incident.
Leicestershire police has set up a police cordon near to Moores Road.
The cordon has been set up on an area of the pavement on the junction of Melton Road and Moores Road.
Leicestershire Live has given updates on this incident and they appear below.
We've also got an app for Android as well as Apple devices which can be tailored to deliver the news and sport that you're interested in.
Police have released a statement and an appeal for information following the assault on Moores Road in Belgrave in the early hours of this morning.
Another picture of the cordon from the scene.
Our reporter on the scene has seen a large pool of blood splashed up the front door of one of the properties is evident behind the cordon.
A police forensic investigation vehicle has just arrived at the scene.
Our reporter on the scene has seen a discarded green medical bag lay inside the cordon next to the remains of the broken bottle.
Police have confirmed that two men are in hospital following an assault in a Leicester street.
At around 12.50am today, officers were called to Moores Road a report that a man had been attacked.
One man in his 30s has been taken to the Queen's Medical Centre in Nottingham and one man also in his 30s was taken to Leicester Royal Infirmary.
We are continuing to establish the full circumstances of the incident and remain in the area this morning carrying out enquiries. Two men have been injured in this incident, one suffering serious injuries, and we urge anyone with any information to contact police.
Police have cordoned off a section of a main city road in the Belgrave area of the city.
The junction of Melton Road and Moores Road outside Sea King frozen sea food suppliers is cordoned off with a PCSO guarding the scene.
Inside the cordon which stretches for 100 yards along Moores Road there are blood stains stretching several yards and a broken bottle.
A picture of the cordon from the scene.


text:
Great British Bake Off fans, the new line-up has been confirmed
The Stylist web team
The Great British Bake Off is set to make the move to Channel 4 this year – and, as we are all too painfully aware, Mary Berry, Mel Giedroyc, and Sue Perkins will not be joining Paul Hollywood in the white tents ever again. Ever.
For many fans, the Queen of Cakes and co had set the bar incredibly high, making them an (almost) impossible act to follow.
But it seems as if GBBO bosses have done their best to replace the gang with a whole host of new faces – some expected (we already guessed that Prue Leith would be jumping into Bezza's shoes), some completely, completely unexpected.
Like, for instance, Noel Fielding, As in, actual Noel Fielding.
Noel Fielding
While he's not the first person that jumps to mind when you say 'cake', Fielding is very keen to be leaping into the big white tent – even going so far as to claim that Bake Off is "one of his favourite shows".
"[This is] a dream come true," he said.
Read more: Mary Berry controversially names her favourite Bake Off contestant
Madness – or, quite possibly, a stroke of genius from show bosses.
Leith, meanwhile, has said she is "so thrilled to be joining... the biggest show on TV".
"I cannot wait to see what the real stars of the show - the bakers - are going to create for us," the restaurateur and food critic said.
The Bake Off production team are reportedly hoping that Leith – who, at 76, is the same age as Berry was when GBBO first launched – will prove to be a "like for like" replacement for our beloved Bezza.
"Both are from similar backgrounds, so hopefully viewers will be accepting and give Prue a chance to win them over," a source told The Sun.
"Following in Mary's footsteps is a big challenge. But the producers are confident that Prue has all the attributes."
And they're not the only ones in the (batter) mix: Sandi Toksvig will be joining Fielding as the show's co-host – making for a very intriguing new sponsored_longform.
"It is an extraordinary honour to part of this national treasure of a show," she said.
Read more: Candice Brown makes surprise career announcement
Extraordinary is definitely the word we'd use, too. Although there's no denying that this new line-up will make for fascinating viewing when the series airs later this year.
Commenting on his new co-stars, Hollywood admitted that he "could not be more delighted by who will be joining me on the show."
And it seems as if his opinion is shared by Channel 4 bosses, who have said that they hope the new stars will bring "plenty of fun and warmth" to the baking show.
"It's just the sort of innovative line-up audiences expect from Channel 4," they added.
Earlier this year, Berry announced that she was quitting the Great British Bake Off when it moved to its new home on Channel 4 – and that her decision had been made out of loyalty to the BBC.
"My decision to stay with the BBC is out of loyalty to them, as they have nurtured me, and the show, that was a unique and brilliant format from day one," the 81-year-old said in a statement.
"What a privilege and honour it has been to be part of seven years of magic in a tent.
"I am just sad for the audience who may not be ready for change, I hope they understand my decision."
Read more: Why #Breadxit is the worst thing to ever happen in the history of television
Mel and Sue added that they would not "follow the dough" to a rival channel.
Hollywood, however, signed a three-year contract with Channel 4, rumoured to be worth £1 million.
Insisting his decision had nothing to do with the money involved, he later said that the show "been a huge part of my life in the past few years and I just couldn't turn my back on all that – the bakers themselves, the bakes, the team that makes it, and of course the tent, the bunting, and who could forget… the squirrels".
The show is confirmed to return to our screens this year – and applications are now being accepted online.
Images: Rex Pictures, Love Productions
Words: Kayleigh Dray
Mel Giedroyc
Paul Hollywoo

redpajama_set_name: RedPajamaCommonCrawl


text:
\section{Introduction}
For a knot $K$ in $S^3$, let $J_{{{\mbox{\tiny K}}}, n}(t)$ denote the \textit{$n$-colored Jones polynomial} of $K$
with the zero framing, normalized so that for the unknot $U$,
\[J_{{{\mbox{\tiny U}}},n}(t) = \frac{t^{2n}-t^{-2n}}{t^2-t^{-2}}.\]
A remarkable result, proved in \cite{GaLe}, asserts  that for every knot $K$, $J_{{{\mbox{\tiny K}}},n}(t)$ satisfies a nontrivial linear recurrence relation.
By defining $J_{{{\mbox{\tiny K}}},-n}(t) := -J_{{{\mbox{\tiny K}}},n}(t)$, one may  treat $J_{{{\mbox{\tiny K}}},n}(t)$ as
 a discrete function  $$J_{{{\mbox{\tiny K}}},-}(t): {\mathbb Z} \to {\mathbb Z}[t^{\pm 1}].$$
 The quantum torus
\[ {\mathcal T} = {\mathbb C}[t^{\pm 1}] \left<{{\mbox{\small L}}}^{\pm 1}, {{\mbox{\small M}}}^{\pm 1} \right> / ({{\mbox{\small L}}} {{\mbox{\small M}}}-t^2{{\mbox{\small M}}} {{\mbox{\small L}}})\]
acts on   the set of discrete functions $f: {\mathbb Z} \to {\mathbb C}[t^{\pm 1}]$ by
\[({{\mbox{\small M}}} f)(n) := t^{2n}f(n), \quad  ({{\mbox{\small L}}} f)(n) := f(n+1).\]
 Then   linear recurrence relations of $J_{{{\mbox{\tiny K}}},n}(t)$
 correspond naturally to
  annihilators of $J_{{{\mbox{\tiny K}}},n}(t)$ in ${\mathcal T}$.
The latter set, which we denote by
  $${\mathcal A}_{{\mbox{\tiny K}}} := \{P \in {\mathcal T} \mid P J_{{{\mbox{\tiny K}}},n}(t) = 0\},$$
  is obviously  a left ideal of ${\mathcal T}$, called the \textit{recurrence ideal} of $K$.
  The result of \cite{GaLe} cited above states that ${\mathcal A}_{{\mbox{\tiny K}}}$
  is not the zero ideal for every knot $K$.

The ring ${\mathcal T}$ can be extended to a principal left ideal domain $\widetilde{\T}$ by adding inverses of polynomials in $t$ and ${{\mbox{\small M}}}$; that is, $\widetilde{\T}$ is the set of Laurent polynomials in ${{\mbox{\small L}}}$ with coefficients rational functions of $t$ and ${{\mbox{\small M}}}$ with a product defined by
\[f(t,{{\mbox{\small M}}}) {{\mbox{\small L}}}^a \cdot g(t,{{\mbox{\small M}}}) {{\mbox{\small L}}}^b = f(t,{{\mbox{\small M}}}) g(t,t^{2a}{{\mbox{\small M}}}){{\mbox{\small L}}}^{a+b}.\]
The left ideal $\widetilde{\A}_{{\mbox{\tiny K}}}=\widetilde{\T} {\mathcal A_{{\mbox{\tiny K}}}}$ is then generated by some nonzero polynomial in $\widetilde{\T}$, and in particular, this generator can be chosen to be in ${\mathcal A}_{{\mbox{\tiny K}}}$ and be of the form
\[\alpha_{{\mbox{\tiny K}}}(t,{{\mbox{\small M}}},{{\mbox{\small L}}}) = \sum_{i=0}^d P_i {{\mbox{\small L}}}^i,\]
with $d$ minimal and with  $P_1,...,P_d \in {\mathbb Z}[t,{{\mbox{\small M}}}]$ being coprime in ${\mathbb Z}[t,{{\mbox{\small M}}}]$. This polynomial $\alpha_{{\mbox{\tiny K}}}$ is uniquely determined up to a sign and is called the (\textit{normalized}) \textit{recurrence polynomial} of $K$.

The $A$-polynomial was introduced in \cite{CCGLS}. For a knot $K$ in $S^3$,
its $A$-polynomial $A_{{\mbox{\tiny K}}}({{\mbox{\small M}}},{{\mbox{\small L}}})\in {\mathbb Z}[{{\mbox{\small M}}},{{\mbox{\small L}}}]$ is a two variable polynomial
with no repeated factors and with relative prime integer coefficients,
which is uniquely associated to $K$ up to a sign. Note that $A_{{\mbox{\tiny K}}}({{\mbox{\small M}}},{{\mbox{\small L}}})$ always contains the factor ${{\mbox{\small L}}}-1$.

The AJ-conjecture was raised  in \cite{G} which states that for every knot $K$, its recurrence polynomial $\alpha_{{\mbox{\tiny K}}}(t,{{\mbox{\small M}}},{{\mbox{\small L}}})$  evaluated at $t = -1$ is equal to the A-polynomial of $K$, up to a factor of a polynomial in ${{\mbox{\small M}}}$. The conjecture is obviously of fundamental importance as it predicts a strong connection
between two important knot invariants derived from very different backgrounds. This is also a very difficult conjecture; so far only torus knots, some $2$-bridge knots and some pretzel knots are
known to satisfy the conjecture  \cite{G} \cite {Ta} \cite{Hikami}
\cite{Le} \cite{LT} \cite{Tran}.

In this paper, we consider the AJ-conjecture for  cabled knots over torus knots. Recall that the set of nontrivial torus knots $T(p,q)$ in $S^3$
can be indexed, in a standard way,  by pairs of relative prime integers
$(p,q)$ satisfying  $|p|>q\geq 2$.
Also recall that an $(r, s)$-cabled knot on a knot $K$ in $S^3$
is the knot which can be embedded in
the boundary torus of a regular neighborhood of $K$ in $S^3$
as a curve of slope $r/s$ with respect to the meridian/longitude coordinates of $K$
satisfying $(r,s)=1$, $s\geq 2$.
Note that $r$ can be any integer relatively prime to $s$.
We have
\begin{thm}\label{main result}
The $AJ$-conjecture holds for each $(r,s)$-cabled knot $C$ over each $(p,q)$-torus knot $T$
if $r$ is not an integer between $0$ and $pqs$.
\end{thm}

A cabling formula for $A$-polynomials of cabled knots
 in $S^3$ is  given in \cite{NZ}. In particular when
 $C$ is  the $(r, s)$-cabled knot over the torus knot $T(p, q)$ in $S^3$,  its
 $A$-polynomial $A_{{\mbox{\tiny C}}}({{\mbox{\small M}}}, {{\mbox{\small L}}})$  is given explicitly as in (\ref{A-poly of C}) below.
For  a pair of relative prime integers $(p,q)$  with $q\geq 2$, define $F_{(p,q)}({{\mbox{\small M}}},{{\mbox{\small L}}}), G_{(p,q)}({{\mbox{\small M}}},{{\mbox{\small L}}})\in {\mathbb Z}[{{\mbox{\small M}}},{{\mbox{\small L}}}]$ to be the associated
 polynomials in variables ${{\mbox{\small M}}}$ and ${{\mbox{\small L}}}$
by:
$$F_{(p, q)}({{\mbox{\small M}}}, {{\mbox{\small L}}}):= \left\{\begin{array}
 {ll}{{\mbox{\small M}}}^{2p}{{\mbox{\small L}}}+1,\;\;&  \mbox{if $q=2$, $p>0$,}\\
{{\mbox{\small L}}}+{{\mbox{\small M}}}^{-2p},\;\;  &\mbox{if $q=2$,  $p<0$,}\\
{{\mbox{\small M}}}^{2pq}{{\mbox{\small L}}}^2-1,\;\;&  \mbox{if $q>2$, $p>0$,}\\
{{\mbox{\small L}}}^2-{{\mbox{\small M}}}^{-2pq},\;\; & \mbox{if $q>2$, $p<0$}\end{array}\right.$$and
$$G_{(p, q)}({{\mbox{\small M}}}, {{\mbox{\small L}}}):= \left\{\begin{array}
 {ll}{{\mbox{\small M}}}^{pq}{{\mbox{\small L}}}-1,\;\;&  \mbox{if  $p>0$,}\\
{{\mbox{\small L}}}-{{\mbox{\small M}}}^{-pq},\;\; & \mbox{if  $p<0$.}\end{array}\right.
$$
Then
\begin{equation}\label{A-poly of C}A_{{\mbox{\tiny C}}}({{\mbox{\small M}}},{{\mbox{\small L}}})=\left\{\begin{array}{ll}
({{\mbox{\small L}}}-1)F_{(r, s)}({{\mbox{\small M}}}, {{\mbox{\small L}}})F_{(p,q)}({{\mbox{\small M}}}^{s^2},{{\mbox{\small L}}}),&\mbox{if $s$ is odd};\\
({{\mbox{\small L}}}-1)F_{(r, s)}({{\mbox{\small M}}}, {{\mbox{\small L}}})G_{(p,q)}({{\mbox{\small M}}}^{s^2},{{\mbox{\small L}}}),&\mbox{if $s$ is even}.
\end{array}\right.
\end{equation}

A cabling formula for the $n$-colored Jones polynomial of
the $(r,s)$-cabled knot $C$ over a knot $K$ is given in \cite{Morton} (see also
\cite{Veen}) which in our normalized form is:
\begin{equation}\label{cabling formula of colored Jones}
 \begin{array}{ll}\displaystyle J_{{{\mbox{\tiny C}}},n}(t) &=
t^{-rs(n^2-1)}\sum_{k=-\frac{n-1}{2}}^{\frac{n-1}{2}}
t^{4rk(ks+1)}J_{{{\mbox{\tiny K}}},2ks+1}(t).\end{array}
\end{equation}
In particular the $n$-colored Jones polynomial of the $(p,q)$-torus knot $T$
 (which is the $(p,q)$-cabled knot over the unknot $U$) is:
\begin{equation}\label{colored Jones of T}
\begin{array}{ll}
J_{{{\mbox{\tiny T}}}, n}(t)&=t^{-pq(n^2-1)}\sum_{k=-\frac{n-1}{2}}^{\frac{n-1}{2}}
t^{4pk(kq+1)}J_{{{\mbox{\tiny U}}},2kq+1}(t)
\\& =t^{-pq(n^2-1)}\sum_{k=-\frac{n-1}{2}}^{\frac{n-1}{2}}
t^{4pk(kq+1)}\frac{t^{4kq+2}-t^{-4qk-2}}{t^2-t^{-2}}.\end{array}\end{equation}

We divide the proof of Theorem \ref{main result} into the following cases:
\begin{enumerate}
\item $s$ is odd and $q>2$;
\item $s$ is odd and $q=2$;
\item $s>2$ is even;
\item $s=2$.
\end{enumerate}
In each case, we will find an annihilator of $J_{{{\mbox{\tiny C}}},n}(t)$ by applying the formulas
(\ref{colored Jones of T}) and (\ref{cabling formula of colored Jones}) (where taking the general knot $K$ to be the $(p,q)$-torus knot $T$), and then proceed to prove
that it is the recurrence polynomial ${\alpha}_{{\mbox{\tiny C}}}(t,{{\mbox{\small M}}},{{\mbox{\small L}}})$ of $C$ when $r$ is not an integer between
$0$ and $pqs$,  making use of the degree formulas given in Section \ref{degree section}. Of course we will also compare  ${\alpha}_{{\mbox{\tiny C}}}(-1,{{\mbox{\small M}}},{{\mbox{\small L}}})$
with $A_{{\mbox{\tiny C}}}({{\mbox{\small M}}},{{\mbox{\small L}}})$ given in (\ref{A-poly of C}) to complete the verification of
the $AJ$-conjecture for $C$.
For convenience, we often get $\alpha_{{\mbox{\tiny C}}}(t,{{\mbox{\small M}}},{{\mbox{\small L}}})$ in the form
$P=\sum_{i=0}^d P_iL^i\in \widetilde{\A}_{{\mbox{\tiny C}}}$, with $d$ minimal and with $P_i\in {\mathbb Q}(t, {{\mbox{\small M}}})$
and with $P(-1,{{\mbox{\small M}}},{{\mbox{\small L}}})\ne 0$. Such $P$ only differs from ${\alpha}_{{\mbox{\tiny C}}}$ by a factor of a rational
 function $f(t,{{\mbox{\small M}}})\in {\mathbb Q}(t, {{\mbox{\small M}}})$ with $f(-1,{{\mbox{\small M}}})\ne 0$
and thus  is clearly as good as the normalized
  recurrence polynomial in verification for the $AJ$-conjecture.
 We often simply call  such $P$ the recurrence polynomial of
 $C$.
 Also notice
  from the formula (\ref{A-poly of C}) that changing the sign of $r$ or $p$ only changes the $A$-polynomial of $C$ up to a power of ${{\mbox{\small M}}}$, so in checking
    that $P(-1,{{\mbox{\small M}}},{{\mbox{\small L}}})=A_{{\mbox{\tiny C}}}({{\mbox{\small M}}},{{\mbox{\small L}}})$  up to a factor of a
   rational function  in ${{\mbox{\small M}}}$ we don't need to worry about the sign of $r$ or $p$.

Further investigation of the $AJ$-conjecture for more general  cabled knots,
such as iterated torus knots and cabled knots over some hyperbolic knots,
are being continued in \cite{R}. In particular for some cabled knots over the figure 8 knot
the $AJ$-conjecture has been verified to be true.

\section{Degrees of $J_{{{\mbox{\tiny T}}},n}(t)$ and $J_{{{\mbox{\tiny C}}},n}(t)$}\label{degree section}

From now on in this paper,  $T$
denotes the $(p,q)$-torus knot and $C$ the $(r,s)$-cabled knot over $T$, with the index convention
given in the introduction.

For a polynomial $f(t)\in {\mathbb Z}[t^{\pm1}]$, let $\ell[f]$ and $\hbar[f]$
 denote the lowest degree and the highest degree of $f$ in $t$ respectively.
Obviously for  $f(t),g(t)\in {\mathbb Z}[t^{\pm1}]$,
$\ell[fg]=\ell[f]+\ell[g]$ and $\hbar[fg]=\hbar[f]+\hbar[g]$.

\begin{lemma}\label{degree of J_T} (1) When $p>q$, $$\begin{array}{l}
\ell[J_{{{\mbox{\tiny T}}},n}(t)]=-pqn^2+pq+\frac{1}{2}(1-(-1)^{n-1})(p-2)(q-2),\\
\hbar[J_{{{\mbox{\tiny T}}},n}(t)]=2(p+q-pq)|n|+2(pq-p-q).\end{array}$$
\newline
(2)
When $p<-q$, $$\begin{array}{l}
\ell[J_{{{\mbox{\tiny T}}},n}(t)]=2(p-q-pq)|n|+2(pq-p+q),\\
\hbar[J_{{{\mbox{\tiny T}}},n}(t)]=-pqn^2+pq+\frac{1}{2}(1-(-1)^{n-1})(p+2)(q-2).\end{array}$$
\end{lemma}

{\noindent{\bf Proof.\hspace{2mm}}} The formula for $\ell[J_{{{\mbox{\tiny T}}},n}(t)]$ in part (1)
is  proved in \cite[Lemma 1.4]{Tran}. The rest of the lemma can be
 proved similarly.
{\hspace{2mm}{\small $\diamondsuit$}}

Note that $r\ne pqs$ since $r$ is relatively prime to $s$.

\begin{lemma}\label{degree of J_C}
(1) When $p>q$, $$\begin{array}{lll}
\ell[J_{{{\mbox{\tiny C}}},n}(t)]&=- p q s^2 n^2 + (2pqs^2-2pqs+2r-2rs)n&\\
&\;\;\;\;+2rs-2r+2pqs-pqs^2+\frac{1}{2}(1-(-1)^{(n-1)s})(p-2)(q-2),
&\mbox{if $r<pqs$},\\&&\\
\ell[J_{{{\mbox{\tiny C}}},n}(t)]&=-rsn^2+rs+
\frac{1}{2}(1-(-1)^{(n-1)})(s-2)(r-pqs)\\
&\;\;\;\;+\frac12[1-(-1)^{(n-1)s}](p-2)(q-2), &
\mbox{if $r>pqs$},
\end{array}$$

$$\begin{array}{lll}\hbar[J_{{{\mbox{\tiny C}}},n}(t)]&=-rsn^2+rs+\frac12(1-(-1)^{n-1})(s-2)(r-2pq+2p+2q),
&\mbox{if $r<0$}.
\end{array}$$
\newline
(2)
When $p<-q$, $$\begin{array}{lll}
\hbar[J_{{{\mbox{\tiny C}}},n}(t)]&=- p q s^2 n^2 + (2pqs^2-2pqs+2r-2rs)n&\\
&\;\;\;\;+2rs-2r+2pqs-pqs^2+\frac{1}{2}(1-(-1)^{(n-1)s})(p+2)(q-2),
&\mbox{if $r>pqs$},\\&&\\
\hbar[J_{{{\mbox{\tiny C}}},n}(t)]&=-rsn^2+rs+
\frac{1}{2}(1-(-1)^{(n-1)})(s-2)(r-pqs)\\
&\;\;\;\;+\frac12(1-(-1)^{(n-1)s})(p+2)(q-2), &
\mbox{if $r<pqs$},
\end{array}$$
$$\begin{array}{lll}\ell[J_{{{\mbox{\tiny C}}},n}(t)]&=-rsn^2+rs+\frac12(1-(-1)^{n-1})(s-2)(r-2pq+2p-2q),
&\mbox{if $r>0$}.
\end{array}$$
\end{lemma}

\begin{proof}
(1)  From the formula (\ref{cabling formula of colored Jones}) for $J_{{{\mbox{\tiny C}}},n}(t)$ (replacing $K$ there by $T$), we can see that
$$\ell[J_{{{\mbox{\tiny C}}},n}(t)]= -rs(n^2-1)+\text{min}\left\{ \ell[J_{{{\mbox{\tiny T}}},2sk+1}(t)]+4rk(ks+1) \mid -\frac{n-1}{2} \leq k \leq \frac{n-1}{2}\right\}.$$
 By Lemma \ref{degree of J_T} (1), we have
\begin{align*}
\ell[J_{{{\mbox{\tiny T}}},2ks+1}(t)]+4rk(ks+1) &= -p q (2 k s+1)^2+pq+\frac{1}{2}(1-(-1)^{2 k s})(p-2) (q-2) +4 k r (k s+1)\\
&=  (4 r s-4 p q s^2)k^2+ (4 r-4 p q s)k+\frac{1}{2} (1-(-1)^{2 k s})(p-2) (q-2).
\end{align*}
When $n$ is odd, $k$ is integer valued  and thus the alternating term vanishes, so the above expression  is quadratic in $k$. When $n$ is even, $k$ is half-integer valued and the alternating term is either always equal to zero
(when $s$ is even) or is always equal to $(p-2)(q-2)$ (when $s$ is odd), and thus the above expression
  is again quadratic in $k$.
So if $r < pqs$, it is minimized at  $k = \frac{n-1}{2}$, which yields the first formula in part (1),  and  if  $r> pqs$, it is minimized at  $k = 0$
when $n$ is odd and at $k=-1/2$ when $n$ is even, which yields the second formula in part (1).

  Similarly to get the third formula in (1), we look at
$$\hbar[J_{{{\mbox{\tiny C}}},n}(t)]= -rs(n^2-1)+\text{max}\left\{ \hbar[J_{{{\mbox{\tiny T}}},2sk+1}(t)]+4rk(ks+1) \mid -\frac{n-1}{2} \leq k \leq \frac{n-1}{2}\right\}.$$
 By Lemma \ref{degree of J_T} (1), we have
$$\begin{array}{l}
\hbar[J_{{{\mbox{\tiny T}}},2ks+1}(t)]+4rk(ks+1)
= 2(p+q-pq)|2ks+1|+2(pq-p-q) +4 k r (k s+1)\\
=\left\{\begin{array}{ll}
  4 r sk^2+ (4ps+4qs-4pqs+4r)k, &\mbox{for non-negative  $k$'s},
  \\
   4 r sk^2+ (-4ps-4qs+4pqs+4r)k+4(pq-p-q), &\mbox{for negative $k$'s.}
\end{array}\right.
\end{array}$$
If $r <0$, it is maximized at  $k = 0$
when $n$ is odd and at $k=-1/2$ when $n$ is even, which yields the third formula in part (1).

Part (2) can be proved similarly.
\end{proof}

\section{Case $s>2$ is odd and $q>2$}\label{section s odd}

\subsection{An Annihilator $P$ of $J_{{{\mbox{\tiny C}}},n}(t)$.}
Define
\begin{align*}
\delta_j &= \frac{t^{2(p+q)(j+1)+2}+t^{-2(p+q)(j+1)+2} -t^{2(q-p)(j+1)-2}-t^{-2(q-p)(j+1)-2}}{t^2-t^{-2}},\\
S_n &= \sum_{k=1}^s t^{-4pqskn+2pqsn+4pqk^2-12pqsk+6pqs} \delta_{s(n+3)-1-2k}.
\end{align*}
By \cite[Lemma 1.1]{Tran}, we have
\begin{equation}\label{relation of J_T(n+2)}J_{{{\mbox{\tiny T}}},n+2}(t) = t^{-4pq(n+1)} J_{{{\mbox{\tiny T}}},n}(t)+t^{-2pq(n+1)} \delta_n.
\end{equation}

Note that (\ref{relation of J_T(n+2)}) is valid for every torus knot (although in \cite{Tran}, only positive
$p$ was considered).
The following two lemmas also hold for general $C$ and $T$ (without restriction on $s$ and $q$)
and they shall also be applied in later sections.

\begin{lemma} \label{J_C formulas}
\begin{align*}
J_{{{\mbox{\tiny C}}},n+2}(t) &=
 t^{-4rs n-4rs} J_{{{\mbox{\tiny C}}},n}(t)+ (t^{2(r-rs)n-2rs+2r-4pqs(n+1)}-t^{2(-r-rs)n-2rs-2r})J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)\\
 &\;\;\;\;+t^{2(r-rs)n-2rs+2r-2pqs(n+1)} \delta_{s(n+1)-1}.
\end{align*}
\end{lemma}

\begin{proof}
We know by the cabling formula (\ref{cabling formula of colored Jones})
\begin{align*}
J_{{{\mbox{\tiny C}}},n+2}(t) &= t^{-rs((n+2)^2-1)} \sum_{k=-\frac{n+1}{2}}^{\frac{n+1}{2}}t^{4rk(ks+1)}J_{{{\mbox{\tiny T}}},2ks+1}(t)\\
&= t^{-rs(n^2+4n+3)} \Biggl(\sum_{k=-\frac{n-1}{2}}^{\frac{n-1}{2}}t^{4rk(ks+1)}J_{{{\mbox{\tiny T}}},2ks+1}(t) + t^{4r\left(\frac{n+1}{2}\right)(\left(\frac{n+1}{2}\right)s+1)} J_{{{\mbox{\tiny T}}},s(n+1)+1}(t) \\
& \;\;\;\;+t^{4r\left(-\frac{n+1}{2}\right)(\left(-\frac{n+1}{2}\right)s+1)} J_{{{\mbox{\tiny T}}},-s(n+1)+1}(t)\Biggr).
\end{align*}
Noting that $J_{{{\mbox{\tiny T}}},-s(n+1)+1}(t) = -J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)$, we have
\begin{align*}
J_{{{\mbox{\tiny C}}},n+2}(t) &=  t^{-rs(n^2+4n+3)} \Biggl(t^{rs(n^2-1)} t^{-rs(n^2-1)}\sum_{k=-\frac{n-1}{2}}^{\frac{n-1}{2}}t^{4rk(ks+1)}J_{{{\mbox{\tiny T}}},2ks+1}(t) \\
&\;\;\;\;+ t^{(n+1)^2 rs+2r(n+1)} J_{{{\mbox{\tiny T}}},s(n+1)+1}(t)-t^{(n+1)^2 rs-2r(n+1)} J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)\Biggr)\\
&= t^{-4rs n-4rs} J_{{{\mbox{\tiny C}}},n}(t)+ t^{2(r-rs)n-2rs+2r}J_{{{\mbox{\tiny T}}},s(n+1)+1}(t)-t^{2(-r-rs)n-2rs-2r}J_{{{\mbox{\tiny T}}},s(n+1)-1}(t).
\end{align*}
Since $J_{{{\mbox{\tiny T}}},s(n+1)+1}(t)$ and $J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)$ are related by equation (\ref{relation of J_T(n+2)}) as
\[J_{{{\mbox{\tiny T}}},s(n+1)+1} = t^{-4pqs(n+1)} J_{{{\mbox{\tiny T}}},s(n+1)-1}+t^{-2pqs(n+1)} \delta_{s(n+1)-1},\]
we have
\begin{align*}
J_{{{\mbox{\tiny C}}},n+2}(t) &=
 t^{-4rs n-4rs} J_{{{\mbox{\tiny C}}},n}(t)+ (t^{2(r-rs)n-2rs+2r-4pqs(n+1)}-t^{2(-r-rs)n-2rs-2r})J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)\\
 &\;\;\;\;+t^{2(r-rs)n-2rs+2r-2pqs(n+1)} \delta_{s(n+1)-1}.
\end{align*}
\end{proof}

\begin{lemma} \label{s peel identity}  For all positive integers $m$, we have
\begin{align*}
J_{{{\mbox{\tiny T}}},n}(t) &= t^{-4pqm(n+1)+4pqm(m+1)} J_{{{\mbox{\tiny T}}},n-2m}(t) + \sum_{k=1}^m t^{(-4pqk+2pq)n+4pqk^2-4pqk+2pq} \delta_{n-2k}
\end{align*}
and in particular, with $s$ any positive integer,
\[J_{{{\mbox{\tiny T}}},s(n+3)-1}(t) = t^{-4pqs^2n-8pqs^2-4pqs} J_{{{\mbox{\tiny T}}},s(n+1)-1}(t) + S_n.\]
\end{lemma}

\begin{proof} We induct on $m$. The base case $m=1$ follows directly from
 equation (\ref{relation of J_T(n+2)}). Then assume the formula holds for some positive integer $m$. Applying equation (\ref{relation of J_T(n+2)}) again yields
\begin{align*}
J_{{{\mbox{\tiny T}}},n}(t) &= t^{-4pqm(n+1)+4pqm(m+1)} J_{{{\mbox{\tiny T}}},n-2m}(t) + \sum_{k=1}^m t^{(-4pqk+2pq)n+4pqk^2-4pqk+2pq} \delta_{n-2k}\\
&= t^{-4pqm(n+1)+4pqm(m+1)} (t^{-4pq(n-2m-1)}J_{{{\mbox{\tiny T}}},n-2m-2}(t)+t^{-2pq(n-2m-1)}\delta_{n-2m-2}) \\
&\quad + \sum_{k=1}^m t^{(-4pqk+2pq)n+4pqk^2-4pqk+2pq} \delta_{n-2k}\\
&= t^{-4pqm(n+1)+4pqm(m+1)-4pq(n-2m-2+1)} J_{{{\mbox{\tiny T}}},n-2m-2}(t)\\
&\quad +t^{-4pqm(n+1)+4pqm(m+1)-2pq(n-2m-1)}\delta_{n-2m-2} + \sum_{k=1}^m t^{(-4pqk+2pq)n+4pqk^2-4pqk+2pq} \delta_{n-2k}.
\end{align*}
If we compare the terms in the summation to the $\delta_{n-2m-2}$ term outside, we can easily see that this is precisely the term where $k = m+1$. So moving it inside, we have
\begin{align*}
J_{{{\mbox{\tiny T}}},n}(t) &= t^{-4pq(m+1)(n+1)+4pq(m^2+m)+4pq(2m+2)} J_{{{\mbox{\tiny T}}},n-2(m+1)}(t) + \sum_{k=1}^{m+1} t^{(-4pqk+2pq)n+4pqk^2-4pqk+2pq} \delta_{n-2k}\\
&= t^{-4pq(m+1)(n+1)+4pq(m+1)(m+2)} J_{{{\mbox{\tiny T}}},n-2(m+1)}(t) + \sum_{k=1}^{m+1} t^{(-4pqk+2pq)n+4pqk^2-4pqk+2pq} \delta_{n-2k}
\end{align*}
as needed. Applying the formula at $s(n+3)-1$  gives the particular equation.
\end{proof}

 We shall now find an annihilator for $J_{{{\mbox{\tiny C}}},n}(t)$.
By Lemma \ref{J_C formulas}, replacing $t^{2n}$ with ${{\mbox{\small M}}}$ gives us
\begin{align*}J_{{{\mbox{\tiny C}}},n+2}(t)&={{\mbox{\small M}}}^{-2rs}t^{-4rs} J_{{{\mbox{\tiny C}}},n}(t)+({{\mbox{\small M}}}^{r-rs-2pqs}t^{-2rs+2r-4pqs}-{{\mbox{\small M}}}^{-r-rs}t^{-2rs-2r})J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)
\\&\;\;\;\;+{{\mbox{\small M}}}^{r-rs-pqs}t^{-2rs+2r-2pqs} \delta_{s(n+1)-1},\end{align*}
and since $J_{{{\mbox{\tiny C}}},n+2}(t) = {{\mbox{\small L}}}^2 J_{{{\mbox{\tiny C}}},n}(t)$, we find
\begin{align*}({{\mbox{\small L}}}^2-{{\mbox{\small M}}}^{-2rs}t^{-4rs})J_{{{\mbox{\tiny C}}},n}(t)
&=({{\mbox{\small M}}}^{r-rs-2pqs}t^{-2rs+2r-4pqs}-{{\mbox{\small M}}}^{-r-rs}t^{-2rs-2r})J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)
\\&\;\;\;\;+{{\mbox{\small M}}}^{r-rs-pqs}t^{-2rs+2r-2pqs} \delta_{s(n+1)-1}.\end{align*}
In this equation, let $$a(t,{{\mbox{\small M}}})={{\mbox{\small M}}}^{r-rs-2pqs}t^{-2rs+2r-4pqs}-{{\mbox{\small M}}}^{-r-rs}t^{-2rs-2r},$$
 which is the coefficient of $J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)$,
then obviously $a(t,{{\mbox{\small M}}})\ne 0$, and we have
\begin{equation}\label{first relation}
\begin{array}{l}a^{-1}(t,{{\mbox{\small M}}})({{\mbox{\small L}}}^2-{{\mbox{\small M}}}^{-2rs}t^{-4rs})J_{{{\mbox{\tiny C}}},n}(t)
\\=J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)
+a^{-1}(t,{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-pqs}t^{-2rs+2r-2pqs}\delta_{s(n+1)-1}.\end{array}
\end{equation}
From Lemma \ref{s peel identity}, we have
$$({{\mbox{\small L}}}^2-t^{-8pqs^2-4pqs}{{\mbox{\small M}}}^{-2pqs^2})J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)=S_n.$$
So multiplying (\ref{first relation}) from the left by $({{\mbox{\small L}}}^2-t^{-8pqs^2-4pqs}{{\mbox{\small M}}}^{-2pqs^2})$
gives
\begin{equation}\label{second relation}\begin{array}{l}
({{\mbox{\small L}}}^2-t^{-8pqs^2-4pqs}{{\mbox{\small M}}}^{-2pqs^2})a^{-1}(t,{{\mbox{\small M}}})({{\mbox{\small L}}}^2-{{\mbox{\small M}}}^{-2rs}t^{-4rs})J_{{{\mbox{\tiny C}}},n}(t)
\\
=S_n
+({{\mbox{\small L}}}^2-t^{-8pqs^2-4pqs}{{\mbox{\small M}}}^{-2pqs^2})a^{-1}(t,{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-pqs}t^{-2rs+2r-2pqs}\delta_{s(n+1)-1}\\
=S_n+a^{-1}(t,t^4{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-pqs}t^{-6rs+6r-6pqs} \delta_{s(n+3)-1} \\
\;\;\;\;-a^{-1}(t,{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-pqs-2pqs^2}t^{-2rs+2r-8pqs^2-6pqs}\delta_{s(n+1)-1}.\end{array}\end{equation}
Let $b(t,{{\mbox{\small M}}})/(t^2-t^{-2})$ denote the right hand side of (\ref{second relation}).
Then $b(t,{{\mbox{\small M}}})$ is a rational function in $t$ and ${{\mbox{\small M}}}$.
We claim that $b\ne 0$, which we show by checking $b(-1,{{\mbox{\small M}}}):=\lim_{t{\rightarrow} -1}b(t,{{\mbox{\small M}}})\ne 0$.
Recall
$$ S_n = \sum_{k=1}^s {{\mbox{\small M}}}^{-2pqsk+pqs}t^{4pqk^2-12pqsk+6pqs} \delta_{s(n+3)-1-2k}.$$
So we have \begin{equation}\label{S limit}\begin{array}{ll}
\lim_{t \to -1} (t^2-t^{-2}) S_n &= \sum_{k=1}^s {{\mbox{\small M}}}^{-2pqsk+pqs}({{\mbox{\small M}}}^{s(p+q)} +{{\mbox{\small M}}}^{-s(p+q)}  - {{\mbox{\small M}}}^{s(q-p)} -{{\mbox{\small M}}}^{-s(q-p)}) \\
&= ({{\mbox{\small M}}}^{s(p+q)} +{{\mbox{\small M}}}^{-s(p+q)}  - {{\mbox{\small M}}}^{s(q-p)} -{{\mbox{\small M}}}^{-s(q-p)}) \frac{{{\mbox{\small M}}}^{pqs} (1-{{\mbox{\small M}}}^{-2pqs^2})}{{{\mbox{\small M}}}^{2pqs}-1}.\end{array}\end{equation}
Also
$$\begin{array}{l}
\lim_{t \to -1} (t^2-t^{-2})\big(a^{-1}(t,t^4{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-pqs}t^{6r-6rs-6pqs} \delta_{s(n+3)-1} \\
\qquad \qquad\qquad \qquad-a^{-1}(t,{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-pqs-2pqs^2}t^{-2rs+2r-8pqs^2-6pqs}\delta_{s(n+1)-1} \big)\\
= a^{-1}(-1,{{\mbox{\small M}}})({{\mbox{\small M}}}^{r-rs-pqs}-{{\mbox{\small M}}}^{r-rs-pqs-2pqs^2})({{\mbox{\small M}}}^{s(p+q)} +{{\mbox{\small M}}}^{-s(p+q)}  - {{\mbox{\small M}}}^{s(q-p)} -{{\mbox{\small M}}}^{-s(q-p)})\\
=\frac{1}{{{\mbox{\small M}}}^{r-rs-2pqs}-{{\mbox{\small M}}}^{-r-rs}}({{\mbox{\small M}}}^{r-rs-pqs}-{{\mbox{\small M}}}^{r-rs-pqs-2pqs^2})({{\mbox{\small M}}}^{s(p+q)} +{{\mbox{\small M}}}^{-s(p+q)}  - {{\mbox{\small M}}}^{s(q-p)} -{{\mbox{\small M}}}^{-s(q-p)}).\end{array}
$$
Summing up the two limits above, we get
$$\begin{array}{l}
b(-1,{{\mbox{\small M}}}):=\lim_{t{\rightarrow} -1}b(t,{{\mbox{\small M}}}) \\= \left( \frac{{{\mbox{\small M}}}^{pqs} (1-{{\mbox{\small M}}}^{-2pqs^2})}{{{\mbox{\small M}}}^{2pqs}-1}+\frac{{{\mbox{\small M}}}^{r-rs-pqs}-{{\mbox{\small M}}}^{r-rs-pqs-2pqs^2}}{{{\mbox{\small M}}}^{r-rs-2pqs}-{{\mbox{\small M}}}^{-r-rs}}\right)
({{\mbox{\small M}}}^{s(p+q)} +{{\mbox{\small M}}}^{-s(p+q)}  - {{\mbox{\small M}}}^{s(q-p)} -{{\mbox{\small M}}}^{-s(q-p)})\\
= \frac{(-{{\mbox{\small M}}}^{pqs-r-rs}+
{{\mbox{\small M}}}^{r-rs+pqs})(1-{{\mbox{\small M}}}^{-2pqs^2})
({{\mbox{\small M}}}^{ps}-{{\mbox{\small M}}}^{-ps})({{\mbox{\small M}}}^{qs}-{{\mbox{\small M}}}^{-qs})}{({{\mbox{\small M}}}^{2pqs}-1)({{\mbox{\small M}}}^{r-rs-2pqs}-{{\mbox{\small M}}}^{-r-rs})}
,\end{array}
$$
which is not zero. So $b \neq 0$ and we conclude that our recurrence (\ref{second relation}) is   inhomogeneous. Therefore,
\begin{align*}
P(t,{{\mbox{\small M}}},{{\mbox{\small L}}}) &= ({{\mbox{\small L}}}-1)b^{-1}(t,{{\mbox{\small M}}})({{\mbox{\small L}}}^2-t^{-8pqs^2-4pqs}{{\mbox{\small M}}}^{-2pqs^2})a^{-1}(t,{{\mbox{\small M}}})({{\mbox{\small L}}}^2-{{\mbox{\small M}}}^{-2rs}t^{-4rs})
\end{align*}
is an annihilator of $J_{{{\mbox{\tiny C}}},n}(t)$ in $\widetilde{\A}_{{\mbox{\tiny C}}}$.

Up to this point all the results above in this section are valid for
general $C$ over $T$.
From now on in this section,  we put in the restriction that  $s$ is odd and $q > 2$.
Once we prove that $P$ is of minimal degree in $L$, it will follow that $P$ is the recurrence polynomial of $J_{{{\mbox{\tiny C}}},n}(t)$ up to normalization. We can check the AJ-conjecture by evaluating $P$ at $t = -1$.
\[P(-1,{{\mbox{\small M}}},{{\mbox{\small L}}}) =b^{-1}(-1,{{\mbox{\small M}}})a^{-1}(-1,{{\mbox{\small M}}})({{\mbox{\small L}}}-1)({{\mbox{\small L}}}^2-{{\mbox{\small M}}}^{-2pqs^2})({{\mbox{\small L}}}^2-{{\mbox{\small M}}}^{-2rs}),\]
which, up to a nonzero rational function in ${{\mbox{\small M}}}$, is equal to the A-polynomial of $C$.

\subsection{$P$ is the Recurrence Polynomial of $C$.}
We now wish to show that the operator $P$ is the recurrence polynomial of $C$, up to normalization. It is enough to show that if an operator $Q = D_4 L^4 + D_3 L^3 + D_2 L^2 + D_1 L + D_0$ is an annihilator of $J_{{{\mbox{\tiny C}}},n}(t)$
with $D_0,...,D_4\in {\mathbb Z}[t^{\pm1},{{\mbox{\small M}}}^{\pm1}]$, then $Q = 0$.

Suppose $Q J_{{{\mbox{\tiny C}}},n}(t) = 0$, that is,
\[D_4 J_{{{\mbox{\tiny C}}},n+4}(t) + D_3 J_{{{\mbox{\tiny C}}},n+3}(t) + D_2 J_{{{\mbox{\tiny C}}},n+2}(t) + D_1 J_{{{\mbox{\tiny C}}},n+1}(t) + D_0 J_{{{\mbox{\tiny C}}},n}(t) = 0.\]
We wish to show that $D_i = 0$ for $i = 0,1,2,3,4$. Applying our Lemma \ref{J_C formulas}, we have
\begin{align*}
0 &= D_4 J_{{{\mbox{\tiny C}}},n+4}(t) + D_3 J_{{{\mbox{\tiny C}}},n+3}(t) + D_2 J_{{{\mbox{\tiny C}}},n+2}(t) + D_1 J_{{{\mbox{\tiny C}}},n+1}(t) + D_0 J_{{{\mbox{\tiny C}}},n}(t)\\
&=D_4\big({{\mbox{\small M}}}^{-2rs}t^{-12rs}J_{{{\mbox{\tiny C}}},n+2}(t)+({{\mbox{\small M}}}^{r-rs-2pqs}t^{-6rs+6r-12pqs}
-{{\mbox{\small M}}}^{-r-rs}t^{-6rs-6r})J_{{{\mbox{\tiny T}}},s(n+3)-1}(t)
\\&\;\;\;\;+{{\mbox{\small M}}}^{r-rs-pqs}t^{-6rs+6r-6pqs} \delta_{s(n+3)-1}\big)\\
&\;\;\;\;+ D_3 \big({{\mbox{\small M}}}^{-2rs}t^{-8rs}J_{{{\mbox{\tiny C}}},n+1}(t)+({{\mbox{\small M}}}^{r-rs-2pqs}t^{-4rs+4r-8pqs}
-{{\mbox{\small M}}}^{-r-rs}t^{-4rs-4r})J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)
\\&\;\;\;\;+{{\mbox{\small M}}}^{r-rs-pqs}t^{4r-4rs-4pqs} \delta_{s(n+2)-1}\big)\\&
\;\;\;\;+D_2J_{{{\mbox{\tiny C}}},n+2}(t) + D_1 J_{{{\mbox{\tiny C}}},n+1}(t) + D_0 J_{{{\mbox{\tiny C}}},n}(t)\end{align*}
\begin{align*}&=(D_4{{\mbox{\small M}}}^{-2rs}t^{-12rs}+ D_2)\big ({{\mbox{\small M}}}^{-2rs}t^{-4rs} J_{{{\mbox{\tiny C}}},n}(t)+({{\mbox{\small M}}}^{r-rs-2pqs}t^{-2rs+2r-4pqs}-{{\mbox{\small M}}}^{-r-rs}t^{-2rs-2r})J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)
\\&\qquad+{{\mbox{\small M}}}^{r-rs-pqs}t^{2r-2rs-2pqs} \delta_{s(n+1)-1}\big)\\
&\quad+D_4\big(({{\mbox{\small M}}}^{r-rs-2pqs}t^{-6rs+6r-12pqs}
-{{\mbox{\small M}}}^{-r-rs}t^{-6rs-6r})J_{{{\mbox{\tiny T}}},s(n+3)-1}(t)
+{{\mbox{\small M}}}^{r-rs-pqs}t^{6r-6rs-6pqs} \delta_{s(n+3)-1}\big)\\&
\quad+ D_3 \big({{\mbox{\small M}}}^{-2rs}t^{-8rs}J_{{{\mbox{\tiny C}}},n+1}(t)+({{\mbox{\small M}}}^{r-rs-2pqs}t^{-4rs+4r-8pqs}
-{{\mbox{\small M}}}^{-r-rs}t^{-4rs-4r})J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)
\\&\qquad+{{\mbox{\small M}}}^{r-rs-pqs}t^{4r-4rs-4pqs} \delta_{s(n+2)-1}\big)\\&
\;\;\;\; + D_1 J_{{{\mbox{\tiny C}}},n+1}(t) + D_0 J_{{{\mbox{\tiny C}}},n}(t)\\
&= (D_0 + D_2 {{\mbox{\small M}}}^{-2rs}t^{-4rs} + D_4 {{\mbox{\small M}}}^{-4rs}t^{-16rs})J_{{{\mbox{\tiny C}}},n}(t) + (D_1  +  D_3{{\mbox{\small M}}}^{-2rs}t^{-8rs}) J_{{{\mbox{\tiny C}}},n+1}(t)\\
&\quad+D_4({{\mbox{\small M}}}^{r-rs-2pqs}t^{-6rs+6r-12pqs}
-{{\mbox{\small M}}}^{-r-rs}t^{-6rs-6r})J_{{{\mbox{\tiny T}}},s(n+3)-1}(t)\\
&\quad+D_3 ({{\mbox{\small M}}}^{r-rs-2pqs}t^{-4rs+4r-8pqs}
-{{\mbox{\small M}}}^{-r-rs}t^{-4rs-4r})J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)\\
&\quad+(D_4{{\mbox{\small M}}}^{-2rs}t^{-12rs}+ D_2)({{\mbox{\small M}}}^{r-rs-2pqs}t^{-2rs+2r-4pqs}-{{\mbox{\small M}}}^{-r-rs}t^{-2rs-2r})J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)\\
&\quad+(D_4{{\mbox{\small M}}}^{-2rs}t^{-12rs}+ D_2){{\mbox{\small M}}}^{r-rs-pqs}t^{2r-2rs-2pqs} \delta_{s(n+1)-1}+D_3{{\mbox{\small M}}}^{r-rs-pqs}t^{4r-4rs-4pqs}\delta_{s(n+2)-1}\\
&\qquad+D_4{{\mbox{\small M}}}^{r-rs-pqs}t^{6r-6rs-6pqs}\delta_{s(n+3)-1},
\end{align*}
and applying Lemma \ref{s peel identity},
\begin{align*}
&= (D_0 + D_2 {{\mbox{\small M}}}^{-2rs}t^{-4rs} + D_4 {{\mbox{\small M}}}^{-4rs}t^{-16rs})J_{{{\mbox{\tiny C}}},n}(t) + (D_1  +  D_3{{\mbox{\small M}}}^{-2rs}t^{-8rs}) J_{{{\mbox{\tiny C}}},n+1}(t)\\
&\quad+\big(D_4({{\mbox{\small M}}}^{r-rs-2pqs}t^{-6rs+6r-12pqs}
-{{\mbox{\small M}}}^{-r-rs}t^{-6rs-6r}){{\mbox{\small M}}}^{-2pqs^2}t^{-8pqs^2-4pqs}\\&
\qquad+(D_4{{\mbox{\small M}}}^{-2rs}t^{-12rs}+ D_2)({{\mbox{\small M}}}^{r-rs-2pqs}t^{-2rs+2r-4pqs}-{{\mbox{\small M}}}^{-r-rs}t^{-2rs-2r})\big)J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)
\\
&\quad+D_3 ({{\mbox{\small M}}}^{r-rs-2pqs}t^{-4rs+4r-8pqs}
-{{\mbox{\small M}}}^{-r-rs}t^{-4rs-4r})J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)\\
&\quad+(D_4{{\mbox{\small M}}}^{-2rs}t^{-12rs}+ D_2){{\mbox{\small M}}}^{r-rs-pqs}t^{2r-2rs-2pqs} \delta_{s(n+1)-1}+D_3{{\mbox{\small M}}}^{r-rs-pqs}t^{4r-4rs-4pqs}\delta_{s(n+2)-1}\\
&\qquad+D_4{{\mbox{\small M}}}^{r-rs-pqs}t^{6r-6rs-6pqs}\delta_{s(n+3)-1}+D_4({{\mbox{\small M}}}^{r-rs-2pqs}t^{-6rs+6r-12pqs}
-{{\mbox{\small M}}}^{-r-rs}t^{-6rs-6r})S_n
\\&= D_4' J_{{{\mbox{\tiny C}}},n}(t) + D_3' J_{{{\mbox{\tiny C}}},n+1}(t) + D_2' J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)+D_1' J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)+D_0'
\end{align*}
We claim that each $D_i' = 0$, and it then follows that each $D_i = 0$. Indeed, it follows easily from $D_3' = D_1' = 0$ that $D_3 = D_1 = 0$. For the rest, it is enough to show that the two linear equations defined by
$D_0'=0$ and $D_2'=0$ are linearly independent (with $D_2$ and $D_4$ as variables).
We can check that the determinant of the linear system is nonzero, and in particular, we multiply by $(t^2-t^{-2})$ and then evaluate at $t=-1$ in order to use equation (\ref{S limit}):
\begin{align*}
& ({{\mbox{\small M}}}^{r-rs-2pqs}-{{\mbox{\small M}}}^{-r-rs})({{\mbox{\small M}}}^{r-3rs-pqs}+{{\mbox{\small M}}}^{r-rs-pqs}+({{\mbox{\small M}}}^{r-rs-2pqs}-{{\mbox{\small M}}}^{-r-rs}){{\mbox{\small M}}}^{pqs}\frac{(1-{{\mbox{\small M}}}^{-2pqs^2})}{{{\mbox{\small M}}}^{2pqs}-1})\\
&\quad ({{\mbox{\small M}}}^{s(p+q)} +{{\mbox{\small M}}}^{-s(p+q)}  - {{\mbox{\small M}}}^{s(q-p)} -{{\mbox{\small M}}}^{-s(q-p)})\\
&-({{\mbox{\small M}}}^{r-rs-2pqs-2pqs^2}-{{\mbox{\small M}}}^{-r-rs-2pqs^2}+{{\mbox{\small M}}}^{r-3rs-2pqs}-{{\mbox{\small M}}}^{-r-3rs}){{\mbox{\small M}}}^{r-rs-pqs}\\
&\quad ({{\mbox{\small M}}}^{s(p+q)} +{{\mbox{\small M}}}^{-s(p+q)}  - {{\mbox{\small M}}}^{s(q-p)} -{{\mbox{\small M}}}^{-s(q-p)})\\
=&\frac{1}{{{\mbox{\small M}}}^{2pqs}-1}({{\mbox{\small M}}}^{s(p+q)} +{{\mbox{\small M}}}^{-s(p+q)}  - {{\mbox{\small M}}}^{s(q-p)} -{{\mbox{\small M}}}^{-s(q-p)}) ({{\mbox{\small M}}}^{r-2pqs}-{{\mbox{\small M}}}^{-r})\\
&\quad \big( {{\mbox{\small M}}}^{-rs}({{\mbox{\small M}}}^{r-rs-pqs}({{\mbox{\small M}}}^{-2rs}+1)({{\mbox{\small M}}}^{2pqs}-1)+({{\mbox{\small M}}}^{r-rs-pqs}-{{\mbox{\small M}}}^{-r-rs+pqs})(1-{{\mbox{\small M}}}^{-2pqs^2}))\\
&\quad -{{\mbox{\small M}}}^{r-rs-pqs}({{\mbox{\small M}}}^{-2pqs^2-rs}+{{\mbox{\small M}}}^{-3rs})({{\mbox{\small M}}}^{2pqs}-1)\big),
\end{align*}
expanding some of the terms to observe cancelation,
\begin{align*}
=&\frac{1}{{{\mbox{\small M}}}^{2pqs}-1}({{\mbox{\small M}}}^{s(p+q)} +{{\mbox{\small M}}}^{-s(p+q)}  - {{\mbox{\small M}}}^{s(q-p)} -{{\mbox{\small M}}}^{-s(q-p)}) ({{\mbox{\small M}}}^{r-2pqs}-{{\mbox{\small M}}}^{-r})\\
&\quad \big( {{\mbox{\small M}}}^{r-4rs+pqs} - {{\mbox{\small M}}}^{r-4rs-pqs}+{{\mbox{\small M}}}^{r-2rs+pqs}-{{\mbox{\small M}}}^{r-2rs-pqs} \\
&\qquad +{{\mbox{\small M}}}^{r-2rs-pqs}-{{\mbox{\small M}}}^{r-2rs-pqs-2pqs^2} -{{\mbox{\small M}}}^{-r-2rs+pqs} + {{\mbox{\small M}}}^{-r-2rs+pqs-2pqs^2} \\
&\qquad +{{\mbox{\small M}}}^{r-2rs-pqs-2pqs^2} + {{\mbox{\small M}}}^{r-4rs-pqs} - {{\mbox{\small M}}}^{r-2rs+pqs-2pqs^2}-{{\mbox{\small M}}}^{r-4rs+pqs} \big)\\
=&\frac{({{\mbox{\small M}}}^{r-2pqs}-{{\mbox{\small M}}}^{-r})}{{{\mbox{\small M}}}^{2pqs}-1}
({{\mbox{\small M}}}^{ps}-{{\mbox{\small M}}}^{-ps})({{\mbox{\small M}}}^{qs}-{{\mbox{\small M}}}^{-qs})
\big(  ({{\mbox{\small M}}}^{-2pqs^2}-1)({{\mbox{\small M}}}^{-r-2rs+pqs} -{{\mbox{\small M}}}^{r-2rs+pqs}) \big)
\end{align*}
which is indeed nonzero.

We now prove that if $r$ is not an integer between $0$ and $pqs$, we have $D_i' = 0$ for each $i = 0,1,2,3,4$ and thus our annihilator $P$ is of minimal ${{\mbox{\small L}}}$-degree.

We say that a function $f:{\mathbb Z} \to {\mathbb Z}$ is a \textit{quasi-polynomial} if there exist periodic functions $a_0, \ldots, a_d$ each with integral period such that
\[f(n) = \sum_{i=0}^d a_i(n) n^i,\]
and $f$ is of degree $d$ if $a_d \neq 0$. In particular, we say $f$ is \textit{quasi-quadratic} if $f$ is a quasi-polynomial of degree 2.

\begin{lemma} \label{degree argument for s odd}(1) When $p > q$ and either $r < 0$ or $r > pqs$, we have $D_i' = 0$ for $i = 0,1,2,3,4$.

(2) When $p < -q$ and either $r > 0$ or $r < pqs$, we have $D_i' = 0$ for $i = 0,1,2,3,4$.
\end{lemma}

\begin{proof} (1) Suppose $p > q$, $r > pqs$, and some $D_i' \neq 0$. Then
there must be another nonzero $D_j'$ such that one of the following equalities hold:
\begin{align*}
\ell[D_4' J_{{{\mbox{\tiny C}}},n}(t)] &= \ell[D_3' J_{{{\mbox{\tiny C}}},n+1}(t)],\\
\ell[D_4' J_{{{\mbox{\tiny C}}},n}(t)] &= \ell[D_2' J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)],\\
\ell[D_4' J_{{{\mbox{\tiny C}}},n}(t)] &= \ell[D_1' J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)],\\
\ell[D_3' J_{{{\mbox{\tiny C}}},n+1}(t)] &= \ell[D_2' J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)],\\
\ell[D_3' J_{{{\mbox{\tiny C}}},n+1}(t)] &= \ell[D_1' J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)],\\
\ell[D_2' J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)] &= \ell[D_1' J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)].
\end{align*}
That is, two of the summands must share a lowest degree, and since $\ell[D_0']$ is only linear in $n$ for large enough $n$ while $\ell[J_{{{\mbox{\tiny C}}},n}(t)]$ and $\ell[J_{{{\mbox{\tiny T}}},n}(t)]$ are quasi-quadratic
by Lemmas \ref{degree of J_C} and \ref{degree of J_T}, we can immediately dispose of the cases
 involving $D_0'$.

\begin{subcase}\label{c,n,n+1-case}$\ell[D_4' J_{{{\mbox{\tiny C}}},n}(t)] = \ell[D_3' J_{{{\mbox{\tiny C}}},n+1}(t)]$:\end{subcase} From the second formula of Lemma \ref{degree of J_C}(1), we have
\begin{align*}
\ell[D_4']-\ell[D_3'] &= \ell[J_{{{\mbox{\tiny C}}},n+1}(t)] - \ell[J_{{{\mbox{\tiny C}}},n}(t)]\\
&= -2rsn-rs-(-1)^n((s-2)(r-pqs)+(p-2)(q-2)),
\end{align*}
but for sufficiently large $n$, $\ell[D_4']-\ell[D_3']$ is a linear function in $n$, while the right hand side is not a polynomial, so we have a contradiction.

\begin{subcase}$\ell[D_4' J_{{{\mbox{\tiny C}}},n}(t)] = \ell[D_2' J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)]$: \end{subcase}From Lemmas \ref{degree of J_C}(1) and \ref{degree of J_T}(1), we have
\begin{align*}
\ell[D_4']-\ell[D_2'] &= \ell[J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)] - \ell[J_{{{\mbox{\tiny C}}},n}(t)]\\
&= s(r-p q s)n^2+(2 p q s-2 p q s^2)n - p q s^2+2p q s- r s
 \\
&\quad - \frac12(1-(-1)^{n-1})(s-2)(r-pqs),
\end{align*}
which is quasi-quadratic, while the left hand side is at most linear, giving us another contradiction.

\begin{subcase}$\ell[D_4' J_{{{\mbox{\tiny C}}},n}(t)] = \ell[D_1' J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)]$: \end{subcase}Here we have
\begin{align*}
\ell[D_4']-\ell[D_1'] &= \ell[J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)]-\ell[J_{{{\mbox{\tiny C}}},n}(t)]\\
&=   s(r - p q s)n^2 + (2 p q s - 4 p q s^2)n -p q(4s^2-4s)- r s\\
&\quad - \frac12(1-(-1)^{n-1})(s-2)(r-pqs)+ (-1)^n (p-2)(q-2)
\end{align*}
again giving us a quasi-quadratic function on the right and a linear function on the left, which is a contradiction.

\begin{subcase}$\ell[D_3' J_{{{\mbox{\tiny C}}},n+1}(t)] = \ell[D_2' J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)]$:\end{subcase} We have
\begin{align*}
\ell[D_3']-\ell[D_2'] &= \ell[J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)] - \ell[J_{{{\mbox{\tiny C}}},n+1}(t)]\\
&= s(r - p q s) n^2 + (2 p q s + 2 r s - 2 p q s^2) n -pq(s^2-2s) \\
 &\quad   - \frac12(1-(-1)^{n-1})(s-2)(r-pqs)+ (-1)^n (p-2)(q-2)
\end{align*}
which is again quasi-quadratic on the right and linear on the left,
 again a contradiction.

\begin{subcase}$\ell[D_3' J_{{{\mbox{\tiny C}}},n+1}(t)] = \ell[D_1' J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)]$:\end{subcase} We have
\begin{align*}
\ell[D_3'] - \ell[D_1'] &= \ell[J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)] - \ell[J_{{{\mbox{\tiny C}}},n+1}(t)]\\
&= (r s - p q s^2) n^2 + (2 p q s + 2 r s - 4 p q s^2) n -pq(4s^2-4s) \\
 &\quad - \frac12(1-(-1)^{n})(s-2)(r-pqs)
\end{align*}
giving us another contradiction.

\begin{subcase}\label{(t,s(n+2),s(n+1))-case}$\ell[D_2' J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)] = \ell[D_1' J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)]$:\end{subcase} This time we have
\begin{align*}
\ell[D_2'] - \ell[D_1'] &= \ell[J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)] - \ell[J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)]\\
&= -2p q s^2 n + 2 p q s - 3 p q s^2 + \frac{1}{2}(-1)^{(n+1)s}(1-(-1)^s)(p-2)(q-2)\\
&=- 2 p q s^2 n + 2 p q s - 3 p q s^2  -(-1)^n(p-2)(q-2)
\end{align*}
which is alternating on the right and eventually linear on the left, which is a contradiction.  This exhausts the possibilities of the case $r > pqs$.

Now assume $p > q$ and $r < 0$. We first consider
 the highest degrees of the summands in the equation
 $$0= D_4' J_{{{\mbox{\tiny C}}},n}(t) + D_3' J_{{{\mbox{\tiny C}}},n+1}(t) + D_2' J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)+D_1' J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)+D_0'$$
 for large positive $n$'s. If $D_3'\ne 0$ or $D_4'\ne 0$, then
 both of them cannot be zero since by Lemma \ref{degree of J_C}(1)
  $\hbar[J_{{{\mbox{\tiny C}}}, n}]$ and $\hbar[J_{{{\mbox{\tiny C}}},n+1}]$ are each quasi-quadratic while
 by Lemma \ref{degree of J_T} (1) $\hbar[J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)]$ and
 $\hbar[J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)]$ are each linear in $n$ (for positive $n$'s), and
  we must have the following

 \begin{subcase}$\hbar[D_4'J_{{{\mbox{\tiny C}}}, n}(t)]=\hbar[D_3'J_{{{\mbox{\tiny C}}},n+1}(t)]$:
 \end{subcase}
 Then we have, by Lemma \ref{degree of J_C}(1),
 $$\begin{array}{ll}
 \hbar[D_4']-\hbar[D_3']&=\hbar[J_{{{\mbox{\tiny C}}}, n+1}(t)]-\hbar[J_{{{\mbox{\tiny C}}},n}(t)]\\
& = -2rs n -rs-(-1)^n(s-2)(r-2pq+2p+2q)
\end{array}$$
which is a linear polynomial for large $n$ on the left but is not a linear polynomial on the right, giving a contradiction.

So both $D_4'$ and $D_3'$ are zero. So we have
 $0=D_2' J_{{{\mbox{\tiny T}}},s(n+1)-1}(t) + D_1' J_{{{\mbox{\tiny T}}},s(n+2)-1}(t) + D_0'$.
We can then analyze the lowest degrees in a similar fashion as above; if one of the $D_i'$ is not zero, we must have
\begin{subcase}$\ell[D_2' J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)] = \ell[D_1' J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)]$:\end{subcase}

This can be treated similarly as Subcase \ref{(t,s(n+2),s(n+1))-case}.
We conclude that each $D_i' = 0$. This completes the proof of part (1).

Part (2) of the lemma can be proved similarly with the use of Lemmas \ref{degree of J_T}(2) and \ref{degree of J_C}(2).
\end{proof}

\begin{rem}\label{subcase remark}{\rm In the proof of Lemma \ref{degree argument for s odd} we used the condition
that $s>2$  odd and $q> 2$ in several subcases. Some of these subcases will disappear
accordingly in later
sections when we impose the condition $s$ odd and $q=2$ or $s>2$ even or $s=2$.}
\end{rem}

\section{Case $s>2$ is odd and $q=2$}
\subsection{An Annihilator $P$ of $J_{{{\mbox{\tiny C}}},n}(t)$.}
Define:
 \[U_n = \sum_{k=1}^s (-1)^{k-1} t^{2psn-4psnk+2pk^2-8psk+2pk+4ps}\frac{t^{4sn+8s-2-4k}-t^{-4sn-8s+2+4k}}{t^2-t^{-2}}.\]
 When $q = 2$, we have by \cite[Lemma 1.5]{Tran} the identity
\begin{equation} \label{q=2 J_T(n+1) relation}
J_{{{\mbox{\tiny T}}},n+1}(t) = -t^{-(4n+2)p} J_{{{\mbox{\tiny T}}},n}(t)+t^{-2pn}\frac{t^{4n+2}-t^{-4n-2}}{t^2-t^{-2}}.
\end{equation}
Note again that (\ref{q=2 J_T(n+1) relation}) is valid for negative $p$ as well.
\begin{lemma} \label{q=2 s peel lemma}When $q=2$, for all positive integers $m$, we have
\[
J_{{{\mbox{\tiny T}}},n}(t) = (-1)^m t^{(-4m n+2m^2)p}J_{{{\mbox{\tiny T}}},n-m}(t) + \sum_{k=1}^m (-1)^{k-1} t^{-(4k-2)pn+(2k^2-2k+2)p}\frac{t^{4n+2-4k}-t^{-4n-2+4k}}{t^2-t^{-2}}
\]
and in particular, when $s$ is odd,
 \[J_{{{\mbox{\tiny T}}},s(n+2)-1}(t) = - t^{-4ps^2n+4ps-6ps^2} J_{{{\mbox{\tiny T}}},s(n+1)-1}(t) + U_n.\]
\end{lemma}
\begin{proof} Apply the relation (\ref{q=2 J_T(n+1) relation}) $m$ times.
\end{proof}

Note that the relation (\ref{first relation}) is valid for general $C$ over $T$.
Specializing it at $q=2$ and $s$ odd, we have
\begin{equation}\label{first relation for q=2}
\begin{array}{l}a^{-1}(t,{{\mbox{\small M}}})({{\mbox{\small L}}}^2-{{\mbox{\small M}}}^{-2rs}t^{-4rs})J_{{{\mbox{\tiny C}}},n}(t)
\\=J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)
+a^{-1}(t,{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-2ps}t^{2r-2rs-4ps}\delta_{s(n+1)-1}.\end{array}
\end{equation}
From  Lemma \ref{q=2 s peel lemma}, we get
\[({{\mbox{\small L}}}+{{\mbox{\small M}}}^{-2ps^2}t^{4ps-6ps^2}) J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)= U_n.\]
Multiplying (\ref{first relation for q=2}) from the left by
$({{\mbox{\small L}}}+{{\mbox{\small M}}}^{-2ps^2}t^{4ps-6ps^2})$ yields
\begin{equation}\label{second relation for q=2}
\begin{array}{l}({{\mbox{\small L}}}+{{\mbox{\small M}}}^{-2ps^2}t^{4ps-6ps^2})a^{-1}(t,{{\mbox{\small M}}})({{\mbox{\small L}}}^2-{{\mbox{\small M}}}^{-2rs}t^{-4rs})J_{{{\mbox{\tiny C}}},n}(t)
\\=U_n+({{\mbox{\small L}}}+{{\mbox{\small M}}}^{-2ps^2}t^{4ps-6ps^2})a^{-1}(t,{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-2ps}t^{2r-2rs-4ps}\delta_{s(n+1)-1}
\\=U_n+a^{-1}(t,t^2{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-2ps}t^{4r-4rs-8ps}\delta_{s(n+2)-1}\\
\quad
+a^{-1}(t,{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-2ps-2ps^2}t^{2r-2rs-6ps^2}\delta_{s(n+1)-1}.\end{array}
\end{equation}

As in Section \ref{section s odd}, let $b(t,{{\mbox{\small M}}})/(t^2-t^{-2})$ denote the right hand side of (\ref{second relation for q=2}).
Then $b(t,{{\mbox{\small M}}})$ is a rational function in $t$ and ${{\mbox{\small M}}}$.
We now show that  $b\ne 0$  by checking $b(-1,{{\mbox{\small M}}}):=\lim_{t{\rightarrow} -1}b(t,{{\mbox{\small M}}})\ne 0$.
Rewrite $U_n$ as a function of $t$ and ${{\mbox{\small M}}}$ by changing $t^{2n}$ to ${{\mbox{\small M}}}$:
$$ U_n =\sum_{k=1}^s (-1)^{k-1} {{\mbox{\small M}}}^{ps-2psk}t^{2pk^2-8psk+2pk+4ps}\frac{{{\mbox{\small M}}}^{2s}t^{8s-2-4k}-{{\mbox{\small M}}}^{-2s}t^{-8s+2+4k}}{t^2-t^{-2}}.$$
So we have \begin{equation}\label{U limit}\begin{array}{ll}
\lim_{t \to -1} (t^2-t^{-2})U_n &=
 \sum_{k=1}^s (-1)^{k-1} {{\mbox{\small M}}}^{ps-2psk}
 ({{\mbox{\small M}}}^{2s}-{{\mbox{\small M}}}^{-2s})=\frac{({{\mbox{\small M}}}^{2s}-{{\mbox{\small M}}}^{-2s}){{\mbox{\small M}}}^{-ps}(1+{{\mbox{\small M}}}^{-2ps^2})}{1+{{\mbox{\small M}}}^{-2ps}}.\end{array}\end{equation}
Also
$$\begin{array}{l}
\lim_{t \to -1} (t^2-t^{-2})\big(a^{-1}(t,t^2{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-2ps}t^{4r-4rs-8ps}\delta_{s(n+3)-1}\\
\qquad \qquad \qquad \qquad +a^{-1}(t,{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-2ps-2ps^2}t^{2r-2rs-6ps^2}\delta_{s(n+1)-1}\big)\\
=\frac{{{\mbox{\small M}}}^{r-rs-2ps}+{{\mbox{\small M}}}^{r-rs-2ps-2ps^2}}{{{\mbox{\small M}}}^{r-rs-4ps}-{{\mbox{\small M}}}^{-r-rs}}({{\mbox{\small M}}}^{s(p+2)} +{{\mbox{\small M}}}^{-s(p+2)}  - {{\mbox{\small M}}}^{s(2-p)} -{{\mbox{\small M}}}^{-s(2-p)}).\end{array}
$$
Summing up the two limits above, we get
$$\begin{array}{l}
b(-1,{{\mbox{\small M}}}):=\lim_{t{\rightarrow} -1}b(t,{{\mbox{\small M}}}) \\=\frac{({{\mbox{\small M}}}^{2s}-{{\mbox{\small M}}}^{-2s}){{\mbox{\small M}}}^{-ps}(1+{{\mbox{\small M}}}^{-2ps^2})}{1+{{\mbox{\small M}}}^{-2ps}}
+\frac{{{\mbox{\small M}}}^{r-rs-2ps}+{{\mbox{\small M}}}^{r-rs-2ps-2ps^2}}{{{\mbox{\small M}}}^{r-rs-4ps}-{{\mbox{\small M}}}^{-r-rs}}({{\mbox{\small M}}}^{s(p+2)} +{{\mbox{\small M}}}^{-s(p+2)}  - {{\mbox{\small M}}}^{s(2-p)} -{{\mbox{\small M}}}^{-s(2-p)})\\
=\frac{({{\mbox{\small M}}}^{2s}-{{\mbox{\small M}}}^{-2s})(1+{{\mbox{\small M}}}^{-2ps^2}){{\mbox{\small M}}}^{-rs-ps}({{\mbox{\small M}}}^r-{{\mbox{\small M}}}^{-r})}
{(1+{{\mbox{\small M}}}^{-2ps})({{\mbox{\small M}}}^{r-rs-4ps}-{{\mbox{\small M}}}^{-r-rs})},\end{array}
$$
which is not zero. So $b \neq 0$ and we conclude that our recurrence (\ref{second relation for q=2}) is   inhomogeneous. Therefore,
\begin{align*}
P(t,{{\mbox{\small M}}},{{\mbox{\small L}}}) &= ({{\mbox{\small L}}}-1)b^{-1}(t,{{\mbox{\small M}}})({{\mbox{\small L}}}+{{\mbox{\small M}}}^{-2ps^2}t^{4ps-6ps^2})a^{-1}(t,{{\mbox{\small M}}})({{\mbox{\small L}}}^2-{{\mbox{\small M}}}^{-2rs}t^{-4rs})
\end{align*}
is an annihilator of $J_{{{\mbox{\tiny C}}},n}(t)$ in $\widetilde{\A}_{{\mbox{\tiny C}}}$.

Once we prove that $P$ is of minimal degree in $L$, it will follow that $P$ is the recurrence polynomial of $J_{{{\mbox{\tiny C}}},n}(t)$ up to normalization. We can check the AJ-conjecture by evaluating $P$ at $t = -1$.
\[P(-1,{{\mbox{\small M}}},{{\mbox{\small L}}}) =b^{-1}(-1,{{\mbox{\small M}}})a^{-1}(-1,{{\mbox{\small M}}})({{\mbox{\small L}}}-1)({{\mbox{\small L}}}+{{\mbox{\small M}}}^{-2ps^2})({{\mbox{\small L}}}^2-{{\mbox{\small M}}}^{-2rs}),\]
which, up to a nonzero factor  in ${\mathbb Q}({{\mbox{\small M}}})$, is equal to the A-polynomial of $C$.

\subsection{$P$ is the Recurrence Polynomial of $C$.}\label{subsection s odd and q=2}
We now wish to show that the operator $P$ is the recurrence polynomial of $C$. It is enough to show that if an operator $Q = D_3 {{\mbox{\small L}}}^3 + D_2 {{\mbox{\small L}}}^2 + D_1 {{\mbox{\small L}}} + D_0$
 with $D_0,...,D_3\in {\mathbb Z}[t^{\pm1},{{\mbox{\small M}}}^{\pm1}]$ is an annihilator of $J_{{{\mbox{\tiny C}}},n}(t)$, then $Q = 0$.

Suppose $Q J_{{{\mbox{\tiny C}}},n}(t) = 0$, that is,
$D_3 J_{{{\mbox{\tiny C}}},n+3}(t) + D_2 J_{{{\mbox{\tiny C}}},n+2}(t) + D_1 J_{{{\mbox{\tiny C}}},n+1}(t) + D_0 J_{{{\mbox{\tiny C}}},n}(t) = 0$.
We wish to show that $D_i = 0$ for $i = 0,1,2,3$. We have by Lemma \ref{J_C formulas}
(specialized at $q=2$)

\begin{align*}
0 &= D_3 J_{{{\mbox{\tiny C}}},n+3}(t) + D_2 J_{{{\mbox{\tiny C}}},n+2}(t) + D_1 J_{{{\mbox{\tiny C}}},n+1}(t) + D_0 J_{{{\mbox{\tiny C}}},n}(t)\\
&= D_3 \big({{\mbox{\small M}}}^{-2rs}t^{-8rs}J_{{{\mbox{\tiny C}}},n+1}(t)+({{\mbox{\small M}}}^{r-rs-4ps}t^{-4rs+4r-16ps}
-{{\mbox{\small M}}}^{-r-rs}t^{-4rs-4r})J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)
\\&\qquad+{{\mbox{\small M}}}^{r-rs-2ps}t^{4r-4rs-8ps} \delta_{s(n+2)-1}\big)\\&
\quad+D_2\big ({{\mbox{\small M}}}^{-2rs}t^{-4rs} J_{{{\mbox{\tiny C}}},n}(t)+({{\mbox{\small M}}}^{r-rs-4ps}t^{-2rs+2r-8ps}-{{\mbox{\small M}}}^{-r-rs}t^{-2rs-2r})J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)
\\&\qquad+{{\mbox{\small M}}}^{r-rs-2ps}t^{2r-2rs-4ps} \delta_{s(n+1)-1}\big)\\
&\quad + D_1 J_{{{\mbox{\tiny C}}},n+1}(t) + D_0 J_{{{\mbox{\tiny C}}},n}(t)\\&=
 (D_0 +D_2{{\mbox{\small M}}}^{-2rs}t^{-4rs})J_{{{\mbox{\tiny C}}},n}(t)+(D_1 +D_3 {{\mbox{\small M}}}^{-2rs}t^{-8rs})J_{{{\mbox{\tiny C}}},n+1}(t)\\
&\quad+D_3 ({{\mbox{\small M}}}^{r-rs-4ps}t^{-4rs+4r-16ps}
-{{\mbox{\small M}}}^{-r-rs}t^{-4rs-4r})J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)\\
&\quad+D_2({{\mbox{\small M}}}^{r-rs-4ps}t^{-2rs+2r-8ps}-{{\mbox{\small M}}}^{-r-rs}t^{-2rs-2r})J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)\\
&\quad+ D_2{{\mbox{\small M}}}^{r-rs-2ps}t^{2r-2rs-4ps} \delta_{s(n+1)-1}+D_3{{\mbox{\small M}}}^{r-rs-2ps}t^{4r-4rs-8ps}\delta_{s(n+2)-1},
\end{align*}
and applying Lemma \ref{q=2 J_T(n+1) relation},
\begin{align*}
&= (D_0 +D_2{{\mbox{\small M}}}^{-2rs}t^{-4rs})J_{{{\mbox{\tiny C}}},n}(t)+(D_1 +D_3 {{\mbox{\small M}}}^{-2rs}t^{-8rs})J_{{{\mbox{\tiny C}}},n+1}(t)\\
&\quad + D_3 ({{\mbox{\small M}}}^{r-rs-4ps}t^{-4rs+4r-16ps}
-{{\mbox{\small M}}}^{-r-rs}t^{-4rs-4r})(-{{\mbox{\small M}}}^{-2ps^2}t^{4ps-6ps^2}J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)+U_n)\\
&\quad + D_2({{\mbox{\small M}}}^{r-rs-4ps}t^{-2rs+2r-8ps}-{{\mbox{\small M}}}^{-r-rs}t^{-2rs-2r})J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)\\
&\quad+ D_2{{\mbox{\small M}}}^{r-rs-2ps}t^{2r-2rs-4ps} \delta_{s(n+1)-1}+D_3{{\mbox{\small M}}}^{r-rs-2ps}t^{4r-4rs-8ps}\delta_{s(n+2)-1}\\
&= (D_0 +D_2{{\mbox{\small M}}}^{-2rs}t^{-4rs})J_{{{\mbox{\tiny C}}},n}(t)+(D_1 +D_3 {{\mbox{\small M}}}^{-2rs}t^{-8rs})J_{{{\mbox{\tiny C}}},n+1}(t)\\
&\quad +\big(D_3 ({{\mbox{\small M}}}^{r-rs-4ps}t^{-4rs+4r-16ps}
-{{\mbox{\small M}}}^{-r-rs}t^{-4rs-4r})(-{{\mbox{\small M}}}^{-2ps^2}t^{4ps-6ps^2})\\&\qquad+ D_2({{\mbox{\small M}}}^{r-rs-4ps}t^{-2rs+2r-8ps}-{{\mbox{\small M}}}^{-r-rs}t^{-2rs-2r})\big)J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)\\
&\quad+ D_2{{\mbox{\small M}}}^{r-rs-2ps}t^{2r-2rs-4ps} \delta_{s(n+1)-1}+D_3\big(
({{\mbox{\small M}}}^{r-rs-4ps}t^{-4rs+4r-16ps}
-{{\mbox{\small M}}}^{-r-rs}t^{-4rs-4r})U_n\\&\qquad+{{\mbox{\small M}}}^{r-rs-2ps}t^{4r-4rs-8ps}\delta_{s(n+2)-1}\big)\\
&=D_3'J_{{{\mbox{\tiny C}}},n}(t)+D_2'J_{{{\mbox{\tiny C}}},n+1}(t)+D_1'J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)+D_0'
\end{align*}
We claim that each $D_i' = 0$, and it then follows as in the previous section that each $D_i = 0$.
We again wish to show that the two linear equations defined by
$D_0'=0$ and $D_1'=0$ are linearly independent (with $D_2$ and $D_3$ as variables).
So let's check the determinant of the linear system, multiplied by $(t^2-t^{-2})$
and then valued at $t=-1$, is nonzero:
\begin{align*}
&({{\mbox{\small M}}}^{r-rs-4ps}
-{{\mbox{\small M}}}^{-r-rs})(-{{\mbox{\small M}}}^{-2ps^2})
{{\mbox{\small M}}}^{r-rs-2ps}({{\mbox{\small M}}}^{s(p+2)} +{{\mbox{\small M}}}^{-s(p+2)}  - {{\mbox{\small M}}}^{s(2-p)} -{{\mbox{\small M}}}^{-s(2-p)})\\
&-({{\mbox{\small M}}}^{r-rs-4ps}-{{\mbox{\small M}}}^{-r-rs})\big(({{\mbox{\small M}}}^{r-rs-4ps}-{{\mbox{\small M}}}^{-r-rs})
\frac{({{\mbox{\small M}}}^{2s}-{{\mbox{\small M}}}^{-2s}){{\mbox{\small M}}}^{-ps}(1+{{\mbox{\small M}}}^{-2ps^2})}{1+{{\mbox{\small M}}}^{-2ps}}\\
&\quad+{{\mbox{\small M}}}^{r-rs-2ps}({{\mbox{\small M}}}^{s(p+2)} +{{\mbox{\small M}}}^{-s(p+2)}  - {{\mbox{\small M}}}^{s(2-p)} -{{\mbox{\small M}}}^{-s(2-p)})\big)\\
=&-({{\mbox{\small M}}}^{r-rs-4ps}
-{{\mbox{\small M}}}^{-r-rs})({{\mbox{\small M}}}^{-2ps^2}+1)\big({{\mbox{\small M}}}^{r-rs-2ps}
({{\mbox{\small M}}}^{s(p+2)} +{{\mbox{\small M}}}^{-s(p+2)}  - {{\mbox{\small M}}}^{s(2-p)} -{{\mbox{\small M}}}^{-s(2-p)})\\
&+({{\mbox{\small M}}}^{r-rs-4ps}-{{\mbox{\small M}}}^{-r-rs})
\frac{({{\mbox{\small M}}}^{2s}-{{\mbox{\small M}}}^{-2s}){{\mbox{\small M}}}^{-ps}}{1+{{\mbox{\small M}}}^{-2ps}}\big)
\end{align*}
\begin{align*}
=&-({{\mbox{\small M}}}^{r-rs-4ps}-{{\mbox{\small M}}}^{-r-rs})({{\mbox{\small M}}}^{-2ps^2}+1)({{\mbox{\small M}}}^{2s}-{{\mbox{\small M}}}^{-2s})\big({{\mbox{\small M}}}^{r-rs-2ps}
({{\mbox{\small M}}}^{ps}-{{\mbox{\small M}}}^{-ps})\\
&+({{\mbox{\small M}}}^{r-rs-4ps}-{{\mbox{\small M}}}^{-r-rs})
\frac{1}{{{\mbox{\small M}}}^{ps}+{{\mbox{\small M}}}^{-ps}}\big)\\
=&-\frac{1}{{{\mbox{\small M}}}^{ps}+{{\mbox{\small M}}}^{-ps}}({{\mbox{\small M}}}^{r-rs-4ps}-{{\mbox{\small M}}}^{-r-rs})({{\mbox{\small M}}}^{-2ps^2}+1)({{\mbox{\small M}}}^{2s}-{{\mbox{\small M}}}^{-2s})\big({{\mbox{\small M}}}^{r-rs-2ps}
({{\mbox{\small M}}}^{2ps}-{{\mbox{\small M}}}^{-2ps})\\
&+({{\mbox{\small M}}}^{r-rs-4ps}-{{\mbox{\small M}}}^{-r-rs})\big)\\
=&-\frac{1}{{{\mbox{\small M}}}^{ps}+{{\mbox{\small M}}}^{-ps}}({{\mbox{\small M}}}^{r-rs-4ps}-{{\mbox{\small M}}}^{-r-rs})({{\mbox{\small M}}}^{-2ps^2}+1)({{\mbox{\small M}}}^{2s}-{{\mbox{\small M}}}^{-2s})({{\mbox{\small M}}}^{r-rs}-{{\mbox{\small M}}}^{-r-rs})
\end{align*}
which is indeed nonzero.

The following lemma shows that each $D_i' = 0$ if $r$ is not a number between $0$ and $pqs$.

\begin{lemma} \label{q=2 min degree lemma} (1) When $p > q = 2$ and either $r < 0$ or $r > pqs$, we have $D_i' = 0$ for $i = 0,1,2,3$.

(2) When $p < -q$ and either $r > 0$ or $r < pqs$, we have $D_i' = 0$ for $i = 0,1,2,3$.
\end{lemma}

\begin{proof} The proof is entirely similar to that of Lemma \ref{degree argument for s odd}
(also cf Remark \ref{subcase remark}).
\end{proof}

\section{Case $s>2$ is even}
\subsection{An Annihilator $P$ of $J_{{{\mbox{\tiny C}}},n}(t)$.}
Recall our definition
\[\delta_j = \frac{t^{2(p+q)(j+1)+2}+t^{-2(p+q)(j+1)+2} -t^{2(q-p)(j+1)-2}-t^{-2(q-p)(j+1)-2}}{t^2-t^{-2}}\]
and further define:
\[V_n = \sum_{k=1}^{s/2}t^{-4pqsnk+2pqsn+4pqk^2-8pqsk+4pqs} \delta_{s(n+2)-1-2k}.\]

\begin{lemma} \label{s/2 peel identity} If $s$ is even, then
\[J_{{{\mbox{\tiny T}}},s(n+2)-1}(t) = t^{-2pqs^2n-3pqs^2+2pqs} J_{{{\mbox{\tiny T}}},s(n+1)-1}(t) + V_n \]
 \end{lemma}
\begin{proof} Apply Lemma \ref{s peel identity}, setting $m = s/2$. \end{proof}

 Lemma \ref{s/2 peel identity} yields  the relation
 $$({{\mbox{\small L}}}-{{\mbox{\small M}}}^{-pqs^2}t^{-3pqs^2+2pqs}) J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)=V_n.$$
 So applying the operator $({{\mbox{\small L}}}-{{\mbox{\small M}}}^{-pqs^2}t^{-3pqs^2+2pqs})$
  to both sides of (\ref{first relation}) gives:
$$\begin{array}{l}({{\mbox{\small L}}}-{{\mbox{\small M}}}^{-pqs^2}t^{-3pqs^2+2pqs})
a^{-1}(t,{{\mbox{\small M}}})({{\mbox{\small L}}}^2-{{\mbox{\small M}}}^{-2rs}t^{-4rs})J_{{{\mbox{\tiny C}}},n}(t)
\\=V_n+({{\mbox{\small L}}}-{{\mbox{\small M}}}^{-pqs^2}t^{-3pqs^2+2pqs})a^{-1}(t,{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-pqs}t^{2r-2rs-2pqs}\delta_{s(n+1)-1}\\
=V_n+a^{-1}(t,t^2{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-pqs}t^{4r-4rs-4pqs}\delta_{s(n+2)-1}
-a^{-1}(t,{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-pqs^2-pqs}t^{2r-2rs-3pqs^2}\delta_{s(n+1)-1}.\end{array}$$
To see this is an inhomogeneous recursion for $J_{{{\mbox{\tiny C}}},n}(t)$, let $b(t,{{\mbox{\small M}}})/(t^2-t^{-2})$ be the right hand side of this equation and check it is nonzero.
As before it suffices to check that $\lim_{t{\rightarrow} -1}b(t,{{\mbox{\small M}}})\ne0$,
considering $V_n$ and ${\delta}_{s(n+k)-j}$'s as functions of $t$ and ${{\mbox{\small M}}}$ (changing $t^{2n}$ to ${{\mbox{\small M}}}$).
We have $$\begin{array}{l}b(-1,{{\mbox{\small M}}})=\lim_{t{\rightarrow}-1}b(t,{{\mbox{\small M}}})\\
=\lim_{t{\rightarrow} -1}(t^2-t^{-2})(V_n+a^{-1}(t,t^2{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-pqs}t^{4r-4rs-4pqs}\delta_{s(n+2)-1}
\\ \qquad -a^{-1}(t,{{\mbox{\small M}}}){{\mbox{\small M}}}^{r-rs-pqs^2-pqs}t^{2r-2rs-3pqs^2}\delta_{s(n+1)-1})\\
=(\sum_{i=1}^{s/2}{{\mbox{\small M}}}^{-2pqsj+pqs}+a^{-1}(-1,{{\mbox{\small M}}})({{\mbox{\small M}}}^{r-rs-pqs}-{{\mbox{\small M}}}^{r-rs-pqs^2-pqs}))\\
\qquad \times ({{\mbox{\small M}}}^{s(p+q)} +{{\mbox{\small M}}}^{-s(p+q)}  - {{\mbox{\small M}}}^{s(q-p)} -{{\mbox{\small M}}}^{-s(q-p)})\\
=\left(\frac{{{\mbox{\small M}}}^{-pqs}-{{\mbox{\small M}}}^{-pqs^2-pqs}}
{1-{{\mbox{\small M}}}^{-2pqs}}+\frac{{{\mbox{\small M}}}^{r-rs-pqs}-{{\mbox{\small M}}}^{r-rs-pqs^2-pqs}}{
{{\mbox{\small M}}}^{r-rs-2pqs}-{{\mbox{\small M}}}^{-r-rs}}\right)
({{\mbox{\small M}}}^{s(p+q)} +{{\mbox{\small M}}}^{-s(p+q)}  - {{\mbox{\small M}}}^{s(q-p)} -{{\mbox{\small M}}}^{-s(q-p)})\\
=\frac{(1-{{\mbox{\small M}}}^{-pqs^2}){{\mbox{\small M}}}^{-rs-pqs}({{\mbox{\small M}}}^r-{{\mbox{\small M}}}^{-r})({{\mbox{\small M}}}^{ps}-{{\mbox{\small M}}}^{-ps})({{\mbox{\small M}}}^{qs}-{{\mbox{\small M}}}^{-qs})}
{(1-{{\mbox{\small M}}}^{-2pqs})({{\mbox{\small M}}}^{r-rs-2pqs}-{{\mbox{\small M}}}^{-r-rs})},\end{array}$$
which is indeed nonzero.
Hence
\begin{align*}
P(t,{{\mbox{\small M}}},{{\mbox{\small L}}}) &= ({{\mbox{\small L}}}-1)b^{-1}(t,{{\mbox{\small M}}})({{\mbox{\small L}}}-{{\mbox{\small M}}}^{-pqs^2}t^{-3pqs^2+2pqs})
a^{-1}(t,{{\mbox{\small M}}})({{\mbox{\small L}}}^2-{{\mbox{\small M}}}^{-2rs}t^{-4rs})
\end{align*}
is an annihilator of $J_{{{\mbox{\tiny C}}},n}(t)$.
Assuming $P$ is of minimal degree in ${{\mbox{\small L}}}$, we can now check the AJ-conjecture by evaluating $P$ at $t = -1$. We have
\[P(-1,{{\mbox{\small M}}},{{\mbox{\small L}}}) =b^{-1}(-1,{{\mbox{\small M}}}) a^{-1}(-1,{{\mbox{\small M}}})({{\mbox{\small L}}}-1)({{\mbox{\small L}}}-{{\mbox{\small M}}}^{-pqs^2})({{\mbox{\small L}}}^2-{{\mbox{\small M}}}^{-2rs}),\]
which agrees with the A-polynomial of $C$,  up to a nonzero factor of a rational function in ${\mathbb Q}({{\mbox{\small M}}})$.

\subsection{$P$ is the Recurrence Polynomial of $C$.}

We now want to show that the operator $P$ is the recurrence polynomial of $C$. It is enough to prove that if  $Q = D_3 L^3 + D_2 L^2 + D_1 L + D_0$ is an element in ${\mathcal A}_{{\mbox{\tiny C}}}$, then $Q = 0$. As in Subsection \ref{subsection s odd and q=2}
we have
\begin{align*}
0 &= D_3 J_{{{\mbox{\tiny C}}},n+3}(t) + D_2 J_{{{\mbox{\tiny C}}},n+2}(t) + D_1 J_{{{\mbox{\tiny C}}},n+1}(t) + D_0 J_{{{\mbox{\tiny C}}},n}(t)\\
&= (D_1  +  D_3{{\mbox{\small M}}}^{-2rs}t^{-8rs}) J_{{{\mbox{\tiny C}}},n+1}(t)+(D_0 + D_2 {{\mbox{\small M}}}^{-2rs}t^{-4rs})J_{{{\mbox{\tiny C}}},n}(t) \\
&\quad+D_3 ({{\mbox{\small M}}}^{r-rs-2pqs}t^{-4rs+4r-8pqs}
-{{\mbox{\small M}}}^{-r-rs}t^{-4rs-4r})J_{{{\mbox{\tiny T}}},s(n+2)-1}(t)\\
&\quad+D_2({{\mbox{\small M}}}^{r-rs-2pqs}t^{-2rs+2r-4pqs}-{{\mbox{\small M}}}^{-r-rs}t^{-2rs-2r})J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)\\
&\quad+ D_2{{\mbox{\small M}}}^{r-rs-pqs}t^{2r-2rs-2pqs} \delta_{s(n+1)-1}+D_3{{\mbox{\small M}}}^{r-rs-pqs}t^{4r-4rs-4pqs}\delta_{s(n+2)-1},
\end{align*}
and applying Lemma \ref{s/2 peel identity},
\begin{align*}
&= (D_1  +  D_3{{\mbox{\small M}}}^{-2rs}t^{-8rs}) J_{{{\mbox{\tiny C}}},n+1}(t)+(D_0 + D_2 {{\mbox{\small M}}}^{-2rs}t^{-4rs})J_{{{\mbox{\tiny C}}},n}(t) \\
&\quad+D_3 ({{\mbox{\small M}}}^{r-rs-2pqs}t^{-4rs+4r-8pqs}
-{{\mbox{\small M}}}^{-r-rs}t^{-4rs-4r})({{\mbox{\small M}}}^{-pqs^2}t^{-3pqs^2+2pqs} J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)+V_n)\\
&\quad+D_2({{\mbox{\small M}}}^{r-rs-2pqs}t^{-2rs+2r-4pqs}-{{\mbox{\small M}}}^{-r-rs}t^{-2rs-2r})J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)\\
&\quad+ D_2{{\mbox{\small M}}}^{r-rs-pqs}t^{2r-2rs-2pqs} \delta_{s(n+1)-1}+D_3{{\mbox{\small M}}}^{r-rs-pqs}t^{4r-4rs-4pqs}\delta_{s(n+2)-1}\\&
= (D_1  +  D_3{{\mbox{\small M}}}^{-2rs}t^{-8rs}) J_{{{\mbox{\tiny C}}},n+1}(t)+(D_0 + D_2 {{\mbox{\small M}}}^{-2rs}t^{-4rs})J_{{{\mbox{\tiny C}}},n}(t) \\
&\quad+\big(D_3 ({{\mbox{\small M}}}^{r-rs-2pqs}t^{-4rs+4r-8pqs}
-{{\mbox{\small M}}}^{-r-rs}t^{-4rs-4r}){{\mbox{\small M}}}^{-pqs^2}t^{-3pqs^2+2pqs}\\&\qquad
+D_2({{\mbox{\small M}}}^{r-rs-2pqs}t^{-2rs+2r-4pqs}-{{\mbox{\small M}}}^{-r-rs}t^{-2rs-2r})\big)J_{{{\mbox{\tiny T}}},s(n+1)-1}(t)\\
&\quad+ D_2{{\mbox{\small M}}}^{r-rs-pqs}t^{2r-2rs-2pqs} \delta_{s(n+1)-1}+D_3({{\mbox{\small M}}}^{r-rs-pqs}t^{4r-4rs-4pqs}\delta_{s(n+2)-1}\\&
\qquad+({{\mbox{\small M}}}^{r-rs-2pqs}t^{-4rs+4r-8pqs}
-{{\mbox{\small M}}}^{-r-rs}t^{-4rs-4r})V_n)\\&
= D_3' J_{{{\mbox{\tiny C}}},n}(t) + D_2' J_{{{\mbox{\tiny C}}},n+1}(t) + D_1' J_{{{\mbox{\tiny T}}},s(n+1)-1}(t) + D_0'
\end{align*}

As in the previous section, we show that  $D_i' = 0$, $i=1,...,3$,
implies $D_i = 0$, $i=0,...,3$.
We just need to show that the two linear equations  defined by
$D_0'=0$ and $D_1'=0$ are linearly independent.
Again we just need check the determinant of the linear system, multiplied by $(t^2-t^{-2})$
and then valued at $t=-1$, is nonzero:
$$\begin{array}{ll}
&\big(({{\mbox{\small M}}}^{r-rs-2pqs}-{{\mbox{\small M}}}^{-r-rs}){{\mbox{\small M}}}^{-pqs^2}{{\mbox{\small M}}}^{r-rs-pqs}
\\&-({{\mbox{\small M}}}^{r-rs-2pqs}-{{\mbox{\small M}}}^{-r-rs})({{\mbox{\small M}}}^{r-rs-pqs} +({{\mbox{\small M}}}^{r-rs-2pqs}-{{\mbox{\small M}}}^{-r-rs})\frac{{{\mbox{\small M}}}^{-pqs}-{{\mbox{\small M}}}^{-pqs^2-pqs}}
{1-{{\mbox{\small M}}}^{-2pqs}})\big)
\\&({{\mbox{\small M}}}^{s(p+q)} +{{\mbox{\small M}}}^{-s(p+q)}  - {{\mbox{\small M}}}^{s(q-p)} -{{\mbox{\small M}}}^{-s(q-p)})\\
=&\frac{({{\mbox{\small M}}}^{r-rs-2pqs}-{{\mbox{\small M}}}^{-r-rs})}{1-{{\mbox{\small M}}}^{-2pqs}}\big({{\mbox{\small M}}}^{r-rs-pqs^2-pqs}(1-{{\mbox{\small M}}}^{-2pqs})
-{{\mbox{\small M}}}^{r-rs-pqs}(1-{{\mbox{\small M}}}^{-2pqs})
\\&-({{\mbox{\small M}}}^{r-rs-2pqs}-{{\mbox{\small M}}}^{-r-rs})({{\mbox{\small M}}}^{-pqs}-{{\mbox{\small M}}}^{-pqs^2-pqs})\big)
({{\mbox{\small M}}}^{s(p+q)} +{{\mbox{\small M}}}^{-s(p+q)}  - {{\mbox{\small M}}}^{s(q-p)} -{{\mbox{\small M}}}^{-s(q-p)})\\
=&({{\mbox{\small M}}}^{r-rs-2pqs}-{{\mbox{\small M}}}^{-r-rs})({{\mbox{\small M}}}^{-pqs^2}-1)({{\mbox{\small M}}}^{r-rs-pqs}-{{\mbox{\small M}}}^{-r-rs-pqs})({{\mbox{\small M}}}^{ps}-{{\mbox{\small M}}}^{-ps})({{\mbox{\small M}}}^{qs}-{{\mbox{\small M}}}^{-qs})
\end{array}$$
which is indeed nonzero.

The following lemma shows that each $D_i' = 0$ if $r$ is not a number between $0$ and $pqs$.

\begin{lemma} (1) When $p > q$ and either $r < 0$ or $r > pqs$, we have $D_i' = 0$ for $i = 0,1,2,3$.

(2) When $p < -q$ and either $r > 0$ or $r < pqs$, we have $D_i' = 0$ for $i = 0,1,2,3$.
\end{lemma}

The proof is similar to that of Lemma \ref{degree argument for s odd}.

\section{Case $s=2$}
\subsection{An Annihilator $P$ of $J_{{{\mbox{\tiny C}}},n}(t)$.}
In this section, we assume that $C$ is  a $(r,2)$-cabled knot over
a torus knot $T=T(p,q)$.
In   $$\begin{array}{ll}\displaystyle J_{{{\mbox{\tiny C}}},n+1}(t) &=
t^{-2r((n+1)^2-1)}\sum_{k=-\frac{n}{2}}^{\frac{n}{2}}
t^{4rk(2k+1)}J_{{{\mbox{\tiny T}}},4k+1}(t),\end{array}$$
let $k=-(j+\frac12)$, then
\begin{equation}\label{s=2 first relation}\begin{array}{ll}\displaystyle J_{{{\mbox{\tiny C}}},n+1}(t) &=
t^{-2r((n+1)^2-1)}\sum_{j=\frac{n-1}{2}}^{-\frac{n+1}{2}}
t^{4r(2j+1)j}J_{{{\mbox{\tiny T}}},-4j-1}(t)\\
 &=-
t^{-2r((n+1)^2-1)}\sum_{j=\frac{n-1}{2}}^{-\frac{n+1}{2}}
t^{4r(2j+1)j}J_{{{\mbox{\tiny T}}},4j+1}(t)\\
&=-t^{-2r((n+1)^2-1)}[t^{2rn(n+1)}J_{{{\mbox{\tiny T}}}, 2n+1}+
\sum_{j=\frac{n-1}{2}}^{-\frac{n-1}{2}}
t^{4r(2j+1)j}J_{{{\mbox{\tiny T}}},4j+1}(t)]
\\
&=-t^{-2r((n+1)^2-1)}[t^{2rn(n+1)}J_{{{\mbox{\tiny T}}}, -2n-1}+
t^{2r(n^2-1)}J_{{{\mbox{\tiny C}}},n}(t)]
\\
&=t^{-2rn}J_{{{\mbox{\tiny T}}}, 2n+1}-
t^{-4rn-2r}J_{{{\mbox{\tiny C}}},n}(t).\end{array}\end{equation}
Turning $t^{2n}$ into ${{\mbox{\small M}}}$ and $J_{{{\mbox{\tiny C}}},n+1}(t)$ into ${{\mbox{\small L}}} J_{{{\mbox{\tiny C}}},n}(t)$, we see that
\[({{\mbox{\small L}}}+{{\mbox{\small M}}}^{-2r}t^{-2r})J_{{{\mbox{\tiny C}}},n}(t) = {{\mbox{\small M}}}^{-r} J_{{{\mbox{\tiny T}}},2n+1}(t),\]
or
\[{{\mbox{\small M}}}^r({{\mbox{\small L}}}+{{\mbox{\small M}}}^{-2r}t^{-2r})J_{{{\mbox{\tiny C}}},n}(t) = J_{{{\mbox{\tiny T}}},2n+1}(t).\]
We now wish to find an inhomogeneous recurrence for $J_{{{\mbox{\tiny T}}},2n+1}(t)$. Recall equation (\ref{relation of J_T(n+2)}):
\[J_{{{\mbox{\tiny T}}},n+2}(t) = t^{-4pq(n+1)} J_{{{\mbox{\tiny T}}},n}(t)+t^{-2pq(n+1)} \delta_n,\]
which implies that
\begin{equation}\label{s=2 second relation}\begin{array}{ll}
J_{{{\mbox{\tiny T}}},2n+3}(t) &= t^{-4pq(2n+2)}J_{{{\mbox{\tiny T}}},2n+1}(t)+t^{-2pq(2n+2)} \delta_{2n+1}\\
&= {{\mbox{\small M}}}^{-4pq}t^{-8pq}J_{{{\mbox{\tiny T}}},2n+1}(t)+{{\mbox{\small M}}}^{-2pq}t^{-4pq} \delta_{2n+1},
\end{array}\end{equation}
and so
\[({{\mbox{\small L}}}-{{\mbox{\small M}}}^{-4pq}t^{-8pq})J_{{{\mbox{\tiny T}}},2n+1}(t) = {{\mbox{\small M}}}^{-2pq}t^{-4pq} \delta_{2n+1}.\]
Letting $b(t,{{\mbox{\small M}}})/(t^2-t^{-2}) = {{\mbox{\small M}}}^{-2pq}t^{-4pq} \delta_{2n+1}$.
Then   $b(t,{{\mbox{\small M}}}) \in {\mathbb Z}[t^{\pm 1},{{\mbox{\small M}}}^{\pm 1}]$, which is obviously non-zero, and  we obtain an operator $P(t,{{\mbox{\small M}}},{{\mbox{\small L}}})$ which annihilates $J_{{{\mbox{\tiny C}}},n}(t)$ given by
\[
P(t,{{\mbox{\small M}}},{{\mbox{\small L}}}) = ({{\mbox{\small L}}}-1)b^{-1}(t,{{\mbox{\small M}}})({{\mbox{\small L}}}-{{\mbox{\small M}}}^{-4pq}t^{-8pq}){{\mbox{\small M}}}^{r}({{\mbox{\small L}}}+{{\mbox{\small M}}}^{-2r}t^{-2r}).
\]

Assuming $P$ has the minimal ${{\mbox{\small L}}}$ degree,
we can check the AJ-conjecture. Evaluating $P(-1,{{\mbox{\small M}}},{{\mbox{\small L}}})$ gives
\[P(-1,{{\mbox{\small M}}},{{\mbox{\small L}}}) =  b^{-1}(-1,{{\mbox{\small M}}}) ({{\mbox{\small L}}}-1)({{\mbox{\small L}}}-{{\mbox{\small M}}}^{-4pq}){{\mbox{\small M}}}^{r}({{\mbox{\small L}}}+{{\mbox{\small M}}}^{-2r}),\]
which is equal to the $A$-polynomial of $C$ up to a nonzero factor in ${\mathbb Q}({{\mbox{\small M}}})$.

\subsection{$P$ is the Recurrence Polynomial of $C$.}

Next we show that the operator $P$ given above is a generator
of the  ideal $\widetilde{\A}_{{\mbox{\tiny C}}}$.
It amounts to show that if an operator $Q=D_2{{\mbox{\small L}}}^2+D_2{{\mbox{\small L}}}+D_0$,
where each $D_j\in {\mathbb Z}[t^{\pm1}, {{\mbox{\small M}}}^{\pm1}]$, is an annihilator
of $J_{{{\mbox{\tiny C}}}, n}(t)$, then $Q=0$.

So suppose that $QJ_{{{\mbox{\tiny C}}},n}(t)=0$, i.e.
\begin{equation}\label{degree at least three}
D_2J_{{{\mbox{\tiny C}}},n+2}(t)+D_1J_{{{\mbox{\tiny C}}},n+1}+D_0J_{{{\mbox{\tiny C}}},n}(t)=0.
\end{equation}
Our goal is to show that $D_i=0$, $i=0,1,2$.

Using (\ref{s=2 first relation}) and (\ref{s=2 second relation}) we can transform  (\ref{degree at least three})
into
$$\label{degree at most 2}
\begin{array}{ll}0&=D_2(t^{-4rn-2r}J_{{{\mbox{\tiny T}}},2n+3}(t)-
t^{-4rn-6r}({{\mbox{\small M}}}^{-r}J_{{{\mbox{\tiny T}}},2n+1}(t)-
t^{-4rn-2r}J_{{{\mbox{\tiny C}}},n}(t)))\\&\quad+D_1(t^{-rn}J_{{{\mbox{\tiny T}}},2n+1}(t)-
t^{-4rn-2r}J_{{{\mbox{\tiny C}}},n}(t))+D_0J_{{{\mbox{\tiny C}}},n}(t)
\\&=(D_2t^{-8rn-8r}-D_1t^{-4rn-2r}+D_0)J_{{{\mbox{\tiny C}}},n}(t)\\&
\quad+D_2t^{-2rn-2r}(t^{-8pqn-8pq}J_{{{\mbox{\tiny T}}},2n+1}(t)+t^{-4pqn-4pq} \delta_{2n+1})
\\&\quad+(-D_2t^{-6rn-6r}+D_1t^{-2rn})J_{{{\mbox{\tiny T}}},2n+1}(t)\\&
=(D_2t^{-8r(n+1)}-D_1t^{-4rn-2r}+D_0)J_{{{\mbox{\tiny C}}},n}(t)\\
&\quad+
(D_2(t^{-2r(n+1)-8pq(n+1)}-t^{-6r(n+1)})
+D_1t^{-2rn})J_{{{\mbox{\tiny T}}}, 2n+1}(t)\\
&\quad+
D_2t^{-2r(n+1)-4pq(n+1)}\delta_{2n+1}
\\&=D_2'J_{{{\mbox{\tiny C}}},n}(t)+D_1'J_{{{\mbox{\tiny T}}},2n+1}(t)+D_0'.
\end{array}$$

If we can show that $D_i'=0$, $i=0,1,2$, then it will follow right away
that  $D_i=0$, $i=0,1,2$.
As  in Lemma \ref{degree argument for s odd}, we can show that $D_i'=0$, $i=0,1,2$, if $r$ is
not an integer between $0$ and $2pq$.

redpajama_set_name: RedPajamaArXiv


Diamond Lake Elementary School is a public elementary school located in Mundelein, IL in the Diamond Lake School District 76. It enrolls 325 students in grades 1st through 12th. Diamond Lake Elementary School is the 1,618th largest public school in Illinois and the 39,346th largest nationally. It has 12.4 students to every teacher.
The form below lets you find Diamond Lake Elementary School alumni info and Diamond Lake Elementary School students.
Out of 2,421 ranked schools in Illinois, Diamond Lake Elementary School is ranked 1,110th for total students on lunch assistance.
The percentage of Diamond Lake Elementary School students on free and reduced lunch assistance (60.3%) is slightly higher than the state average of 52.9%. This may indicate that the area has a higher level of poverty than the state average.


text:
One Land, Five Nations is a documentary (in two parts) by Mahmoud Shoolizadeh. It introduces the cultural and social and geographical life of the people in Faridan in Isfahan Province of Iran. In this region, five different Nations: Armenians, Lors, Georgian, Turkish and Persian are living peacefully together, celebrating their own languages, customs and cultures.
This film participated in The first short films festival in Tehran, Iran, where it became candidate for the best documentary film and best photography in 1990

Technical specifications and Film crew

One Land, Five Nations

16 mm, 45 & 35mins, Documentary in two parts, Iran, 1989
Researcher, Script writer and Director: Mahmoud Shoolizadeh,
Photograph: Mohammad Abedi,
Edit : marziyeh Soleymani,
Music: Saeed Sharifiyan
Producer: Javad Peyhani ( I.R.I.B, Isfahan )

1989 films
Iranian documentary films
1989 documentary films

redpajama_set_name: RedPajamaWikipedia


Buying travel insurance for a trip who's primary destination is a restricted nation can be difficult. Certain countries are subject to sanctions by the U.S. Government thru the Office of Foreign Assets control. These sanctions make it difficult for an insurance company to insure travelers visiting such countries and in some cases not even possible. The legal issues are daunting and many insurers simply are unwilling to or unable to provide coverage.
The list of restricted countries varies by travel insurance company. Some will insure to a broader range of countries on the OFAC list.
American Modern restricts purchase for: Afghanistan, Algeria, Myanmar, Burundi, Central African Republic, Chad, Cuba, Democratic Republic of Congo, Eritrea, Guinea, Haiti, Iran, Iraq, Ivory Coast, Kenya, Lebanon, Liberia, Libya, Mali, Mauritania, Niger, Nigeria, North Korea, Pakistan, Palestinian Authority, Somalia, South Sudan, Sudan, Syria, Tunisia, Yemen, Zimbabwe.
Nationwide restricts purchase for: Albania, Belarus, Bosnia-herzegovina, Myanmar, Central African Republic, Cuba, Democratic Republic of Congo, Iran, Iraq, Ivory Coast, Kosovo, Lebanon, Liberia, Libya, Macedonia, Montenegro, North Korea, Palestinian Authority, Somalia, Sudan, Syria, Yemen, Zimbabwe.
Old Republic restricts purchase for: Albania, Belarus, Bosnia-Herzegovina, Cuba, Democratic Republic of Congo, Iran, Iraq, Ivory Coast, Kosovo, Liberia, Macedonia, Myanmar, North Korea, Palestinian Authority, Somalia, Sudan, Syria, Yemen, Zimbabwe.
Arch and US Fire restrict purchase for: Cuba, Iran, Iraq, North Korea, Palestinian Authority, Sudan, Syria.
What is the US OFAC and what does it do?
The United States Office of Foreign Assets Control administers and enforces economic sanctions programs primarily against countries and groups of individuals, such as terrorists and narcotics traffickers. The sanctions can be either comprehensive or selective, using the blocking of assets and trade restrictions to accomplish foreign policy and national security goals.
Basically the US government has levied some level of economic sanctions against the country, and prohibits US citizens from doing business with them. This includes restricting travel to many of these countries. It's not prudent for any company to ignore the OFAC regulations. The fines for violations can be substantial. Depending on the program, criminal penalties can include fines ranging from $50,000 to $10,000,000 and imprisonment ranging from 10 to 30 years for willful violations. Depending on the program, civil penalties range from $250,000 or twice the amount of each underlying transaction to $1,075,000 for each violation.
Does this mean you could go to jail for visiting one of these restricted countries? Probably not, in most cases, but you should consult with an attorney before traveling to a sanctioned country to be sure. What is clear though, is that our insurance partners will not make payments to a medical facility, travel supplier, or emergency evacuation supplier that is located in an OFAC restricted country. Some of our partners may reimburse a traveler who has already made payments in an OFAC restricted country.
We set up a warning if you select one of the OFAC restricted nations as a primary destination during enrollment that lets you know if we can't sell you the insurance plan you have selected for that trip. This saves people the trouble of filling out the entire form, entering a credit card, and then having the transaction be rejected.


text:
Retailers cashing in on Wii shortages by forcing bundles; Reggie not amused
Contributor	13 years
I was recently at Fry's buying, I don't know, something. They had a stack of hard-to-find Wiis right up front, a great big pallet. I thought about grabbing one, then realized it cost $329.
Knowing that the retail price of a Wii is $250, I asked what was up, and an anxious employee told me the Wiis were only available in bundles that included three additional controllers and nunchucks, a game, and some pretty useless stuff. He said it was over $400 worth of merchandise, to which I said, "I live alone. Just me. I just want a Wii."
And he said I couldn't have one. I thought this to be dishonest and a lousy way to make a buck, and now it seems our favorite corporate goon ever, Reggie Fils-Aime, agrees. While Nintendo doesn't have a way of stopping them — retailers can re-package, or make "kits", with anything they want to — he made it clear that Nintendo does not approve of the practice.
We're for the idea of Nintendo withholding shipments of Wiis to these nefarious villians, and we're hoping they get reallocated to more scrupulous retailers.
Reggie: Retailers have been told we are not fans of forced Wii bundles [MaxCon]


D-line Tabletop Autoclaves are designed for the needs of modern clinics where autoclaves are expected to operate non-stop every workday.
- Door remains closed during the drying stage to maintain sterility and ensure efficient drying (included with EA/EKA models). A high efficiency air pump circulates filtered air through the chamber for fast drying.


Congratulations to our Recipe Contest Winner, Katie!
Out of all the recipes we received, this dish stood out the most. We love the combination of sweetness form the sweet potatoes and carrots, and the subtle smokiness of the Bohea. Makes a perfect side dish or a hearty meal served with brown rice on a chilly day. Thanks Katie! We hope you enjoy your new teapot with flowering teas!
Heat the water and steep the Bohea to make a strong tea. Set aside.
In 3 quart saucepan, heat oil over medium heat. Add onion and cook, stirring until tender (about 5 min). Stir in ginger, garlic, cumin, and ground red pepper. Cook for about 1 minute. Stir in sweet potato, carrot, lentils, broth, salt, and brewed tea.
Heat to boiling on high heat and cook for 10 min. Reduce heat, cover and simmer, stirring, until lentils are just tender (20-25 min).


Are you looking for the map of Fredrikstad? Find any address on the map of Fredrikstad or calculate your itinerary to and from Fredrikstad, find all the tourist attractions and Michelin Guide restaurants in Fredrikstad. The ViaMichelin map of Fredrikstad: get the famous Michelin maps, the result of more than a century of mapping experience.


text:
Q: xcode iOS check url prefix every time I am at start in Xcode , i need a short script which verifies my url prefix
in php it would be something like this
<?php
    $prefix1 = "loc:";
    if (Webview == $prefix1)
    {
        echo "Url Prefix is loc:";
    }
?>

A: NSURL *url = ...;
if ([[url scheme] isEqualToString:@"loc"]) {
    NSLog(@"Url Prefix is loc:");
}

A: There is a function available in NSString class - hasPrefix method
if ([myString hasPrefix:@"http"])

The other way is available in NSURL class - Scheme property
NSURL* url = [NSURL URLWithString:@"http://www.google.com"];

Then
url.scheme means, http

Hope this helps.

A: If you are 100% sure you have a valid URL, use
NSURL* myURL = [NSURL URLWithString: @"http://example.com"];
if( [myURL.scheme isEqualToString: @"http"] )
{
    // do something here.
}

If this URL is entered by the user, use NSURLComponents instead, which will cope better with partial or invalid URLs, and give less surprising results with that kind of data.

redpajama_set_name: RedPajamaStackExchange


text:
Compact Series
SCH Series
Find a Charger
Commercial Charging
Home Charging
Fleet Charging
EV Charge Times
Why Sun Country
WLGHP
Ontario Program
BC Incentive Program
The Longest Greenest Highway Project
The Sun Country Highway team set out on November 17th, 2012 from St. John's, NL to demonstrate the longest network of free EV Charging stations in the world.
The 10,125 km project was completed in 35 days upon arrival in Victoria, BC while commemorating the 50th Anniversary of the opening of the Trans Canada Highway and the 100th Anniversary of Thomas Wilby's 1st Motor Car Tour Across Canada. The SCH Trans Canada Highway network, consists of over 80 free public access charging stations that are located at restaurants, hotels, tourist destinations, municipalities and various other sustainability focused businesses across Canada.
The SCH team chose an all electric Tesla Roadster due to showcase EV technology and organized media interviews at each charging station.
This amazing project which was completed without the use of tax payer money to promote electric vehicles as a sustainable means of transportation and, as a key enabler of economic development opportunity.
Longest Greenest Highway Project Highlights
https://www.youtube.com/watch?v=SR9qP6lfWPc
Experience the highlights of the Sun Country Highway team on the Longest Greenest Highway Project. From coast to coast through rain, sleet and snow – our all EV fleet drove over 3000km to promote the EV movement in Canada.
https://www.youtube.com/watch?v=RaAAfDCcHDM
CHCH Sun Country
Highway Interview
Follow Sun Country
Learn more about our latest adventures
Bringing Sustainability to Life
Empowering individuals to lead global change by adopting models of economic, social and environmental sustainability.
Sun Country Highway © 2017
Sign up for updates from Sun Country
Contact Us Home Charging
Business Charging


text:
Mary Landrieu's 'Peak Education' Economic Illiteracy
January 30th, 2012 MacAoidh
We could do another post on the idiocy of the state's teacher unions in their opposition to Gov. Bobby Jindal's education reform package, and we could do something on Jindal's counterbarrage calling for Michael Walker-Jones, executive director of the Louisiana Association of Educators, to resign after the latter suggested poor people aren't capable of picking schools for their kids.
But when we found this, it just jumped out as blog material.
U.S. Sen. Mary Landrieu sought to pour some cold water on one of the central proposals in Gov. Bobby Jindal's education reform agenda Monday, using state data to show a huge gap between the number of students who would technically qualify for the governor's proposed private school voucher program and the number of seats that may actually exist in the state's private schools.Under Jindal's plan, about 380,000 students would qualify to receive state aid for tuition at a private or religious school, Landrieu pointed out.
But even if every private school in the state could immediately grow their enrollment by 10 percent to accommodate an influx of voucher recipients, there would only be about 8,000 seats available. She based her estimate on figures from the state Department of Education that show about 80,000 private school students in Louisiana overall.
Vouchers, Landrieu said, "cannot be the centerpiece of our reforms based on the reality of these numbers."
Landrieu, it should be pointed out, isn't actually all that bad on education. She's more or less a fan of charter schools, for example, and she's been supportive of Jindal's call for urgency in school reform.
But the voucher criticism as expressed today is just painfully stupid.
Does Landrieu really believe that when the state issues school vouchers covering the cost, essentially, of public school – which is somewhere between $10,500 and $13,000 per student per year, depending on your source – that there won't be an increase in the supply of spots in private schools? Particularly when those vouchers are available to some 380,000 students?
The point is to create a market for education. The point is to incentivize entrepreneurs in the educational field to get into that market and try all kinds of things which might well work better than the crappy public schools those 380,000 kids are stuck in. And the point is that the parents of those kids in those crappy schools, rather than politicians, would get to make the decision where those kids and the tuition dollars which follow them go – once the politicians have proven themselves to be failures in operating public schools.
Because that's where the voucher program kicks in. It applies to kids in schools which score C, D or F, and that's a bit more than half of the state's 700,000 K-12 kids. If your school stinks, Jindal wants to give you a chance to try a private school.
In all likelihood that doesn't mean Newman, St. Martin's, Country Day or Episcopal. It means Catholic school, since Catholic schools are a lot cheaper per student than public schools are.
But when there's no more room at the current private-school inn, guess what happens?
Somebody comes along and builds more private schools. That's called the marketplace.
Let's say they build an interstate exit out in the sticks. Invariably, somebody will build a gas station there. And inevitably, that gas station will get pretty busy pretty quickly. Maybe even so busy that folks will sometimes be stuck lining up for gas there.
Guess what happens in no time…
Another gas station.
This plays out the same way with fast food restaurants, with bars, with theme parks (feel free to count how many they've got in Orlando now since Disney built the first one there) and lots of other kinds of business establishments.
Does anybody really think the marketplace will operate differently somehow where schools are involved?
Of course not. The minute Jindal's voucher program goes statewide America's entrepreneurial class will descend on Louisiana armed with every possible business model in an effort to land those 380,000 kids and their parents as clients. Because doing it better, cheaper means the potential for profit and success in any business.
The majority of the charter schools currently in Louisiana are non-profits, and charters go through the government – whether local school boards, or, increasingly in the current political landscape, the state Board of Elementary and Secondary Education. And that responsible government agency can pull a charter if they don't think the methods or results are worthy of taxpayer dollars.
Which is a perfectly fine idea and in New Orleans it works like a charm. After Hurricane Katrina the school system was largely dissolved and better than 70 percent of the public school kids in the city now attend charters. Now, after six years of building a charter school-based market there, two-thirds of the folks agree that the schools are better than they were before the storm.
But vouchers go even further than that.
Because with vouchers, there are no politicians around to pull the plug and there is no board to grant a charter, or not. With vouchers the final arbiter is the consumer. Nobody wants to go to a bad school.
The voucher program in New Orleans, limited though it might be, is even more popular than the charter schools are. This came out on Thursday…
The Louisiana Federation for Children today praised newly-released results of a recent parental satisfaction survey showing overwhelmingly high satisfaction rates from parents with children participating in the New Orleans voucher program.
The survey, conducted by direct mail in December 2011 by the Black Alliance for Educational Options (BAEO), reported an overwhelming satisfaction rate – 93.4 percent – among parents with children in the Student Scholarships for Educational Excellence program, which allows qualifying families to send their children to participating private schools of their choice. The percentage includes those parents who reported being "satisfied" or "very satisfied" with their child's school.
Why are they satisfied? Because they directly picked their school. It's not rocket science; it's the market at work.
There is no better way to insure quality of service than competition among suppliers of that service. The voucher program makes a growth industry out of education in Louisiana thanks to the creation of a sizable market from what is essentially thin air.
It's pretty simple for most of us. Open up a significant demand, and you'll see a significant increase in supply. Basic economics.
But Mary Landrieu has practically zero private-sector experience. She got elected to the state legislature in 1980, when she was 25, and she's been an elected official since. So it's no surprise that Mary Landrieu can't envision education becoming a service industry like insurance or medicine or IT or staffing or food service; she has zero frame of reference to see something like that.
And because she has zero frame of reference, she offers up something akin to the Peak Oil prophecies of the 1960's and 1970's applied to education. We'll never have enough private school spots to cover the kids who want them, so a voucher plan won't work. We'll never have them because there's a finite amount of classroom space in Louisiana. We just can't kill dinosaurs fast enough to train private-school teachers or build schoolhouses, you see. Or something like that.
This is stupidity. It's breathtakingly vapid to come out of the mouth of a U.S. Senator. But it's what we have, and it's emblematic of why our schools are what they are.
Interested in more news from Louisiana? We've got you covered! See More Louisiana News
Previous Article Retirement Reform Is Risky Business
Next Article Video From The Louisiana Democratic Party's Meandering Misery Tour


Transcript: Hey guys, it's Ben Lai from Sales Ethos. Today is part one of three on the topic "What to look for when recruiting sales people." Following on from my interview with Tommy Sim the other day, I thought I'd provide my different perspective on what I would look for in the ideal candidate.
Look, there are lots of different things that are going to contribute towards the success of a salesperson. But if you can identify just a few of the top traits, then you're going to go a long way in terms of the hiring process.
So the first thing that I would look for in hiring a salesperson is the willingness to learn. The willingness to learn is an important trait because it demonstrates that they are humble… That they can accept constructive feedback… That they're constantly looking for ways to improve themselves. As the saying goes, you hire for attitude, and you train for skills. A lack of experience, or a lack of skills can be made up for very easily. But to get a person to change their attitude is incredibly difficult.
So a willingness to learn is the best way and the easiest way to make sure you're covered in this area. As a result of having this trait, they're going to be extremely easy to coach. It's going to be easy to give them feedback, and you know that the potential on their performance is going to be unlimited.
Now, how do you go about looking for and finding this trait in people? The kinds of questions that I would ask in the interview would be "What was the last book that you read?" If they can't answer that, you know you're in a bit of trouble there. The other questions that you can ask or you can look on their CV is to look at what courses they have attended. What self-development activities have they done?
If they have no evidence of that, then you know you're in trouble. But if on the other hand, you know they've read several books, and they can describe them in detail to you, then you know that they have this important trait of being willing to learn.
So I'd love to hear your comments. That's just my first thought on what I would look for in an ideal sales candidate. But what are the traits that you would look for? What is the top and the most important one? Leave a comment below. Don't forget to like and subscribe, and remember that integrity plus skills equals success.
Sales Ethos helps companies increase revenue through sales training in Melbourne. We achieve results through sales coaching, customer service training, sales management training, and sales process consulting. Ben Lai is a sales trainer and business coach in Melbourne, and offers public sales training courses in Melbourne.
This entry was posted in Sales Management Training. Bookmark the permalink.
Or, subscribe for free weekly sales training videos!
Opt out any time. Your information will never be shared or sold to a 3rd party.