$N$-tuples $(x_1,\dots,x_N)$ and $(K_1,\dots,K_N)$ (thus deviating from the
convention in the rest of the paper where boldface denotes matrices) and
$\bx^\bK/\bK!$ for the divided power $\prod x_j^{K_j}/K_j!\,$. To prove the theorem,
we have to bound both the denominators of the numerical coefficient~$c(T)$ and the
further denominators coming from the Gaussian averaging

We begin with the latter question.  For this we recall first that the Gaussian
average $\langle f\rangle_Q$ is given by $e^{\D_Q}(f)\bigl|_{x=0}$ for any power
series~$f$, where $\D_Q$ is the Laplacian associated to~$Q$, and hence is equal to
$\D_Q^\ell(f)/\ell!$ if $f$ is a homogeneous polynomial of degree $2\ell$. (For
polynomials of odd degree it of course vanishes trivially.)  The Laplacian $\D_Q$
is a quadratic polynomial in the derivatives $\pt_i=\pt/\pt x_i$, and we can enlarge
the ring~$R$ by adjoining to it the coefficients of this polynomial, so since the
image of any divided factorial under any product $\pt_1^{\ell_1}\cdots\pt_N^{\ell_N}$
is an integer, we then certainly have that the Gaussian integral
$\langle\bx^\bK/\bK!\,\rangle$ is $1/\ell!$ times an element of~$R$. Unfortunately
it turns out that this estimate is not good enough for our purposes, and we have
to work a little harder.

Writing $\D_Q$ as an $R$-linear combination of binomials $\pt_i\pt_j$ with
$1\le i\le j\le N$, and applying the multinomial theorem, we see that
$\D_Q^\ell/\ell!$ is an $R$-linear combination of terms
$\prod_{i\le j}(\pt_i\pt_j)^{m_{ij}}/m_{ij}!$ with $m_{ij}\in\Z_{\ge0}$. Define an even
symmetric $N\times N$ matrix $M$ by setting $M_{ij}=M_{ji}=m_{ij}$ for $i<j$ and
$M_{ii}=2 m_{ii}$. Then $\prod_{i\le j}(\pt_i\pt_j)^{m_{ij}}=\prod_j\pt_j^{K_j}$
with $\bK=M\mb1$, where $\mb1$ is the vector consisting of~$N$~1's, and this
sends $\bx^\bK/\bK!$ to~1 and all other monomials to~0.
It follows that a universal denominator of $\langle\bx^\bK/\bK!\,\rangle$ is
the number
\D(\bK) \;:=\; \text{l.c.m.}\Biggl\{ \prod_{1\le i\le N}(M_{ii}/2)!
\prod_{1\le i<j\le N}M_{ij}!
\;\,\Big|\;\text{$M=M^t\in M_{N,N}(\Z_{\ge0})$ even, $M\mb1=\bK$}\Biggr\}\;.
Notice that this does divide~$\ell!$, because $\ell$ is the sum of the diagonal
$M_{ii}/2$ and of the $M_{ij}$ with $i<j$, so that this statement refines the bound
given above, but $\D(\bK)$ is in general much smaller than~$\ell!$ and this
improvement will be needed for the proof. A usually sharper multiplicative upper
bound for $\D(\bK)$ is the largest integer~$S$ whose square divides the product of
the~$K_j!$ (the proof of this also uses only the integrality of multinomial
coefficients), and then of course $\D(\bK)$ also divides the g.c.d.~of $\ell!$ and
$S$, which in general is smaller than either one. (For instance, for $\bK=(6,9,9,10)$
we have $\ell!=355687428096000$, $S=3135283200$, and $(\ell!,S)=S/3$.)
Either of these two latter upper bounds would be sufficient for our proof, but in
fact there is an easy upper bound that is stronger than either one of them and is
extremely sharp (in particular, it
is equal to $\D(\bK)$ for all $\bK$ with $N\le4$ and $\max(K_j)\le 30$), namely
\D^*(\bK) \;:=\; \prod_{\text{$p$ prime}} p^{\delta_p(\K)} \qquad \text{with} \qquad
\delta_p(K_1,\dots,K_N) \;:=\; \sum_{s\ge1}
\biggl[\frac12\sum_{j=1}^N\biggl[\frac{K_j}{p^s}\biggr]\biggr]\;.
To show that $\D(\bK)$ divides~$\D^*(\bK)$, we need
$\sum_iV_p(M_{ii}/2)+\sum_{i<j}V_p(M_{ij})\le\delta_p(\bK)$ for every prime~$p$ and
every even symmetric matrix~$M$ with non-negative entries and row sums~$\bK$, where
$V_p(m):=v_p(m!)$ for any~$m\ge0$ denotes the largest power of~$p$ dividing~$m!$.
In view of the standard formula $V_p(m)=\sum_{s\ge1}[m/p^s]$, it suffices for this
to show that $\sum_i[M_{ii}/2q]+\sum_{i<j}[M_{ij}/q]\le\frac12\sum_j[K_j/q]$
for every prime power~$q$, and this follows immediately from the obvious facts
$[x/2q]=[[x/2]/q]$ and $[x]+[y]+\cdots \le[x+y+\cdots]$ valid for arbitrary real
numbers $x,\,y,\dots$.

Now going back to our main problem, we now see that it suffices to show that the
product $\D^*(\bK)\,c(T)$ has denominator dividing~$D_n$ for all terms~$T$ as above,
with $\bK=(K_1,\dots,K_N)$ and~$n$ defined by~\eqref{nAndK}. We will prove this
one prime at a time (ignoring the prime~2), which is natural in view of the fact
that the upper bound~$\D^*(\bK)$ is defined by its prime power decomposition. To do
this, we will split both the term~$T$ and the corresponding numerical
coefficient~$c(T)$, and also each of the $N$ factors $T_j$ and $c(T_j)$ of which
they are comprised, as the product of four factors in a way depending on the
prime~$p$ being studied, labelled ``$s$" (terms with $r=0$ and $k$ smaller than~$p$),
``$b$" (terms with $r=0$ and $k$ bigger than or equal to~$p$),
``1"~(terms with $r=1$), and ``$\ge2$" (terms with $r\ge2$), with a similar splitting
of the individual weights~$K_j$ into the sum of four pieces
$K_j^{(s)}=\sum_{3\le k<p}\l_j(0,k)\,k$, $K_j^{(b)}=\sum_{k\ge p}\l_j(0,k)\,k$, $
K_j^{(1)}=\sum_{k\ge1}\l_j(1,k)\,k$, and $K_j^{(\ge2)}=\sum_{r\ge2,\,k\ge0}\l_j(r,k)\,k$.
We also decompose the number~$n$ (the expoent of~$h$ in~$T$) in~\eqref{nAndK} as
$\ell+n'-t$ with
\ell\,:=\,\frac12\,\sum_{j=1}^NK_j\,,   \qquad
n'\,:=\sum_{\substack{1\le j\le N \\ r\ge2,\;k\ge0\substack}}\l_j(r,k)\,(r-1)\,, \qquad
t\,:=\sum_{\substack{1\le j\le N \\k\ge3\substack}}\l_j(0,k)\,.
and also split $t$ as $t^{(s)}+t^{(b)}$ according as $3\le k\le p-1$ or $k\ge p$ in
the summation, with each of $t^{(s)}$ and $t^{(b)}$ splitting into the sum over
$1\le j\le N$ of pieces $t_j^{(s)}$ and $t_j^{(b)}$ in the obvious way.

The numerical coefficient $c(T)$ can be decomposed as
\be
\label{cTdecomp}
c(T) \= \prod_{1\le j\le N}\frac{K_j!}{P_j(T)}\;\cdot\;
\prod_{\substack{1\le j\le N \\ r\ge2,\;k\ge0\substack}}
\frac1{\l_j(r,k)!}\,\biggl(\frac{B_r}{r!}\biggr)^{\l_j(r,k)}
\ee
P_j(T) \= \prod_{0\le r\le1,\,k\ge1}\l_j(r,k)!\,k!^{\l_j(r,k)} \;\cdot\;
\prod_{r\ge2,\,k\ge0}k!^{\l_j(r,k)}\,,
which we can split up further as $P_j^{(s)}(T)P_j^{(b)}(T)P_j^{(1)}(T)P_j^{(\ge2)}(T)$.
The reason that we have included the factor $\l_j(r,k)$ into the definition of
$P_j(T)$ for $0\le r\le1$ but not for~$r\ge2$ is that $\l!k!^\l$ divides $(k\l)!$
for all $\l\ge0$ if $k$ is strictly positive but not if $k=0$, and the terms with
$r\ge2$ can have~$k=0$.   Then the product $P_j^{(b)}(T)P_j^{(1)}(T)P_j^{(\ge2)}(T)$
divides $(K_j-K_j^{(s)})!$, while the first factor $P_j^{(s)}(T)$ divides $t_j^{(s)}!$
up to a $p$-adic unit because here $k$ is always less than~$p$ and therefore $k!$
is not divisible by~$p$. (Here we have made repeated use of the integrality of
multinomial coefficients.) On the other hand, by Lemma~\ref{LemmaB} below
and the submultiplicativity of $D_n$, the second factor in~\eqref{cTdecomp} has
denominator dividing~$D_{n'}$. Putting this all together, we deduce that $c(T)$
is $G(T)/D_{n'}$ times a $p$-adic integer for every~$p$ (always different from~2
and not dividing the denominators of the elements of~$R$), where
$G(T)=\prod_{j=1}^N\bigl(K_j!/t_j^{(s)}!\,(K_j-K_j^{(s)})!\bigr)$.  Using the
submultiplicativity of $D_n$ again, this reduces the problem to showing that
$\delta_p(\bK)\le v_p\bigl(D_{\ell-t}\,G(T)\bigr)$ for each~$p$, and in view of
the definitions of $\delta_p(\bK)$ and $D_n$ and of the above-mentioned formula
$V_p(m)=\sum_{s\ge1}[m/p^s]$ for the $p$-adic valuation of factorials, this in turn
will follow if we can show that
\be
\label{Final}
\biggl[\frac12\sum_{j=1}^N\biggl[\frac{K_j}q\biggr]\biggr] \quad\le\quad
\biggl[\frac{\ell-t}{q^*}\biggr]
\+  \sum_{j=1}^N \biggl(\biggl[\frac{K_j}q\biggr] \m \biggl[\frac{t_j^{(s)}}q\biggr]
\m \biggl[\frac{K_j-K_j^{(s)}}q\biggr] \biggr)
\ee
for each prime power $q=p^s$ with $s\ge1$, where $q^*:=p^{s-1}(p-2)$.  % =(1-2/p)\,q$.

For this final step, we first note that
\frac{\ell-t}{q^*} \= \sum_{j=1}^N
\frac{K_j^{(s)}+K_j^{(b)}+K_j^{(1)}+K_j^{(\ge2)}-2t_j^{(s)}-2t_j^{(b)}}{2q^*}
\;\ge\; \sum_{j=1}^N\frac{K_j-2t_j^{(s)}}{2q}
since $(K_j^{(b)}-2t_j^{(b)})/q^*\ge (1-2/p)\,K_j^{(b)}/q^*=K_j^{(b)}/q$
\,(because $k\ge p$ in the terms defining $K_j^{(b)}$ and $t_j^{(b)}$) and~$q^*<q$.
Using that $[\frac x{2q}]=[\frac12[\frac xq]]$ we deduce that
\biggl[\frac{\ell-t}{q^*}\biggr] \quad\ge\quad
\biggl[\frac12\sum_{j=1}^N\biggl[\frac{K_j-2t_j^{(s)}}{q}\biggr]\biggr]
and hence (since $x\le y$ certainly implies $[x/2]\le[y/2]$) the
inequality~\eqref{Final} will follow if we have the inequality
\biggl[\frac{K_j}q\biggr] \quad\le\quad \biggl[\frac{K_j-2t_j^{(s)}}{q}\biggr]
\+  2\, \biggl(\biggl[\frac{K_j}q\biggr] \m \biggl[\frac{t_j^{(s)}}q\biggr]
\m \biggl[\frac{K_j-K_j^{(s)}}q\biggr] \biggr)
for every $1\le j\le N$. But this inequality is trivial, since
$K_j-K_j^{(s)}\le K_j-3t_j^{(s)}\le K_j-2t_j^{(s)}$ (because every $k$ in the
definition of $K_j^{(s)}$ is $\ge3$) and
+2\Bigl[\frac{t_j^{(s)}}q\Bigr]$. This completes the proof of
Theorem~\ref{thm.denom} modulo that of the following lemma.
\end{proof}

\begin{lemma}
\label{LemmaB}
For any integers $r\ge2$ and $\l\ge0$ we have
\be
\label{BernEstimate}
\frac1{\l!}\, \Bigl(\frac{B_r}{r!}\Bigr)^\l \;\in\;\frac1{D_{\l(r-1)}}\,\Z\,.
\ee
\end{lemma}

\begin{proof}
We prove this one prime at at time. By well-known results of
von Staudt and Clausen, the Bernoulli number $B_r$ ($r>0$) has $p$-adic valuation
$-1$ if $(p-1)|r$ and $B_r/r$ is $p$-integral if $p-1$ does not divide~$r$. From this
we deduce that the $p$-adic valuation of the denominator of $B_r/r!$ is bounded
above by $\bigl[\frac r{p-1}\bigr]$, which is $\le\bigl[\frac {r-1}{p-2}\bigr]$
since $\frac{r-1}{p-2}\ge\frac r{p-1}$ if $r\ge p-1$ and both expressions
vanish otherwise. The $p$-adic valuation of the denominator of
$\frac1{\l!}\bigl(\frac{B_r}{r!}\bigr)^\l$ is therefore bounded above by
$V_p(\l)+\l\bigl[\frac{r-1}{p-2}\bigr]$.  On the other hand, from the definition
of~$D_n$ we have $v_p(D_n)=m+V_p(m)=V_p(pm)$, where $m=\bigl[\frac n{p-2}\bigr]\,$.
We must therefore show that
V_p(\l)\+\l\,\biggl[\frac{r-1}{p-2}\biggr] \quad\le\quad
for all $\l\ge0$ and $r\ge2$. For this we make a case distinction: if $\,2\le r<p-1$
\text{LHS} \= V_p(\l) \= V_p\biggl(p\,\biggl[\frac\l p\biggr]\biggr) \;\le\;
V_p\biggl(p\,\biggl[\frac\l{p-2}\biggr]\biggr) \;\le\; \text{RHS}\,,
while if $r\ge p-1$ then we set $h=\bigl[\frac{r-1}{p-2}\bigr]\ge1$ and have
\text{LHS} \= V_p(\l)\+\l h \;\le\; \l h\+V_p(\l h) \= V_p(p\l h) \;\le\; \text{RHS}
because $[\l x]\ge \l\,[x]$ for any positive real number~$x$.
\end{proof}

\smallskip

We end this subsection with several further observations concerning the denominators
and integrality properties of the coefficients of our divergent power series. The
first is that the bound in Theorem~\ref{thm.denom} is not only sharp in the strong
sense that it is best possible for \emph{every} integer~$n\ge0$ and not merely that
it is attained for some~$n$, but that this optimality is reached in two very
different extreme ways: in the above lemma if $r=p-1$ and $\l\ge0$ is arbitrary
(in which case both sides of~\eqref{BernEstimate} have the same value~$V_p(p\l)$)
and again in the calculation~\eqref{Final} in the case when only $K_j^{(b)}$ occurs
and all $k_i$ are~equal to $p$, so that $K_j^{(b)}$ is exactly $p\,t_j^{(b)}$ (in
other words whenever the dominating contribution in~\eqref{term} comes from the
terms with $(r,k)=(p-1,0)$ or~$(0,p)$). The fact that two completely different
mechanisms lead to the same function $n\mapsto D_n$ suggests that this function
may be a more fundamental one than appears at first sight and may have a broader
domain of applicability.

The second observation is that the universal denominator statement given by
Theorem~\ref{thm.denom} can be sharpened by considering the series at the logarithmic
level, or equivalently, by studying the denominators of the contributions from
connected rather than from all Feynman diagram. This was motivated by the observation
of Peter Scholze that the logarithm of the series $\Phi_0^{(4_1)}(h)$ in~\eqref{as41},
which we had calculated up to order $\,\O(h^{150})$, had coefficients with smaller
denominators than those of the series itself.  Specifically, he found experimentally
that the first occurrence of $p^k$ for small primes $p$~($\ne2,\,3$) and $k\ge1$ in
$\log\Phi_0^{(4_1)}(h)$ occurred for the coefficient of $h^n$ with $n=k(p-1)-1$
rather than $n=k(p-2)$ as for the power series $\Phi_0^{(4_1)}(h)$ itself.
At first sight this statement seems to contradict the intuition mentioned at the
beginning of the section that the arithmetic of the series~$\Phi_\a(h)$ is much
simpler if one does not take their logarithms. But in fact both statements are true!
The reason is that in general $\Phi_\a(h)$ is a linear combination of finitely
many power series correspoding to the stationary points of the function being
integrated (specifically, there are~$c^M$ such series, where $c$ is the denominator
of~$\a$ and $M$ can be taken to be the number of tetrahedra in a triangulation of
the knot complement), and it is not reasonable to take logarithms of sums. But
for~$\a=0$ there is only one summand, so here it is reasonable to take the logarithm,
and for general~$\a$ the logarithm of each of the finitely many summands of
$\Phi_a(h)$ coming from the contribution to the state integral near an individual
stationary point is indeed simpler than this summand itself, because it corresponds
to a sum over only connected rather than over all Feynman diagrams. The final
statement is given in the following theorem.

\begin{theorem}
\label{thm.denomc}
For each integer $n\ge1$ define
\be
\label{Dnc}
D_n\conn \=  \prod_{\text{$p$  prime}} \,p^{[(n+1)/(p-1)]} \,\;.
\ee
Then the coefficient of $h^n$ in $\log\Phi_0^{(K,\s)}(h)$ for any knot $K$ and
any $\s\in\calP_K$ has denominator dividing $D_n\conn$ apart from a finite set
of primes depending only on $K$ and~$\s$. More generally, for any $\a\in\Q$ we
have $\Phi_\a^{(K,\s)}(h)\in R\otimes\exp\bigl(\sum_{n\ge1} R\,h^n/D_n\conn\bigr)
\subset R_D[[h]]$.
\end{theorem}

The first few values of $D_n\conn$ are given by
%% see: knots-perturbative-series-modularity/notes/note_universal_denominator.txt
\be
\label{Dnconnvalues}
\begin{aligned}
& \quad 2, 12, 24, 720, 1440, 60480, 120960, 3628800, 7257600, 479001600, \
958003200, \\ & \quad 2615348736000, 5230697472000, \dots
\end{aligned}
\ee
This sequence too appears in the the online-encyclopedia of integer
sequences~[66] under the name~\texttt{A091137} and with the formula given
above, and coincides with the denominator of the Todd polynomials given in
Lemma 1.5.2 and 1.7.3 of Hirzebruch's book~[51]
without proof and quoted from the~[2] paper.

The proof of the above theorem (which actually implies Theorem~\ref{thm.denom})
is similar to the proof of Theorem~\ref{thm.denom} and is ommitted.

\noindent
Note that the ``connected denominators" $D_n\conn$ are considerably smaller than
the ``additive denominators"~$D_n\,$: $\,D_n/n!$ is an integer growing exponentially
like $(44.621\cdots+\text o(1))^n$, as already mentioned, while $D_n\conn/(n+2)!$
is an integer of subexponential growth.

The final remarks concern the relation of the above results with the known integrality
properties of elements of the Habiro ring.  The proof of Theorem~\ref{thm.denom} as
given above only works for the power series $\Phi^{(\s)}_\a$ with $\s\ne\s_0$, because
the perturbative expansion does not apply to the case~$\s=\s_0$.  However, as we
know, this remaining case is actually simpler because it belongs to the Habiro ring
and therefore satisfies $\Phi_0^{(\s_0)}(h)\in\Z[[e^h-1]]$, and more generally
$\Phi_\a^{(\s_0)}(h)\in\Z(\e(\a))[[\e(\a)e^{-h}-1]]$. The corresponding property no
longer holds for $\s\ne\s_0$, even for $\a=0$ and the figure~8 knot, and in some sense
should not even be expected, because the ``natural" invariant here is the completed
power series $\Phih_0^{(4_1)}$, which contains a transcendental factor $e^{\V(4_1)/h}$.
However, if we consider the product
$\sqrt{3}\Phi(h)\Phi(-h)=-\sqrt{-3}\Phi^{(1)}(h)\Phi^{2)}((h)$, which would be
unchanged if we replaced the power series by their completions, then we \emph{do}
find experimentally that it belongs to the ring~$\Z[1/3][[e^{-h}-1]]$, i.e.~after the
change of variables from $q=e^{-h}$ to $q=1+x$ it becomes a 3-integral power series
in~$x$. We expect, and have checked numerically, that the same should be true for
$5_2$ if one multiplies all three series $\Phi_0^{(5_2,\s_i)}$ ($i=1,2,3$), and for
$(-2,3,7)$ for both products of three series corresponding to the two number fields
$\Q(\xi$ and $\Q(\eta)$ corresponding to this knot. These properties are explained
by the properties of Habiro rings for general number fields as being developed

Finally, we found experimentally that we can obtain a power series that is already
integral (away from 2 and~3) in~$e^{-h}-1$ from $\Phi(h)=\Phi^{(4_1,\s_1)}(h)$
\emph{without} multiplying it by $\Phi(-h)=-i\Phi^{(4_1,\s_2)}(h)$ if we multiply
instead by
$\mathcal E^{(4_1,\s_1)}(h)
:=\exp\bigl(-\sum_{r\ge1}^\infty \frac{|B_{r+1}|}{(r+1)!}C_rh^r\bigr)$ with $C_r$ as
in Section~\ref{sub.proof41.0}.  (Notice that this implies the previous statement
becomes $\mathcal E(h)\mathcal E(-h)=1$.) We expect that there will be a similar
correction factor $\mathcal E^{(K,\s)}(h)$ for any $\Phi_\a^{(K,\s)}(h)$ and that
the corrected $\Phi$-series can be seen as the $h$-deformed versions of the units
constructed in~[10], and hope to study this too in~[36].

\section{Numerical aspects}
\label{sec.compKash}

In this section we describe how the power series whose existence is predicted
by the modularity conjecture can be computed effectively via a
numerical computation of the Kashaev invariant, extrapolation, and
recognition of algebraic numbers in a known number field.  We also
describe other methods that are applicable to the power series $\Phi_\a^{(K,\s)}(h)$
for $\s$ different from~$\s_1$, as well as the smooth truncation
methods of ``evaluating" factorially divergent power series at non-zero arguments
that were discussed in~\ref{sub.4.3}.
%  to high precision.  using among other things
The actual numerical data for several knots will then be presented in the appendix.

\subsection{Computing the power series $\Phi_\a^{(K,\s)}(h)$}
\label{sub.comp2}
In this subsection we explain the various methods that can be used to compute the
coefficients of the power series $\Phi_\a^{(K,\s)}(h)$ for all $\a\in\Q$ and
$\s\in\calP_K$ numerically and then exactly as algebraic numbers.

\smallskip\noindent{\bf Step 1: Computing the colored Jones polynomials.}
To compute the Kashaev invariant $\la K \ra_N$ of a knot~$K$, we use
the Murakami--Murakami formula $\la K \ra_N=J_N(e^{2 \pi i/N})$, where $J_n(q)=J_{K,n}(q)$
is the $n$-th colored Jones polynomial, together with a theorem of T.T.Q.~L\^ e and the
first author~[34] that asserts the existence of a recursion relation for
the polynomials $J_n(q)$. This reduces the problem to that of
computing/guessing this recursion relation concretely for a given knot.
This in turn has been solved for several knots in joint work of the first author,
X.~Sun and C.~Koutschan~[37, 32, 33]. The solution required a modulo $p$
computation of the $N$-th colored Jones polynomial (for several primes $p$
and several thousand values of $N$), together with a careful guess of the
supporting coefficients of such a recursion. In particular, the recursion
was computed in~[37] for the twist knots $K_p$ with $|p|\le15$, was
guessed in~[32] for the $(-2,3,3+2p)$ pretzel knots with $|p|\le5$,
and was computed (when $p=2$) or guessed (when $p=3,4,5$) in~[33]
for the double twist knots $K_{p,p}$ with $2\le p\le5$.

\smallskip\noindent{\bf Step 2: Computing the Kashaev invariant for large~$N$.}
In order to get numerical information about the asymptotics
of the Kashaev invariant $\la K \ra_N$, we ned to be able to compute it
numerically to high precision for large values of $N$, say of the order
of~$N=5000$. Although both the Kashaev invariant and the colored Jones
polynomial are given by finite-dimensional terminating $q$-hypergeometric sums,
and the latter have been programmed in \texttt{Mathematica}~[3],
these programs can only give the value of the colored Jones polynomial
and of the Kashaev invariant for modest values of $N$, say, up to~$N=20$,
which is far less than we need for the numerical extrapolation. By using the
recursion, we can compute $J_n(\z_N)$ numerically to high precision for
$N$ large and $0<n<N$.  (This is far faster than computing the colored Jones
polynomials $J_n(q)$ for these values of~$n$ and substituting $q=\z_N$ at
the end.)  However, this does not give the Kashaev invariant
$\la K \ra_N=J_{K,N}(\z_N)$ because the recursion gives $P(q,q^n)J_n(q)$
as a linear combination of a bounded number of previous values $J_{n-i}(q)$,
where $P(q,x)$ is a fixed polynomial that is always divisible by~$1-x$.
To overcome this, we use the recursion relation and its first $r$ derivatives,
where $(1-x)^r\|P(q,x)$, to get a recursion for the length-$r$ vector
$(J_n(q),J_n'(q),\dots,J_n^{(r)}(q))$. We can the use the recursion to compute
the whole vector numerically for~$q=\z_N$ and~$0<n<N$, and the single value
$J_n(q)$ for~$n=N$. It follows that

\begin{proposition}
\label{prop.ckashaev}
The Kashaev invariant $\la K \ra_N$ of a knot $K$ can be computed numerically
in $O(N)$~steps.
\end{proposition}

\noindent
This linear-time algorithm, which seems to be new even for the Kashaev invariant,
can be used equally well to compute $J(\g N)$ for~$\g=\sma abcd\in\SL_2(\Z)$ fixed and
$N$ tending to infinity, or even $J(\g X)$ with~$\g=\sma abcd$ fixed and $X$~tending to
infinity with fixed fractional part, since this simply the value of $J_n(\g X)$ with $n$
equal to the denominator of $\g X$ and can be calculated by the same trick.

\noindent{\bf Step 3: Computing $\Phi_\a^{(K,\s_1)}(h)$.}
Once we know how to compute $J(\g N)$ for large integers~$N$ (or even $J(\g X)$
for large~$X$ with bounded denominator), we can we can obtain the first few
coefficients of the power series $\Phi^{(K,\s_1)}_{\a}(h)$ numerically for a fixed
rational number~$\a=\g(\infty)$ by combining the Quantum Modularity
Conjecture~\eqref{QMCa} (or~\eqref{QMCb})
together with the extrapolation method of the second author (as described in detail
in~[45] or the appendix of~[39]) or the closely related
Richardson transform~\cite[Chpt.~8]{Bender}.  This is quite effective and gives, for
instance, the first hundred coefficients of the series~\eqref{as41} or forty
coefficients of the series~\eqref{as52} in only a few minutes of computing time.

\noindent {\bf Step 4: Recognizing the coefficients exactly.}
Given that the coefficients of $\Phi^{(K,\s_1)}_{\a}(h)$
are conjecturally algebraic
numbers in a specific number field, we can then test numerically by using
the standard LLL (Lenstra-Lenstra-Lovasz) algorithm to approximate the
numerically computed coefficients by rational linear combinations of
a basis of this field.  If this works to high precision with coefficients
that are not too large and have only small primes in the denominator, then
we have considerable confidence that the approximate equality is an exact one.
The method is self-verifying in the sense that the success at each
stage requires the correctness of the answer guessed at the previous stage.

\noindent{\bf Step 5: Computing $\Phi_\a^{(K,\s_0)}(h)$.}
In this step, we explain how to compute the expansion of the element
of the Habiro ring at a root of unity, in linear time. More precisely,
we have:

\begin{proposition}
\label{prop.hab}
The series $\Phi^{(K,\s_0)}_\a(h) + O(h)^{N+1}$ is computable in $O(N)$ steps.
\end{proposition}
This follows from the fact that the expansion of the Habiro element at
$q=\z_\a e^h$ up to $O(h)^{N+1}$ requires $cN$ terms of the
cyclotomic polynomial of $K$, which is a linear combination of the
first $cN$ colored Jones polynomials of $K$. An alternative formula,
inspired by Mahler's ideas on $p$-adic interpolation, gives an
following expansion of the Habiro element:
\be
\label{hab-mahler}
\Phi^{(K,\s_0)}_\a(h) = \sum_{k=1}^{cn} \widehat{J}^K_n(\z_\a e^h) + O(h^{n+1})
\ee
where $c$ is the denominator of $\a$, $\z_\a=\e(\a)$ and
\be
\label{tJ}
\widehat{J}^K_n(q) = \sum_{k=1}^n (-1)^{n-k} \binom{n}{k}_q
q^{\frac{k(k-1)}{2}} J^K_{n-k+1}(q) q^{-\frac{n(n+1)}{2}}
\ee
and $\binom{n}{k}_q = (q;q)_n/((q;q)_k (q;q)_{n-k})$ is the $q$-binomial
symbol and $(q;q)_n=(1-q)(1-q^2)\dots (1-q^n)$ is the $q$-Pochhammer
symbol. The right hand side of~\eqref{hab-mahler} gives
a well-defined formal power series since $\widehat{J}^K_{cn}(\z_\a e^h)$ lies
in $h^n \BQ[[h]]$.

%% see pari files: qKnots/qSeries/237pretzel/Habiro
%% and:            qKnots/qSeries/ColoredJones/ColoredJonesAllKnots

\noindent{\bf Step 6: Computing the remaining power series.}
Once we have the leading term in the original QMC, we can obtain the numerical terms
by successively subtracting the corrections coming from the values of~$\s$ different
from~$\s_1$ as explained in subsections~\ref{sub.4.1} (using ``optimal truncation")
and~\ref{sub.4.3} (using the more precise ``smooth truncation" described
in~\ref{sub.optimal} and in more detail in~[42]), where numerical
examples were also given.

\noindent{\bf Step 7: Alternative methods.}
In Steps 2 and~3 we discussed how to obtain the coefficients of $\Phi_\a^{(K,\s_1)}(h)$
from the original QMC together with the high-speed high-precision computation of
Jones polynomials at roots of unity and numerical extrapolation; in Steps~4 and~5
we explained how to get the series $\Phi_\a^{(K,\s)}(h)$ for $\s$ Galois conjugate
to~$\s_1$ and for $\s=\s_0$, respectively; and in Step~6 we described how to obtain
the remaining series by using the \emph{refined} quantum modularity conjecture
together with optimal truncation. However, there are at least two other ways to get
these other series that are of interest and are sometimes more efficient.

The first way is to use the formal Gaussian integration of Section~\ref{sec.perturbation}
and~[14, 15].  This method uses exact arithmetic and allows the computation of
few terms $A^{(K,\s)}_\a(k)$ (in practice $k \leq 5$) as exact algebric numbers for
many knots (such as those with ideal triangulations with up to 15 ideal tetrahedra).
See also~[35].

The other, which is completely different, is based on the asymptotics near roots of
unity of the holomorphic functions in the upper half plane (generalized Nahm sums)
that we study in the companion paper~[41]. Since Nahm sums converge
quadratically, the values of those functions at $\tau=\a+i/N$ can be computed in
$\,\text O(\sqrt{N})$ steps and after extrapolation give a numerical computation of
the algebraic numbers $A^{(K,\s)}_\a(k)$. This method, when applicable, is not only
much faster (time $\,\text O(\sqrt N)$ rather than $\,\text O(N)$), but also has
the major advantage of allowing the simultaneous numerical computation of
$A^{(K,\s)}_\a(k)$ for all $\a$ of a fixed denominator and for all $\s$ in a Galois
orbit which (after multiplication by $D_k$ and by a suitable $S$-unit)
reduces the problem of recognizing the list of coefficients $A^{(K,\s)}_\a(k)$ as
algebraic numbers to the problem of recognizing numerically computed \emph{integers},
albeit of growth $k!^2 C^k$. This allowed us to compute, for instance,
100~coefficients of the series $\Phi^{(K,\s)}_\a(h)$ for the $5_2$ knot for $\a=0$
and $a=1/2$ and for all three representations $\s$ in the Galois orbit of the
geometric representation,
and it allowed us to compute 37 terms of the series of the $(-2,3,7)$
pretzel knot for $\a=0$ and for both Galois orbits (each of size 3) of
nontrivial representations $\s$. It is remarkable that this method allows
the computation of series for representations not Galois conjugate to the
geometric one, though not for the trivial representation~$\s_0$.

{\bf Orientation conventions.} Finally, we have to discuss an annoying technical
point, namely, the choice of a consistent set of conventions for the two
classical invariants (namely the trace field and the complex volume of
a knot) and the two quantum invariants (namely the colored Jones polynomial
and the Kashaev invariant of a knot). These conventions are especially
important since the tables of knots rarely distinguish a knot from its
mirror, and (for instance) the name $5_2$ of the unique hyperbolic
5-crossing knot does not convey this distinction.

On the other hand, replacing a knot $K$ by its mirror $K^*$ reverses the
orientation of the knot complement $M_K = S^3 \ssm K$, which implies that
\begin{itemize}
\item
$F_{K^*} = \overline{F_K}$ and $\tV(K^*) =\overline \tV(K)$.
\item
$J_{K^*,N}(q)=J_K(q^{-1})$ and $\la K^* \ra_N=\overline{\la K \ra_N}$ and
\end{itemize}
Thus, a random orientation convention for $K$ might not match the asymptotics
whose coefficients are in a fixed subfield of the complex numbers, and
not on its complex conjugate subfield.

The Jones (hence, also the colored Jones) polynomial
$J_{K}(q) \in \BZ[q^{\pm 1}]$ of a knot (or a link) $K$ is uniquely determined
by the following skein-relation~[54]

\be
\label{jones}
q \, J_{\left(\includegraphics[height=0.1in]{figures/Rp.pdf} \right)}(q)
-q^{-1} J_{\left(\includegraphics[height=0.1in]{figures/Rn.pdf} \right)}(q) =
J_{\left(\includegraphics[height=0.1in]{figures/Rs.pdf} \right)}(q),
\qquad J_{\text{unknot}}(q)=1 \,.
\ee
In particular, for the left-hand trefoil $3_1$, we have:
$J_{3_1}(q)= -q^{4} + q^{3} + q^{1}$. Moreover, the colored Jones polynomial
$J_{K,N}(q) \in \BZ[q^{\pm 1}]$ is normalized to be $1$ at the unknot, and
to equal to the Jones polynomial when $N=2$.

%% see programs/snappy/note.normalization.jones.polynomial.pdf

Note that the \texttt{SnapPy} program~[12] for computing the Jones
polynomial agrees with~\eqref{jones}, whereas the \texttt{Mathematica}
program \texttt{KnotAtlas}~[3] polynomial differs by replacing $q$
by $1/q$:

\medskip

\begin{center}
\begin{tabular}{|l|l|} \hline
\verb|L = Link(braid_closure=[-1,-1,-1])| &
\verb|Jones[BR[2, {-1, -1, -1}]][q]|\\
\verb|L.jones_polynomial()| &
\verb|-q^-4 + q^-3 + q^-1|
\\
\verb|-q^4 + q^3 + q^1|     &  \\ \hline
\end{tabular}
\end{center}

\medskip

We will be using the consistent orientation convention for $M_K$ of the
\texttt{SnapPy} program (when $K$ is given as the closure of a braid, or
via a planar projection, or via an augmented DTcode or Gauss code),
which has the added advantage that it also gives shapes of tetrahedra
corresponding to the hyperbolic structure (exactly or numerically),
as well as the trace field (exactly) and the complex volume $\tV(K)$
(exactly or numerically).

\subsection{Optimal truncation and smoothed optimal truncation}
\label{sub.optimal}

One of the main numerical aspects concerns smoothed optimal translation which
was originally an appendix to an earlier draft of this paper but has now been
relegated to an independent publication ([42]) because the methods
are applicable to many problems outside the realm of quantum topology.

We already explained in Section~\ref{sub.4.1} the naive optimal truncation
$\Phi(h)^\opt$ of a factorially divergent series $\Phi(h)=\sum_{n=0}^\infty A_nh^n$,
defined simply as $\sum_{n=0}^NA_nh^n$ where $N$ is the approximate
value of~$n$ at which the term $A_nh^n$ takes on its minimum absolute value,
given explicitly by $N=|B/h|$ if $A_n$ grows like $n!\,B^{-n}$.  The idea of
smoothed optimal truncation is very simply to replace $\Phi(h)^\opt$ by a
``smoothed" version $\Phi(h)^\smo$ which is defined as $\Phi(h)^\opt+\ve_N(h)$
where the exponentially small correction term $\ve_N(h)$ depends on the cutoff
parameter~$N$ in such a way that $\Phi(h)^\smo$ does not jump when one changes $N$
by~1. This means simply that we require $\ve_{N-1}(h)-\ve_N(h)=A_nh^n$.  Of course, if
we knew how to solve this equation exactly, then the function $\Phi(h)^\smo$
would be completely independent of~$N$ and would give us a canonical way to
lift the power series $\Phi(h)$ to an actual function.  This is not the case,
but if $A_n$ has a known asymptotic expansion, which is true for all of the
series in this paper (see Section~\ref{sub.CoeffA}) then we can define
$\e_N(h)$ asymptotically as the product of $e^{-N}$ and a power series in~$1/n$
chosen in such a way that the desired equality  $\ve_{N-1}(h)-\ve_N(h)=A_nh^n$
is true asymptotically to all orders in~$1/n$.  The details of how to do this\
if $A_n$ has the asymptotic form $B^{-n}\,\sum_{\ell=0}^\infty c_\ell\,\G(n+\k-\ell)$
for some real number~$\k$ and some numerical coefficients~$c_\ell$ (as in
equations~\eqref{AnFirst}, \eqref{BnFirst} or \eqref{CoeffAsymp}) are given
in~[42] and will not be repeated here. The only thing that is of
importance to us here is that the result of the smoothing gives an evaluation of
$\Phi(h)$ that is independent of all choices (and hence gives a predicted ``right"
definition of the corresponding function) up to an error that is exponentially small
with a better exponent than that given by naive optimal truncation.  Specifically,
the new error is $e^{-N}\,(1+|C|)^{-N}$ rather than simply $e^{-N}$ as before if, as
is always the case for us, the coefficients $c_\ell$ themselves grow factorially
like $\ell!\,C^{-\ell}$.  Examples of this dramatic numerical improvement were given
in Section~\ref{sub.4.3}, where in one case the error in evaluating a series
$\Phi(h)$ was of the order of~$10^{-29}$ using naive optimal truncation but of the
order of~$10^{-56}$ using smoothed optimal truncation.

\newpage

\part*{Appendix: Numerical data for five sample knots}
\label{sec.num}

In this section we present numerical data that support the quantum modularity
conjecture for a choice of knots. Initially, we hoped that pairs
of geometrically similar knots (that have identical trace fields and
equal elements in the Bloch group, modulo torsion---henceforth called
``sisters'') might have identical or nearly identical
series~$\Phi^{(K,\s_1)}_\a$. With this in mind, and having already performed the
computations for the $4_1$ knot, we were led to consider its sister,
the \texttt{m003} census manifold. The latter is not a knot complement
(it is the complement of a knot in a lens space), but its 5-fold cyclic
cover is the complement of the (twisted) $5$-chain link, with a computable
Kashaev invariant, to be compared with the $5$-th power of the Kashaev
invariant of the $4_1$ knot. No relation between the series $\Phi^{(K,\s_1)}_\a$
was found for this pair, but the units $\ve(K)_\a$ did match (up to roots
of unity). We then tried the $5_2$ knot, whose sister is the
$(-2,3,7)$ pretzel knot. Here again the series  $\Phi^{(K,\s_1)}_\a$ were
different, but the units $\ve(K)_\a$ matched. The final
example of the $6_1$ knot was chosen because its Bloch group
has rank~$2$ and its $\SL_2(\BC)$-character variety is more complicated,
making the verification of the QMC, the Galois invariance
property~\eqref{cunit} and the match with the unit of~[10]
more subtle, but again all three were verified numerically. For this knot
we did not make any ``sister'' computations.

Recall the coefficients $A^{(K,\s)}_\a(k)$ of the power series
$\Phi^{(K,\s)}_\a(h)$ are algebraic numbers. In this section,
we present the numerically obtained data for $A^{(K,\s)}_\a(k)$
written in the form
\be
\label{PhiA}
A^{(K,\s)}_\a(k) \= C^{(K,\s)}_\a \tiA^{(K,\s)}_{\a}(k)\,,
% \Phi^{(K,\s)}_\a(h)
% \= C^{(K,\s)}_\a \sum_{k=0}^\infty A^{(K,\s)}_{\a}(k) \, h^k
\ee
where $C^{(K,\s)}_\a=\mu_{\s,\a}\cdot(\ve_{\s,\a})^{\frac{1}{c}} \cdot
\delta_\s^{-\frac{1}{2}}$ is given in~\eqref{Phiprec2}.
Note however that $C^{(K,\s)}_\a$ and $\tiA^{(K,\s)}_{\a}(k)$ are not
canonically defined numbers, only their product is. (Since we are focusing
on the geometric representation $\s_1$, and we are fixing the knot $K$, we
omit the superscript $(K,\s)$ from the notation in the right hand side of
the above equation.) We will further specify a choice of an algebraic number
$\lambda_c$ such that $\lambda_c^k \, D_k \, \tiA_{\a}(k) \in \calO_{F(\z_\a)}$
is an algebraic integer, where $c$ is the denominator of $\a$, $F$ is the
trace field of the knot and $D_n$ is
the universal denominator~\eqref{Dn}. Using a basis of free abelian groups
$\calO_{F(\z_\a)}$ we can represent the above algebraic integers by lists of
integer numbers.

\subsection*{The figure eight knot}
\label{sub.41}

%% The files Dropbox/qKnots have a local copy in programs/pari/zagier/qKnots

%% pari files: Dropbox/qKnots/Kashaev/Fig8/Fig8Modularity
%% pari files: Dropbox/qKnots/Kashaev/5link/Modularity

In this section we discuss the numerical aspects of the quantum modularity
conjecture for the simplest hyperbolic $4_1$ knot, for which we currently
know how to prove the modularity conjecture.
(The proof was presented in Section~\ref{sec.QMC41}.)
Needless to say, the numerically obtained results agree with the exact
computation of the expansion coefficients given in Section~\ref{sec.QMC41}.
Some information about the numerical aspects of the Kashaev invariant
of the $4_1$ knot were already presented in the introduction, but we give some
additional data (e.g., for the expansion near seventh roots of unity),
since this is the most accessible knot numerically and also to illustrate
the formulas occuring in the proof.

Since the knot is fixed in this section, and so is the
geometric representation $\s_1$, we will suppress them
from the notation. As mentioned in Section~\ref{sec.QMCK}, the
trace field of the $4_1$ knot is $\BQ(\sqrt{-3})$ and the torsion
is $\delta = \sqrt{-3}$. Some terms of the series $\Phi_\a(h)$
when $c=1$ or $c=5$ were given in
Equations~\eqref{as41} and~\eqref{at5}, respectively.
The numerical method allows us to compute and identify
the power series $\Phi_{a/c}(h)$ to any desired precision. Although
the coefficients of the series $\Phi_{a/c}(h)$, divided by the
constant term, is an element of the number field $\BQ(\z_{3c})$ (when
$c$ is coprime to $3$) or $\BQ(\z_c)$ when $3$ divides $p$, a judicious
choice of the constant $C_{a/c}$, combined with the Galois invariance
of the coefficients allows us to list the coefficients $\tiA_{a/p}(k)$ for
$p \neq 3$ prime and for $a=1,\dots,p-1$ by giving a $p-1$ tuple of elements
in the trace field $\BQ(\z_3)=\BQ(\sqrt{-3})$ of the knot. Furthermore,
since the knot is amphicheiral, it follows that $\tiA_{a/p}(k)$ is real or purely
imaginary (for $k$ odd or even, respectively), and combined with the above
discussion, allows to list the vector of coefficients
$(\tiA_{1/p}(k),\dots,\tiA_{(p-1)/p}(k))$ by a $(p-1)$-dimensional vector of rational
numbers. Our numerical extrapolation method allows us to compute this
tuple efficiently, and what is more, our code is self-correcting in
several ways: if a wrong denominator for $\tiA_{\a}(k)$ is guessed
for some $k$, its factorization in primes involves prime larger than $k+1$,
and the precision of the computation drops in the next $k+1$ step by
a factor of two. As a result, we were able to compute 100 terms of the
series $\Phi_0(h)$ when $c=1$, and the results agree with the
computations given in~[16] as well as computations obtained by
a different method by the first author.

In addition to this, we computed the constant term $\Phi_\a(0)$ for all
$\a$ with denominator a prime less than 100, and confirmed that its norm
agrees with the predictions of Section 4.1 of~[15] for $c \leq 19$.
We also computed 20 terms of the series $\Phi_\a(h)$ for all $\a$ with
denominator a prime less than 100.

To present a sample of our computations, we start with the special case of
$c=1,2,3,6$ where the $c$-th root of unity is in the trace field
$\BQ(\sqrt{-3})$. With the choice of
% I divided k_c=LL[c] in pari file by c because of
% (X+d/c)*L[c]=(cX+d)*L[c]/c
% L[1]=12, L[2]=48, L[3]=36, L[6]=144.
%$\lambda_1=1$, $\lambda_2=2$, $\lambda_3=1$, $\lambda_6=3$,
$C_\a$ and $\lambda_\a$
% I divide the C_a in the pari file by c^{3/2} because of X+d/c vs c*X+d
\begin{center}
\def\arraystretch{1.2}
\begin{tabular}{|c|c|c|cc|cc|}
\hline
& $0$  & $1/2$ & $1/3$ & $2/3$ & $1/6$ & $5/6$ \\
\hline
$C_\a$ & $3^{-{1/4}}$ & $3^{1/4}$ & $2 \cdot 3^{-1/12}$ &
& $2^2 \cdot 3^{1/12}$ & $3^{17/12}$ \\
\hline
$\lambda_c$ & $72 \sqrt{-3}$ & $72 \sqrt{-3}$ &
\multicolumn{2}{c|}{$24 \sqrt{-3}$}
& \multicolumn{2}{c|}{$36 \sqrt{-3}$} \\
\hline
\end{tabular}
\end{center}
it turns out that $\lambda_c^k D_k \tiA_{\a}(k)$ are integers are given by

%% sample table with multilines and multicolumns
%\multirow{3}{*}{A} & \multicolumn{2}{c|}{User B} & %
%    \multicolumn{2}{c|}{User C} & \multirow{3}{*}{D}\\
% & \multicolumn{2}{c|}{Value} & \multicolumn{2}{c|}{Value} & \\
% & B1 & B2 & C1 & C2 & \\
% & & & & & \\
% & & & & & \\
%% THIS IS SAMPLE

%% data computed in:
%% programs/pari/zagier/qKnots/Kashaev/Fig8/AsympWithDenom.stavros
%% and saved in:
%% programs/pari/zagier/qKnots/Kashaev/Fig8/A1vector, A2vector
%% A31vector, A32vector, A4vector, A5vector, A7vector etc

%% vector(8,n,A1[n]) == [1, 11, 697, 724351, 278392949, 244284791741, 1140363907117019, 212114205337147471]
%% vector(7,n,A2[n]) == [1, 41, 12625, 48022429, 72296210981, 252636824949503, 4705529517313888499]
%% vector(7,n,A31[n]) == [1, 37, 7785, 21535937, 24220768661, 63245072194611, 881169308393417291]
%% vector(7,n,A32[n]) == [1, 25, 6449, 18981677, 21569737445, 56749680285647, 795019302903681491]
%% vector(7,n,A4[n]) == [[1, 2], [365, 370], [311785, 420386], [4219048201, 6325027802], [24805519728725, 38098972914250], [340419470401244075, 531593492940700894], [25036998069742932352139, 39557220304220645794918]]
%% vector(7,n,A61[n]) == [1, 579, 1224117, 39903107571, 535664049856461, 16693882665527364525, 2781195926314033920017079]
%% vector(7,n,A65[n]) == [1, 201, 782865, 29648832381, 412895509718949, 13164162601119392223, 2226683577626946593056179]
%% vector(7,n,A5[n]) == [[-1, -4, -4, -6], [-55, -5140, -7660, -9690], [-7586065, -48629140, -58401700, -81382470], [-1066837647875, -5818148628500, -6620399493500, -9407838821250], [-51952598327049125, -274293246490488500, -309180073069692500, -440171876888046750], [-5814113396376116334625, -29960825153926862627500, -33500926926525556664500, -47835527737950677253750], [-3440238162871379293314141875, -17444570215945410898913817500, -19394322331421620186055667500, -27754399327684612597988351250]]
%% vector(7,n,A7[n]) == [[-20, 7, 2, 5, -14, -8], [-98140, 8267, -19670, 27937, -39214, -16576], [-2199415652, 426208447, -172006030, 524259533, -1237405358, -619260152], [-676432728043100, 166452454682479, -15638648253886, 168799271208365, -406506539584838, -215671594628336], [-86350611733284233860, 22591735955847949331, -702673247614974230, 21808440520527403561, -52829131820839184902, -28340444966866544008], [-25671367091358132079572196, 6928168872402051353797277, 10873595841062215161670, 6492789075493742592974935, -15896921084389159954206466, -8579075179324647599719264], [-40778149057063373929933607755660, 11230283616311811275545576789037, 254227567976955185183216119222, 10324895672816198996021631076855, -25455304659369651131719478440714, -13795408035312995480440595565448]]

\begin{center}
\def\arraystretch{1.3}
\begin{tabular}{|c|c|c|cc|cc|}
\hline
& $\lambda_1^k D_k \tiA_{0}(k)$  & $\lambda_2^k D_k \tiA_{1/2}(k)$ &
$\lambda_3^k D_k \tiA_{1/3}(k)$ & $\lambda_3^k D_k \tiA_{2/3}(k)$ &
$\lambda_6^k D_k \tiA_{1/6}(k)$ & $\lambda_6^k D_k \tiA_{5/6}(k)$ \\
\hline
$0$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
\hline
$1$ & $11$ & $41$ & $37$ & $25$ & $579$ & $201$ \\
\hline
$2$ & $697$ & $12625$ & $7785$ & $6449$ & $1224117$ & $782865$\\
\hline
$3$ & $724351$ & $48022429$ & $21535937$ & $18981677$ & $39903107571$ & $29648832381$ \\
\hline
$4$ & $278392949$ & $72296210981$ & $24220768661$ & $21569737445$ & $535664049856461$ & $412895509718949$
\\
\hline
$5$ & $244284791741$ & $252636824949503$ & $63245072194611$ & $56749680285647$ & $16693882665527364525$
& $13164162601119392223$
\\
\hline
\end{tabular}
\end{center}

When $c=4$ and $a= \pm 1 \bmod 4$, with the choice
$C_{a/4}=\pm (3 (2 \pm \sqrt{3}))^{-1/4}$ and $\lambda_4=6\sqrt{-3}$,
we can write
\lambda_4^k \, D_k \, \tiA_{\pm 1/4}(k) = \tiB_{1/4}(k) \pm \tiB_{-1/4}(k) i
where $B_4(k)=(\tiB_{1/4}(k),\tiB_{-1/4}(k)) \in \BZ^2$ with the first
six values are given by

\begin{center}
\begin{tabular}{|c|l|}
\hline
$B_4(0)$ & $\la 1, 2 \ra$ \\
\hline
$B_4(1)$ & $\la 365, 370 \ra$ \\
\hline
$B_4(2)$ & $\la 311785, 420386 \ra$ \\
\hline
$B_4(3)$ & $\la 4219048201, 6325027802 \ra$ \\
\hline
$B_4(3)$ & $\la 24805519728725, 38098972914250 \ra$ \\
\hline
$B_4(4)$ & $\la 340419470401244075, 531593492940700894 \ra$ \\
\hline
$B_4(5)$ & $\la 25036998069742932352139, 39557220304220645794918 \ra$ \\
\hline
\end{tabular}
\end{center}

Finally, when $c=p$ is a prime different from $3$, we found
out for that for the primes less than 100, the
constant $C_\a$ of~\eqref{PhiA} can be taken to be
C_{\a} = 3^{(-2 \pm 1)/4} \, p^{1/2}\, (\ve_\a)^{1/p}
\qquad \text{for} \qquad p=\pm 1 \bmod 6
\,
\ve_\a = \prod_{|k| \leq \frac{p-1}{2}} (\ve(p'k\a))^k, \qquad
p'=\mp 1/4 \bmod p,
\qquad \ve(x)=2 \cos 2 \pi(x-1/3) \,.
Note that the unit $\ve_\a$ in $\BQ(\z_{3p})$
that appears in the choice of $C_\a$ agrees, up to $p$-th powers of units,
with the theoretically computed unit from Equation~\eqref{tag8} (for $r=0$)
below. With the above choice of $C_\a$, the numbers $A_{\a}(k)$ lie in the
field $\BQ(\z_{3p})$, satisfy the Galois invariance described in detail
in the introduction, and this allows them to be expressed in terms of vectors
$B_p(k)=\la \tiB_{1/p}(k),\dots,\tiB_{(p-1)/p}(k) \ra \in \BZ^{p-1}$
as follows:
\lambda_p^k \, D_k \, \tiA_{a/p}(k) = \sum_{b=1}^{p-1} \eta(ab/p) \tiB_{b/p}(k),
\qquad \eta(x) = 2 \sin(2 \pi (x-1/3)) \,.
where $\lambda_p = 3p^2 \sqrt{-3}/2$. The vectors $B_p(k)$ for $k \leq 20$ and
$p$ a prime less than $100$ were numerically obtained and for $p=5$ and
$p=7$ are given by

\begin{center}
\begin{tabular}{|c|l|}
\hline
$B_5(0)$ & $\la -1, -4, -4, -6 \ra$ \\
\hline
$B_5(1)$ & $\la -55, -5140, -7660, -9690 \ra$ \\
\hline
$B_5(2)$ & $\la -7586065, -48629140, -58401700, -81382470 \ra$ \\
\hline
$B_5(3)$ & $\la -1066837647875, -5818148628500, -6620399493500, -9407838821250 \ra$ \\
\hline
$B_5(4)$ & $\la -51952598327049125, -274293246490488500, -309180073069692500, -440171876888046750 \ra$ \\
\hline
$B_5(5)$ & $\la -5814113396376116334625, -29960825153926862627500, -33500926926525556664500, -47835527737950677253750 \ra$
\\
\hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{|c|l|}
\hline
$B_7(0)$ & $\la -20, 7, 2, 5, -14, -8 \ra$ \\
\hline
$B_7(1)$ & $\la -98140, 8267, -19670, 27937, -39214, -16576 \ra$ \\
\hline
$B_7(2)$ & $\la -2199415652, 426208447, -172006030, 524259533, -1237405358, -619260152 \ra$ \\
\hline
$B_7(3)$ & $\la -676432728043100, 166452454682479, -15638648253886, 168799271208365, -406506539584838, -215671594628336
\ra$ \\
\hline
$B_7(4)$ & $\la -86350611733284233860, 22591735955847949331, -702673247614974230, 21808440520527403561, $ \\
& $ -52829131820839184902, -28340444966866544008 \ra$ \\
\hline
$B_7(5)$ & $\la -25671367091358132079572196, 6928168872402051353797277, 10873595841062215161670, $ \\
& $ 6492789075493742592974935, -15896921084389159954206466, -8579075179324647599719264 \ra$ \\
\hline
\end{tabular}
\end{center}

%% 100 values of the rational numbers $(72 \sqrt{-3})^k \tiA_{0}(k)$
%% were computed in: qKnots/Kashaev/Fig8/Near1
%% and saved in:     qKnots/Kashaev/Fig8/A1vector

%% The numbers $B_{p}(k)$ are computed in:
%%   qKnots/Kashaev/Fig8/pLimit/AsympWithDenom.stavros
%%   qKnots/Kashaev/Fig8/pLimit/ScalarAsymptotics
%%   qKnots/Kashaev/Fig8/pLimit/VectorAsymptotics
%%   and saved in: qKnots/Kashaev/Fig8/A1vector, A2vector
%%   A31vector, A32vector, A4vector, A5vector, A7vector etc

%% Cleanest files:
%%   qKnots/Kashaev/Fig8/pLimit
%%   qKnots/Kashaev/Fig8/Uniform5

%% Earlier files:
%%   qKnots/Kashaev/Fig8/Fig8new
%%   qKnots/Kashaev/Fig8/Fig8Modularity
%%   qKnots/Kashaev/Fig8/Fig8and6j.stavros2

\subsection*{The sister of the figure eight knot}
\label{sub.s41}

Its quotient by $\BZ/5\BZ$, which
is a knot in the lens space $L(5,1)$ rather than the 3-sphere,
is the sister of the $4_1$ knot, with the same trace field and same
Bloch group invariant. We therefore
expect to find similarities between the asymptotic power series
associated to $\K1$ and to~$\K2$.

Next, we discuss the case of the sister of the $4_1$ knot, the manifold
\texttt{m003} in the hyperbolic knot census~[12], which is not
the complement of a knot in the 3-sphere, but is the complement of a
nullhomologous knot in the Lens space $L(5,1)$. This complicates things
since the sister knot has no Jones (hence, no colored Jones) polynomial,
and although it has a Kashaev invariant, a formula for it is not available
to us. However, the $5$-fold cyclic cover of the sister of the
$4_1$ knot is the 5-chain link $L$ in $S^3$ (denoted by $10^5_3$ and also
by \texttt{L10n113}). This is a famous link because virtually every census
manifold is a Dehn filling on it~[17]. The link $L$
has a colored Jones polynomial
$J_{L,N}(q)$ (with all components colored by the same $N$-dimensional
representation) with a formula available from~[70]
and a Kashaev invariant. More precisely, we have:

%% pari formula taken from qKnots/Kashaev/5link/Modularity

% {JJC(N) = X=x^N; CC=vector(N); s=0; for(n=0,N-1, c=X^(-n)/prod(j=1,n,1-x^j)*
%   sum(k=0,N-n-1,X^(-k)*prod(j=k+1,n+k,(1-x^(N-j))*(1-x^j)));
%     s += (x^(n+1)-x^(-n))* c^2*subst(c,x,1/x)^3); -1/(1-X)*s}

%{\obeylines\font\small=cmr10 scaled \magstephalf\parindent=0pt\hangafter=0\raggedright\tt
%\hskip0pt  $\{$ JJC(N) = X=x\^{\kern.2pt}N; CC=vector(N); s=0; for(n=0,N-1, c=X\^{\kern.2pt}(-n)/prod(j=1,n,1-x\^{\kern.2pt}j)*
%\hskip0pt sum(k=0,N-n-1,X\^{\kern.2pt}(-k)*prod(j=k+1,n+k,(1-x\^{\kern.2pt}(N-j))*(1-x\^{\kern.2pt}j)));
%\hskip0pt s += (x\^{\kern.2pt}(n+1)-x\^{\kern.2pt}(-n))* c\^{\kern.2pt}2*subst(c,x,1/x)\^{\kern.2pt}3); -1/(1-X)*s $\}$ \par\ \leftskip=0pt }

J_{L,N}(q) \,= -\frac{1}{1-q^N}
\sum_{n=0}^{N-1} (q^{n+1}-q^{-n})c(N,n)(q)^2 c(N,n)(q^{-1})^3 \,
c(N,n)(q)=\frac{q^{-Nn}}{(q;q)_n}\sum_{k=0}^{N-n-1} q^{-Nk}
\prod_{j=k+1}^{n+k} (1-q^{N-j})(1-q^j) \,.
\noindent
The above formula is $O(N^2)$ can can be rewritten in terms of an $O(N)$
formula that has a recursion relation. However the latter has the
disadvantage that the middle term of the
summand ($k=N/2$) now vanishes when evaluated at $\e(1/N)$. To overcome
this, we compute the sum from both sides by differentiation. Having done so,
we tested the QMC and no suprises were found. We computed 10 terms of the
series $\Phi^{L}_\a(h)$ when $\a=0$ (given below)
and 8 terms when $\a=1/2$.
% app(X) = C0*exp(a*X)/X^2.5
% Z6=.5+sqrt(-.75); V=imag(polylog(2,Z6)); a=5*V/Pi + 3*I*Pi/4; c0=2*Pi*I/432;
% C0= Z6/2^3.5/3^.25

We now give the data for $\a=0$. The trace field of $L$ is $\BQ(\sqrt{-3})$,
same as for the $4_1$ knot. The complex volume of $L$ is given by
\V(L)=5 \, \V(4_1) -3\pi^2 %+ \frac{3}{2} \pi i
and its torsion is given by
\delta(L) = 2^7 \sqrt{-3} \,.
Since $L$ is a link, in~\eqref{K1}, we should replace the exponent
$3/2$ by $5/2$. With these changes, and with the notation of
Equation~\eqref{PhiA} we get algebraic integers
$12^k D_k A^{L}_{0}(k)$ in the ring~$\BZ[\sqrt{-3}]$ and the first
10 are given as follows.

\def\arraystretch{1.3}
\begin{array}{|c|l|} \hline
k & 12^{k} \, D_k \, \tiA^{K_2}_{0}(k) \\ \hline
0 & 1
\\ \hline
1 & -115 \sqrt{-3} + 279
\\ \hline
2 & -49050 \sqrt{-3} + 53286
\\ \hline
3 & -112270440 \sqrt{-3} + 163969920
\\ \hline
4 & -131463532440 \sqrt{-3} + 2948624280
\\ \hline
5 & 4388324675760 \sqrt{-3} - 163377997734672
\\ \hline
6 & -155232475000358400 \sqrt{-3} + 1614884631367642560
\\ \hline
7 & -456051590815208713920 \sqrt{-3} - 409415976078904226880
\\ \hline
8 & 1201424680509251029718400 \sqrt{-3} - 2426468490157451971144320
\\ \hline
9 & 280843674420360230423881689600 \sqrt{-3} + 767958533539384912591107225600
\\ \hline
\end{array}
%% pari files: qKnots/Kashaev/5link/Modularity.DEEXPONENTIATED

% [1,
% -115*t + 279,
% -49050*t + 53286,
% -112270440*t + 163969920,
% -131463532440*t + 2948624280,
% 4388324675760*t - 163377997734672,
% -155232475000358400*t + 1614884631367642560,
% -456051590815208713920*t - 409415976078904226880,
% 1201424680509251029718400*t - 2426468490157451971144320,
% 280843674420360230423881689600*t + 767958533539384912591107225600]

% Z6=.5+sqrt(-.75); V=imag(polylog(2,Z6)); a=5*V/Pi + 3*I*Pi/4; c0=2*Pi*I/432;
% C0= Z6/2^3.5/3^.25; S=quadgen(-12); a1=(279-115*S)/2; a2=(26643-24525*S)/4;
% a3 = (1366416-935587*S)/2; a4 = (24571869-1095529437*S)/16;
% a5 = (-3403708286139 + 91423430745*S)/560;  a6 = 3*(186907943445329 - 17966721643560*S)/560;
% a7 = 3*(-6769444048923681 - 7540535562420779*S)/160; a8 = 9*(-46806876739148379073 + 23175630411058083135*S)/8960;
% a9 = (740700746083511682668892 + 270875457581365962985997*S)/49280;  aa = [1,a1,a2,a3,a4,a5,a6,a7,a8,a9];
% app(X) = C0*exp(a*X)/X^2.5
% P(n,d) =  sum(j=0,d,aa[j+1]*(Pi*I/216/n)^j)
% try(d,r=6) = dv=diffm(vector(L,n,(H+n)^(r+d)*(v[n]-P(H+n,d-1))),r)/r!/c0^d; l=length(dv); C=dv[l]; print(size(dv[l-5]/C-1)); C
% TRY(d,R=14) = for(r=2,R,print1(r,"  ");try(d,r))

However, we failed to find any relation between the series for the
$4_1$ knot and for the 5-fold cover of its sister.

\subsection*{The $5_2$ knot}
\label{sub.52}

The pair of the $4_1$ knot and its sister from the previous section in
unsatisfactory in two ways. For once, the Quantum Modularity Conjecture is
proven for the $4_1$ knot.
Moreover, the sister of $4_1$ (and its 5-fold cover) is not a knot.
The next next simplest hyperbolic knot after~$4_1$ is the $5_2$ knot,
whose sister is the mirror of the $(-2,3,7)$ pretzel knot. Sister (or
geometrically similar) knots have a decomposition into a finite number of
congruent ideal tetrahedra, hence they have the same trace field
and equal elements in the Bloch group, modulo torsion.

A formula for the Kashaev invariant of the $5_2$ knot
was given in~\cite[Eqn.2.3]{K97},
\be
\label{K52}
\J^{5_2}(x)=\sum_{m=0}^{N-1} \sum_{k=0}^m
%b_{m,k}(\zeta_N), \qquad b_{m,k}(q)=
q^{-(m+1)k} \frac{(q;q)_m^2}{(q^{-1};q^{-1})_k}, \qquad q=\e(x)
\ee
where $N$ is the denominator of $x \in \BQ$. After multiplication of
the above by $\e(x)$, it agrees with the evaluation of the colored
Jones polynomial $J_{5_2,N}(\e(x))$, where the Jones polynomial of $5_2$
is $J_{5_2}(q)=q - q^2 + 2 q^3 - q^4 + q^5 - q^6$.
%% see: mathematica/zagier/Prepare.Recursion.For.Kashaev.Pari.nb
%% see: mathematica/kashaev/Kashaev.52.nb
The formula~\eqref{K52} allows a computation of the Kashaev invariant
in~$O(N^2)$ steps, and a simplification of it was found by one of the
\J^{5_2}(x)=\sum_{m=0}^{N-1} (q;q)_m^2 %\text{Conj}
\Bigl(
(q^{-1};q^{-1})_m \sum_{k=0}^m \frac{q^{-k^2}}{(q^{-1};q^{-1})_k^2} \Bigr),
\qquad q=\e(x)
that allows an $O(N)$-step computation of the Kashaev invariant.
An alternative computation of the latter in $O(N)$-steps can be performed
using the recursion relation for the colored Jones polynomial of

As mentioned in Section~\ref{sub.p}, the trace field of $5_2$ is $F=\BQ(\xi)$
\be
\label{a52}
\xi^3-\xi^2+1\=0, \qquad \xi = 0.877438833\cdots \;-\;
\ee
%It will be useful to introduce the primes $\p_1$ and $\p_2$ of norm $23$
%\p_1\=3\xi-2\,,\qquad \p_2\=3\xi+1\,,\qquad 23\=-\p_1^2\,\p_2\,,
%\qquad \l\=\xi\, \p_1^3\;.

%% The data was computed in the clean pari file:
%% qKnots/Kashaev/52knot/52expansions.NEWER
%% includes de-exponentiation of power series
%% For the complex volumes, see /don-own-cloud/Knots/NewHabiro/52Complete
The trace field has three embeddings labeled by $\s_j$ for $j=1,2,3$
(as discussed in Section~\ref{sub.p}) and their volumes are given
% see don-own-cloud/Knots/NewHabiro/52Complete
% and don-own-cloud/Knots/Check-52-237-PhiSeries
%V  = [-1.11345455247 + 0.E-519*I, 3.02412837651 + 2.82812208833*I, 3.02412837651 - 2.82812208833*I, 0]

\begin{align*}
\V(\s_1) &= -3 R(\xi_1)+\frac{2 \pi^2}{3}
= 3.0241283 \ldots + 2.8281220 \ldots \, i
\\
\V(\s_2) &= -3 R(\xi_2)+\frac{2 \pi^2}{3}
= 3.0241283 \ldots -2.8281220 \ldots \, i
\\
\V(\s_3) &= 3 R(\xi_3/(1-\xi_3))-\frac{\pi^2}{3} = -1.1134545 \ldots
\end{align*}
where $R(x)$ denotes the Rogers dilogarithm defined by
\be
\label{Rogers}
R(x) \= \text{Li}_2(x) \+ \frac12\,\log(x)\,\log(1-x)\,-\,\frac{\pi^2}6
\qquad \text{for \
\ee
The torsion of the $5_2$ knot is given by $\delta(5_2)=3\xi-2$.

%\tV(5_2) &=%-3/2/Pi/I (polylog(2,a)+log(a) log(1-a)/2) - I Pi/3
%-\frac{3}{2 \pi i} R(\xi) -\frac{\pi i}{12} =
%  0.450109610\dots
%- 0.481304979\dots i
%%&\= -1/\sqrt{\p_1}\=-0.5233766\cdots\,-\, 0.3958279 \cdots\,i
%\delta(5_2) &\= 3\xi-2 \,.

% In the data presented below, we use a slightly different normalization
% of $\Phi^K_{a/c}$ (that reflects the difference between $cX+d$ and $X+d/c$),
% \J^{K^*}\Bigl(\frac{aX+b}{cX+d}\Bigr) \;\sim\;
% (2\pi/h)^{3/2}\, \J^{K^*}(X) \, e^{\tV(K)(X+d/c)}\,
% \ti\Phi^{K}_{a/c}(h), where h=\frac{2\pi i}{X+c/d}
%defined by
%\Phi^K_{a/c}(h)=(i c)^{-3/2} \ti\Phi^K_{a/c}(ch)
%\ti\Phi^K_{a/c}(h) = \ti \mu_{8c} \cdot \delta(K)^{-\frac12}
%\cdot \ve(K)_{a/c}^{\frac{1}{c}} \cdot \ti\phi^K_{a/c}(h)

\noindent
$\bullet$ Modularity at 0 :
% asymptotics of: rat(X) = f(1/X)/f(-X)
% \J^{5_2}(1/X)/\J^{5_2}(-X) as $X\to\infty$ with bounded denominator,
% \J^{5_2}(1/X)/\J^{5_2}(-X) = \J^{5_2^*}(-1/X)/\J^{5_2^*}(X)
We choose $\ve(5_2)_0=1$, and with the notation of Equation~\eqref{PhiA}
the first eleven terms are given as follows:

\def\arraystretch{1.3}
\begin{array}{|c|l|} \hline
k & (2^3 \xi^5 (3\xi-2)^3)^k \, D_k \, \tiA^{5_2}_{0}(k) \\ \hline
0 & 1
\\ \hline
1 & -12 \xi^2 + 19 \xi - 86
\\ \hline
2 & -1343 \xi^2 - 12052 \xi + 14620
\\ \hline
3 & 1381097 \xi^2 + 36300408 \xi - 10373787
\\ \hline
4 & -939821147 \xi^2 - 7647561573 \xi - 5587870829
\\ \hline
5 & 114451233224986 \xi^2 - 51239666382079 \xi - 6305751988731
\\ \hline
6 & -2263527400987641127 \xi^2 - 631762147829071739 \xi - 1298875409805289208
\\ \hline
7 & -757944502306007361580 \xi^2 + 1425054483652604079482 \xi + 2654782623273180246011
\\ \hline
8 & 16785033822956024557916646 \xi^2 - 2226340168480665471705515 \xi
\\ &
- 14930684354870794067096358
\\ \hline
9 & -3735848035153601836654158090473 \xi^2 - 3510831690088210470322102227368 \xi
\\ &
- 449224959824265576892987954854
\\ \hline
10 & -34345984964128841574873487072878291 \xi^2 + 25085231887789675521906921078089414 \xi
\\ &
+ 52364404634270110370401111089362065
\\ \hline
\end{array}
% C(5_2)=2 \pi i \tV(5_2)
% \J^{5_2}(1/X)/\J^{5_2}(-X) as $X\to\infty$ with bounded denominator,
% \J^{5_2}(1/X)/\J^{5_2}(-X) = \J^{5_2^*}(-1/X)/\J^{5_2^*}(X)
% we have
%\J^{5_2}(1/X)/\J^{5_2}(-X) \sim -\delta_{5_2}^{-1/2} (2 \pi /h)^{3/2}
%e^{C(5_2)/h} \Phi_{5_2,0}(h),
%\qquad (\, h \= \frac{2\pi i}{X}\,)\,,
% $\gamma=\sma 0 1 {-1} 0$, $\gamma(Y)=-1/Y$ so $Y=-X \to -\infty$
% and $\J^{5_2}(-1/Y)/\J^{5_2}(Y) \sim -\delta_{5_2}^{-1/2} (2 \pi /h)^{3/2}
% e^{C(5_2)/h} \Phi_{5_2,0}(h)$ with $h=-\frac{2\pi i}{Y}$, that's
% why I take the mirror knot.

\noindent
$\bullet$ Modularity at 1/2 :
% asymptotics of: rat2(X) = f((X+1)/(2*X+1))/f(-X)
%as $X\to\infty$ with bounded denominator, we have
%\J^{5_2^*}\Bigl(\dfrac{-X-1}{2X+1}\Bigr)/\J^{5_2^*}(X) =
%B(5_2)_{1/2} \;(2\pi/h)^{3/2}\, e^{C(5_2)/h} \phi^{5_2}_{1/2}(h)
%\qquad (\,h \= \frac{2\pi i}{X+\frac12}\,)\,,
We choose $\ve(5_2)_{1/2} =\xi^{-5}$ and with the notation of
Equation~\eqref{PhiA} the first six
terms are given as follows:
% CONVERSION from pari series of 2*Pi*I/(X+d/c) to series in 2*Pi*I/(cX+d)
% * Multiply the k-th term of pari by c^k because
% 2*Pi*I/(X+d/c) = c*2*Pi*I/(cX+d)
% * Multiply the B term by c^{-3/2}.

\def\arraystretch{1.3}
\begin{array}{|c|l|} \hline
k & (2 \, \xi^5 (3\xi-2)^3)^k \, D_k \, \tiA^{5_2}_{1/2}(k) \\ \hline
0 & \xi + 2
\\ \hline
1 & 307 \xi^2 - 138 \xi - 628
\\ \hline
2 & -573109 \xi^2 - 168712 \xi + 457975
\\ \hline
3 & 2096955561 \xi^2 + 5077310601 \xi + 1165885531
\\ \hline
4 & 6470888990010 \xi^2 - 5414463743327 \xi - 10380246225743
\\ \hline
5 & 289484322041800655 \xi^2 - 138373378538474483 \xi - 156775910252412286
\\ \hline
\end{array}

\noindent
$\bullet$ Modularity at 1/3 :
Here, the constant term $\Phi^{5_2}_{1/3}(0)$ was numerically computed to
high precision
% B33 = e(-2/9) * p7 *sqrt(-1)/sqrt(p1)
% b30 = p7*p43; \Phi^{5_2}_{1/3}(0) = B33*b30
\Phi^{5_2}_{1/3}(0)=
-1.3478490468923913068\dots - 1.5706460265356353326\dots i
but was not initially recognized. To identify it,
we used the formula~\eqref{K52} for the Kashaev invariant and performed
a theoretical computation analogous to the constants $S(\a)$ and $E_0(\a)$
(given in~\eqref{tag8} and~\eqref{tag10}) of the $4_1$ knot which produced
the primes
\p_7 = (\xi^2-1) \z_6 - \xi + 1, \qquad \p_{43} = 2\xi^2-\xi - \z_6
of norm 7 and 43 respectively in the number field $F_3=\BQ(\xi,\z_3)$.
Note that the same primes appear in~\cite[Sec.6.2]{DG2}.
In addition, the above constant involves $\delta(5_2)^{-1/2}$ and a number
whose third power is in $F_3$. After some experimentation, we concluded that
\Phi^{5_2}_{1/3}(0)= \e(1/36) %\e(-2/9) \, i \,
\frac{1}{\sqrt{3\xi-2}} \, \p_7^2 \, \p_{43} \,.
It follows that a representative of the unit at $\a=1/3$ is given by
\be
\label{unit.51}
\ve(5_2)_{1/3} \=\e(1/12) \,. %\e(-2/3)
\ee
It was a bit of a surprise to find that the unit is torsion although the
Bloch group of $F_{6_1}$ has rank 1--on the other hand
$3$ (as well as $2$ and a few other primes) are exceptional ones in the

Once the constant term was recognized, it turnt out that we needed to
separate one factor of $\p_7$ in the constant term $\Phi^{5_2}_{1/3}(0)$
from the remaining terms, in order to avoid spurious denominators.
With the choice of $C_{1/3}=\e(1/36) (3\xi-2)^{-\frac{1}{2}} \p_7 \, \p_{43}$
and the notation of~\eqref{PhiA}, the first seven terms were found

\def\arraystretch{1.3}
\begin{array}{|c|l|l|} \hline
k & (\xi^5 (3\xi-2)^3)^k \, D_k \, \tiA^{5_2}_{1/3}(k) \\ \hline
0 & (-\xi^2 + 2 \xi - 2) \z_6 + (2 \xi^2 - 4 \xi)
\\ \hline
1 & (717 \xi^2 - 822 \xi + 947) \z_6 + (-2226 \xi^2 + 1856 \xi + 106)
\\ \hline
2 & (-680145 \xi^2 + 1283633 \xi - 1844797) \z_6 + (4731470 \xi^2 - 1215426 \xi + 785050)
\\ \hline
3 & (-4879664798 \xi^2 - 15547118437 \xi + 26771206405) \z_6
\\ &
+ (-20691193336 \xi^2 - 35194065214 \xi - 73160959238)
\\ \hline
4 & (237593851209955 \xi^2 - 123624865686699 \xi + 65688152000880) \z_6
\\ &
+ (-455730563794746 \xi^2 + 258640669065738 \xi + 244974132213716)
\\ \hline
5 & (-8559119253981428654 \xi^2 + 9164193255880569642 \xi - 8506396294603249043) \z_6
\\ &
+ (-8914434881967188748 \xi^2 - 7549553228397039176 \xi + 21232362162256499338)
\\ \hline
6 & (1206971041591026374138836 \xi^2 - 1471979903142920023426465 \xi + 1526039068996370402375484) \z_6
\\ &
+ (2034143372251380409655636 \xi^2 + 5390411863643322238842526 \xi - 935392258601663466664696)
\\ \hline
\end{array}
where $\z_6=\e(1/6)$. % s=\xi, t=\z_6

\subsection*{The $(-2,3,7)$ pretzel knot}
\label{sub.237}

Next, we discuss the case of a sister the $5_2$ knot, namely the
(mirror of) the $(-2,3,7)$ pretzel knot.
% $(-2,3,7)$-pretzel knot (\texttt{K12n242}).
Note that the trace fields of $5_2$ and $(-2,3,7)$ coincide,
which allow us to use the notation of~\eqref{a52}.

% pari file: programs/pari/zagier/atlanta.11/PRETZEL5
% pari file: programs/pari/zagier/MPI.11/52KNOT.2/PRETZEL5
% pari file: programs/pari/zagier/qKnots/Kashaev/237pretzel/PRETZEL7.DEEXPONENTIATED

Unlike the case of the $4_1$ and $5_2$ knots, the Kashaev invariant of
$(-2,3,7)$ can only be computed via the recursion of the colored Jones
polynomial which was guessed in~[32], with the convention that the
Jones polynomial of $(-2,3,7)$ is given by
$J^{(-2,3,7)}(q)=q^{-5} + q^{-7} - q^{-11} + q^{-12} - q^{-13}$. The above
inhomogeneous recursion has order 6, maximal degree $(6, 58, 233)$
with respect to the shift variable, the $q^n$ and the $q$ variables,
and contains a total of
90 terms, which can be found in~[24].
% see Mathematica file: mathematica/koutschan/pretzel/AqPolyPretzelKnots.nb
In contrast, the $A$-polynomial of the $(-2,3,7)$ knot has maximal degree
$(6,55)$ with respect to the $(L,M)$ variables and contains 12 terms.
In addition, we multiply the Kashaev invariant of $(-2,3,7)$ by $q^{-4}$.

Since $(-2,3,7)$ is a sister of the $5_2$ knot, they have a common
trace field $\BQ(\xi)$ given in~\eqref{a52}.
The trace field has three embeddings labeled by $\s_j$ for $j=1,2,3$
(as discussed in Section~\ref{sub.p}) and their complex volumes are given
% pari file: don-own-cloud/Knots/Check-52-237-PhiSeries
% pari file: programs/pari/zagier/qKnots/Kashaev/237pretzel/PRETZEL.EVENNEWER.stavros
\begin{align*}
\V(\s_1) &= -3 R(\xi_1) + \frac{\pi^2}{3} = 4.6690624 \ldots + 2.8281220 \ldots i
\\
\V(\s_2) &= -3 R(\xi_2) + \frac{\pi^2}{3} = 4.6690624 \ldots - 2.8281220 \ldots i
\\
\V(\s_3) &= 3 R(\xi_3/(\xi_3-1)) + \frac{\pi^2}{3} = 0.5314795 \ldots
\end{align*}
The torsion of $(-2,3,7)$ are given by
$\delta((-2,3,7)) = -2 (3\xi-2) \xi^{-2}$.

%f(x) \= q^4\,K(q)\,, \quad K(q)\
%=\text{Kashaev invariant for $(-2,3,7)$ pretzel}
%\tV((-2,3,7)) & \= \frac{1}{2 \pi}(
%R(1-\xi^2)\,-\,\pi \,i\,\log(\xi)\,+\,2\,R(1-\xi)\,+\,\pi^2 )
%= 0.450109610\dots - 0.743104367\dots i
%\delta((-2,3,7)) & = -2 (3\xi-2) \xi^{-2}
%%\delta((-2,3,7))^{-1/2} & = \xi/\sqrt{-2\,\p_1}
%%\=-0.0300721\cdots\,-\, 0.5332066\cdots\,i

%and are related to the complex volume and the torsion of $5_2$ by:
%\tV((-2,3,7))= \tV(5_2) - \pi i/12,
%\qquad \delta((-2,3,7)) = -2 \xi^{-2} \delta(5_2) \,.

%% for computation of $\delta$ and $\epsilon_{a/c}$ for $c=0,1$ \
%% for the 5_2 and (-2,3,7) pretzel knot see:
%% programs/mathematica/dimofte/S1/from.dimofte/S0123.All.Knots.nb

%% $$
%% C \,-\,\frac{7\pi^2}{6}\=- 6.8454\cdots + 2.82812\cdots \,i \=
%% \text{value given by snap for $(-2,3,7)$ pretzel}
%% $$
%% sage: from snappy import *
%% sage: M=Manifold("pretzel-237.tri")
%% sage: HTLinkExteriors.identify(M)
%% K12n242(0,0)
%% sage: M=Manifold("K12n242")
%% sage: M.volume()
%% 2.8281220883
%% sage: M.chern_simons()
%% 0.236537467
%% sage: p=M.ptolemy_variety(2,obstruction_class="1")
%% sage: p
%% Ptolemy Variety for K12n242, N = 2, obstruction_class = 1
%% sage: sols=p.compute_solutions()
%% sage: len(sols)
%% 1
%% sage: sol=sols[0]
%% sage: sol.number_field()
%% x^3 - x^2 + 1
%% sage: z=sol.cross_ratios()
%% sage: z
%% {'z_0000_0': Mod(x^2, x^3 - x^2 + 1),
%% 'z_0000_1': Mod(x, x^3 - x^2 + 1),
%% 'z_0000_2': Mod(x^2 - x + 1, x^3 - x^2 + 1),
%% 'zp_0000_0': Mod(x^2 - x + 1, x^3 - x^2 + 1),
%% 'zp_0000_1': Mod(x^2, x^3 - x^2 + 1),
%% 'zp_0000_2': Mod(x, x^3 - x^2 + 1),
%% 'zpp_0000_0': Mod(x, x^3 - x^2 + 1),
%% 'zpp_0000_1': Mod(x^2 - x + 1, x^3 - x^2 + 1),
%% 'zpp_0000_2': Mod(x^2, x^3 - x^2 + 1)}
%% sage: pari.set_real_precision(100)
%% 15
%% sage: sol.complex_volume_numerical()[2]
%% vc = 2.828122088330783162763898809276634942770981317300649477043520327258802548322471630936947017929999107 - 0.2657397571871512132248791115974493844534495601428615629678158152310972325202080011607886429390444443*I
%% VC-conj(vc)=-Pi^2/2

%% 100 coefficients of A^{sigma}_{0}(k) for all 6 sigma were
%% computed in: qKnots/qSeries/237pretzel/hSeries.13
%% saved in:    qKnots/qSeries/237pretzel/c37
%% using the 6 q-series
%% to convert the 6-vector of c37 into numbers,
%% use:         qKnots/qSeries/237pretzel/6x7matrix.stavros

\noindent
$\bullet$ Modularity at 0 :  Using the notation of~\eqref{PhiA}, we write
((2 \xi^2 - 6)^3/\xi^5)^k \, D_k \, \tiA^{(-2,3,7),\s_1}_{0}(k) = (1,\xi,\xi^2)\cdot B^{(-2,3,7),\s_1}_{0}(k)
where $B^{(-2,3,7),\s_1}_{0}(k) \in \BZ^3$ is a vector of integers with the first
11 values given by

\begin{array}{|c|l|} \hline
k & B^{(-2,3,7),\s_1}_{0}(k) \\ \hline
0 &  \la 1, 0, 0\ra
\\ \hline
1 &  \la -33, 128, -90\ra
\\ \hline
2 &  \la 79245, -104172, 50944\ra
\\ \hline
3 &  \la 333329999, -597644460, 317584318\ra
\\ \hline
4 &  \la -12580573862099, 16160668928488, -9152599685200\ra
\\ \hline
5 &  \la 275061075796915969, -366241217321535656, 209464837107544698\ra
\\ \hline
6 &  \la -21464059785100413194817, 28432876033981872108244, -16179201892533998639888\ra
\\ \hline
7 &  \la 39552725057509518276438631, -52341801268123421363828580, 29838036942620515077356206\ra
\\ \hline
8 &  \la 249767901145868199725688538645, -330081248453503483229302323376, 187971265625750854805584690976\ra
\\ \hline
9 &  \la -3700925786017810109833640742259950499, 4903075033684898536256604949931358320,
\\ &
\\ \hline
10 &  \la 392518725914904741935043787434245408953117, -519977480066306945985500543478969169892188,
\\ &
\\ \hline
\end{array}

\noindent
$\bullet$ Modularity at 1/2 :
% Constant term of Phi_{1/2}(0)=-sqrt(a^7*8/p1)
% \= -0.7369478\cdots \,-\, 2.9451833\cdots\,i
% need to multiply each term by 2^k because of 1/(X+1/2)=2/(2X+1).
If we choose $\ve_{1/2}((-2,3,7))=2 \xi^5 $, with the notation
of~\eqref{PhiA}, the first four terms are given by
%% pari file: programs/pari/zagier/qKnots/Kashaev/237pretzel/PRETZEL7.DEEXPONENTIATED
\def\arraystretch{1.3}
\begin{array}{|c|l|} \hline
k & (4 \, \xi (3\xi-2)^3)^k \, D_k \, \tiA^{((-2,3,7),\s_1)}_{1/2}(k) \\ \hline
0 & 1
\\ \hline
1 & -225 \xi^2 + 404 \xi - 249
\\ \hline
2 & 87535 \xi^2 - 158073 \xi + 123948
\\ \hline
3 & 1981731163 \xi^2 - 3465695160 \xi + 2508787814
\\ \hline
\end{array}

\noindent
$\bullet$ Modularity at 1/3 : Here the constant term and the next two
coefficients of the power series $\Phi^{(-2,3,7)}_{1/3}(h)$
$\Phi^{(-2,3,7)}_{2/3}(h)$ were computed to high precision, and using as a guidance
the appearance of primes of norm 373 (conjectured in~\cite[Sec.6.2]{DG}),
we identified the constant terms
\Phi^{(-2,3,7)}_{1/3}(0) \= \e(2/9) \sqrt{-\frac{27}{2 (3\xi-2)}} \p_{373},
\qquad
\Phi^{(-2,3,7)}_{2/3}(0) \= \e(5/9) \sqrt{-\frac{27}{2 (3\xi-2)}} \p'_{373},
where $\p_{373}=\xi^2+2 \xi \z_6 +1$ and $\p'_{373}=\xi^2+2 \xi (1-\z_6) +1$
are primes in $\BQ(\xi,\z_6)$ of norm $373$.
%% see: programs/pari/zagier/qKnots/Kashaev/237pretzel/UnitNew
%% however, we found there the complex conjugate of the above expression.
It follows that the unit at $\a=1/3$ is given by
\be
\label{237unit}
\ve((-2,3,7))_{1/3} = \e(2/3) \,.
\ee
The units of $5_2$ and $(-2,3,7)$ at $\a=1/3$ match up to a $24$-th root of

As mentioned in Section~\ref{sub.p}, the $(-2,3,7)$ pretzel knot
has 6 parabolic nonabelian representations that come in two Galois orbits
of size 3 each: one is defined over the trace field (the cubic field of
discriminant $-23$ discussed above), and another defined over the real
field $\BQ(\eta)$, the abelian field of discriminant 49. At first glance,
the latter three parabolic representations (which are $\SL_2(\BR)$
representations of zero volume) are not seen by the Kashaev invariant. Yet,
one can detect them using the asymptotics of the coefficients of the former
three representations as explained in Section~\ref{sub.optimal}.

In the subsequent paper~[41], we used the 6 pairs of $q$-series
associated the $(-2,3,7)$ pretzel knot and their asymptotics to compute 37
terms of all six series $\Phi^{((-2,3,7),\s_j)}_{0}(h)$ for $j=1,\dots,6$.
Below, we give the first 11 terms of the series associated to the abelian
number field $\BQ(\eta)$ given in Section~\ref{sub.p}.
%% C7=[23,-13,11]*Pi*I/84; CC = concat(C23, C7); V=I*Pi*CC; 2*V
%% e7 = vector(3,j,2*cos(2*Pi*j/7));
%% Y = concat(vector(3,j,(a[j]^2 - 3)^3/a[j]^5),[7,7,7]);
Consider the embeddings $\s_{3+j}$ of the above field for $j=1,2,3$
given in Section~\ref{sub.p} which send $\eta$ to $2\cos(2\pi j/7)$
and let
%Let $\eta=2\cos(2\pi/7)$ be a generator of the ring of integers of the
%number field $\BQ(\cos(2\pi/7))$ and
$C^{((-2,3,7),\s_{3+j})}_0 = \sqrt{(\eta_j-2)/14}$. The complex volumes
of $\s_{3+j}$ are given by
\V(\s_4)= -\frac{1}{21} \pi^2, \qquad \V(\s_5)= \frac{1}{14} \pi^2, \qquad
\V(\s_6)= -\frac{1}{42} \pi^2
and the torsion equals to $\delta((-2,3,7), \s_{3+j})=14/(\eta_j-2)$.
Using the notation of~\eqref{PhiA}, we write
7^k \, D_k \, \tiA^{((-2,3,7),\s_{3+j})}_{0}(k)
= (1,\eta_j,\eta_j^2)\cdot B^{((-2,3,7),\s_{3+j})}_{0}(k)
where $B^{((-2,3,7),\s_{3+j})}_{0}(k) \in \BZ^3$ is a vector of integers with
the first 11 values given by

\begin{array}{|c|l|} \hline
k & B^{((-2,3,7),\s_{3+j})}_{0}(k) \\ \hline
0 &  \la 1, 0, 0\ra
\\ \hline
1 & \la 43, 0, -21\ra
\\ \hline
2 & \la 3928, 63, -1491\ra
\\ \hline
3 & \la -9658210, -2570400, 8759835\ra
\\ \hline
4 & \la -12802855375, 9661452255, 660110430\ra
\\ \hline
5 & \la -42833879089694, 5736063757095, 23026249581258\ra
\\ \hline
6 & \la -360522404258392495, -58094689166990595, 278695629206010765\ra
\\ \hline
7 & \la 108480519886094978165, 114336214602228319050, -161431920455740612440\ra
\\ \hline
8 & \la 420957357301236147078125, -601694281205047856100870, 211820529501946639071105\ra
\\ \hline
9 & \la 276051903390093831791757795950, -105329146895536652560323534375, -93062298372659896456977171525\ra
\\ \hline
10 & \la 3837169849511929903158156720021580, 1712034755788650551262940860512280,
\\ &
\ra
\\ \hline
\end{array}

\subsection*{The $6_1$ knot}
\label{sub.61}

%% Recursion relation prepared with:
%% zagier/Prepare.Recursion.For.Kashaev.Pari.nb
%% zagier/README.Prepare.Recursion.For.Kashaev.Pari

%% Dropbox/qKnots/Kashaev/61knot/61modularity, 61modularity.NEWER

%% Technically, the asymptotics at $\a=0$ up to second order were
%% given by the paper with Tudor and matched the numerical values.

% sage: L=Link('6_1').mirror()
% sage: L.braid_word()
% [1, 2, 3, 2, -4, -1, -3, 2, -3, 4, -3, 2]
% sage: L.jones_polynomial()
%q^-4 - q^-3 + q^-2 - 2*q^-1 + 2 - q + q^2
% sage: M=L.exterior()
% sage: M.complex_volume()
% 3.16396322888315 - 3.07886290184117*I
% sage: K=M.tetrahedra_field_gens()
% sage: K.find_field(prec=200, degree=10, optimize=True)
% (Number Field in z with defining polynomial x^4 + x^2 - x + 1,
% <ApproxAN: 0.547423794586 + 0.58565197969*I>,
% [2*z^3 + 2*z^2 + 3*z, z^3 + z^2 + z + 1, 1/8*z^3 + 3/8*z^2 + 1/4*z + 5/8,
% z^2])

% pari:
% VC = 3.16396322888314398399101 - 16.23833543662698312044*I
% cs = - 3.07886290184117*I
% (imag(VC)-cs)/Pi^2
% -1.333333333333333
% imag(VC)-(cs-4*Pi^2/3) \\ -1.6286 E-15

In this section, we look at one further knot (this time without a sister), the
$6_1$ knot for two reasons. Firstly, the trace field
is $F_{6_1} = \BQ(\xi)$, a number field of discriminant 257 (a prime) where
\xi^4 + \xi^2 - \xi + 1=0, \qquad \xi=0.5474\dots + 0.5856\dots i
The trace field has two complex embeddings, hence its Bloch group has rank
two, giving a nontrivial test for the unit $\ve(6_1)_\a$. Secondly,
the $\SL_2(\BC)$ character variety (and the corresponding $A$-polynomial)
is a curve whose quotient modulo the involution
$\iota:(M,L)\mapsto (M^{-1},L^{-1})$ is not a rational curve.
It was observed by Borot that his recent work with Eynard
[7] suggested a mechanism (based on the {\em topological recursion})
that could explain at least a weak part of the Modularity Conjecture, namely
that the asymptotics of $\J^K(\ep)$ (as $\ep$ tends to zero through rational
numbers with bounded denominators), is always given by the same series
$\Phi^K_0(\ep)$ up to a constant factor, not predicted by their model.
However, Borot could make this argument precise only in the case where the
space of holomorphic differentials of the corresponding spectral curve
was anti-invariant under the involution
$\iota:(M,L)\mapsto (M^{-1},L^{-1})$. This condition is equivalent to
the statement to the rationality of the quotient of the spectral curve by
$\iota$. This led us to conduct a final experiment for the $6_1$ knot.
The question here was whether the Modularity Conjecture itself might fail,
or had to be modified in the context where the argument based on the work
of Borot-Eynard no longer applied. Fortunately, however, we found no anomalies.

To fix conventions, the $6_1$ knot is the closure of the braid word
{\tiny $[1, 2, 3, 2, -4, -1, -3, 2, -3, 4, -3, 2]$} where $j$ (respectively,
$-j$) corresponds to the standard generator $s_j$ (respective, $s_j^{-1}$)
of the braid group in 4 stands, and
%$s_1 s_2 s_3 s_2 s_4^{-1} s_1^{-1} s_3^{-1} s_2 s_3^{-1} s_4 s_3^{-1} s_2$
with Jones polynomial $q^{-4} - q^{-3} + q^{-2} - 2 q^{-1} + 2 - q + q^2$.
The complex volume is given by:
% see pari file don-own-cloud/Knots/Check-52-237-PhiSeries
% and pari/zagier/qKnots/Kashaev/61knot/61modularity.DEEXPONENTIATED
\V(6_1)=  -(2 R_1+R_2+R_3) - \frac{4}{3} \pi^2
= -3.0788629 \ldots + 3.1639632 \ldots i
\begin{align*}
R_1 &= R(-\xi^2) - \frac{1}{2} \pi i \log(-\xi^2) + \pi i \log(1+\xi^2) \\
R_2 &= R(1-\xi^3)- \pi i \log(1-\xi^3) + \frac{1}{2} \pi i \log(\xi^3) \\
R_3 &= R(1-\xi) \,.
\end{align*}
The torsion, a prime in $F_{6_1}$ of norm 257 is given by
\delta(6_1)= 1+\xi+4 \xi^2+\xi^3 \,.

Here, the Kashaev invariant of the $6_1$ knot was computed not
using the formula given in Equation~(24) of~[56] (since the
latter is an $O(N^3)$ computation), but it was computed in $O(N)$ steps
using the recursion relation of the colored Jones polynomial.
The rather complicated inhomogeneous recursion has order $4$,
has maximal degree $(4,15,31)$ with respect to the shift variable, the
$q^n$ and the $q$ variables, and contains a total of 346 terms, which
can be found in~[37, 23].
% see Mathematica file: mathematica/sun/APolyTwistKnots/AqPolyTwistKnots.nb
In contrast, the $A$-polynomial of the $6_1$ knot has maximal degree
$(4,8)$ with respect to the $(L,M)$ variables and contains 21 terms.
Due to the complexity of the recursion, we were forced to use precision
3000 in \texttt{pari} when $N=1000$. The first three coefficients of
$\Phi^{6_1}_0(h)$ were numerically computed at $\a=0$, and using the
prediction of~[14] and the notation of`\eqref{Phiprec2},
those algebraic numbers were identified as follows:
\begin{align*}
\Phi^{6_1}_0(h) & \= \frac{1}{\sqrt{\delta}} \Bigl( 1
+ \frac{194 \xi^3 - 331 \xi^2 + 207 \xi - 245}{2^3\cdot 3 \cdot \delta(6_1)^3} h
\\
& \quad
+ \frac{-154734 \xi^3 - 34354 \xi^2 + 127399 \xi - 119864}{2^7 \cdot 3^2
\cdot \delta(6_1)^6}h^2
+ O(h^3) \Bigr) \,.
\end{align*}

% A0 = (1+a+4*a^2+a^3)/2;   \\  = 0.487465... + 1.738045...*I
% AA = - sqrt(I/2/A0);  AA - A    \\  = 5.843 E-13 + 9.118 E-14*I,
% so still 14 digits accuracy

% for(M=1,2,print(M,"  ",factor(1/content(polcoeff(p,M)))))
% 1  [2, 3; 3, 1]
% 2  [2, 7; 3, 2]

% for(M=1,2,print(M,"  ",polcoeff(p,M)/content(polcoeff(p,M))))
% 1  194*t^3 - 331*t^2 + 207*t - 245
% 2  -154734*t^3 - 34354*t^2 + 127399*t - 119864

\bibliographystyle{plain}
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\end{thebibliography}

\end{document}