text:
\section{Introduction}
\par Let ${\mathfrak H}$ be the complex upper half plane and let $\Gamma_1(N)$
be a congruence subgroup of $SL_2({\mathbb Z})$ whose elements are
congruent to $\left(\smallmatrix 1&*\\0&1 \endsmallmatrix\right) \mod N \, (N=1,2,\cdots)$. Since
the group $\Gamma_1(N)$ acts on ${\mathfrak H}$ by linear fractional
transformations, we get the modular curve
$X_1(N)=\Gamma_1(N)\backslash{\mathfrak H}^*$, as a projective closure
of the smooth affine curve $\Gamma_1(N)\backslash{\mathfrak H}$, with
genus $g_{1,N}$.
 Here, ${\mathfrak H}^*$ denotes the union of ${\mathfrak H}$ and $\mathbb
 P^1({\mathbb Q})$.
 \par Ishida and Ishii showed in \cite{Ishida-Ishii}
 that for $N\ge 7$, the function field $K(X_1(N))$ is generated
 over ${\mathbb C}$ by the modular functions $X_2(z,N)^{\epsilon_N
 \cdot N}$ and $X_3(z,N)^N$, where
 $X_r(z,N)=e^{2\pi i \frac{(r-1)(N-1)}{4N}} \prod_{s=0}^{N-1}
 \frac{K_{r,s}(z)}{K_{1,s}(z)}$ and $\epsilon_N$ is $1$ or $2$
 according as $N$ is odd or even. Here, $K_{r,s}(z)$ is a Klein
 form of level $N$ for integers $r$ and $s$ not both congruent to
 $0 \mod N$.
 On the other hand, since the genus $g_{1,N}=0$ only for the
eleven cases $1\le N \le 10$ and $N=12$ (\cite{Miyake},
\cite{Chang2}), the function field $K(X_1(N))$ in this case is a
rational function field ${\mathbb C}(j_{1,N})$ for some modular
function $j_{1,N}$ (Table 3, Appendix).
 \par The element $\left(\smallmatrix 1&1 \\ 0&1 \endsmallmatrix\right)$ of $\Gamma_1(N)$ takes
 $z$ to $z+1$, and in particular a modular function $f$ in
 $K(X_1(N))$ is periodic. Thus it can be written as a Laurent series in
 $q=e^{2\pi i z}$ ($z\in {\mathfrak H}$), which is called a $q$-{\it series} \, (or
 $q$-{\it expansion}) \, of $f$. We call $f$ {\it normalized} \, if
 its $q$-series starts with $q^{-1}+0+a_1 q+a_2 q^2+\cdots$. By a
 {\it Hauptmodul} $t$ we mean the normalized generator of
 a genus zero function field $K(X_1(N))$ and
 we write $t=q^{-1}+0+\sum_{k\ge 1} H_k q^k $ for its $q$-series.
\par For a Fuchsian group $\Gamma$, let $\overline{\Gamma}$ denote
the inhomogeneous group of $\Gamma$ ($=\Gamma/\pm I$). Let
$\Gamma_0(N)$ be the Hecke subgroup given by $\{ \left(\smallmatrix a&b\\c&d \endsmallmatrix\right)
\in SL_2({\mathbb Z}) \mid c\equiv 0 \mod N \}$. Also, let $t={\mathcal
N}
 (j_{1,N})$ be the
Hauptmodul of $\Gamma_1(N)$ and $X_n(t)$ be a unique polynomial in
$t$ of degree $n$ such that $X_n(t)-\frac 1n q^{-n}$ belongs to
the maximal ideal of the local ring ${\mathbb C}[[q]]$. Polynomials
with this property are known as the Faber polynomials
(\cite{Duren}, Chapter 4). Write $X_n(t)=\frac 1n q^{-n} +
\underset{m\ge 1}{\sum} H_{m,n} q^m$.
 \par When
 $\overline{\Gamma}_1(N)=\overline{\Gamma}_0(N)$, ${\mathcal N}
 (j_{1,N})$ becomes a replicable function, that is, it satisfies
 the following replication formulae
 \begin{equation*}
  H_{a,b}=H_{c,d} \text{ \hskip0.17cm \hskip0.17cm
 whenever $ab=cd$ and
 $(a,b)=(c,d)$} \tag{$*$}
 \end{equation*} (\cite{ACMS}, \cite{Cummins-Norton}, \cite{Koike}).
  Given a replicable function $f$
  the {\it $n$-plicate} of $f$ is defined iteratively by
 $$ f^{(n)}(nz)= -\sideset{}{'}\sum_{\smallmatrix ad=n \\ 0\le b < d
 \endsmallmatrix}
   \, f^{(a)}\left(\frac{az+b}{d}\right)+nX_n(f) $$
 where the primed sum means that the term with $a=n$ is omitted (\cite{Cummins-Norton}).
  We call $f$ {\it completely replicable} if $f$ is a replicable
 function with rational integer coefficients and has only a finite
 number of distinct replicates, which are themselves replicable functions.
 According to \cite{ACMS} there are, excluding the trivial cases $q^{-1}+aq$,
 326 completely replicable functions of which 171 are monstrous
 functions, i.e.,  modular
 functions whose $q$-series coincide with the Thompson series
 $T_g(q)=\sum_{n\in {\mathbb Z}} {\rm Tr}(g| V_n)q^n$ for some element
 $g$ of the monster simple group $M$ whose order is approximately
 $8\cdot 10^{53}$. Here we observe that $V=\underset{n\in {\mathbb Z}}{\bigoplus} V_n$
 is the infinite dimensional graded representation of $M$
 constructed by Frenkel {\it et al.} (\cite{Frenkel1},
 \cite{Frenkel2}).
 Furthermore, in \cite{Cummins-Norton} Cummins and Norton showed that
 if $f$ is replicable, it can be
 determined only by the 12 coefficients of its first 23 ones.
 \par If $\overline{\Gamma}_1(N)\neq
 \overline{\Gamma}_0(N)$, unlike those replicable functions mentioned above,
 we show in \S 3 that the Fourier coefficients of $X_n(t)$ with $t={\mathcal N}(j_{1,N})$
 ($N \neq 7, 9$)
  satisfy a twisted
 formula (\ref{super}) by a character $\psi$ (see Corollary
 \ref{Repli2}). Here we note that when we work with the
 Thompson series, it is reduced to replication formulae in $(*)$ by
 viewing $\psi$ as the trivial character. Thus in this sense it gives
 a more general class of modular functions, which we
 propose to call ${\mathcal N}(j_{1,N})$ a {\it super-replicable}
 function.
 \par There would be certain similarity between some of
 replicable functions and super-replicable ones as follows.
 We derived in \cite{Kim-KooA}
 the following self-recursion formulas for the Fourier
 coefficients of ${\mathcal N}(j_{1,N})$ without the aid
 of its 2-plicate when
 $N=2,6,8,10,12$:  for $k\ge 1$,
\begin{align*}
H_{4k-1} &= \frac{H_{2k-1}}{2} + 2\sum_{1\le j\le k-1}
H_{2j}H_{4k-2j-2}
          +\alpha\cdot H_{4k-2} - \frac{{H_{2k-1}}^2}{2}
      -\sum_{1\le j\le 2k-2} H_j H_{4k-j-2} \\
H_{4k}   &= -\beta\cdot H_{4k-2} - \sum_{1\le j < 2k-1} H_j
H_{2(2k-j-1)}\\ H_{4k+1} &= \frac{H_{2k}}{2} + 2\sum_{1\le j <k}
H_{2j}H_{4k-2j}
          +\alpha\cdot H_{4k} + \frac{{H_{2k}}^2}{2}
      -\sum_{1\le j <2k} H_j H_{4k-j} \\
H_{4k+2} &= -\beta\cdot H_{4k} - \sum_{1\le j < 2k} H_j
H_{2(2k-j)}
\end{align*}
where $\alpha=-{\mathcal N}(j_{1,N})(\frac{1+N/2}{N})$ and
      $\beta =-{\mathcal N}(j_{1,N})(\frac{1}{N/2})$.
 Furthermore, we verified in \cite{Kim-KooB} that the above recursion
 can be also applied to 14 monstrous functions of even levels
 (including $\mathcal
 N(j_{1,2})$ and ${\mathcal N}(j_{1,6})$)
  which are Thompson series of type
 $2B$, $6C$, $6E$, $6F$, $10B$, $10E$, $14B$, $18C$, $18D$, $22B$,
 $30C$, $30G$, $42C$, $46AB$ (these are all replicable functions)
 and one monster-like function of type $18e$
 (for the definition of monster-like
 function, we refer to \cite{Ferenbaugh}).
 Therefore the Hauptmoduln mentioned above which have
 self-recursion formulas can be determined just by the first four
 coefficients $H_1$, $H_2$, $H_3$ and $H_4$ without the aid of
 2-plicate. What is more interesting would be the fact that there seems
 to be a connection between super-replicable functions and
 infinite dimensional Lie superalgebras. That is, considering the
 arguments from Borcherds \cite{Borcherds}, Kang \cite{Kang} and
 Koike \cite{Koike} we believe that the super-replication formulae
 in (\ref{super}) might suggest the existence of certain
 infinite dimensional Lie superalgebra whose denominator identity
 implies such formulae.

 \par Lastly, as an application of super-replication formulae
  we consider the following periodically
 vanishing property.
 Many of monstrous functions, for example, Thompson series of type
 $4B$, $4C$, $4D$, $6F$, $8B$, $8C$, $8D$, $8E$, $8F$, $9B$, etc
 have periodically vanishing properties among the Fourier coefficients
 (see the
 Table 1 in \cite{Mckay-Strauss}). This result must be known to
 experts, but we could not find a reference. Hereby we describe it
 in Theorem \ref{VVV}.
 Meanwhile, as for the case of super-replicable functions, we see from the
 Appendix, Table 4 that only the Haupmodul
 ${\mathcal N}(j_{1,12})$ seems to have such property.
 To this end, we shall first derive in \S 4
 an identity (\ref{twisted-replication}) which
 is analogous to the ``$2^k$-plication formula" (\cite{Ferenbaugh2}, \cite{Koike})
 satisfied
 by replicable functions. And, combining this with the super-replication formulae
  we are able to verify that the Fourier
 coefficients $H_m$ of ${\mathcal N}(j_{1,12})$ vanish whenever $m\equiv
 4 \mod 6$ (Corollary \ref{Replication3}).

 \par Through the article we adopt the following notations:
 \par\noindent $\bullet$ \, $S_{\Gamma_1(N)}$ \hskip0.17cm the set of
 $\Gamma_1(N)$-inequivalent cusps
 \par\noindent $\bullet$ \, $q_h=e^{2\pi iz/h}, \hskip0.17cm z\in {\mathfrak H}$
 \par\noindent $\bullet$ \, $f|_{\left(\smallmatrix a&b \\ c&d
 \endsmallmatrix\right)}=f\left(\frac{az+b}{cz+d}\right)$
 \par\noindent $\bullet$ \,
     $f(z)=g(z)+O(1) \hskip0.17cm \text{ means that $f(z)-g(z)$ is bounded as $z$
                    goes to $i\infty$. }$
\section{Hauptmodul of $\Gamma_1(12)$}
 \par
 In this section we investigate the generalities of the modular
 function $j_{1,12}$ which is under primary consideration and
 construct the Hauptmodul ${\mathcal N}(j_{1,12})$. We also examine
 some number theoretic property of ${\mathcal N}(j_{1,12})$. As for
 more arithmetic properties, we refer to \cite{Hong-Koo}.
\begin{lem}
Let $\frac ac$ and $\frac{a'}{c'}$ be fractions in lowest terms.
Then $\frac ac$ is $\Gamma_1(N)$-equivalent to $\frac{a'}{c'}$ if
and only if \hskip0.17cm $\pm\left(\smallmatrix a' \\ c' \endsmallmatrix\right) \equiv
           \left(\smallmatrix a+nc \\ c \endsmallmatrix\right) \mod N$
           for some $n\in{\mathbb Z}$.
       \label{Equiv2}
\end{lem}
\begin{pf} Straightforward.
\end{pf}
Using the above lemma \, we can check that the cusps
$$ 0, \, 1/2,  \, 1/3, \, 1/4, \, 1/5, \, 1/6, \, 1/8, \,
   1/9, \, \infty, \, 5/12 $$
are $\Gamma_1(12)$-inequivalent. But from \cite{Chang2} we know
that the cardinality of $S_{\Gamma_1(12)}$ is 10, whence
$$ S_{\Gamma_1(12)}
   = \{0, \, 1/2, \, 1/3, \, 1/4, \, 1/5, \, 1/6, \, 1/8, \,
       1/9, \, \infty, \, 5/12 \}. $$
For later use we are in need of calculating the widths of the
cusps of $\Gamma_1(12)$.
\begin{lem} Let \, $a/c\in {\mathbb P}^1({\mathbb Q})$ be a cusp where $(a,c)=1$. Then
            the width of \, $a/c$ \, in $X_1(N)$ is given by $N/(c,N)$
        if $N\neq 4$.
        \label{Width}
\end{lem}
\begin{pf} If $N\mid 4$, the statement is obvious.
Hence, we assume that $N$ does not divide 4, i.e., $N\neq 1,2,4$.
First, choose $b$ and $d$ such that $\left(\begin{smallmatrix} a&
b \\ c& d \end{smallmatrix}\right)\in SL_2({\mathbb Z})$. Let $h$ be
the width of the cusp $a/c$. Then $h$ is the smallest positive
integer such that
$$ \begin{pmatrix} a&b \\ c&d \end{pmatrix}
   \begin{pmatrix} 1&h \\ 0&1 \end{pmatrix}
   {\begin{pmatrix} a&b \\ c&d \end{pmatrix}}^{-1} \in \pm \Gamma_1(N). $$
Thus we have
$$ \begin{pmatrix} 1-cah & * \\ -c^2h & 1+cah \end{pmatrix} \in \pm \Gamma_1(N). $$
If $\begin{pmatrix} 1-cah & * \\ -c^2h & 1+cah \end{pmatrix}$ is an element of
$-\Gamma_1(N)$, by taking trace $2\equiv -2 \mod N$; hence $N\mid
4$. Thus when $N\neq 1,2,4$, $\begin{pmatrix} 1-cah & * \\ -c^2h & 1+cah \end{pmatrix}
\in \Gamma_1(N)$. This condition is equivalent to saying that
$$ h\in \frac{N}{(c^2,N)} {\mathbb Z} \, \bigcap \,
        \frac{N}{(ca,N)} {\mathbb Z} \, = \, \frac{N}{(c,N)} {\mathbb Z}. $$
\end{pf}
We then have the following table of inequivalent cusps of
$\Gamma_1(12)$:
\begin{center}
{\bf Table 1}. \par
\begin{tabular}{|r|r|r|r|r|r|r|r|r|r|r|} \hline
 cusp & $\infty$ & $0$ & $1/2$ & $1/3$ & $1/4$ & $1/5$
         & $1/6$ & $1/8$ & $1/9$ & $5/12$ \\ \hline
 width &   1      &  12  &    6   &   4  &   3   &    12
          &   2  &    3   &   4  &   1    \\ \hline
\end{tabular}
\end{center}
Recall the Jacobi theta functions $\theta_2,\theta_3,$ and
$\theta_4$ defined by
\begin{align*}
    \theta_2(z) & =  \sum_{n\in{\mathbb Z}} q_2^{(n+\frac{1}{2})^2} \\
    \theta_3(z) & =  \sum_{n\in{\mathbb Z}} q_2^{n^2} \\
    \theta_4(z) & =  \sum_{n\in{\mathbb Z}} (-1)^nq_2^{n^2}
\end{align*}
   for $z\in{\mathfrak H}$.
We have the following transformation formulas (\cite{Rankin}
pp.218-219).
\begin{align}
       \theta_2(z+1) &= e^{\frac{1}{4}\pi i}\theta_2(z) \label{T1} \\
       \theta_3(z+1) &= \theta_4(z) \label{T2} \\
       \theta_4(z+1) &= \theta_3(z)  \label{T3} \\
       \theta_2\left(-\frac{1}{z}\right)
       &= (-iz)^{\frac{1}{2}}\theta_4(z) \label{I1} \\
       \theta_3\left(-\frac{1}{z}\right)
       &= (-iz)^{\frac{1}{2}}\theta_3(z)  \label{I2} \\
       \theta_4\left(-\frac{1}{z}\right)
       &= (-iz)^{\frac{1}{2}}\theta_2(z). \label{I3}
\end{align}
\begin{lem} Let $k$ be an odd positive integer and
       $N$ be a multiple of $4$. Then for
       $F(z)\in M_{\frac k2}(\widetilde{\Gamma}_0(N))$
       and $m\ge 1$,
       $F(mz)\in M_{\frac k2}(\widetilde{\Gamma}_0(mN),\chi_m)$
       with $\chi_m(d)=\left(\frac{m}{d}\right)$ and $(d,m)=1$.
       \label{half2}
\end{lem}
\par\smallskip\noindent{\bf{\it Proof.}}\hskip0.4cm
\cite{Shimura2}, Proposition 1.3.
\par\noindent
Put $j_{1,12}(z)=\theta_3(2z)/\theta_3(6z)$.
\begin{theo}
(a) $\theta_3(2z)\in M_{\frac 12}(\widetilde{\Gamma}_0(4))$ and
    \, $\theta_3(6z)\in M_{\frac 12}(\widetilde{\Gamma}_0(12),\chi_3)$.
\par\noindent
(b) $K(X_1(12))$ is equal to \, ${\mathbb C}(j_{1,12}(z))$.
    $j_{1,12}$ takes
    the following value at each cusp:
    $j_{1,12}(\infty)=1$, $j_{1,12}(0)=\sqrt{3}$,
    $j_{1,12}(\frac12)=0$ (a simple zero),
    $j_{1,12}(\frac13)=i$, $j_{1,12}(\frac14)=\sqrt{3} i$,
    $j_{1,12}(\frac15)=-\sqrt3$,
    $j_{1,12}(\frac16)=\infty$ (a simple pole),
    $j_{1,12}(\frac18)=-\sqrt3 i$,
    $j_{1,12}(\frac19)=-i$,
    $j_{1,12}(\frac{5}{12})=-1$.
\label{Haupt2}
\end{theo}
\begin{pf} For the first part, we recall that (\cite{Koblitz}, p.184)
$$ \theta_3(2z)\in M_{\frac 12}(\widetilde{\Gamma}_0(4)).$$
Then by Lemma \ref{half2} we immediately get that
$$ \theta_3(6z)\in M_{\frac 12}(\widetilde{\Gamma}_0(12),\chi_3).$$
By the assertion (a), it is clear that $j_{1,12}(z)\in
K(X_1(12)).$ Thus for (b), it is enough to show that \,
$j_{1,12}(z)$ has only one simple zero and one simple pole on the
curve $X_1(12)$. As is well-known, $\theta_3(z)$ never vanishes
on ${\mathfrak H}$. Hence we are forced to investigate the zeroes and
poles of \, $j_{1,12}$ at each cusp of $\Gamma_1(12)$. Let $S=\left(\smallmatrix
0 & -1 \\ 1 & 0 \endsmallmatrix\right)$ and $T=\left(\smallmatrix 1 & 1 \\ 0 & 1 \endsmallmatrix\right)$.
\par\noindent
(i) $s=\infty$:
   $$ j_{1,12}(\infty)=\lim_{z\to i\infty}\frac{\theta_3(2z)}{\theta_3(6z)}
              =\lim_{q\to 0}\frac{1+2q+2q^4+\cdots}{1+2q^3+2q^{12}+\cdots}
              =1. $$
\par\noindent
(ii) $s=0$:
 \begin{align*}
   j_{1,12}(0)  &=\lim_{z\to i\infty}
             {\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{ \hskip0.17cm S} \\
           &=\lim_{z\to i\infty}
         \frac{\sqrt{-i\frac z2} \hskip0.17cm \theta_3(\frac z2)}
                      {\sqrt{-i\frac z6} \hskip0.17cm \theta_3(\frac z6)}
            \hskip0.17cm \hskip0.17cm \text{ by (\ref{I2})}  \\
               &=\sqrt{3}.
 \end{align*}
\par\noindent
(iii) $s=\frac12$:
 We Observe that $(ST^{-2}S)\infty=\left(\smallmatrix -1&0\\-2&-1 \endsmallmatrix\right)\infty=\frac12$.
 \par\noindent \hskip0.17cm Considering the identities
 \begin{align*}
           \theta_3(2z)^2|_{[S]_1}
           &= z^{-1}\theta_3\left(-\frac{2}{z}\right)^2
           = z^{-1}\lbrace\left(-\frac{iz}{2}\right)
           ^{\frac{1}{2}}\theta_3\left(\frac{z}{2}\right)\rbrace^2
           \text{ \hskip0.17cm by (\ref{I2})} \\
           &= -\frac i2 \theta_3\left(\frac{z}{2}\right)^2 \\
           \theta_3(2z)^2|_{[ST^{-2}]_1}
           &= -\frac{i}{2}\theta_3\left(\frac{z}{2}\right)^2|_{[T^{-2}]_1}
               = -\frac{i}{2}\theta_4\left(\frac{z}{2}\right)^2
               \text{ \hskip0.17cm by (\ref{T3}) }\\
           \theta_3(2z)^2|_{[ST^{-2}S]_1}
           &= -\frac{i}{2}\theta_4\left(\frac{z}{2}\right)^2|_{[S]_1}
           = -\frac{i}{2}z^{-1}\{(-2iz)^{\frac{1}{2}}\theta_2(2z)\}^2
           \text{ \hskip0.17cm by (\ref{I3})}\\
           &= -\theta_2(2z)^2,
     \end{align*}
     we get that
     \begin{align*}
           \theta_3(2z)^2|_{s=\frac12}
           &=\lim_{z\to i\infty}\theta_3(2z)^2|_{[ST^{-2}S]_1}
              =\lim_{z\to i\infty}-\theta_2(2z)^2   \\
           &=\lim_{z\to i\infty} -2^2 \hskip0.17cm q_2(1+q^2+q^6+q^{12}+\cdots)^2
              \hskip0.17cm \text{ since } \theta_2(z)= 2q_8(1+q+q^3+\cdots) \\
           &=0 \text{\hskip0.17cm (a triple zero). }
     \end{align*}
     On the other hand
     \begin{align*}
           \theta_3(2z)^2|_{\left(\smallmatrix -1&0\\-2&-1 \endsmallmatrix\right)_1}
       & =\frac{1}{\sqrt3}\, \theta_3(2z)^2|_{\left(\smallmatrix 3&0\\0&1 \endsmallmatrix\right)_1
       \left(\smallmatrix -1&0\\-2&-1 \endsmallmatrix\right)_1}
         =\frac{1}{\sqrt3}\, \theta_3(2z)^2
              |_{\left(\smallmatrix 1&1\\0&1 \endsmallmatrix\right)_1\left(\smallmatrix -1&0\\-2&-1 \endsmallmatrix\right)_1
          \left(\smallmatrix 1&-1\\0&3 \endsmallmatrix\right)_1} \\
       & =\frac{1}{\sqrt3}\, \theta_3(2z)^2|_{\left(\smallmatrix -1&0\\-2&-1 \endsmallmatrix\right)_1
       \left(\smallmatrix 1&-1\\0&3 \endsmallmatrix\right)_1}
         =-\frac{1}{\sqrt3}\, \theta_2(2z)^2
              |_{\left(\smallmatrix 1&-1\\0&3 \endsmallmatrix\right)_1} \\
       & =-3^{-1} \, \theta_2\left(2\cdot \frac{z-1}{3}\right)^2
         =-3^{-1} e^{-\frac{\pi i}{3}} \, q_6(1+\cdots)^2,
     \end{align*}
 so that $\theta_3(6z)^2$ has a simple zero at $\frac12$. Thus $j_{1,12}^2
 =\frac{\theta_3(2z)^2}{\theta_3(6z)^2}$ has a double zero at $\frac12$, whence
 $j_{1,12}$ has a simple zero at $\frac12$.

\par\noindent
(iv) $s=\frac 13$: $(ST^{-3}S) \infty=\frac13$.
 First we recall that
  ${\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{ \hskip0.17cm S}
   =\sqrt3 \, \frac{\theta_3(\frac z2)}
                      {\theta_3(\frac z6)}
            \hskip0.17cm \hskip0.17cm \text{ by (\ref{I2})}.  $
 Observe that $\theta_2(2z)=\frac12 (\theta_3(\frac z2)-\theta_4(\frac z2))$
             and $\theta_3(2z)=\frac12 (\theta_3(\frac z2)+\theta_4(\frac z2))$.
 From these identities we can write
      \, $\theta_3(\frac z2)=\theta_2(2z)+\theta_3(2z)$ and
      \, $\theta_3(\frac z6)=\theta_2(\frac23 z)+\theta_3(\frac23 z)$.
 Then we have that
 \begin{align*}
   {\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{ \hskip0.17cm ST^{-3}}
    &= \sqrt{3}\cdot{\frac{\theta_2(2z)+\theta_3(2z)}{\theta_2(\frac23 z)+\theta_3(\frac23 z)}
               \vline}_{ \hskip0.17cm T^{-3}}   \\
    &=\sqrt{3}\cdot\frac{(e^{-\frac{\pi i}{4}})^6 \theta_2(2z)+\theta_3(2z)}
                 {(e^{-\frac{\pi i}{4}})^2 \theta_2(\frac23 z)+\theta_3(\frac23 z)}
       \text{ \hskip0.17cm by (\ref{T1}), (\ref{T2}) and (\ref{T3}),}\\
    &=\sqrt3\cdot\frac{i\theta_2(2z)+\theta_3(2z)}{-i\theta_2(\frac23 z)+\theta_3(\frac23 z)}
 \end{align*}
 and
 \begin{align*}
   {\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{ \hskip0.17cm ST^{-3}S}
  &= \sqrt3\cdot
    {\frac{i\theta_2(2z)+\theta_3(2z)}{-i\theta_2(\frac23 z)+\theta_3(\frac23 z)}\vline}
    _{\hskip0.17cm S} \\
  &= \sqrt3\cdot\frac{i\theta_4(\frac z2)+\theta_3(\frac z2)}
     {-i\sqrt3\theta_4(\frac32 z)+\sqrt3\theta_3(\frac32 z)}
     \text{ \hskip0.17cm by (\ref{I1}) and (\ref{I2}) }
 \end{align*}
 which goes to $\frac{i+1}{-i+1}=i$ as $z\to i\infty$, so that
 $$ j_{1,12}\left(\frac13\right)=i.$$

\par\noindent
(v) $s=\frac14$: $\left(\smallmatrix 1&0\\4&1 \endsmallmatrix\right) \infty= \frac14$. In this
case we use the following well-known fact from \cite{Koblitz}
p.148 : For $\gamma\in\Gamma_0(4)$ and $z\in{\mathfrak H}$,
$$ \Theta(\gamma z)=\left(\frac cd \right)\sqrt{\left(\frac{-1}{d}\right)}^{-1}
                    \sqrt{cz+d} $$
where $\Theta(z)=\theta_3(2z)$. Then
\begin{align*}
         {\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{ \hskip0.17cm \left(\smallmatrix 1&0\\4&1 \endsmallmatrix\right)}
    &= \frac{\Theta(\left(\smallmatrix 1&0\\4&1 \endsmallmatrix\right) z)}{\Theta(\left(\smallmatrix 3&0\\0&1 \endsmallmatrix\right) \left(\smallmatrix 1&0\\4&1 \endsmallmatrix\right) z)}
     =\frac{\Theta(\left(\smallmatrix 1&0\\4&1 \endsmallmatrix\right) z)}{\Theta(\left(\smallmatrix 3&2\\4&3 \endsmallmatrix\right) \left(\smallmatrix 1&-2\\0&3 \endsmallmatrix\right) z)}\\
    &=\frac{\sqrt{4z+1} \hskip0.17cm \Theta(z)}
       {\left(\frac43\right) i^{-1}\sqrt{4\cdot\frac{z-2}{3}+3}\hskip0.17cm \Theta(\left(\smallmatrix 1&-2\\0&3 \endsmallmatrix\right) z)}\\
    &=\sqrt3 i \, \frac{\Theta(z)}{\Theta\left(\frac{z-2}{3}\right)}
\end{align*}
which tends to $\sqrt3 i$ when $z$ goes to $i\infty$. Therefore
 $$ j_{1,12}\left(\frac14\right)=\sqrt3 i.$$
\par\noindent
(vi) $s=\frac 15$: Because $\left(\smallmatrix 5&1 \\ 24&5 \endsmallmatrix\right) \left(\smallmatrix 0&-1 \\ 1&0
\endsmallmatrix\right)$ sends $\infty$ to $\frac 15$,
 \begin{align*}
   j_{1,12}\left(\frac15\right)  &=\lim_{z\to i\infty}
             {\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{
     \left(\smallmatrix 5&1 \\ 24&5 \endsmallmatrix\right) \left(\smallmatrix 0&-1 \\ 1&0 \endsmallmatrix\right)} \\
           &=\lim_{z\to i\infty}
         -{\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{\left(\smallmatrix 0&-1\\1&0
               \endsmallmatrix\right)}
            \hskip0.17cm \hskip0.17cm \text{ by (a)}  \\
               &=-j_{1,12}(0)=-\sqrt{3}.
 \end{align*}
(vii) $s=\frac16$: Observe that $(ST^{-6}S)\infty=\frac16$.
 \par\noindent \hskip0.17cm Considering the identities
 \begin{align*}
   {\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{ \hskip0.17cm S}
    &= \sqrt{3}\cdot\frac{\theta_3(\frac z2)}{\theta_3(\frac z6)}
       \text{ \hskip0.17cm by } (\ref{I2})\\
   {\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{ \hskip0.17cm ST^{-6}}
    &= \sqrt{3}\cdot{\frac{\theta_3(\frac z2)}{\theta_3(\frac z6)}
               \vline}_{ \hskip0.17cm T^{-6}}
        =\sqrt{3}\cdot\frac{\theta_4(\frac z2)}{\theta_4(\frac z6)}
       \text{ \hskip0.17cm by  (\ref{T2}) and (\ref{T3})} \\
   {\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{ \hskip0.17cm ST^{-6}S}
    &= \sqrt{3}\cdot{\frac{\theta_4(\frac z2)}{\theta_4(\frac z6)}
               \vline}_{ \hskip0.17cm S}
        =\sqrt{3}\cdot\frac{\sqrt{-2iz} \hskip0.17cm \theta_2(2z)}
                   {\sqrt{-6iz} \hskip0.17cm \theta_2(6z)}
       \text{ \hskip0.17cm by } (\ref{I3}),
 \end{align*}
 we have that
 \begin{align*}
         j_{1,12}\left(\frac16\right)
     & =\lim_{z\to i\infty}
        {\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{ \hskip0.17cm ST^{-6}S}
           =\lim_{z\to i\infty}
        \frac{\theta_2(2z)}{\theta_2(6z)} \\
     & =\lim_{z\to i\infty}
        \frac{2\,q_8^2(1+q^2+q^6+\cdots)}{2\,q_8^6(1+q^6+q^{18}+\cdots)}
        \hskip0.17cm\q \text{ since $\theta_2(z)=2\, q_8(1+q+q^3+\cdots)$.}
 \end{align*}
 Thus by Table 1, $j_{1,12}$ has a simple pole at $\frac16$.
\par\noindent
(viii) $s=\frac 18$: Because $\left(\smallmatrix -5&1 \\ -36&7 \endsmallmatrix\right) \left(\smallmatrix 1&0 \\
4&1 \endsmallmatrix\right)$ sends $\infty$ to $\frac 18$,
 \begin{align*}
   j_{1,12}\left(\frac18\right)  &=\lim_{z\to i\infty}
             {\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{
     \left(\smallmatrix -5&1 \\ -36&7 \endsmallmatrix\right) \left(\smallmatrix 1&0 \\ 4&1 \endsmallmatrix\right)} \\
           &=\lim_{z\to i\infty}
         -{\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{\left(\smallmatrix 1&0\\4&1
               \endsmallmatrix\right)}
            \hskip0.17cm \hskip0.17cm \text{ by (a)}  \\
               &=-j_{1,12}\left(\frac14\right)=-\sqrt{3}i.
 \end{align*}
\par\noindent
(ix) $s=\frac 19$: Because $\left(\smallmatrix -5&2 \\ -48&19 \endsmallmatrix\right) \left(\smallmatrix 1&0 \\
3&1 \endsmallmatrix\right)$ sends $\infty$ to $\frac 19$,
 \begin{align*}
   j_{1,12}\left(\frac19\right)  &=\lim_{z\to i\infty}
             {\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{
     \left(\smallmatrix -5&2 \\ -48&19 \endsmallmatrix\right) \left(\smallmatrix 1&0 \\ 3&1 \endsmallmatrix\right)} \\
           &=\lim_{z\to i\infty}
         -{\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{\left(\smallmatrix 1&0\\3&1
               \endsmallmatrix\right)}
            \hskip0.17cm \hskip0.17cm \text{ by (a)}  \\
               &=-j_{1,12}\left(\frac13\right)=-i.
 \end{align*}
\par\noindent
(x) $s=\frac{5}{12}$: \hskip0.17cm $\left(\smallmatrix 5 &2 \\ 12&5 \endsmallmatrix\right) \cdot
\infty=\frac{5}{12}$.
 \begin{align*}
   j_{1,12}\left(\frac{5}{12}\right)  &=\lim_{z\to i\infty}
             {\frac{\theta_3(2z)}{\theta_3(6z)}\vline}_{
     \left(\smallmatrix 5&2 \\ 12&5  \endsmallmatrix\right)} \\
           &=\lim_{z\to i\infty}
         -\frac{\theta_3(2z)}{\theta_3(6z)}
            \hskip0.17cm \hskip0.17cm \text{ by (a)}  \\
               &=-1.
 \end{align*}
\end{pf}
We will now construct the Hauptmodul ${\mathcal N}(j_{1,12})$ from
the modular function $j_{1,12}$ mentioned in Theorem \ref{Haupt2}.
\begin{align*}
\frac{2}{j_{1,12}(z)-1}
  & =\frac{2\hskip0.17cm\theta_3(6z)}{\theta_3(2z)-\theta_3(6z)}
   =\frac{2(1+2q^3+2q^{12}+2q^{27}+\cdots)}
      {2q-2q^3+2q^4+2q^9-2q^{12}+\cdots} \\
  & =\frac1q+q+q^2+q^3-q^6-q^7-q^8-q^9+q^{11}+2q^{12}+\cdots,
\end{align*}
which is in \, $q^{-1}{\mathbb Z}[[q]]$. From the uniqueness of the
normalized generator it follows that ${\mathcal
N}(j_{1,12})=\frac{2}{j_{1,12}-1}$.
 By Theorem \ref{Haupt2}-(b) we have the following
table:
\begin{center}
{\bf Table 2}. Cusp values of $j_{1,12}$ and ${\mathcal N}(j_{1,12})$
\par
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline
 $s$ & $\infty$ & $0$ & $1/2$ & $1/3$ & $1/4$ & $1/5$
                      & $1/6$ & $1/8$ & $1/9$ & $5/12$  \\ \hline
 $j_{1,12}(s)$ & $1$ & $\sqrt3$ & $0$ & $i$ & $\sqrt3 i$ & $-\sqrt3$
                      & $\infty$ & $-\sqrt3 i$ & $-i$ & $-1$ \\ \hline
${\mathcal N}(j_{1,12})(s)$ & $\infty$ & $\sqrt3+1$ & $-2$ & $-1-i$ &
$\frac{-1-\sqrt3 i}{2}$ & $1-\sqrt3$
                      & $0$ & $\frac{-1+\sqrt3 i}{2}$ & $-1+i$ & $-1$ \\ \hline
\end{tabular}
\end{center}
\vskip0.5cm
\begin{theo}
       Let $d$ be a square free positive integer and \, $t=N(j_{1,N})$
       be the normalized generator of $K(X_1(N))$.
       Let $s$ be a cusp of \, $\Gamma_1(N)$
       whose width is $h_s$. If $t\in q^{-1}\mathbb Z[[q]]$ and
       $\prod_{s\in S_{\Gamma_1(N)}
        \smallsetminus \{\infty\}} (t(z)-t(s))^{h_s} $ is a polynomial in
       $\mathbb Z[t]$, then $t(\tau)$ is an algebraic integer
       for \, $\tau\in{\mathbb Q}(\sqrt{-d})\cap\mathfrak H$.
       \label{Integer}
\end{theo}
\begin{pf} Let $j(z)=\dfrac1q+744+196884q+\cdots$, the elliptic
 modular function. It is well-known that $j(\tau)$ is an algebraic
 integer for $\tau\in{\mathbb Q}(\sqrt{-d})\cap\mathfrak H$ (\cite{Lang},
 \cite{Shimura1}). For algebraic proofs, see \cite{Deuring},
 \cite{Neron}, \cite{Serre} and \cite{Silverman}. Now, we view $j$
 as a function on the modular curve $X_1(N)$. Then $j$ has a pole
 of order $h_s$ at the cusp $s$. On the other hand, $t(z)-t(s)$
 has a simple zero at $s$. Thus $$ j\times \prod_{s\in
 S_{\Gamma_1(N)}
      \smallsetminus \{\infty\}} (t(z)-t(s))^{h_s} $$
 has a pole only at $\infty$ whose degree is \,
 $\mu_N=[\overline{\Gamma}(1):
 \overline{\Gamma}_1(N)]$,
 and so by the Riemann-Roch Theorem it is a monic polynomial in \,
 $t$ \, of degree \, $\mu_N$ which we denote by $f(t)$. Since
 $\prod_{s\in S_{\Gamma_1(N)}
      \smallsetminus \{\infty\}} (t(z)-t(s))^{h_s} $
      is a polynomial in $\mathbb Z[t]$ and
$j$, $t$ have integer coefficients in the $q$-expansions, $f(t)$
is a monic polynomial in ${\mathbb Z}[t]$ of degree $\mu_N$. This
shows that $t(\tau)$ is integral over ${\mathbb Z}[j(\tau)]$.
Therefore $t(\tau)$ is integral over $\mathbb Z$
 \, for $\tau\in{\mathbb Q}(\sqrt{-d})\cap\mathfrak H$.
\end{pf}
\begin{cor} For \, $\tau\in{\mathbb Q}(\sqrt{-d})\cap\mathfrak H$,
${\mathcal N}(j_{1,12})(\tau)$ is an algebraic integer.
\end{cor}
\begin{pf}
${\mathcal N}(j_{1,12})$ has integral Fourier coefficients. And by
Table 1 and 2,
$$ \prod_{s\in S_{\Gamma_1(12)
      \smallsetminus \{\infty\}}} (t(z)-t(s))^{h_s}
   = (t^2-2t-2)^{12} \, (t+2)^6 \, (t^2+2t+2)^4 \, (t^2+t+1)^3 \,
     t^2 \, (t+1) \hskip0.17cm \in \mathbb Z[t].  $$
Now the assertion is immediate from Theorem \ref{Integer}.
\end{pf}

\section{Super-replication formulae}
\par
Let $\Delta^n$ be the set of $2\times 2$ integral matrices $\left(\smallmatrix
a&b
\\ c&d \endsmallmatrix\right)$ where $a\in 1+N{\mathbb Z}, \, c\in N{\mathbb Z}$, and
$ad-bc=n$. Then $\Delta^n$ has the following right coset
decomposition: (See \cite{Koblitz}, \cite{Miyake},
\cite{Shimura1})
\begin{equation}
\Delta^n=\bigcup_{\smallmatrix a|n \\ (a,N)=1 \endsmallmatrix}
     \bigcup_{i=0}^{\frac{n}{a}-1} \Gamma_1(N)\sigma_a
          \begin{pmatrix} a&i \\ 0&\frac{n}{a} \end{pmatrix}   \label{Coset}
\end{equation}
where $\sigma_a\in SL_2({\mathbb Z})$ such that $\sigma_a\equiv \left(\smallmatrix
a^{-1}&0 \\ 0&a \endsmallmatrix\right) \mod N$. Let $f(z)$ be a modular function
with respect to $\Gamma_1(N)$. For brevity, let us call it {\it
$f(z)$ is on} $\Gamma_1(N)$.  For $f\in K(X_1(N))$ we define an
operator $U_n$ and $T_n$ by
\begin{align*}
 f|_{U_n} &=n^{-1} \sum_{i=0}^{n-1} f|_{\left(\smallmatrix 1&i\\0&n \endsmallmatrix\right)} \\
 \intertext{and}
 f|_{T_n} &= n^{-1} \sum_{\smallmatrix a\mid n \\ (a,N)=1
\endsmallmatrix}
   \sum_{i=0}^{\frac na-1} f|_{\sigma_a \left(\smallmatrix a& i \\ 0& \frac na \endsmallmatrix\right)}.
\end{align*}
\begin{lem}
For $f\in K(X_1(N))$ and $\gamma_0\in \Gamma_0(N)$,
$(f|_{T_n})|_{\gamma_0}=(f|_{\gamma_0})|_{T_n}$ for any positive integer $n$.
In particular, $f|_{T_n}$ is again on $\Gamma_1(N)$.
\label{Hecke}
\end{lem}
\begin{pf}
First we claim that
$$ \Delta^n\gamma_0=\gamma_0\Delta^n \text{ \hskip0.17cm
 for }\gamma_0=\left(\smallmatrix x&y\\z&w\endsmallmatrix\right)\in\Gamma_0(N).$$
Let $\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)\in \Delta^n$. Then
 $$
{\gamma_0}^{-1}\left(\smallmatrix a&b\\c&d \endsmallmatrix\right) \gamma_0 = \left(\smallmatrix w& -y\\ -z&x\endsmallmatrix\right)
\left(\smallmatrix a&b\\c&d \endsmallmatrix\right) \left(\smallmatrix x&y\\z&w\endsmallmatrix\right) \equiv \left(\smallmatrix 1&*\\ 0&d \endsmallmatrix\right) \mod
N.
 $$
Hence \, ${\gamma_0}^{-1}\Delta^n\gamma_0 \subset \Delta^n$ \, so
that \, $\Delta^n\gamma_0\subset \gamma_0\Delta^n$. By the same
argument we can show the reverse inclusion. We note that
\begin{align*}
\Delta^n\gamma_0 & =
     \bigcup_{\smallmatrix a|n \\ (a,N)=1 \endsmallmatrix}
     \bigcup_{i=0}^{\frac{n}{a}-1} \Gamma_1(N)\sigma_a
          \begin{pmatrix} a&i \\ 0&\frac{n}{a} \end{pmatrix}  \gamma_0 \\
\intertext{ and }
\gamma_0\Delta^n & =
     \bigcup_{\smallmatrix a|n \\ (a,N)=1 \endsmallmatrix}
     \bigcup_{i=0}^{\frac{n}{a}-1} \gamma_0\Gamma_1(N)\sigma_a
          \begin{pmatrix} a&i \\ 0&\frac{n}{a} \end{pmatrix}  \\
 & = \bigcup_{\smallmatrix a|n \\ (a,N)=1 \endsmallmatrix}
     \bigcup_{i=0}^{\frac{n}{a}-1} \Gamma_1(N)\gamma_0\sigma_a
          \begin{pmatrix} a&i \\ 0&\frac{n}{a} \end{pmatrix}
      \text{ \hskip0.17cm because $\Gamma_1(N)\triangleleft\Gamma_0(N)$.}
\end{align*}
Here we note that $\sigma_a\left(\smallmatrix a&i \\ 0&\frac{n}{a} \endsmallmatrix\right)
\gamma_0$'s are the matrices appearing in the definition of
$(f|_{T_n})|_{\gamma_0}$ and $\gamma_0\sigma_a\left(\smallmatrix a&i \\
0&\frac{n}{a} \endsmallmatrix\right)$'s are those appearing in the definition of
$(f|_{\gamma_0})|_{T_n}$. Now the assertion follows.
\end{pf}
For a positive integer $N$ with $g_{1,N}=0$, we let $t$ (resp.
$t_0$) be the Hauptmodul of $\Gamma_1(N)$ (resp. $\Gamma_0(N)$).
And, we write $X_n(t)=\frac 1n q^{-n} + \sum_{m\ge 1} H_{m,n} q^m$
and $X_n(t_0)=\frac 1n q^{-n} + \sum_{m\ge 1} h_{m,n} q^m$.
\begin{lem}
 For positive integers $m$ and $n$, $H_{m,n}=H_{n,m}$ and
 $h_{m,n}=h_{n,m}$. \label{CCC}
\end{lem}
\begin{pf}
 Let $p=e^{2\pi i y}$ and $q=e^{2\pi i z}$ with $y,z\in {\mathfrak
 H}$. Note that $X_n(t)$ can be viewed as the coefficient of
 $p^n$-term in $-\log p - \log (t(y)-t(z))$ (\cite{Norton}). Thus
 $H_{m,n}$ becomes the coefficient of $p^n q^m$-term of
 \begin{align*}
  & -\log p - \log (t(y)-t(z)) \\
  = & -\log (1-p/q)+\log (p^{-1} - q^{-1}) - \log (t(y)-t(z)) \\
  = & \sum_{i\ge 1} \frac 1i (p/q)^i - F(p,q)
 \end{align*}
 where $F(p,q)=\log \left(
  \frac{p^{-1}-q^{-1}+\sum_{i\ge 1} H_i (p^i -q^i)}{p^{-1}-q^{-1}}
  \right)$.
 We then come up with $F(p,q)=F(q,p)$, which implies that
 $H_{m,n}=H_{n,m}$. Similarly if we
 work with $t_0$ instead of $t$, the identity $h_{m,n}=h_{n,m}$ follows.
 \end{pf}
\begin{theo}
For positive integers $n$ and $l$ such that $(n,N)=(l,n)=1$,
$$ X_l(t)|_{T_n} = X_{ln}(t)|_{\sigma_n} + c $$
where $c$ is a constant. In particular,
$$ t|_{T_n} = X_{n}(t)|_{\sigma_n} + c. $$
\label{Hecke2}
\end{theo}
\begin{pf}
Since $X_l(t)$ has poles only at $\Gamma_1(N)\infty$, the poles
of $X_l(t)|_{T_n}$ can occur only at $\left(\smallmatrix a&i \\ 0&\frac na
\endsmallmatrix\right)^{-1} {\sigma_a}^{-1} \Gamma_1(N)\infty$ where $a$ and $i$ are
the indices appearing in the definition of $T_n$. On the other
hand, we have
\begin{align*}
\left(\smallmatrix a&i \\ 0&\frac na \endsmallmatrix\right)^{-1} {\sigma_a}^{-1}\Gamma_1(N)\infty
  &=n^{-1}\left(\smallmatrix \frac na&-i \\ 0&a \endsmallmatrix\right){\sigma_a}^{-1}\Gamma_1(N)\infty \\
  & = \left(\smallmatrix \frac na&-i \\ 0&a \endsmallmatrix\right){\sigma_a}^{-1}\Gamma_1(N)\infty.
\end{align*}
Let $\gamma$ be an element in $\Gamma_1(N)$.
Then
\begin{align*}
\left(\smallmatrix \frac na&-i \\ 0&a \endsmallmatrix\right){\sigma_a}^{-1}\gamma\infty
& \equiv \left(\smallmatrix \frac na&-i \\ 0&a \endsmallmatrix\right)
  \left(\smallmatrix a&0\\0&a^{-1}\endsmallmatrix\right)\left(\smallmatrix 1&*\\0&1\endsmallmatrix\right)\infty \mod N \\
& \equiv \left(\smallmatrix n&*\\0&1 \endsmallmatrix\right) \infty = \frac{n+Nm}{Nk}
\end{align*}
for some $k,m\in {\mathbb Z}$. If there exists an integer $x>1$ with
$x\mid (n+Nm,Nk)$, then $x$ must divide $\det\left[\left(\smallmatrix \frac
na&-i\\ 0&a \endsmallmatrix\right) {\sigma_a}^{-1}\gamma\right]=n$. In this case
$x\nmid N$ because $(n,N)=1$. Therefore $\left(\smallmatrix a&i \\ 0&\frac na
\endsmallmatrix\right)^{-1} {\sigma_a}^{-1}\gamma\infty$ is of the form
$\gamma_0\infty$ for some $\gamma_0\in \Gamma_0(N)$. Now we
conclude that $X_l(t)|_{T_n}$ can have poles only at
$\gamma_0\infty$ for some $\gamma_0\in\Gamma_0(N)$. By Lemma
\ref{Hecke},
\begin{align*}
 & n(X_l(t)|_{T_n})|_{\gamma_0} = n(X_l(t)|_{\gamma_0})|_{T_n} \\
= & \sum_{a\mid n }
   \sum_{i=0}^{\frac na-1} (X_l(t)|_{\gamma_0})
   |_{\sigma_a \left(\smallmatrix a& i \\ 0& \frac na \endsmallmatrix\right)}
= \sum_{a\mid n }
   \frac na {(X_l(t)|_{\gamma_0})
   |_{\sigma_a}}|_{U_\frac na} (az) \\
=& \sum_{\smallmatrix a\mid n \\ a\neq n \endsmallmatrix}
   \frac na {(X_l(t)|_{\gamma_0})
   |_{\sigma_a}}|_{U_\frac na} (az) + X_l(t)|_{\gamma_0\sigma_n}(nz).
\end{align*}
Here we note that
$$ \sum_{\smallmatrix a\mid n \\ a\neq n \endsmallmatrix}
   \frac na {(X_l(t)|_{\gamma_0})
   |_{\sigma_a}}|_{U_\frac na} (az) = O(1).$$
 In fact, if $\gamma_0\sigma_a \notin\pm\Gamma_1(N)$, it is clear
 that ${(X_l(t)|_{\gamma_0})
   |_{\sigma_a}}|_{U_\frac na}$ has a holomorphic $q$-expansion.
   Otherwise
$$ X_l(t)|_{\gamma_0\sigma_a U_\frac na} = X_l(t)|_{U_\frac na} =
(l^{-1}q^{-l}+\text{ terms of positive degree })|_{U_\frac na} =
O(1)
$$ because $(l,n)=1$. Now we have
\begin{equation}
 n(X_l(t)|_{T_n})|_{\gamma_0}
= X_l(t)|_{\gamma_0\sigma_n}(nz) + O(1).
\label{A}
\end{equation}
This implies that $X_l(t)|_{T_n}$ has a pole at $\gamma_0\infty$
if and only if $\gamma_0\sigma_n \in \pm\Gamma_1(N)$, that is,
$\gamma_0\in \pm\Gamma_1(N){\sigma_n}^{-1}$. Hence $X_l(t)|_{T_n}$
has poles only at cusps $\Gamma_1(N){\sigma_n}^{-1}\infty$. In
this case we derive from (\ref{A})
\begin{equation*}
  n(X_l(t)|_{T_n})|_{{\sigma_n}^{-1}}
= X_l(t)|_{{\sigma_n}^{-1}\sigma_n}(nz)+ O(1) = l^{-1}q^{-ln} + O(1).
\end{equation*}
Thus \, $(X_l(t)|_{T_n})|_{{\sigma_n}^{-1}} =(nl)^{-1}q^{-ln} +
O(1)$ \, and \, $(X_l(t)|_{T_n})|_{{\sigma_n}^{-1}}$ \, has poles
only at cusps $\sigma_n
\Gamma_1(N){\sigma_n}^{-1}\infty=\Gamma_1(N)\infty.$ On the other
hand, $X_{ln}(t)$ has poles only at $\Gamma_1(N)\infty$ too and
$X_{ln}(t)=(ln)^{-1}q^{-ln} + O(1).$ Therefore
$(X_l(t)|_{T_n})|_{{\sigma_n}^{-1}} = X_{ln}(t)+c$ for some
constant $c$. Then we have $X_l(t)|_{T_n} =
X_{ln}(t)|_{\sigma_n}+c$, as desired.
\end{pf}
\begin{cor} Let $N$ be a positive interger such that the genus
$g_{1,N}$ is zero and
$[\overline{\Gamma}_0(N):\overline{\Gamma}_1(N)]\le 2$. For
positive integers $n,l$ and $m$ such that $(n,N)=(l,n)=1$, we have
$$ \sum_{\smallmatrix e\mid (m,n)\\ e>0 \endsmallmatrix}
   e^{-1}\left\{ \psi (e)
      \left( 2 H_{\frac{mn}{e^2},l}-h_{\frac{mn}{e^2},l}\right)
       +h_{\frac{mn}{e^2},l}\right\}
      = \psi (n)(2H_{m,ln}-h_{m,ln})+h_{m,ln} $$
where $\psi: \, ({\mathbb Z}/N{\mathbb Z})^\times \to \{ \pm 1\}$ is a
character defined by
$$ \psi(e)= \begin{cases} 1, & \text{ if } e\equiv \pm 1 \mod N \\
-1, & \text{otherwise.} \end{cases} $$ \label{RRR1}
\end{cor}
\begin{pf}
It follows from Theorem \ref{Hecke2} that
\begin{equation}
 X_l(t)|_{T_n} = \sum_{\smallmatrix e\mid n\\ e>0 \endsmallmatrix}
   e^{-1} \left(X_l(t)|_{\sigma_e}\right)|_{U_\frac ne}(ez)
   = X_{ln}(t)|_{\sigma_n}+ \text{ constant. } \label{BB}
\end{equation}
 Note that for each positive integer r,
$$ X_r(t)|_{\sigma_e}+X_r(t)=
\begin{cases} 2 X_r(t), &  \text{ if } e\equiv \pm 1 \mod N \\
               X_r(t_0)+ \text{ constant }, &  \text{ otherwise.}
\end{cases} $$
In the above when $e$ is not congruent to $\pm 1 \mod N$, $
X_r(t)|_{\sigma_e}+X_r(t)$ is on $\Gamma_0(N)$ and has poles only
at $\Gamma_0(N)\infty$ with $r^{-1}q^{-r}$ as its pole part, which
guarantees the above equality. We then have $$
X_r(t)|_{\sigma_e}=\frac12
  \{\psi(e)(2X_r(t)-X_r(t_0))+X_r(t_0)\} + \text{ constant.} $$
Now (\ref{BB}) reads
\begin{gather*}
\sum_{\smallmatrix e\mid n \\ e>0 \endsmallmatrix}
 e^{-1}\cdot \frac 12
  \{\psi(e)(2X_l(t)-X_l(t_0))+X_l(t_0)\}|_{U_\frac ne}(ez) \\
 = \frac 12
  \{\psi(n)(2X_{ln}(t)-X_{ln}(t_0))+X_{ln}(t_0)\}+ \text{constant.}
\end{gather*}
Comparing the coefficients of $q^m$-terms on both sides, we get
the corollary.
\end{pf}
\begin{cor}
 Let $N$ be a positive integer such that the genus $g_{1,N}$ is zero
 and $[\overline{\Gamma}_0(N)
 :\overline{\Gamma}_1(N)]\le 2$. For positive integers $a,b,c,d$ with
 $ab=cd$, $(a,b)=(c,d)$ and $(b,N)=(d,N)=1$,
 \begin{equation}
 H_{a,b}=\psi(bd) H_{c,d} + \frac{(1-\psi(bd))}{2}h_{c,d}.
 \label{super}
 \end{equation}
 \label{Repli2}
\end{cor}
\begin{pf}
In Corollary \ref{RRR1} we take $n=b$, $l=1$ and $m=a$. Then it
follows from the conditions and the replicability of $h_{m,n}$
that
$$ \psi(b)(2H_{a,b}-h_{a,b})=\psi(d)(2H_{c,d}-h_{c,d}). $$
Now the assertion follows.
\end{pf}
\begin{cor}
Let $N$ be a positive integer with $g_{1,N}=0$ and
$[\overline{\Gamma}_0(N):\overline{\Gamma}_1(N)]=2$. If $(mn,N)=1$
and $mn\not\equiv \pm 1 \mod N$, then $h_{m,n}=2H_{m,n}$.
\end{cor}
\begin{pf}
In Corollary \ref{Repli2} we take $a=m, b=n$ and $c=n, d=m$.
 The condition that $\psi(mn)=-1$ implies
\begin{align*}
 H_{m,n}  &= -H_{n,m}+h_{n,m} \\
          &= -H_{m,n}+h_{m,n} \text{ \hskip0.17cm by Lemma \ref{CCC}. }
\end{align*}
This proves the corollary.
\end{pf}

\section{Vanishing property in the Fourier coefficients of ${\mathcal N}(j_{1,12})$}
\par
 As mentioned in the introduction, many of the Thompson series
 have periodically
 vanishing properties among the Fourier coefficients. Now we will
 give a more theoretical explanation for these phenomena.
 \par Let $T_g$ be the Thompson series
 of type $g$ and $\Gamma_g$ be its corresponding genus zero group.
 To describe $\Gamma_g$ we are in need of some notations.
 Let $N$ be a positive integer and $Q$ be any Hall divisor of
$N$, that is, $Q$ be a positive divisor of $N$ for which $(Q,
N/Q)=1$. We denote by $W_{Q,N}$ a matrix $\left(\smallmatrix Qx & y \\ Nz & Qw
\endsmallmatrix\right)$ with $\det W_{Q,N}=Q$ and $x,y,z$ and $w\in \mathbb Z$, and
call it an Atkin-Lehner involution. Let $S$ be a subset of Hall
divisors of $N$ and let $\Gamma=N+S$ be the subgroup of
$PSL_2(\mathbb R)$ generated by $\Gamma_0(N)$ ($=\{ \left(\smallmatrix a&b\\c&d \endsmallmatrix\right)
\in SL_2(\mathbb Z) \, | \, c\equiv 0 \mod N \}$) and all
Atkin-Lehner involutions $W_{Q,N}$ for $Q\in S$.
 For a positive divisor $h$ of $24$, let
$n$ be a multiple of $h$ and set $N=nh$. When $S$ is a subset of
Hall divisors of $n/h$, we denote by $\Gamma_0(n|h)+S$ the group
generated by $\left(\smallmatrix h&0 \\ 0&1 \endsmallmatrix\right)^{-1} \circ \Gamma_0(n/h)
\circ\left(\smallmatrix h&0 \\ 0&1 \endsmallmatrix\right) $ and $\left(\smallmatrix h&0 \\ 0&1 \endsmallmatrix\right)^{-1} \circ
W_{Q,n/h} \circ \left(\smallmatrix h&0 \\ 0&1 \endsmallmatrix\right)$ for all $Q\in S$. If there
exists a homomorphism $\lambda$ of $\Gamma_0(n|h)+S$ into $\mathbb
C^*$ such that
 \begin{align}
 & \lambda(\Gamma_0(N))=1, \label{H1} \\
 &\lambda(\left(\smallmatrix 1&1/h\\ 0&1 \endsmallmatrix\right))=e^{-2\pi i/h}, \label{H2} \\
 & \lambda(\left(\smallmatrix 1&0\\ n&1 \endsmallmatrix\right))= \begin{cases} e^{2\pi i/h} & \text{ \hskip0.17cm if $n/h\in
   S$}, \\ e^{-2\pi i/h} & \text{ \hskip0.17cm if $n/h \notin S$,}
     \end{cases} \label{H3} \\
 & \lambda \text{ is trivial on all Atkin-Lehner involutions of
    } \Gamma_0(N) \text{ in } \Gamma_0(n|h)+S, \label{H4}
 \end{align}
  then we let $n|h+S$ be the kernel of $\lambda$ which is
 a subgroup of $\Gamma_0(n|h)+S$ of index $h$.
 Ferenbaugh (\cite{Ferenbaugh}) found out a necessary and
 sufficient condition for the homomorphism $\lambda$ to exist and
 calculated the genera of groups of type $n|h+S$. All the genus
 zero groups of type $n|h+S$ are listed in \cite{Ferenbaugh},
 Table 1.1 and 1.2.
 Now we have the
 following theorem.
 \begin{theo} Write $T_g(q)=q^{-1}+\sum_{m\ge 1} c_g(m) q^m$.
 \par\noindent
 (i) If $\Gamma_g=\Gamma_0(N)$ and $h$ is the largest
 integer such that $h\mid 24$ and $h^2 \mid N$, then
 $ c_g(m)=0 \text{ unless } m\equiv -1 \mod h. $
 \par\noindent
 (ii) If $\Gamma_g=N+S$ and there exists a prime $p$ such that
 $p^2|N$ and $p\nmid Q$ for all $Q\in S$, then
 $ c_g(m)=0 \text{ whenever } m\equiv 0 \mod p. $
 \par\noindent
 (iii) If $\Gamma_g=n|h+S$, then
 $ c_g(m)=0 \text{ unless } m\equiv -1 \mod h. $
 \label{VVV}
 \end{theo}
 \begin{pf} (i) Note that $\left(\smallmatrix 1& h^{-1}\\ 0&1 \endsmallmatrix\right) $ belongs to
 the normalizer of $\Gamma_0(N)$. Thus $T_g|_{\left(\smallmatrix 1& h^{-1}\\ 0&1
 \endsmallmatrix\right)}$ has poles only at $\infty$, which is a simple pole.
 This enables us to write $T_g|_{\left(\smallmatrix 1& h^{-1}\\ 0&1
 \endsmallmatrix\right)}=c\cdot T_g$ for some constant $c$. By comparing the
 coefficients of $q^m$-terms on both sides, the assertion follows.
 \par\noindent
 (ii) From Corollary 3.1 \cite{Koike} it follows that $T_g|_{U_p}=0$.
 Thus
 (ii) is clear.
 \par\noindent
 (iii) Considering the identity in \cite{Koike}, p.27 we have
 $T_g(z+1/h)=e^{-2\pi i/h}\cdot T_g(z)$. Then
 $T_g(q)=q^{-1}+\sum_{0<l\in {\mathbb Z}} c_g(lh-1) q^{lh-1}$, which
 implies (iii).
 \end{pf}
 Unlike the cases of Thompson series, when we handle the
 super-replicable function ${\mathcal N}(j_{1,12})$ we can not
 directly use the ingredients adopted in Theorem \ref{VVV}.
 Therefore we start with
 \begin{lem} For $\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)\in \Gamma_0(12)$,
 $j_{1,12}|_{\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)}
 =\left(\frac{3}{d}\right)j_{1,12}$. Here $\left(\frac{\cdot}{\cdot}\right)$
 denotes the generalized quadratic residue symbol.
 \label{Haupt}
 \end{lem}
 \begin{pf} Immediate from Lemma \ref{half2} and Theorem
 \ref{Haupt2}.
 \end{pf}
 We fix $N=12$ and let $t$ denote the Hauptmodul ${\mathcal N}(j_{1,12})$
 in what follows.

\begin{lem} For $\left(\smallmatrix a&b \\ c&d \endsmallmatrix\right) \in\Gamma_0(24)$,
$$ \left(t|_{U_2}\right)_\chi|_{\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)}
= (-1)^\frac{c}{24}\cdot \left(\frac{3}{a+\frac c4}\right)
  \cdot\left(t|_{U_2}\right)_\chi$$
where $\chi=\left(\frac{-1}{\cdot}\right)$ is the Jacobi symbol
and $\left(t|_{U_2}\right)_\chi$ is the twist of $t|_{U_{2}}$ by
$\chi$ (\cite{Koblitz}, p.127). \label{AAA}
\end{lem}
\begin{pf}
From \cite{Koblitz}, p.128 we observe that
\begin{align}
\left(t|_{U_2}\right)_\chi &=\frac{1}{\sqrt{-4}}
\left(t|_{U_2}\left(z+\frac14\right)-t|_{U_2}\left(z+\frac34\right)\right)
\label{DD} \\
&=\frac{1}{\sqrt{-4}} \sum_{i\in ({\mathbb Z}/4{\mathbb Z})^\times}
\left(\frac{-1}{i}\right) \left(t|_{U_2}+\frac12\right)|_{\left(\smallmatrix
4&i\\0&4 \endsmallmatrix\right)}. \notag
\end{align}
It then follows from \cite{Kim-KooA}, Corollary 28 \, that
$$ \left(t-t\left(\frac{1}{6}\right)\right)\times t|_{U_2} =H_2.
$$ If we compare the coefficients of $q$-term on both sides, we get
$H_4-t\left(\frac{1}{6}\right)\cdot H_2 = 0$. And, substituting
$H_2=1$ and $H_4=0$ we get
 \, $t\left(\frac{1}{6}\right) = 0$.
Now
\begin{equation}
t|_{U_2}=\frac1t=\frac{j_{1,12}-1}{2}.
\label{KK}
\end{equation}
 Then for $\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)\in\Gamma_0(12)$
\begin{align}
\left(t|_{U_2}+\frac12\right)|_{\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)}
&= \frac12 j_{1,12}|_{\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)}
 =\frac12\cdot\left(\frac{3}{d}\right)j_{1,12}
  \text{ \hskip0.17cm by Lemma \ref{Haupt}}
  \label{EE} \\
&=\left(\frac{3}{d}\right)\left(t|_{U_2}+\frac12\right)
 =\left(\frac{3}{a}\right)\left(t|_{U_2}+\frac12\right) \notag
\end{align} since $ad\equiv 1 \mod 12$.
Let $\left(\smallmatrix a&b\\c&d \endsmallmatrix\right) \in\Gamma_0(24)$. For $i=1,3$, we consider
$\left(\smallmatrix 4&i\\0&4 \endsmallmatrix\right)\left(\smallmatrix a&b\\c&d \endsmallmatrix\right) = \left(\smallmatrix 4a+ic&4b+id\\4c&4d
\endsmallmatrix\right)$. Since $(4a+ic,4c)$ divides $\det\left(\smallmatrix 4a+ic&4b+id\\4c&4d
\endsmallmatrix\right)$, we must have $(4a+ic,4c)=4$; hence $(a+ic/4,c)=1$. Thus we
can choose integers $x_i$ and $y_i$ such that $\gamma_i=\left(\smallmatrix
a+ic/4&x_i\\c&y_i \endsmallmatrix\right)\in SL_2({\mathbb Z})$. Write
${\gamma_i}^{-1}\sm4&i\\0&4 \endsmallmatrix\right)\left(\smallmatrix a&b\\c&d \endsmallmatrix\right) = \left(\smallmatrix
4&z_i\\0&4\endsmallmatrix\right)$ for some integer $z_i$. Then we have
$$ \sm4&i\\0&4 \endsmallmatrix\right)\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)
 = \left(\smallmatrix a+ic/4&x_i\\c&y_i \endsmallmatrix\right)\left(\smallmatrix 4&z_i\\0&4\endsmallmatrix\right)$$
for some integer $z_i$. Comparing (1,2)-component on both sides we get
$$ 4b+id=(a+ic/4)z_i + 4x_i.$$ Thus $id\equiv(a+ic/4)z_i \mod 4$. Then
\begin{align*}
z_i & \equiv (a+ic/4)id \mod 4 \text{ \hskip0.17cm \hskip0.17cm because \hskip0.17cm $n^2\equiv
1 \mod 4$
       \hskip0.17cm for every odd integer $n$} \\
    & \equiv aid+i^2\cdot\frac c4\cdot d \mod 4 \equiv i + \frac c4\cdot d
       \mod 4 \text{ \hskip0.17cm \hskip0.17cm due to \, $ad\equiv 1 \mod 4$} \\
    & \equiv i+6c_1d \mod 4, \text{\hskip0.17cm \hskip0.17cm where we write $c=24c_1$.}
\end{align*}
Therefore we derive that for $\left(\smallmatrix a&b\\c&d \endsmallmatrix\right) \in\Gamma_0(24)$,
\begin{align*}
\sqrt{-4}\cdot\left(t|_{U_2}\right)_\chi|_{\left(\smallmatrix a&b\\c&d\endsmallmatrix\right)} &=
\sum_{i\in ({\mathbb Z}/4{\mathbb Z})^\times} \left(\frac{-1}{i}\right)
\left(t|_{U_2}+\frac12\right)|_{\left(\smallmatrix 4&i\\0&4 \endsmallmatrix\right)\left(\smallmatrix a&b\\c&d
\endsmallmatrix\right)}
\text{ \hskip0.17cm by (\ref{DD})} \\
&= \sum_{i\in ({\mathbb Z}/4{\mathbb Z})^\times}
\left(\frac{-1}{i}\right) \left(t|_{U_2}+\frac12\right)|_
{\left(\smallmatrix a+ic/4&x_i\\c&y_i \endsmallmatrix\right)\left(\smallmatrix 4&z_i\\0&4 \endsmallmatrix\right)} \\
&= \sum_{i\in ({\mathbb Z}/4{\mathbb Z})^\times}
\left(\frac{-1}{i}\right)\left(\frac{3}{a+ic/4}\right)
\left(t|_{U_2}+\frac12\right)|_{\left(\smallmatrix 4&z_i\\0&4 \endsmallmatrix\right)} \text{ \hskip0.17cm by
(\ref{EE}).}
\end{align*}
Now if we set $c=24c_1$ as before, then we have
\begin{align*}
& \hskip0.17cm \sqrt{-4}\cdot\left(t|_{U_2}\right)_\chi|_{\left(\smallmatrix a&b\\c&d\endsmallmatrix\right)}
\\ &= \sum_{i\in ({\mathbb Z}/4{\mathbb Z})^\times}
\left(\frac{-1}{i}\right)\left(\frac{3}{a+6c_1i}\right)
\left(t|_{U_2}+\frac12\right)|_{\left(\smallmatrix 4&i+6c_1d\\0&4 \endsmallmatrix\right)}  \\ &=
\left(\frac{3}{a+6c_1}\right)\sum_{i\in ({\mathbb Z}/4{\mathbb
Z})^\times} \left(\frac{-1}{i}\right)
\left(t|_{U_2}+\frac12\right)|_{\left(\smallmatrix 4&i+6c_1d\\0&4 \endsmallmatrix\right)}  \\ &
\text{ \hskip0.17cm \hskip0.17cm since $a+6c_1i\equiv a+6c_1 \mod 12$} \\ &=
\left(\frac{3}{a+6c_1}\right)\sum_{i\in ({\mathbb Z}/4{\mathbb
Z})^\times}
\left(\frac{-1}{i}\right)\left(\frac{-1}{i+6c_1d}\right)
\left(\frac{-1}{i+6c_1d}\right)
\left(t|_{U_2}+\frac12\right)|_{\left(\smallmatrix 4&i+6c_1d\\0&4 \endsmallmatrix\right)}  \\ &=
(-1)^{c_1}\cdot\left(\frac{3}{a+6c_1}\right)\cdot\sqrt{-4}\cdot
\left(t|_{U_2}\right)_\chi
\end{align*}
because $\left(\frac{-1}{i}\right)\left(\frac{-1}{i+6c_1d}\right)
=\left(\frac{-1}{i^2+6c_1id}\right)=\left(\frac{-1}{1+2c_1id}\right)
=(-1)^{c_1id}=(-1)^{c_1}$ and $i+6c_1d$ runs over $({\mathbb Z}/4\mathbb
Z)^\times$. This completes the lemma.
\end{pf}
\begin{lem}
 (i) For each $k\ge 1$, we put $g(z)=(-1)^{k-1}
 \cdot \frac 12\cdot \left(X_{2^k}(t)(z)-X_{2^k}(t)\left(z+\frac 12\right)\right).$
  Then $g$ belongs to $K(X_1(24))$.
 \par\noindent
 (ii) For $\left(\smallmatrix a&b \\ c&d \endsmallmatrix\right) \in\Gamma_0(24)$ and $k\ge 1$, $$
 \left(t|_{U_{2^k}}\right)_\chi|_{\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)} =
 (-1)^\frac{c}{24}\cdot \left(\frac{3}{a+\frac c4}\right)
  \cdot\left(t|_{U_{2^k}}\right)_\chi.$$ In particular,
  $\left(t|_{U_{2^k}}\right)_\chi$ lies in $K(X_1(24))$.
 \label{Replication5}
\end{lem}
\begin{pf}
First we note that for $n\mid N^\infty$, $T_n=U_n$. Here, by
$n\mid N^\infty$ we mean that $n$ divides some power of $N$. To
show $g\in K(X_1(24))$, we observe that
$$ g=(-1)^{k-1}
\left(X_{2^k}(t)(z)-X_{2^k}(t)|_{U_2}(2z)\right).$$ Then using
Lemma \ref{Hecke} we obtain that $g\in K(X_1(24))$. For $\left(\smallmatrix a&b
\\ c&d \endsmallmatrix\right) \in\Gamma_0(12)$,
\begin{align}
\left(t|_{U_{2^k}}+\frac12\right)|_{\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)} &
=\left(t|_{U_{2}}+\frac12\right)|_{U_{2^{k-1}}\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)}
 =\left(t|_{U_{2}}+\frac12\right)|_{\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)
U_{2^{k-1}}}
   \text{ \hskip0.17cm by Lemma \ref{Hecke}}
   \label{FF} \\
& =\left(\frac 3a\right)\left(t|_{U_{2}}+\frac12\right)|_{U_{2^{k-1}}}
   \text{ \hskip0.17cm by (\ref{EE})} \notag \\
& =\left(\frac 3a\right)\left(t|_{U_{2^k}}+\frac12\right). \notag
\end{align}
Now we can proceed in the same manner as in the proof of Lemma
\ref{AAA}.
\end{pf}
\begin{lem}
For $k\ge 1$,
\begin{align*}
& (i) \hskip0.17cm \left(t|_{U_{2^k}}\right)|_{\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right) } =
\begin{cases}
\frac12 {q_2}^{-1}+O(1), & \text{ if $k=1$} \\
O(1), & \text{ otherwise. }
\end{cases}
\\
& (ii) \hskip0.17cm \left(t|_{U_{2^k}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) } =
\begin{cases}
O(1), & \text{ if $k=1$} \\
\frac14 {q_4}^{-1}+O(1), & \text{ if $k=2$ } \\
-\frac i8 {q_2}^{-1}+O(1), & \text{ if $k=3$ } \\
(-1)^{k-1}\cdot\frac{i}{2^k}\cdot{q}^{-2^{k-4}}+O(1), & \text{ if $k\ge 4$. }
\end{cases}
\end{align*}
\label{Replication6}
\end{lem}
\begin{pf}
(i) First, $t|_{\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right)}$ is holomorphic at $\infty$
because $t$ has poles only at the cusps $\Gamma_1(12)\infty$. Now
for $k\ge 1$,
\begin{align*}
\left(t|_{U_{2^k}}\right)|_{\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right) }
&=\left(\left(t|_{U_{2^{k-1}}}\right)|_{U_2}\right)|_{\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right) }
=\frac12\left(t|_{U_{2^{k-1}}}\right)|_{\left(\smallmatrix 1&0\\0&2 \endsmallmatrix\right) \left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right)}
+\frac12\left(t|_{U_{2^{k-1}}}\right)|_{\left(\smallmatrix 1&1\\0&2 \endsmallmatrix\right) \left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right)}
\\
&=\frac12\left(t|_{U_{2^{k-1}}}\right)|_{\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right) \left(\smallmatrix 1&0\\0&2 \endsmallmatrix\right)}
+\frac12\left(t|_{U_{2^{k-1}}}\right)|_{\left(\smallmatrix 7&4\\12&7 \endsmallmatrix\right) \left(\smallmatrix 1&-1\\0&2 \endsmallmatrix\right)}
\\
&=\frac12\left(t|_{U_{2^{k-1}}}\right)\left(\frac z2\right)
+\frac12\left(\left(t|_{\left(\smallmatrix 7&4\\12&7 \endsmallmatrix\right)}\right)|_
{U_{2^{k-1}}}\right)|_{\left(\smallmatrix 1&-1\\0&2 \endsmallmatrix\right)}
\text{ \hskip0.17cm by Lemma \ref{Hecke} } \\
&=\begin{cases}
\frac12 {q_2}^{-1}+O(1), & \text{ if $k=1$} \\
O(1), & \text{ otherwise. }
\end{cases}
\end{align*}
\par\noindent
(ii) We observe that $t|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right)}\in O(1)$. And for
$k\ge 1$,
\begin{align}
\left(t|_{U_{2^k}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) }
&=\left(\left(t|_{U_{2^{k-1}}}\right)|_{U_2}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) }
\label{HH} \\
& =\frac12\left(t|_{U_{2^{k-1}}}\right)|_{\left(\smallmatrix 1&0\\0&2 \endsmallmatrix\right) \left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right)}
+\frac12\left(t|_{U_{2^{k-1}}}\right)|_{\left(\smallmatrix 1&1\\0&2 \endsmallmatrix\right) \left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right)}
\notag \\
&=\frac12\left(t|_{U_{2^{k-1}}}\right)|_{\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right) \left(\smallmatrix 1&0\\0&2 \endsmallmatrix\right)}
+\frac12\left(t|_{U_{2^{k-1}}}\right)|_{\left(\smallmatrix 2&1\\3&2 \endsmallmatrix\right) \left(\smallmatrix 2&0\\0&1 \endsmallmatrix\right)}
\notag
\end{align}
has a holomorphic Fourier expansion if $k=1$. Thus we suppose
$k\ge 2$. We then derive that
\begin{align*}
\left(t|_{U_{2^{k-1}}}\right)|_{\left(\smallmatrix 2&1\\3&2 \endsmallmatrix\right) }
&=\left(\left(t|_{U_{2^{k-2}}}\right)|_{U_2}\right)|_{\left(\smallmatrix 2&1\\3&2 \endsmallmatrix\right) } \\
&=\frac12\left(t|_{U_{2^{k-2}}}\right)|_{\left(\smallmatrix 1&0\\0&2 \endsmallmatrix\right) \left(\smallmatrix 2&1\\3&2 \endsmallmatrix\right)}
+\frac12\left(t|_{U_{2^{k-2}}}\right)|_{\left(\smallmatrix 1&1\\0&2 \endsmallmatrix\right) \left(\smallmatrix 2&1\\3&2 \endsmallmatrix\right)}
\\
&=\frac12\left(t|_{U_{2^{k-2}}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) \left(\smallmatrix 2&1\\0&1 \endsmallmatrix\right)}
+\frac12\left(t|_{U_{2^{k-2}}}\right)|_
{\left(\smallmatrix 11&-1\\12&-1 \endsmallmatrix\right) \left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right)\left(\smallmatrix 1&1\\0&2 \endsmallmatrix\right)}
\\
&=\frac12\left(t|_{U_{2^{k-2}}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) \left(\smallmatrix 2&1\\0&1 \endsmallmatrix\right)}
+\frac12\left(t|_{U_{2^{k-2}}}\right)|_{\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right) \left(\smallmatrix 1&1\\0&2 \endsmallmatrix\right)}.
\end{align*}
If we substitute the above into (\ref{HH}), for $k\ge 2$,
\begin{align}
 & \hskip0.17cm  \left(t|_{U_{2^{k}}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) } \label{II}\\
 & =\frac12\left(t|_{U_{2^{k-1}}}\right)|_{\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right)
   \left(\smallmatrix 1&0\\0&2 \endsmallmatrix\right) }
  +\frac14\left(t|_{U_{2^{k-2}}}\right)|_{\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right) \left(\smallmatrix 1&1\\0&2 \endsmallmatrix\right) \left(\smallmatrix 2&0\\0&1 \endsmallmatrix\right) }
  +\frac14\left(t|_{U_{2^{k-2}}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) \left(\smallmatrix 2&1\\0&1 \endsmallmatrix\right) \left(\smallmatrix 2&0\\0&1 \endsmallmatrix\right) }
   \notag \\
 & =\frac12\left(t|_{U_{2^{k-1}}}\right)|_{\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right)
   \left(\smallmatrix 1&0\\0&2 \endsmallmatrix\right) }
 +\frac14\left(t|_{U_{2^{k-2}}}\right)|_{\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right) \left(\smallmatrix 2&1\\0&2 \endsmallmatrix\right) }
 +\frac14\left(t|_{U_{2^{k-2}}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) \left(\smallmatrix 4&1\\0&1 \endsmallmatrix\right) }
\notag
\end{align}
When $k=2$,
\begin{align*}
& \hskip0.17cm  \left(t|_{U_{4}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) }
=\frac12\left(t|_{U_{2}}\right)|_{\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right)}\left(\frac
z2\right)
 +\frac14 t|_{\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right) }\left(z+\frac12\right)
 +\frac14 t|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) }\left(4z+1\right)  \\
& = \frac14 {q_4}^{-1}+O(1) \text{ \hskip0.17cm\q by (i). }
\end{align*}
If $k=3$,
\begin{align*}
& \hskip0.17cm  \left(t|_{U_{8}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) }
=\frac12\left(t|_{U_{4}}\right)|_{\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right)}\left(\frac
z2\right)
 +\frac14\left(t|_{U_{2}}\right)|_{\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right) } \left(z+\frac12\right)
 +\frac14\left(t|_{U_{2}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) }(4z+1) \\
& = \frac18 e^{-\pi i\left(z+\frac12\right)} + O(1) =
-\frac{i}{8}{q_2}^{-1}+O(1) \text{ \hskip0.17cm\q by (i) and the case $k=1$
in (ii). }
\end{align*}
For $k\ge 4$, we will show by induction on $k$ that
\begin{equation}
\left(t|_{U_{2^{k}}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) }
   = (-1)^{k-1}\cdot \frac{i}{2^k}\cdot q^{-2^{k-4}}+O(1).
   \label{JJ}
\end{equation}
First we note that by (i) and (\ref{II})
$$\left(t|_{U_{2^{k}}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) }
  =\frac14\left(t|_{U_{2^{k-2}}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) }(4z+1)+O(1). $$
If $k=4$,
\begin{align*}
\left(t|_{U_{2^{4}}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) }
  &=\frac14\left(t|_{U_{2^{4-2}}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) }(4z+1)+O(1)\\
&=\frac{1}{16}e^{-\frac{\pi i}{2}(4z+1)}+O(1)
  \text{ \hskip0.17cm \hskip0.17cm by the case $k=2$} \\
&= (-1)^{4-1}\cdot \frac{i}{2^4}\cdot q^{-2^{4-4}}+O(1).
\end{align*}
Thus when $k=4$, (\ref{JJ}) holds. Meanwhile, if $k=5$ then we get
that
\begin{align*}
\left(t|_{U_{2^{5}}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) }
  &=\frac14\left(t|_{U_{2^{5-2}}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) }(4z+1)+O(1)\\
&=-\frac{i}{32}e^{-\pi i(4z+1)}+O(1)
  \text{ \hskip0.17cm \hskip0.17cm by the case $k=3$} \\
&= (-1)^{5-1}\cdot \frac{i}{2^5}\cdot q^{-2^{5-4}}+O(1).
\end{align*}
Therefore in this case (\ref{JJ}) is also valid. Now for $k\ge 6$,
\begin{align*}
\left(t|_{U_{2^{k}}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) }
  &=\frac14\left(t|_{U_{2^{k-2}}}\right)|_{\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right) }(4z+1)+O(1)\\
&=\frac{1}{4}\cdot (-1)^{k-2-1}\cdot \frac{i}{2^{k-2}}\cdot
 e^{-2\pi i(4z+1)\cdot 2^{k-2-4}}+O(1) \\
& \text{ \hskip0.17cm \hskip0.17cm \, by induction hypothesis for $k-2$} \\
&= (-1)^{k-1}\cdot \frac{i}{2^k}\cdot q^{-2^{k-4}}+O(1).
\end{align*}
This proves the lemma.
\end{pf}
\begin{theo} For $k\ge 1$,
\begin{equation}
 \left(t|_{U_{2^k}}\right)_\chi(z)
=(-1)^{k-1}\cdot \frac12\cdot \left(
X_{2^k}(t)(z)-X_{2^k}(t)\left(z+\frac12\right)\right).
\label{twisted-replication}
\end{equation}
This identity twisted by a character $\chi$ is analogous to the
``$2^k$-plication formula" (\cite{Ferenbaugh2}, \cite{Koike})
satisfied by the Hauptmodul of $\Gamma_0(N)$. \label{CC}
\end{theo}
\begin{pf}
Put $g(z)= (-1)^{k-1}\cdot \frac12\cdot \left(
X_{2^k}(t)(z)-X_{2^k}(t)\left(z+\frac12\right)\right)$ as before.
 We see by Lemma \ref{Replication5} that both $\left(t|_{U_{2^k}}\right)_\chi$
 and $g$ sit in $K(X_1(24))$.
 For $\left(t|_{U_{2^k}}\right)_\chi = g$, we will show that
 $\left(t|_{U_{2^k}}\right)_\chi - g$ has no poles in ${\mathfrak H}^*$.
 Recall that $$ \left(t|_{U_{2^k}}\right)_\chi
 =\frac{1}{\sqrt{-4}} \sum_{i\in ({\mathbb Z}/4{\mathbb Z})^\times}
\chi(i)\left(t|_{U_{2^k}}\right)|_{\left(\smallmatrix 4&i\\0&4 \endsmallmatrix\right)}
 =\frac{1}{\sqrt{-4}}
\sum_{i\in ({\mathbb Z}/4{\mathbb Z})^\times}\sum_{j=0}^{2^k-1}
\chi(i)t|_{\left(\smallmatrix 1&j\\0&2^k \endsmallmatrix\right)\left(\smallmatrix 4&i\\0&4 \endsmallmatrix\right)}.
$$
Since $t$ has poles only at $\Gamma_1(12)\infty$,
$\left(t|_{U_{2^k}}\right)_\chi$ can also have poles only at
\begin{align*}
\left(\smallmatrix 4&i\\0&4 \endsmallmatrix\right)^{-1}\left(\smallmatrix 1&j\\0&2^k \endsmallmatrix\right)^{-1}\Gamma_1(12)\infty
& = 16^{-1}\cdot 2^{-k}\cdot
  \left(\smallmatrix 4&-i\\0&4 \endsmallmatrix\right)\left(\smallmatrix 2^k&-j\\0&1 \endsmallmatrix\right)\Gamma_1(12)\infty \\
& =
  \left(\smallmatrix 2^{k+2}&-4j-i\\0&4 \endsmallmatrix\right)\Gamma_1(12)\infty.
\end{align*}
Let $\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)\in \Gamma_1(12)$. Then
$$ \left(\smallmatrix 2^{k+2}&-4j-i\\0&4 \endsmallmatrix\right)\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)\infty
   =\left(\smallmatrix 2^{k+2}a-4jc-ic & * \\ 4c&* \endsmallmatrix\right) \infty
   =\frac{2^{k+2}a-4jc-ic}{4c}. $$
Observe that $(2^{k+2}a-4jc-ic,4c)\mid 2^k \cdot 16=2^{k+4}$.
Write $(2^{k+2}a-4jc-ic,4c)=2^l$ for some integer $l\ge 0$. Then
   $$s=\frac{(2^{k+2}a-4jc-ic)/2^l}{4c/2^l} $$
is in lowest terms. Since $12\mid c$, $s$ is of the form
$s=\frac{n}{3m}$ for some integers $m$ and $n$. We assume
$(3m,n)=1$. Here we consider two cases.
\par\noindent
(i) $2^2\nmid m$: \par Choose integers $x$ and $y$ such that $\left(\smallmatrix
n&x\\ 3m&y\endsmallmatrix\right)\in SL_2({\mathbb Z})$ and consider
\begin{align*}
\left(t|_{U_{2^k}}\right)_\chi|_{\left(\smallmatrix n&x\\3m&y \endsmallmatrix\right) }
&=\frac{1}{\sqrt{-4}}
\left(t|_{U_{2^k}\left(\smallmatrix 4&1\\0&4 \endsmallmatrix\right)\left(\smallmatrix n&x\\3m&y \endsmallmatrix\right)}
     -t|_{U_{2^k}\left(\smallmatrix 4&3\\0&4 \endsmallmatrix\right)\left(\smallmatrix n&x\\3m&y \endsmallmatrix\right)}\right) \\
&=\frac{1}{\sqrt{-4}}
\left(t|_{U_{2^k}\left(\smallmatrix 4n+3m&*\\12m&* \endsmallmatrix\right)}
     -t|_{U_{2^k}\left(\smallmatrix 4n+9m&*\\12m&* \endsmallmatrix\right)}\right).
\end{align*}
Since $(3m,n)=1$ and $2^2\nmid m$, we can write $ \left(\smallmatrix
4n+3m&*\\12m&* \endsmallmatrix\right)=\gamma_0 U_1$ and $ \left(\smallmatrix 4n+9m&*\\12m&*
\endsmallmatrix\right)={\gamma_0}' U_2$ where both $\gamma_0$ and ${\gamma_0}'$ are
in $\Gamma_0(12)$ and $U_1$, $U_2$ are upper triangular matrices.
Then by (\ref{FF}), $\left(t|_{U_{2^k}}\right)_\chi|_{\left(\smallmatrix
n&x\\3m&y \endsmallmatrix\right)} \in O(1)$. Hence, if $2^2\nmid m$ then
$\left(t|_{U_{2^k}}\right)_\chi$ is holomorphic at the cusp
$s=\frac{n}{3m}$.
\par\noindent
(ii) $2^2\mid m$: \par If $2^3\mid m$, then $s$ is of the form
$s=\left(\smallmatrix n&x\\3m&y \endsmallmatrix\right)\infty$ with $\left(\smallmatrix
n&x\\3m&y\endsmallmatrix\right)\in\Gamma_0(24)$. Thus by Lemma \ref{Replication5},
$$ \left(t|_{U_{2^k}}\right)_\chi|_{\left(\smallmatrix n&x\\3m&y \endsmallmatrix\right)}
= (-1)^\frac{3m}{24}\cdot \left(\frac{3}{n+3m/4}\right)
  \cdot\left(t|_{U_{2^k}}\right)_\chi\in O(1).$$
As for the other cases, if we use Lemma \ref{Equiv2}, it is easy
to see that $s$ is equivalent to $\frac{1}{12}$ or $\frac{5}{12}$
under $\Gamma_1(24)$.
\par
Thus we conclude that $\left(t|_{U_{2^k}}\right)_\chi$ can have
poles only at $\frac{1}{12}$, $\frac{5}{12}$ under
$\Gamma_1(24)$-equivalence. Next, let us investigate the poles of
$g$. Recall that $X_{2^k}(t)$ has poles only at
$\Gamma_1(12)\infty$. Therefore $g$ can have poles only at $\left(\smallmatrix
2&i\\0&2 \endsmallmatrix\right)^{-1}\Gamma_1(12)\infty$ for $i=0,1$. And, for $\left(\smallmatrix
a&b\\c&d \endsmallmatrix\right)\in\Gamma_1(12)$,
\begin{align*}
\left(\smallmatrix 2&i\\0&2 \endsmallmatrix\right)^{-1}\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)\infty & = \left(\smallmatrix 2&-i\\0&2
\endsmallmatrix\right)\left(\smallmatrix a&b\\c&d \endsmallmatrix\right)\infty
 = \frac{2a-ic}{2c}
 = \frac{a-ic/2}{c} \text{ \hskip0.17cm \hskip0.17cm in lowest terms. }
\end{align*}
Hence by Lemma \ref{Equiv2}, $g$ can have poles only at
$\frac{1}{24}$, $\frac{5}{24}$, $\frac{7}{24}$, $\frac{11}{24}$,
$\frac{1}{12}$, $\frac{5}{12}$ under $\Gamma_1(24)$-equivalence.
At $\frac{1}{24}$, $\frac{5}{24}$, $\frac{7}{24}$ and
$\frac{11}{24}$, it is easy to check that $g$ is holomorphic. For
example, at $\frac{5}{24}$,
\begin{align*}
g|_{\left(\smallmatrix 5&1\\24&5 \endsmallmatrix\right)}
& = (-1)^{k-1}\cdot \frac12\cdot
\left( X_{2^k}(t)|_{\left(\smallmatrix 5&1\\24&5 \endsmallmatrix\right)}
  -X_{2^k}(t)|_{\left(\smallmatrix 2&1\\0&2 \endsmallmatrix\right)\left(\smallmatrix 5&1\\24&5 \endsmallmatrix\right)}\right) \\
& = (-1)^{k-1}\cdot \frac12\cdot
\left( X_{2^k}(t)|_{\left(\smallmatrix 5&1\\24&5 \endsmallmatrix\right)}
  -X_{2^k}(t)|_{\left(\smallmatrix 34&7\\48&10 \endsmallmatrix\right)}\right) \\
& = (-1)^{k-1}\cdot \frac12\cdot
\left( X_{2^k}(t)|_{\left(\smallmatrix 5&1\\24&5 \endsmallmatrix\right)}
  -X_{2^k}(t)|_{\left(\smallmatrix 17&12\\24&17 \endsmallmatrix\right)\left(\smallmatrix 2&-1\\0&2 \endsmallmatrix\right)}\right) \\
&  \in O(1) \text{ \hskip0.17cm since $\frac{5}{24},
  \frac{17}{24}\notin\Gamma_1(12)\infty$. }
\end{align*}
Now it remains to show that $\left(t|_{U_{2^k}}\right)_\chi-g$ has
no poles at the cusps equivalent to $\frac{1}{12},\frac{5}{12}$
under $\Gamma_1(24)$. At $\frac{1}{12}$,
\begin{align}
\left(t|_{U_{2^k}}\right)_\chi|_{\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right) }
&=\frac{1}{\sqrt{-4}}
\left(t|_{U_{2^k}\left(\smallmatrix 4&1\\0&4 \endsmallmatrix\right)\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right)}
     -t|_{U_{2^k}\left(\smallmatrix 4&3\\0&4 \endsmallmatrix\right)\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right)}\right)
     \label{GG} \\
&=\frac{1}{\sqrt{-4}}
\left(t|_{U_{2^k}\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right)\left(\smallmatrix 16&1\\0&1 \endsmallmatrix\right)}
     -t|_{U_{2^k}\left(\smallmatrix 5&-1\\6&-1 \endsmallmatrix\right)\left(\smallmatrix 8&1\\0&2 \endsmallmatrix\right)}\right) \notag \\
&=\frac{1}{\sqrt{-4}}
\left(t|_{U_{2^k}\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right)\left(\smallmatrix 16&1\\0&1 \endsmallmatrix\right)}
     -t|_{U_{2^k}\left(\smallmatrix 11&-1\\12&-1 \endsmallmatrix\right)\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right)
                 \left(\smallmatrix 8&1\\0&2 \endsmallmatrix\right)}\right)
\notag \\
&=\frac{1}{\sqrt{-4}}
\left(t|_{U_{2^k}\left(\smallmatrix 1&0\\3&1 \endsmallmatrix\right)}(16z+1)
     -t|_{U_{2^k}\left(\smallmatrix 1&0\\6&1 \endsmallmatrix\right)}\left(4z+\frac12\right)\right).
\notag
\end{align}
By (\ref{GG}) and Lemma \ref{Replication6}, we have the following:
\begin{align*}
\text{\noindent If $k=1$, } \hskip2cm
\left(t|_{U_{2}}\right)_\chi|_{\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right) }
& = \frac{1}{\sqrt{-4}}\cdot
   \left(-\frac12 e^{-\pi i\left(4z+\frac12\right)}+O(1)\right) = \frac{1}{4}q^{-2}+O(1).
\end{align*}
\begin{align*}
\text{\noindent If $k=2$, } \hskip2cm
\left(t|_{U_{2^2}}\right)_\chi|_{\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right) }
& = \frac{1}{\sqrt{-4}}\cdot
   \left(\frac14 e^{-\frac{\pi i}{2}\left(16z+1\right)}+O(1)\right)
  = -\frac{1}{8}q^{-4}+O(1).
\end{align*}
\begin{align*}
\text{\noindent If $k=3$, } \hskip2cm
\left(t|_{U_{2^3}}\right)_\chi|_{\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right) }
& = \frac{1}{\sqrt{-4}}\cdot
   \left(-\frac i8 e^{-\pi i\left(16z+1\right)}+O(1)\right) = \frac{1}{16}q^{-8}+O(1).
\end{align*}
\begin{align*}
\text{\noindent If $k\ge 4$, } \hskip2cm
\left(t|_{U_{2^k}}\right)_\chi|_{\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right) }
& = \frac{1}{\sqrt{-4}}\cdot
   \left((-1)^{k-1}\cdot \frac{i}{2^k}\cdot
   e^{-2\pi i\cdot 2^{k-4}\cdot \left(16z+1\right)}+O(1)\right) \\
&=  (-1)^{k-1}\cdot \frac{1}{2^{k+1}}\cdot q^{-2^k}+O(1).
\end{align*}
Observe that the identities for $k=1,2$ and $3$ are the same as
the last one when $k\ge 4$. Hence we conclude that for all $k\ge
1$,
$$\left(t|_{U_{2^k}}\right)_\chi|_{\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right) }
 = (-1)^{k-1}\cdot \frac{1}{2^{k+1}}\cdot q^{-2^k}+O(1).$$
On the other hand,
\begin{align*}
g|_{\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right)}
& = (-1)^{k-1}\cdot \frac12\cdot
\left( X_{2^k}(t)|_{\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right)}
  -X_{2^k}(t)|_{\left(\smallmatrix 2&1\\0&2 \endsmallmatrix\right)\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right)}\right) \\
& = (-1)^{k-1}\cdot \frac12\cdot
\left( X_{2^k}(t)
  -X_{2^k}(t)|_{\left(\smallmatrix 7&4\\12&7 \endsmallmatrix\right)\left(\smallmatrix 2&-1\\0&2 \endsmallmatrix\right)}\right) \\
& = (-1)^{k-1}\cdot \frac{1}{2^{k+1}}\cdot q^{-2^k}+O(1).
\end{align*}
Thus
$$\left(\left(t|_{U_{2^k}}\right)_\chi-g\right)|_
  {\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right) }\in O(1).$$
At $\frac{5}{12}$, we see that $\left(\smallmatrix 5&2\\12&5 \endsmallmatrix\right)=\left(\smallmatrix
-19&2\\-48&5 \endsmallmatrix\right)\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right)$ sends $\infty$ to
$\frac{5}{12}$.  Then
\begin{align*}
 \left(t|_{U_{2^k}}\right)_\chi|_{\left(\smallmatrix 5&2\\12&5 \endsmallmatrix\right)}
&= \left(t|_{U_{2^k}}\right)_\chi|_
  {\left(\smallmatrix -19&2\\-48&5 \endsmallmatrix\right)\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right)} \\
&= (-1)^\frac{-48}{24}\cdot \left(\frac{3}{-19+\frac{-48}{4}}\right)
  \cdot\left(t|_{U_{2^k}}\right)_\chi|_{\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right)}
  \text{\hskip0.17cm by Lemma \ref{Replication5}} \\
&= (-1)\cdot \left(t|_{U_{2^k}}\right)_\chi|_{\left(\smallmatrix 1&0\\12&1 \endsmallmatrix\right) }
 = (-1)^{k}\cdot \frac{1}{2^{k+1}}\cdot q^{-2^k}+O(1).
\end{align*} And
\begin{align*}
g|_{\left(\smallmatrix 5&2\\12&5 \endsmallmatrix\right)}
& = (-1)^{k-1}\cdot \frac12\cdot
\left( X_{2^k}(t)|_{\left(\smallmatrix 5&2\\12&5 \endsmallmatrix\right)}
  -X_{2^k}(t)|_{\left(\smallmatrix 2&1\\0&2 \endsmallmatrix\right)\left(\smallmatrix 5&2\\12&5 \endsmallmatrix\right)}\right) \\
& = (-1)^{k-1}\cdot \frac12\cdot
\left( X_{2^k}(t)|_{\left(\smallmatrix 5&2\\12&5 \endsmallmatrix\right)}
  -X_{2^k}(t)|_{\left(\smallmatrix 11&-1\\12&-1 \endsmallmatrix\right)\left(\smallmatrix 2&1\\0&2 \endsmallmatrix\right)}\right) \\
& = (-1)^{k}\cdot \frac{1}{2}\cdot X_{2^k}(t)|_{\left(\smallmatrix 2&1\\0&2 \endsmallmatrix\right)}+O(1)
= (-1)^{k}\cdot \frac{1}{2^{k+1}}\cdot q^{-2^k}+O(1).
\end{align*}
This implies that
$$\left(\left(t|_{U_{2^k}}\right)_\chi-g\right)|_
  {\left(\smallmatrix 5&2\\12&5 \endsmallmatrix\right) }\in O(1),$$
 from which the theorem follows.
\end{pf}
Now, we are ready to show periodically vanishing property of
${\mathcal N}(j_{1,12})$.
\begin{cor}
As before we let $t$ be the Hauptmodul of $\Gamma_1(12)$ and
write $X_n(t)=\frac 1n q^{-n}+\sum_{m\ge 1}H_{m,n} q^m$. Then
we have \par (i)
$ H_{m,2^k} = (-1)^{k-1}\left(\frac{-1}{m}\right)H_{2^km,1}
\text{ \hskip0.17cm \hskip0.17cm for odd $m$.} $
\par (ii)
$H_m=0$ whenever $m\equiv 4 \mod 6$, and $m=5$.
\label{Replication3}
\end{cor}
\begin{pf}
First if we compare the coefficients of $q^m$-terms on both sides
of the identity in Theorem \ref{CC}, we get (i).
 We see from the Appendix, Table 4 that $H_5=0$. On the other
hand, by the super-replication formula (Corollary \ref{Repli2})
it follows that for $m$ relatively prime to 12,
$$ H_{m,2^k}=H_{2^k,m}=
\begin{cases} H_{2^km,1}, & \text{ if } m\equiv \pm 1 \mod 12 \\
-H_{2^km,1}, & \text{ if } m\equiv \pm 5 \mod 12
\end{cases} $$
because $h_{2^km,1}=0$ in this case (\cite{Koike}, Corollary 3.1).
Then $H_{2^km,1}=0$ when $k$ is odd and $m\equiv 5 \mod 6$, or
$k$ is even and $m\equiv 1 \mod 6$. It is easy to see that
\begin{align*}
& \{ 2^km \, | \, k, m\ge 1, \,
                \text{$k$ odd, $m\equiv 5 \mod 6$} \} \cup
 \{ 2^km \, | \, k, m\ge 1, \,
                \text{$k$ even, $m\equiv 1 \mod 6$} \}  \\
=& \{ l\in{\mathbb Z} \, | \, l\equiv 4 \mod 6 \}.
\end{align*}
This proves (ii).
\end{pf}

 \vskip1cm \centerline{\bf Appendix. \hskip0.17cm Fourier coefficients of the
 Hauptmodul ${\mathcal N} (j_{1,N})$}
\par We shall make use of the following modular forms to construct $j_{1,N}$.
 For $z\in {\mathfrak H}$,
\par\noindent$\bullet$
$\eta(z)$ \hskip0.17cm \hskip0.17cm the Dedekind eta function
\par\noindent$\bullet$
$G_2(z)$ \hskip0.17cm \hskip0.17cm Eisenstein series of weight 2
\par\noindent$\bullet$
$G_2^{(p)}(z)=G_2(z)-pG_2(pz)$ for each prime $p$
\par\noindent$\bullet$
$E_2(z)= G_2(z)/(2\zeta(2))$ \hskip0.17cm \hskip0.17cm normalized Eisenstein series
of weight 2
\par\noindent$\bullet$
$E_2^{(p)}(z)=E_2(z)-pE_2(pz)$ for each prime $p$
\par\noindent$\bullet$
${\mathcal P}_{N,{\bf a}}(z)=\mathcal
P_{L_z}\left(\frac{a_1z+a_2}{N}\right) $ \hskip0.17cm \hskip0.17cm $N$-th division
value of ${\mathcal P}$ where ${\bf a}=(a_1,a_2), L_z={\mathbb Z} z+{\mathbb
Z}$ and ${\mathcal P}_L(\tau)$ is the Weierstrass ${\mathcal P}$-function
(relative to a lattice $L$)
\par Now we get the following tables due to \cite{Kim-Koo2}-\cite{Kim-Koo4}:
\vskip0.5cm
\begin{center}
 {\bf Table 3}. Hauptmoduln $\mathcal N (j_{1,N})$
                       \vskip0.2cm
\begin{tabular}
{|c||c|c|}\hline $N$ & $j_{1,N}$ & $\mathcal N (j_{1,N})$ \\
\hline\hline 1 & $j(z)$ & $j(z)-744$ \\ \hline 2   &
$\theta_2(z)^8/\theta_4(2z)^8$ & $256/j_{1,2}+24$ \\ \hline 3   &
$E_4(z)/E_4(3z)$ & $240/(j_{1,3}-1)+9$  \\ \hline 4   &
$\theta_2(2z)^4/\theta_3(2z)^4$ & $16/j_{1,4}-8$ \\ \hline 5   &
$\frac{4\eta(z)^5/\eta(5z)+E_2^{(5)}(z)}{\eta(5z)^5/\eta(z)}$ &
      $-8/(j_{1,5}+44)-5$  \\ \hline
6   &
$\frac{G_2^{(2)}(z)-G_2^{(2)}(3z)}{2G_2^{(2)}(z)-G_2^{(3)}(z)}$ &
      $2/(j_{1,6}-1)-1$  \\ \hline
7   & $\frac{\mathcal P_{7,(1,0)}(7z)-\mathcal P_{7,(2,0)}(7z)}
            {\mathcal P_{7,(1,0)}(7z)-\mathcal P_{7,(4,0)}(7z)}$ &
      $-1/(j_{1,7}-1)-3$  \\ \hline
8   & $\theta_3(2z)/\theta_3(4z)$ & $2/(j_{1,8}-1)-1$ \\ \hline
9   & $\frac{\mathcal P_{9,(1,0)}(9z)-\mathcal P_{9,(2,0)}(9z)}
            {\mathcal P_{9,(1,0)}(9z)-\mathcal P_{9,(4,0)}(9z)}$ &
      $-1/(j_{1,9}-1)-2$  \\ \hline
10   & $\frac{\mathcal P_{10,(1,0)}(10z)-\mathcal P_{10,(2,0)}(10z)}
             {\mathcal P_{10,(1,0)}(10z)-\mathcal P_{10,(4,0)}(10z)}$ &
      $-1/(j_{1,10}-1)-2$  \\ \hline
12   & $\theta_3(2z)/\theta_3(6z)$ & $2/(j_{1,12}-1)$ \\ \hline
\end{tabular}
\end{center}
 \vskip0.5cm
 When $N=1,2,3,4,6$, ${\mathcal N}(j_{1,N})$ becomes a Thompson
 series $T_g$ with $\Gamma_g=\Gamma_0(N)$. Hence, if $N=4$,
 ${\mathcal N}(j_{1,4})$ has periodically vanishing property
 by Theorem \ref{VVV}-(i).
 Otherwise, the Fourier coefficients of ${\mathcal N}(j_{1,N})$ do not
 vanish (see \cite{Mckay-Strauss}, Table 1).
 Therefore, we consider only the following cases $N$ for which
 $\overline{\Gamma}_1(N)\neq \overline{\Gamma}_0(N)$.
 \vskip0.5cm
 \begin{center} {\bf Table 4}. Fourier coefficients $H_m$
                       of ${\mathcal N}(j_{1,N})$ for $1\le m \le 60$
                       \vskip0.2cm
\begin{tabular} {|c||c|c|c|c|c|c|} \hline   & ${\mathcal N}(j_{1,5})$ & ${\mathcal N}(j_{1,7})$
                                            & ${\mathcal N}(j_{1,8})$ & ${\mathcal N}(j_{1,9})$
                                            & ${\mathcal N}(j_{1,10})$ & ${\mathcal N}(j_{1,12})$\\
 \hline\hline
 $H_{1}$ &     10  &       4 &       3 &        2 &        2 &        1 \\ \hline
 $H_{2}$ &      5  &       3 &       2 &        2 &        1 &        1 \\ \hline
 $H_{3}$ &    -15  &       0 &       1 &        1 &        1 &        1 \\ \hline
 $H_{4}$ &    -24  &      -5 &      -2 &       -1 &        0 &        0 \\ \hline
 $H_{5}$ &     15  &      -7 &      -4 &       -2 &       -1 &        0 \\ \hline

 $H_{6}$ &     70  &      -2 &      -4 &       -3 &       -2 &       -1 \\ \hline
 $H_{7}$ &     30  &       8 &       0 &       -2 &       -2 &       -1 \\ \hline
 $H_{8}$ &   -125  &      16 &       6 &        1 &       -1 &       -1 \\ \hline
 $H_{9}$ &   -175  &      12 &       9 &        4 &        1 &       -1 \\ \hline
 $H_{10}$&     95  &      -7 &       8 &        6 &        3 &        0 \\ \hline

 \end{tabular}
 \vskip0.1cm
 \begin{tabular} {|c||c|c|c|c|c|c|} \hline
 & ${\mathcal N}(j_{1,5})$ & ${\mathcal N}(j_{1,7})$
                                            & ${\mathcal N}(j_{1,8})$ & ${\mathcal N}(j_{1,9})$
                                            & ${\mathcal N}(j_{1,10})$ & ${\mathcal N}(j_{1,12})$\\
 \hline\hline

 $H_{11}$&    420  &     -29 &      -1 &        5 &        4 &        1 \\ \hline
 $H_{12}$&    180  &     -35 &     -12 &        1 &        4 &        2 \\ \hline
 $H_{13}$&   -615  &     -10 &     -20 &       -5 &        1 &        2 \\ \hline
 $H_{14}$&   -826  &      37 &     -16 &      -11 &       -2 &        2 \\ \hline
 $H_{15}$&    410  &      70 &       1 &      -12 &       -6 &        1 \\ \hline

 $H_{16}$&   1760  &      53 &      22 &       -7 &       -8 &        0 \\ \hline
 $H_{17}$&    705  &     -21 &      38 &        3 &       -7 &       -2 \\ \hline
 $H_{18}$&  -2415  &    -106 &      30 &       15 &       -3 &       -3 \\ \hline
 $H_{19}$&  -3100  &    -126 &       1 &       22 &        4 &       -4 \\ \hline
 $H_{20}$&   1530  &     -38 &     -40 &       19 &       10 &       -4 \\ \hline

 $H_{21}$&   6270  &     119 &     -64 &        5 &       14 &       -2 \\ \hline
 $H_{22}$&   2460  &     226 &     -52 &      -15 &       12 &        0 \\ \hline
 $H_{23}$&  -8090  &     164 &      -2 &      -32 &        6 &        3 \\ \hline
 $H_{24}$& -10174  &     -70 &      68 &      -36 &       -6 &        5 \\ \hline
 $H_{25}$&   4840  &    -326 &     107 &      -22 &      -16 &        7 \\ \hline

 $H_{26}$&  19570  &    -378 &      88 &        8 &      -22 &        6 \\ \hline
 $H_{27}$&   7500  &    -106 &      -2 &       40 &      -20 &        4 \\ \hline
 $H_{28}$& -24360  &     353 &    -112 &       58 &       -8 &        0 \\ \hline
 $H_{29}$& -30024  &     652 &    -180 &       50 &        8 &       -4 \\ \hline
 $H_{30}$&  14130  &     469 &    -144 &       12 &       26 &       -8 \\ \hline

 $H_{31}$&  55970  &    -189 &       3 &      -41 &       34 &      -10 \\ \hline
 $H_{32}$&  21155  &    -885 &     182 &      -84 &       31 &       -9 \\ \hline
 $H_{33}$& -67380  &   -1015 &     292 &      -93 &       12 &       -6 \\ \hline
 $H_{34}$& -81926  &    -290 &     228 &      -54 &      -14 &        0 \\ \hline
 $H_{35}$&  37895  &     910 &       4 &       22 &      -41 &        6 \\ \hline

 $H_{36}$& 148410  &    1664 &    -286 &      103 &      -54 &       12 \\ \hline
 $H_{37}$&  55305  &    1179 &    -452 &      148 &      -47 &       14 \\ \hline
 $H_{38}$&-174500  &    -483 &    -356 &      124 &      -20 &       14 \\ \hline
 $H_{39}$&-209577  &   -2205 &      -4 &       32 &       23 &        8 \\ \hline
 $H_{40}$&  96025  &   -2492 &     440 &      -96 &       61 &        0 \\ \hline

 $H_{41}$& 371620  &    -692 &     686 &     -200 &       84 &      -10 \\ \hline
 $H_{42}$& 137160  &    2212 &     544 &     -219 &       72 &      -18 \\ \hline
 $H_{43}$&-427665  &    3998 &      -5 &     -128 &       31 &      -22 \\ \hline
 $H_{44}$&-508800  &    2809 &    -668 &       46 &      -32 &      -20 \\ \hline
 $H_{45}$& 230670  &   -1120 &   -1044 &      231 &      -90 &      -12 \\ \hline

 $H_{46}$& 885070  &   -5119 &    -816 &      330 &     -122 &        0 \\ \hline
 $H_{47}$& 323605  &   -5754 &       5 &      275 &     -107 &       15 \\ \hline
 $H_{48}$&-1001340 &   -1598 &     996 &       67 &      -44 &       26 \\ \hline
 $H_{49}$&-1181123 &    4992 &    1563 &     -216 &       45 &       33 \\ \hline
 $H_{50}$& 531545  &    8968 &    1210 &     -439 &      133 &       29 \\ \hline

 \end{tabular}
 \vskip0.1cm
 \begin{tabular} {|c||c|c|c|c|c|c|} \hline
 & ${\mathcal N}(j_{1,5})$ & ${\mathcal N}(j_{1,7})$
                                            & ${\mathcal N}(j_{1,8})$ & ${\mathcal N}(j_{1,9})$
                                            & ${\mathcal N}(j_{1,10})$ & ${\mathcal N}(j_{1,12})$\\
 \hline\hline

 $H_{51}$& 2022670 &    6251 &       6 &     -477 &      174 &       19 \\ \hline
 $H_{52}$&  734130 &   -2506 &   -1464 &     -275 &      154 &        0 \\ \hline
 $H_{53}$&-2253515 &  -11285 &   -2276 &      107 &       61 &      -20 \\ \hline
 $H_{54}$&-2639348 &  -12579 &   -1768 &      501 &      -68 &      -37 \\ \hline
 $H_{55}$& 1178880 &   -3455 &      -8 &      708 &     -192 &      -45 \\ \hline

 $H_{56}$& 4456650 &   10812 &    2128 &      590 &     -254 &      -42 \\ \hline
 $H_{57}$& 1606500 &   19278 &    3284 &      146 &     -220 &      -26 \\ \hline
 $H_{58}$&-4901250 &   13362 &    2552 &     -447 &      -90 &        0 \\ \hline
 $H_{59}$&-5703676 &   -5278 &      -9 &     -911 &      100 &       27 \\ \hline
 $H_{60}$& 2532720 &  -23765 &   -3056 &     -987 &      272 &       52 \\ \hline
 \end{tabular}
 \end{center}

redpajama_set_name: RedPajamaArXiv


Does coaching work? That is a question I'm sure you are asking yourself and you should be asking yourself. A major component of coaching is helping you discover what gives you the greatest return on your investment of time, energy, effort and resources. This is a portion of just one study done on the ROI of professional coaching.
Coaching was a very effective developmental tool for the leadership development participants, producing financial and intangible benefits for the business.
Decision-making, team performance and the motivation of others were enhanced.
Three-quarters (77%) of the respondents indicated that coaching had a significant or very significant impact on at least one of nine business measures.
Employee satisfaction and customer satisfaction were considered significant sources of intangible benefits.
Respondents cited were work output (30%) and work quality (40%) improvements as a result of coaching.


Check out the students that were members of Mahopac Gymnastics. Visit the registration page if you would like to be listed.
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1. Review your business insurance coverage. Don't just automatically write a check to renew your insurance policies when they come due. Instead, you should sit down with your insurance agent every year. Review your business operations, focusing on any changes. Discuss types of risk that could arise. And ask about new developments in business insurance.
2. Look at your business tax strategy. Consider adjusting taxable earnings for the year, perhaps by accelerating expenses or delaying income at year-end. If you're a cash-basis taxpayer, you could boost 2017 deductions by declaring and paying bonuses in December rather than in early January. Also, you may be able to defer invoices or make early purchases to reduce your 2017 tax bill.
4. Determine your marketing effectiveness. Are your current methods and channels working well, or are you simply doing what you've always done?
7. Update your personal estate planning (if needed). If you're a business owner, your company is likely to be a significant part of your estate. Your company, your personal circumstances and the tax laws are continually changing. You should take time each year to make sure your plans are current.


This full-line pet specialty store has something for every pet, be it scaled, furry or feathered.
Ryan Oaks, general manager of Mini-Critters in Sioux Falls, S.D.
Ryan Oaks: Raw and freeze-dried foods are hot right now. Answers Pet Food, Stella & Chewy's, Rawbble and Primal. From a top-selling standpoint, it is always dog toys and treats—keeping our pets busy and fed!
Oaks: Answers Pet Food's Fermented Fish Stock. It smells fishy—pun intended—but it is an amazing product for dogs' joints, hypertension and kidney relief.
PPN: What industry trends have you been keeping your eye on most closely? Oaks: Raw dog and cat foods and grain-free kibble.
PPN: What business challenge are you currently in the process of tackling? Oaks: The first challenge is always attracting new customers to our store and setting us apart from the competition with our knowledge and customer satisfaction.
Oaks: To expand our self-serve dog wash customer base and, as always, to continue to expand our education on dog and cat foods.
Oaks: Face-to-face interaction. Social media is a good tool, but nothing takes the place of one-on-one interaction.
In her new role, Brecher will provide strategic leadership managing global brand initiatives for Whimzees and serve as a key part of the WellPet marketing team.
TropiClean Pet Products has named LANDON HOBSON as the new CEO of Cosmos Corp. Cosmos Corp., the manufacturer of TropiClean in Wentzville, Mo., oversees the brand on the pet specialty side and the Naturel Promise Pet Products brand for food, drug and mass.
Newbigging joins TropiClean's team with 24 years of experience in the pet industry. Prior to joining TropiClean, Newbigging was the key account manager and export manager for Marcam Pet Nutrition.
ty of California at Davis.


I'll install netbsd on an old computer, but I am sure I'll have a hard time to get wireless internet working in a way or another. I figured I could do that easily if I managed to install things for this computer, on another one, the one I am using now, by crosscompiling. And that it would be a good training, isn't it ?
For now, if pkg_add and so on are recognized, I still can't "pkg_add pkgin" or any software: it says it doesn't know that package. How come ? I see it, it's there.
In regards to installing pkgin, the variable you need to set is PKG_PATH. It is the URL/path to the binary NetBSD package repository you wish to use. Globally, there are at least 8 http/ftp servers. There are some guidlines on setting up pkgin on NetBSD that come up readily on an Internet Search. Use search terms: NetBSD pkgin. pkgin configuration is specified in /usr/pkg/etc/pkgin which should set a default that is linked to the current -stable branch.
I currently have 2 old PC's running NetBSD 7.0.2 patch with WPA2 encrypted pci wireless cards. It is dependent on your hardware - some wireless chips work better than others.
I use the NetBSD wiki guidelines to set up although I have to manually run dhclient the first time.
Last edited by shep; 11th January 2017 at 10:31 PM.
pkg_add: no pkg found for 'pkgin', sorry.
As for the wireless of course I know it is possible and I know too that the problem are drivers. Idem for X, which is the very reason I stopped using NBSD last time. But it was a fun time, I hope it will now too.
edit: Knowing both english and french, I happen to forget when I use one or the other ! sorry !
Last edited by spermwhale_warrior; 11th January 2017 at 08:56 PM.
sperwhale_warrior, per the forum rules, English only please.
If you click on the two links above, you will see that your link is a deadend. Clicking on the second link shows a list of packages one of which is pkgin.
Packages are specific to the NetBSD Release and the pkgsrc version. Every quarter a new version of pkgsrc is released. These are the currently available pkgsrc versions for the amd64 architecture. I copied them from http://ftp.netbsd.org/pub/pkgsrc/packages/NetBSD/amd64/.
pkgsrc2016Q4 was just released and many packages do not presently build. I would recommend retrieving pkgin from the 7.0_2016Q3 repository. When pkgsrc_2016Q4 is deemed ready, the developers will change the links so that you can easily update. Presently 7.0 -> 7.0_2016Q3. When Q4 is ready the link will be 7.0 -> 7.0_2016Q4.
Is it because my tree version is too recent ? I do have the library libbzip2.
Last edited by spermwhale_warrior; 11th January 2017 at 11:34 PM.
If your network connection is not established, pkgin will not be found. Can you ping www.google.com or www.netbsd.org?
PING google.fr (216.58.211.195) 56(84) bytes of data.
Is there a means to run bmake in verbose mode ?
bmake is used when building packages on a NON NetBSD system. When building on a NetBSD system make(1) is used. See Using pkgsrc on non-netbsd systems.
The other thing that can cause you trouble is to mix repository binaries that were built with one version of pkgsrc with a different version of pkgsrc.
1) Use precompiled binaries (packages). Packages built with pkgsrc_2016Q3 have just finished several months of fixing build failures - it is as good as it's going to get.
3) I use NetBSD 7 but most builds are initiated with NetBSD 6.1 The 7.1 release candidate will have some teething problems and I would avoid it until you are comfortable with setting up a system from stable sources. Using a release candidate adds the extra variable of code in testing - did it fail because of something you did or is it just buggy?
Because one of my goal is to learn to use pkgsrc anywhere and get used to crosscompiling, I'll download an older and more stable tree version. Otherwise, I knew that bmake is for non-netbsd system. I doubt that pkg_add non working at all, is because of the "testing part" of the packages, it is too big a bug. I mean, if it is, I don't imagine how is the 'current' branch.
That tree would be mostly devoid of bugs http://ftp.netbsd.org/pub/pkgsrc/stable/pkgsrc.tar.bz2 is that right ?
Last edited by spermwhale_warrior; 12th January 2017 at 08:14 PM.
That tree would be mostly devoid of bugs...is that right ?
Patching and bug fixes continue after a release. If you look at the time and date stamps for pkgsrc_2016Q3, you will see entries that occurred several months after the release of 2016Q3.
A useful mailing list to follow is Pkgsrc bulk build reports. You can scroll back through the archives to get a sense of what is "stable".
If you goal is to install a working Xfce4 desktop, at this time your highest likelyhood of success is with 2016Q3. If you want to help the NetBSD pkgsrc community debug their build scripts, attention is now focused on 2016Q4.


text:
Q: Android app language localization I have developed an app. Now I want to support other regional languages in it. But now the app is almost ready, changing each java file to set the texts in the textview for the corresponding language is a big pain. Is there any other simple way to do that? Like making a different string.xml file and doing some stuff, then all the views will get changed???

A: Create alternative resources
A large part of localizing an app is providing alternative text for different languages. In some cases you also provide alternative graphics, sounds, layouts, and other locale-specific resources.
An app can specify many res// directories, each with different qualifiers. To create an alternative resource for a different locale, you use a qualifier that specifies a language or a language-region combination. (The name of a resource directory must conform to the naming scheme described in Providing Alternative Resources, or else your app cannot compile.)

Example:

Suppose that your app's default language is English. Suppose also that you want to localize all the text in your app to French, and most of the text in your app (everything except the app's title) to Japanese. In this case, you could create three alternative strings.xml files, each stored in a locale-specific resource directory:

*

*res/values/strings.xml Contains English text for all the strings
that the app uses, including text for a string named title.

*res/values-fr/strings.xml Contain French text for all the strings,
including title.

*res/values-ja/strings.xml Contain Japanese text for
all the strings except title.

If your Java-based code refers to R.string.title, here is what happens at runtime:
If the device is set to any language other than French, Android loads title from the res/values/strings.xml file.
If the device is set to French, Android loads title from the res/values-fr/strings.xml file.
Notice that if the device is set to Japanese, Android looks for title in the res/values-ja/strings.xml file. But because no such string is included in that file, Android falls back to the default, and loads title in English from the res/values/strings.xml file.

A: If you properly used resource references (like R.string.hello) You can create another language configuration and it should be properly loaded. If you "hard coded" strings within JAVA code with literals like "hello world" You have quite a bit of refactoring to do.

redpajama_set_name: RedPajamaStackExchange


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Welcome to my web-site. I live in Thrussington, a small village in Leicestershire, where I have lived with my wife for nearly forty years.
For over thirty of those years I have been a full time potter, and for the last ten years or so have concentrated on making ceramic birds. My birds are what I would call decorative art, they are not meant to be a true representation, but are a more stylised version.Each bird is individualy hand made so no two will be the same.
I do hope that you enjoy browsing my site and like the birds that I make.