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\begin{center}
{\bf A MODULAR IDENTITY FOR THE RAMANUJAN IDENTITY MODULO 35}
\vskip 20pt
{\bf Adrian Dan Stanger}\\
{\smallit Department of Mathematics, Brigham Young University, Provo, Utah 84602, USA}\\
{\tt adrian@math.byu.edu}
\end{center}
\vskip 30pt \centerline{\smallit Received:  3/20/02, Revised:
6/27/02, Accepted: 6/27/02, Published:  6/28/02} \vskip 30pt

\centerline{\bf Abstract}


\noindent  In
Rademacher's proof \cite{Rad} of the
Ramanujan identities modulo 5 and 7, 
 the function  $$ L_{k} (z) = \eta(kz)
\sum_{\mu=0}^{k-1} \left\{\frac{1}{\eta \left(\frac{z+24\mu}{k}
\right) } \right\}$$ appears.  This is a modular function on $
\Gamma_0(k)$, provided that $(k,6) = 1$.  The basic identities
\begin{gather*} p(5n + 4) \equiv 0 \bmod 5, \\
  p(7n + 5) \equiv 0 \bmod 7,  \end{gather*} (as well as some generalizations
  to higher powers of 5 and 7) are proved by examining the coefficients of the expansions of $L_5$ and $L_7$
  at infinity.  The two identities above immediately imply the following identity:
  $$ p(35n + 19) \equiv 0 \bmod 35.  $$  Rademacher asked if there was a corresponding modular identity.
  Such an identity is derived in this paper. Also, another identity is obtained from the expansion
  of $L_{35}$ at 0.

  \pagestyle{myheadings} \markright{\smalltt INTEGERS: \smallrm
ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 1
(2002), \#A09\hfill}

\thispagestyle{empty} \baselineskip=15pt \vskip 30pt


\section*{\normalsize 1.  Generalities}
\subsection*{\normalsize 1.1  Basics}
Throughout, $ p(n) $ refers to the unrestricted partition
function.  Euler proved the identity $$ \sum_{n=0}^{\infty} p(n)
q^n = \prod_{m=1}^{\infty} \frac{1}{1-q^m}. $$  We will see that
this ties the partition function intimately to certain types of
modular functions (and forms).
 Let $ \Gamma(1) = \mathrm{PSL}_{2}(\mathbb{Z}) $, which
is
 the special linear group modulo its center $(\pm I).$  $\Gamma(1)$
 is generated by the two transformations $$ S = \begin{pmatrix}
 0 & 1 \\
 -1 & 0
 \end{pmatrix},
 \ \ \ T = \begin{pmatrix}
 1 & 1 \\
 0 & 1
 \end{pmatrix}.$$ We'll have need also for the product $TST,$ which we will call
 $W$. Then in $\Gamma(1)$, we have:
 $$ W = \begin{pmatrix}
        1 & 0\\
        1 & 1
        \end{pmatrix}. $$
   $\Gamma(1) $ acts faithfully on the upper half plane $ \mathcal{H}=\{z \in \mathbb{C}: \Im(z)>0 \}$ via
 linear fractional transformation.  Throughout, we will identify a
 linear fractional transformation and its matrix.

We let $ q = e ^{2 \pi i z}$ throughout. In what follows,
Dedekind's function $\eta(z)$ will be used, so we remind the
reader of some known facts. Recall
 $ \eta (z) = e^{\frac{\pi i z}{12}}\prod_{n=1}^{\infty} \left(1-e^{2\pi
 i n z}\right)$, and note that $\eta(z)$ is periodic of period 24.  It is pole-free and zero-free throughout $
 \mathcal{H}.$
  It is a modular form of weight one half, and
 has a multiplier which is a $24^{th}$ root of unity in its
 transformation equation. The root of unity $\varepsilon(a,b,c,d)$ is determined by the following
formulas (provided $ c \ne 0$):  \begin{multline} \label{root}
 \varepsilon(a,b,c,d) =
\exp\left\{\pi i\left(\frac{a+d}{12c} + s(-d,c)\right) \right\},
\\
 s(h,k) = \sum_{r=1}^{k-1} \frac{r}{k}
\left(\frac{hr}{k}-\left[\frac{hr}{k}\right]-\frac{1}{2}\right),
\end{multline}
 where [ ] is the greatest integer function. For properties of the Dedekind $s(h,k)$ sum see \cite{RWded, RGded}.
 We will often write just
 $\varepsilon$ in place of the more cumbersome
 $\varepsilon(a,b,c,d)$.  Thus, for any $ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma(1) $,
 $\eta (Az) = \varepsilon (cz + d)^{1/2} \eta (z).$
 There is a relationship between $ \eta $ and $ p $.  Namely,
 $$ \sum_{n=0}^{\infty}  p(n) q^n = \frac{q^{\frac{1}{24}}}{\eta(z)}. $$   Also,  $ L_k$ has $q$-expansion given by:
\begin{multline}\label{q-expansion}  L_k (z) = kq^{1 + \left[\frac{k}{24}\right
]}\prod_{m=1}^{\infty} (1-q^{km}) \sum_{n=0}^{\infty} p(kn +
\Delta) q^n,\ \\   \Delta \ \mathrm{least \ positive \ solution \
of} \ 24n \equiv  1 \bmod k.  \end{multline} It is clear that
$L_k$ has all its poles located at the cusps of $ \Gamma_0 (k). $
Note that when $ k = 35 \ (\mathrm{resp.} \ 5, 7)$, $\Delta = 19 \
(\mathrm{resp.} \ 4, 5)$.  This explains the form for the
identities listed earlier. That $L_k(z)$ is a modular function on
$\Gamma_0(k)$ may be seen as follows.  One can show that
$\frac{\eta(kz)}{\eta \left(\frac{z}{k}\right)}$ is a modular
function on $\Gamma_0^0(k)$ using known facts about Dedekind sums
(see particularly Theorems 17 and 18 of \cite{RWded}).  The
matrices $\begin{pmatrix} 1 & 24 \mu \\ 0 & 1 \end{pmatrix},
 (0 \le \mu \le k-1)$ form a set of coset representatives for $\Gamma_0^0(k)$ in
$\Gamma_0(k).$ Therefore, when we sum over this set of coset
representatives to obtain $L_k(z),$ we obtain a modular function
on $\Gamma_0(k).$

Let $\omega \in \mathbb{Q}\bigcup\{i\infty\}$ be a cusp for
$\Gamma_0(k).$ We will let $H_{\omega} $ be the set of modular
functions on $\Gamma_0(k)$ having poles only at $\omega$. It is
with this space of functions that we will work.  In particular we
will consider functions in $H_{\omega}$, where $\omega=0$ or
$i\infty.$

 We mention also the Fricke involution $S_k(z) = \frac{-1}{kz}.$
 Let $\mathcal{F}$ denote the set of modular functions on $\Gamma_0(k),$ and $f
 \in \mathcal{F}.$
$S_k(z)$ is an operator from $\mathcal{F}$ to itself.  Moreover,
it interchanges the polar orders of $f$ at $i \infty$ and $0$ in
the appropriate uniformizing variables (i.e, the expansion of
$f(S_k(z))$ at $i\infty$ has the same order in the variable $q$,
as the expansion of $f(z)$ at zero has in the variable
$q^{\frac{1}{k}}).$ For the rest of section 1, let $k=pQ,$ where
$p$ and $Q$ are two distinct primes strictly greater than 3.  Then
$S_k(z)$ interchanges the polar orders of $f$ at the cusps
$\frac{1}{p}$ and $\frac{1}{Q}$ as well.



\subsection*{\normalsize 1.2  Evaluation at $0$} Pursuant to our task, we need to
determine the order of $L_k$ at each of the cusps. We determine
the expansion about $0$.  To do so we subject $L_k$ to the
transformation $S_k$ and then compute the expansion as usual.

$$L_k(S_k z) = \eta \left[ \begin{pmatrix}
                        0 & -1 \\
                        1 & 0
                        \end{pmatrix}
                         z \right ]
            \sum_{\mu = 0}^{k-1} \left\{\eta \left[\begin{pmatrix}
                                                    24\mu k & -1 \\
                                                    k^2   &   0
                                                    \end{pmatrix} z \right] \right\}^{-1}. $$
We need to factor the inner matrix as the product of a unimodular
matrix and an upper triangular matrix.  Upon factoring and
applying the transformation identities we obtain: $$ L_k(S_k z) =
\frac{1}{\sqrt{k}} \, \eta\left(z\right) \sum_{\mu=0}^{k-1}
\frac{1}{\varepsilon(\frac{24\mu}{\delta}, b, \frac{k}{\delta}, d)
\, {\delta}^{1/2} \, \eta\left(\delta^2 z + \frac{\delta
y}{k}\right)}, $$ where $\delta = (\mu, k)$, and the matrix inside
factors as:
$$
\begin{pmatrix}
                                        24\mu / \delta &  b\\
                                        k/ \delta &      d
                                        \end{pmatrix}
                                        \begin{pmatrix}
                                        \delta k & y \\
                                        0  & k/ \delta
                                        \end{pmatrix},$$ for some
integers $b,d$ and $y$ such that

\begin{gather*}  \frac{24\mu}{\delta}y + \frac{bk}{\delta} = -1, \\
 \frac{24\mu}{\delta}d - \frac{bk}{\delta} = 1. \end{gather*}
 Together, the two equations yield that $y=-d$ (in the case $\mu =
 0$ this is clear). Thus we obtain the following expression:
\begin{equation} \label{at zero}  L_k(S_k z) = \frac{1}{\sqrt{k}}
\prod_{m=1}^{\infty} \left(1-q^m\right) \sum_{\mu=0}^{k-1} \left\{
\varepsilon \,\delta^{\frac{1}{2}} \, e^{\frac{\pi i y
\delta}{12k}} \, q^{\frac{\delta^2 - 1}{24}} \prod_{n=1}^{\infty}
\left(1-e^{2 \pi i \delta yn} q^{\delta^2n} \right)\right\}^{-1}.
\end{equation} Now, since $\mu = 0$ is the only term of the sum in
which $\delta=k $, and this yields the largest negative power of
$q$, we see that $\ord_0(L_k) \ge (1-k^2)/24$. We will return to
equation~\ref{at zero} later.

\subsection*{\normalsize 1.3  Evaluation at all other cusps}
Since k is square-free, all the other cusps of $\Gamma_0(k)$ occur
at the points $\{1/l | \ 1< l < k, \ l|k\,\}$, in other words, at
$\frac{1}{p}$ and $\frac{1}{Q}.$ To determine the order of $L_k$
at $1/l$, we subject it to the transformation $W^{-l}$ and then
evaluate. Thus, we obtain

\begin{gather*}
 L_k \left(W^{-l} z\right) = \eta \left(k W^{-l} z \right) \sum_{\mu = 0}^{k-1}
\left\{\eta \left(\frac{W^{-l} z + 24\mu}{k}\right)\right\}^{-1} \\
\quad = \eta \left[\begin{pmatrix}
                    k/l & 1 - k/l  \\
                    -1   &     1
                    \end{pmatrix}
                    \begin{pmatrix}
                    l &  k/l - 1 \\
                    0  & k/l
                    \end{pmatrix}
                    z \right] \sum_{\mu = 0}^{k-1} \left\{\eta
                    \left[ \begin{pmatrix}
                            1 - 24l\mu  & 24\mu \\
                            -k l & k
                            \end{pmatrix}
                            z \right] \right\}^{-1}.
\end{gather*}

Now, let $\delta = (1 - 24\mu l , k/l)$.  Then the matrix inside
factors as $$ \begin{pmatrix}
               (1 - 24\mu l)/\delta  & b \\
               (-k l)/\delta   & d
               \end{pmatrix}
               \begin{pmatrix}
               \delta  & y \\
               0   & k/\delta
               \end{pmatrix}, $$ for integers $b,d$ and $y$ such that
\begin{gather*}
\frac{1 - 24\mu l }{\delta}\,d + \frac{bk}{\delta}\,l = 1, \\
\frac{1 - 24\mu l }{\delta}\, y + \frac{bk}{\delta} = 24\mu, \\
\frac{-kl}{\delta}\,y + \frac{kd}{\delta} = k.
\end{gather*}

The final equation above yields $ -l y + d =\delta$.  Using this
fact we see that
\begin{gather*}
L_k(W^{-l}(z)) = e^{-\frac{k+l}{12l} \pi i} \sqrt{l/k} \,\, \eta
\left(\frac{l^2 z - l + k}{k}\right) \sum_{\mu=0}^{k-1}
\left\{\varepsilon \,\delta^{1/2} \,\eta \left(\frac{\delta^2 z +
\delta y}{k}\right)\right\}^{-1} \  \\ = e^{-\frac{k + l}{12l} \pi
i} \sqrt{l/k} \prod_{m=1}^{\infty} \left(1 - \rho^{(k-l)n}
q^{\frac{l^2}{k} n}\right) \times \\ \sum_{\mu=0}^{k-1}
\left\{\varepsilon \, \delta^{1/2} \rho^{\frac{dy+k-l}{24}}
q^{\frac{\delta^2-l^2}{24k}}
\prod_{n=1}^{\infty}\left(1-\rho^{\delta y n}
q^{\frac{\delta^2}{k} n} \right) \right\}^{-1}, \ \ \left(\rho =
e^{\frac{2 \pi i}{k}}\right).
\end{gather*}

\vskip 30pt

\section*{\normalsize 2.  Specialization to k = 35}
\subsection*{\normalsize 2.1  Obtaining an explicit polynomial identity at $i\infty$}
Now, we take $k=35$ for the remainder of this paper. We see from
the results above that $L_{35} (z)$ has a zero of order $2$ at
$i\infty$ in the variable $q$, a pole of order $51$ at $0$ in the
variable $q^{\frac{1}{35}}$ , is holomorphic at $1/5$ in
$q^{\frac{1}{7}},$ and has a zero of order at least $1$ at $1/7$
in the variable $q^{\frac{1}{5}}.$ We refer the reader to
\cite{NewClass} for an explicit construction of the basis
functions with which we will work, and for an account of the
functions $A_9$ and $B_8$ (as well as other similar functions used
in deriving the basis of functions at infinity). In summary, the
basis functions $B_4, B_5, B_6, $ and $B_7$ all have poles only at
$ i \infty$ in the same order as the value of their respective
subscript. Similarly, $A_9$ and $B_8$ have poles of order 9 and 8,
respectively, at $i \infty$, and $A_9$ has zeroes of orders 5, 1,
and 3 at $0, 1/5$ and $1/7$ respectively, while $B_8$ has zeros of
orders 6 and 2 at 0 and $1/7$ respectively.

Suffice it to say that $G(z) =  A_9 (z) B_8^8 (z) L_{35} (z)$ is
pole-free in $\mathcal{H}$ and has order $-71$ at $i\infty$, $2$
at $0$, $0$ (or possibly greater) at $1/5$, and $20$ at $1/7$.
 So we have: \begin{theorem} $G(z) \in \mathbb{C}[B_4,
B_5, B_6, B_7].$ \end{theorem}

The proof follows from the work in the previous section and the
fact that the $B_i$ do form a polynomial basis for the set of
modular functions in $H_{i\infty}$.

Since the order at $i\infty$ is known, the polynomial is
determined, using Mathematica to calculate the coefficients. Among
the possible ways of choosing an appropriate form, it was
determined to use $B_7$ to keep the total degree (as a polynomial
in the $B_i$) as small as possible.  For instance the terms
matching $q^{-65}, \ q^{-64}, \ q^{-63} $ are $B_7^8 B_5 B_4, \
B_7^8B_4^2, \ B_7^9$ respectively. Call the polynomial that we
obtain $P_{\infty}$. Then $L_{35} (z) = A_9^{-1}B_8^{-8}
P_{\infty}$.
\begin{theorem} \label{polynomial} Let $c_i$ be the coefficient on
the term of $P_{\infty}$ corresponding to $q^i$.  Then the
polynomial $P_{\infty}$ is determined by the coefficients in the
appendix in table 1, where $c_i$ is the given $i^{th}$ coefficient
multiplied by $1225$ $(={35}^2)$.
\end{theorem}
\begin{cor} $p(35n+19)\equiv 0 \bmod 35. $ \end{cor}
\begin{rem} This identity also shows how the genus dramatically
affects a polynomial identity associated with such an identity for
the partition function. For the cases $k=5$ and $k=7$, where the
genus is $0$, Rademacher obtains a linear polynomial and a
quadratic polynomial, respectively, in the respective basis
function.  Here, the genus is 3 and a polynomial of degree $11$ in
the \emph{four} basis functions is obtained.  In a similar manner,
modular identities for the congruences for 55 and 77 may surely be
constructed. It is expected that more basis functions and a
higher-degree polynomial will be necessary.
\end{rem}
\subsection*{\normalsize 2.2  An Identity at $0$}
We return to equation~\ref{at zero}. Recalling the connection with
the partition function $p$, and separating into cases depending on
$\delta$, we have : $$ L_k(S_k(z)) = \sum_{\delta | 35}
\frac{q^{\frac{1-\delta^2}{24}}}{\sqrt{k}} \sum_{\mu = 0}^{34}
\varepsilon^{-1} \delta ^{-\frac{1}{2}} \sum_{n=0}^{\infty}
e^{\frac{(24 n -1)\delta y \pi i}{12k}} p(n) q ^{\delta^2 n}
\prod_{m=1}^{\infty} (1-q^m). $$ When we group the outer sum into
terms by $\delta$, we obtain Gauss sums for the Jacobi-Legendre
symbol of the appropriate modulus. We have the following terms:
\begin{gather*} \delta = 35: \ \ \ \  \frac{1}{35q^{51}}
\prod_{m=1}^{\infty} (1-q^m) \sum_{n=0}^{\infty} p(n)q^{1225n}, \\
\delta = 7: \ \ \ \  - \frac{1}{7q^2} \prod_{m=1}^{\infty} (1-q^m)
\sum_{n=0}^{\infty} \left ( \frac{n-4}{5} \right ) p(n)q^{49n},
\\ \delta = 5: \ \ \ \  - \frac{1}{5q} \prod_{m=1}^{\infty}
(1-q^m) \sum_{n=0}^{\infty} \left ( \frac{n-5}{7} \right )
p(n)q^{25n},
\\ \delta = 1: \ \ \ \ \prod_{m=1}^{\infty} (1-q^m)
\sum_{n=0}^{\infty} \left ( \frac{n-19}{35} \right ) p(n)q^n,
\end{gather*}
where (-) indicates the Jacobi-Legendre symbol.

Now, $L_{35}(S_{35} z)$ has order $-51$ (resp. $ 2, 1, -1$) at $ i
\infty$ (resp. $ 0, 1/5, 1/7$). For the last two points we really
only know that the order at each of these points is greater than
or equal to the given value. Nevertheless, this is sufficient to
determine that the function $ h(z) = B_8 (z) L_{35} (S_{35} z) $
has its only pole at $ i \infty $, and the order of the pole there
is $ 59.$ Thus, $ h(z) \in \mathbb{C}[B_4, B_5, B_6, B_7].$

To clear fractions, multiply by $ 35 $ and then note $35
L_{35}(S_{35} z) = B_8^{-1} P_0 $, where $P_0$ is a polynomial to
be determined in the four basis functions.  So $$ q^{-51} -5
q^{-2} -7 q^{-1} + 35 + ... = P_0 B_8^{-1}\Sum{n} p(n)q^n. $$ This
yields the following:
\begin{theorem}
$ 35 L_{35} (S_{35} z)$ is a polynomial in the basis functions $
B_4, B_5, B_6, B_7 $, with integral coefficients given in table 2
of the appendix.
\end{theorem}

\begin{cor} \begin{gather*} \Sum{n}\left\{ p(n)q^{1225n}-5\left(\frac{n-4}{5}\right)p(n)q^{49(n+1)}   \right.  \\
\left. - 7\left(\frac{n-5}{7}\right)p(n)q^{25(n+2)}+35\left(\frac{n-19}{35}\right)p(n)q^{n+51} \right\} \\
 = \ q^{51}P_0B_8^{-1}\Sum{n}p(n)q^n. \end{gather*} \end{cor}

I would like to thank my doctoral advisor, Morris Newman, who
suggested this problem to me, and Bruce Berndt and the referee for
their helpful comments. \vskip 30pt

\section*{\normalsize 3.  Appendix: tables}
\begin{longtable}{|c|r|c|r|}
\caption[]{Coefficients for $P_{\infty}$} \\ \hline  i &
\multicolumn{1}{c|}{coefficient} & i &
\multicolumn{1}{c|}{coefficient} \\ \hline \endfirsthead
\caption[]{\emph continued}\\ \hline  i &
\multicolumn{1}{c|}{coefficient} & i &
\multicolumn{1}{c|}{coefficient} \\ \hline \endhead \hline
\multicolumn{4}{r} {\emph {continued on next page}} \endfoot
\hline
\endlastfoot
 -69 & 14 & -33 & -10569223238228513913405 \\ -68 & 10487 & -34 & -29246449555394872646896 \\
  -67 & 1050000 & -33 & -30752043774402282243297 \\
   -66 & 37305717 &  -32 & -10500761656485195685244 \\
-65 & 689742738  &  -31 & -63994584518030323125 \\
 -64 & 6710739396 &  -30 & 747396610368371686739 \\
-63& 66723377634 &  -29 & 41866861538739825786329 \\
 -62 & 306529775888 & -28 & -154825275874629072148111 \\
  -61& 944992220254 & -27 & -330675045299039974754087 \\
   -60 & 8032835276603 & -26 & -330675045299039974754087 \\
   -59 & 41786228768706 & -25 & -195434688782712582767044 \\
    -58 & 163674979606784 & -24 & 106519373950966952221 \\
      -57 & 201541582793635 & -23 & -4489388038461810926594 \\
-56 & 1192164476376773 & -22 & 93815803283069768407781 \\
 -55 & 2133885512377322 & -21 & -868861692593019875600320 \\
  -54 & -224422822302317 & -20 & -780240652273282530768564 \\
   -53 & 9311212319487067 & -19 & -903341606561540466301311 \\
    -52 & 53104246090438488 & -18 & -575916410654649291605538 \\
    -51 & 174875061214811942 & -17 & -72387387334271714225 \\
     -50 & 59790513579741686 & -16 & 5279005749531844180716 \\
     -49 & 418363890733605649 & -15 & -329709784193546752041430 \\
-48 & 1123804424169272601 & -14 & -754715959977583205060136 \\ -47
& -1849756259047888932 & -13 & 2673409023878484316383343 \\ -46 &
 -3961615256484350591 & -12 &
3038512747510492176975342 \\ -45 & 1372328949965814202 & -11 &
1778681601887609961535840 \\ -44 & 19742808525931603996 & -10 &
16760790778989818745 \\ -43 & 2177933533113105157 & -9 &
-1894749705906830722098 \\ -42 & 23516820211582241283 & -8 &
189938163107999640927156 \\ -41 & 123390318981474451231 & -7 &
4789820626890790384933368 \\-40 & -271841378745461882544 & -6 &
-1532454919226226734377599 \\ -39 & -777621797568622505476 & -5 &
-1735248899286586237586336 \\ -38 & -673143178730309518022 & -4 &
-997109474512591275870705 \\ -37 & 268511875093390021788 & -3 & 0
\\ -36 & 10323721732692166606 & -2 & 0 \\ -35 & 334287613007887666292 & -1 & 0\\ -34 &
3963341600197861903691 & 0 & -3000539263195134467759624
\\ \hline
\end{longtable}

\begin{center}
\begin{longtable}{|c|r|c|r|} \caption[]{Coefficients for $P_0$}\\ \hline i &
\multicolumn{1}{c|}{coefficient} & i &
\multicolumn{1}{c|}{coefficient} \\ \hline \endfirsthead
\caption[]{\emph continued} \\ \hline i &
\multicolumn{1}{c|}{coefficient} & i &
\multicolumn{1}{c|}{coefficient} \\ \hline \endhead \hline
\multicolumn{4}{r} {\emph {continued on next page}} \endfoot
\hline
\endlastfoot
 -59 & 1 & -29 & 101856\\ -58 & -5 & -28 & -1088550
\\ -57 & 13 & -27 & -779536
\\ -56 & -24 & -26 & -868518\\ -55 & -124 & -25 & -638720
\\ -54 & -167 & -24 & 183785
\\  -53 & -100 & -23 & 310027 \\ -52 & 186 & -22 & -1009582 \\ -51 & 82 & -21 & -5533 \\ -50 & -784 & -20 &
8917982 \\  -49 & 1899 &  -19 & 10010374 \\  -48 & 7839 & -18 &
6789719
\\ -47 & 9103 & -17 & -21976
\\  -46 & 5400 & -16 & 113545 \\  -45 & -531 & -15& 35220 \\  -44 & 2196 & -14 &
21729982 \\  -43 & -786 & -13 & -703140
\\ -42 & 13705 & -12 & -920456 \\ -41 & -1006 & -11 & -843856 \\ -40 & -4690 & -10 & -236844 \\  -39 & 1882 & -9 &
-376949 \\ -38 & -6611 & -8 & 1365595 \\ -37 & -16039  & -7 &
-4046123
\\ -36 & 44380 & -6 & -12312578
\\ -35 & -84846 & -5 & -13942378
\\ -34 & -393717 & -4 & -9425016 \\ -33 & -427964 & -3 & 0 \\ -32 & -282747 & -2 & 0 \\ -31 & -17208 & -1 & 0
\\
-30 & -53917 & 0 & -30151150 \\ \hline
\end{longtable}
\end{center}


\nocite{Ap41, ApIntro}
\bibliographystyle{alpha}
\begin{thebibliography}{New57}

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\end{thebibliography}


\end{document}

