text:
\section{Introduction}\label{Intro}
At an odd prime $p$, the 2-line
$\operatorname{Ext}^{2,*}$ of the Adams-Novikov
spectral sequence
\[
\operatorname{Ext}^{s,t}:=\operatorname{Ext}^{s,t}_{BP_*BP}(BP_*,BP_*)
\Rightarrow(\pi_{t-s}S^0)_{(p)}
\]
is generated additively by distinguished
classes $\beta_{i/j,k}\in\operatorname{Ext}^{2,
2i(p^2-1)-2j(p-1)}$
of order $p^k$ for certain ordered
triples of positive integers
$(i,j,k)$ \cite{MRW}.
These generators
are known collectively as the
$p$-primary divided beta family.
Behrens \cite{Beh:Cong}
has shown that, for
$p\geq5$, each
$\beta_{i/j,k}$ gives rise to
a modular form $f_{i/j,k}$
over the integers
satisfying conditions
formulated in terms of
$i$, $j$, $k$, and $p$,
including both a congruence
and non-congruence condition.
The proof that these
modular forms exist uses
homotopical properties of spectra
related to the $K(2)$-local
sphere \cite{Beh:Mod}
and ultimately does not explicitly identify the $f_{i/j,k}$.
The purpose of this paper is to
compute the modular forms
$f_{i/j,1}$ associated to order
$p$ divided beta family elements
$\beta_{i/j,1}$
in several cases, including
a complete computation of
such forms at the prime 5.

Behrens' theorem
\cite[Theorem~1.3]{Beh:Cong}
says the following in the case $k=1$.
Let $M_t$
be the space of weight $t$ modular forms
over $\mathbb Z$ for the full
modular group $\Gamma_0(1)=SL_2(\mathbb Z)$,
and for
a positive integer $N$, let $M_t(N)$
be the space of weight $t$ modular
forms over $\mathbb Z$
for the congruence subgroup
$\Gamma_0(N)\subset \Gamma_0(1)$
\cite{Lang:Mod}.
Given an order $p$ divided beta
family element $\beta_{i/j,1}$
for $p\geq5$,
there exists a modular form
$f_{i/j,1}\in M_{i(p^2-1)}$ satisfying
the following conditions:
\begin{enumerate}
\item[(C1)] the Fourier expansion $f_{i/j,1}(q)\in\mathbb Z[[q]]$
is not congruent to 0 mod $p$,
\item[(C2)] the integer $12\cdot\operatorname{ord}_qf_{i/j,1}(q)$ is either
greater than $(p^2-1)i-(p-1)j$, or equal to $(p^2-1)i-(p-1)j-2$,
\item[(C3)] there does not exist $g\in M_t$ for $t<(p^2-1)i$ such that
$f_{i/j,1}(q)\equiv g(q)\mod p$, and
\item[(C4)]\label{C4} for every prime $\ell\neq p$,
there exists $g\in
M_{(p^2-1)i-(p-1)j}(\ell)$
such that $(L_{\ell}f_{i/j,1})(q)\equiv g(q)\mod
p$
\end{enumerate}
where
$L_N:M_t\to M_t(N)$ is a linear operator
to be defined in Section \ref{ModularForms}.

There is ambiguity inherent in the computation
of these modular forms.  For example,
if a modular form $f$ satisfies
conditions (C1) through (C4) for given
$i$, $j$, and $p$, then so does
$cf+h$, where $c\in\mathbb Z$ with $(c,p)=1$
and $h$ is a modular form of the
same weight and $q$-order as $f$ such that
$h(q)\equiv 0\mod p$.  The precise extent to which
a given $f_{i/j,k}$ fails to be unique will be the
subject of future work.   The goal
here is to produce {\em at least one} candidate
for any $f_{i/j,1}$ that we study, and
an assertion of the form $f_{i/j,1}=f$
is understood to mean that $f$ is one of
many possible
choices for $f_{i/j,1}$ in $M_{i(p^2-1)}$.

Our results generalize sample
computer-aided computations of Behrens and Laures \cite{Beh:Lau}.
To state their results, and our
generalizations thereof, requires
additional notation.  In the divided
beta family, we write $\beta_{i/j}$ for
$\beta_{i/j,1}$, we write $\beta_i$ for $\beta_{i/1}$,
and we abbreviate the corresponding $f_{i/j,k}$ similarly.
In the
space of modular forms, $\Delta\in M_{12}$ is the
Ramanujan delta function and $E_4\in M_4$ is the
weight 4 Eisenstein series,
whose definitions we will recall
in Section \ref{ModularForms}.
Behrens and Laures computed the following
modular forms at the prime 5:
\begin{align}\label{powersofD}
\begin{split}
f_1=\Delta^{2},\quad
f_2=\Delta^{4},\quad
f_3=\Delta^{6},\quad
f_4=\Delta^{8},\quad
f_{5/5}=\Delta^{10},
\end{split}
\end{align}
and
\begin{equation}\label{FTFTN}
f_{25/29}=\Delta^{50}+4\Delta^{42}E_4^{24}
+3\Delta^{41}E_4^{27}.
\end{equation}
We shall see that the striking
pattern of powers of $\Delta$
in \eqref{powersofD}
does indeed persist at the
prime 5 and elsewhere.

The formula for $f_{25/29}$
in \eqref{FTFTN} is
the catalyst for our 5-primary computations.
This is despite the fact that
the corresponding element
$\beta_{25/29}$ in homotopy theory is actually
not a member of the 5-primary divided beta family
(see Lemma \ref{betas}), but rather
can be properly interpreted as an element
of a different, closely related
spectral sequence (see \cite{Beh:Lau}).
The fact that $\Delta^{50}+4\Delta^{42}E_4^{24}
+3\Delta^{41}E_4^{27}$ is indeed
a modular form satisfying conditions
(C1) through (C4) in the case $i=25$,
$j=29$, and $p=5$, is instructive.
For example, while $\beta_{25/29}$
is not a 5-primary divided beta family element,
$\beta_{50/29}$ is, and
the way $f_{25/29}$
is built as the sum of a power of $\Delta$ and two
``correction terms'' serves as a model for how to
compute $f_{50/29}$.  In fact,
it is by identifying appropriate
analogs of these ``correction terms''
that we are able to compute
all modular forms $f_{i/j}$ at the prime 5
that are not simply powers of $\Delta$.

Before stating the main theorem
let us define, for $n\geq2$ and $r\geq1$,
modular forms
\begin{align*}
C_{0,n,r}&=4r\Delta^{42\cdot5^{n-2}+2(r-1)5^n}E_4^{24\cdot5^{n-2}},\\
D_{0,n,r}&=3r\Delta^{41\cdot5^{n-2}+2(r-1)5^n}E_4^{27\cdot5^{n-2}}
\end{align*}
in $M_{24r\cdot5^{n}}$,
so that $C_{0,2,1}+D_{0,2,1}=4\Delta^{42}E_4^{24}
+3\Delta^{41}E_4^{27}$
is precisely the summand appearing
in \eqref{FTFTN}.
Moreover, for $n\geq3$, $1\leq m\leq n-2$, and $r\geq1$, define
\begin{align*}
C_{m,n,r}&=3r\Delta^{8\cdot5^{n-1}
	+2\cdot5^{n-m-2}+2(r-1)5^n}
E_4^{6\cdot5^{n-1}-6\cdot5^{n-m-2}},\\
D_{m,n,r}&=r\Delta^{8\cdot5^{n-1}
	+5^{n-m-2}+2(r-1)5^n}
E_4^{6\cdot5^{n-1}-3\cdot5^{n-m-2}}
\end{align*}
in $M_{24r\cdot5^n}$.  Finally,
given a prime $p$ and
any integer $n\geq0$, define
\[
a_n=\begin{cases}
0,&\text{if }n=0,\\
p^n+p^{n-1}-1,&\text{if }n\geq1.
\end{cases}
\]

\begin{thm}\label{main_5}
Given a 5-primary divided beta family
element $\beta_{i/j}$ of order 5,
where $i=r\cdot5^n$ with $(r,5)=1$,
the corresponding modular form
is $f_{i/j}=\Delta^{2r\cdot5^n}$
except when the inequalities $r>1$ and $5^n+1\leq j\leq a_n$
hold simultaneously.  For such $r$ and $j$, there is a positive
integer $u$, $1\leq u\leq n-1$, such that
\[
5^n+5^{n-1}-5^{n-u}+1\leq j\leq5^n+5^{n-1}-5^{n-u-1}
\]
in which case
\[
f_{i/j}=\begin{cases}
\Delta^{2r\cdot5^n}+
\displaystyle\sum_{m=0}^{u-1}
(C_{m,n,r}+D_{m,n,r})
&\text{if}\quad j>5^n+5^{n-1}-5^{n-u}+2\cdot5^{n-u-1},\\
&\, \\
\Delta^{2r\cdot5^n}+\displaystyle\sum_{m=0}^{u-2}
(C_{m,n,r}+D_{m,n,r})+C_{u-1,n,r}
&\text{if}\quad j\leq5^n+5^{n-1}-5^{n-u}+2\cdot5^{n-u-1}.\\
\end{cases}
\]
\end{thm}
In the case $r>1$, Theorem \ref{main_5} shows that
the modular forms $f_{i/j}$ can all be recovered
from $f_{i/a_n}$ by trimming off some number
of summands depending on the size of $j$.
Example \ref{1250} below clearly exhibits this phenomenon.
It is therefore worth noting that the formula
for $f_{i/a_n}$ in Theorem \ref{main_5} has a recursive
interpretation.
\begin{cor}\label{main_5_cor}
For $n\geq1$ and $r\geq2$, $f_{r\cdot5^{n+1}/a_{n+1}}=
(f_{r\cdot5^n/a_n})^5
+C_{n-1,n+1,r}+D_{n-1,n+1,r}$.
\end{cor}
The formulas for a given $f_{i/a_n}$
obtained from Theorem \ref{main_5}
and Corollary \ref{main_5_cor}, respectively, will differ
in $M_{24i}$, but they will be equivalent
modulo 5.
\begin{exmp}\label{preview}
To preview the concepts
and computational methods behind the
proof of Theorem \ref{main_5},
let us employ them to outline a
re-derivation of Equation \eqref{FTFTN}.
Since $f_{25/29}\in M_{600}$, it will
follow from Proposition \ref{basis}
and conditions (C1) and (C3)
that
\[
f_{25/29}=\Delta^{50}+\sum_{m=1}^{50}c_m\Delta^{50-m}
E_4^{3m}
\]
for integers $c_i\in\mathbb Z$.
The demand on $q$-order given by
condition (C2), together with Proposition
\ref{setup}(c),
will imply that $c_{10}=c_{11}=\cdots
=c_{50}=0$.

To verify
condition (C4) at a prime $\ell\neq5$,
Proposition \ref{Serre} will imply
that it suffices to show
$L_{\ell}f_{25/29}$ is
divisible by $E_4^{29}$ in
$M_*(\ell)_{\mathbb Z/5}$.
Proposition \ref{rigidity} will imply
that it suffices to do this for
the case $\ell=2$.
In $M_*(2)_{\mathbb Z/5}$, we
shall see that $L_2\Delta^{50}$ is divisible
by $E_4^{25}$ by Lemma
\ref{L2ONDELTA}.
The $E_4$-divisibility
of $L_2\Delta^{50-m}E_4^{3m}$ for $1\leq m\leq9$
can be computed directly in a similar
manner using Lemma \ref{L2TERM}:\vskip.1truein
\setlength{\tabcolsep}{2pt}
\begin{center}
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
$m$&1&2&3&4&5&6&7&8&9\\\hline
term&$L_2\Delta^{49}E_4^3$&$L_2\Delta^{48}E_4^6$&
$L_2\Delta^{47}E_4^9$&$L_2\Delta^{46}E_4^{12}$&$L_2\Delta^{45}E_4^{15}$
&$L_2\Delta^{44}E_4^{18}$&$L_2\Delta^{43}E_4^{21}$&$L_2\Delta^{42}E_4^{24}$
&$L_2\Delta^{41}E_4^{27}$\\\hline
$E_4$-div.&$E_4^3$&$E_4^7$&$E_4^9$&$E_4^{13}$&$E_4^{15}$&
$E_4^{19}$&$E_4^{21}$&$E_4^{25}$&$E_4^{27}$
\end{tabular}
\end{center}\vskip.1truein
Thus, for $L_2f_{25/29}$ to have the required
$E_4$-divisibility,
we much choose $c_1=c_2=\cdots=c_7=0$.
From there, setting $c_8=4$ makes the term
\[
L_2(\Delta^{50}+c_8\Delta^{42}E_4^{24})
\]
divisible by $E_4^{27}$ in
$M_*(2)_{\mathbb Z/5}$, while all other choices
for $c_8$ (modulo 5) keep its $E_4$-divisibility
at $E_4^{25}$.  Subsequently setting
$c_9=3$ makes
\[
L_2(\Delta^{50}+c_8\Delta^{42}E_4^{24}
+c_9\Delta^{41}E_4^{27})
\]
divisible by $E_4^{29}$ in
$M_*(2)_{\mathbb Z/5}$, and no other choice
of $c_9$ (modulo 5) accomplishes this.
\end{exmp}
\begin{exmp}\label{1250}
Consider the conclusion of
Theorem \ref{main_5} in the
case $n=4$ and $r=2$.
The order 5 divided beta family elements $\beta_{i/j}$
with $i=2\cdot5^4=1250$ are
\[
\{\beta_{1250/j}:1\leq j\leq749\text{ and }
j\neq5,10,15,\ldots,125\}
\]
(see Lemma \ref{betas}).  By Theorem \ref{main_5},
$f_{1250/j}=\Delta^{2500}$ except when $626\leq j\leq749$,
in which case $f_{1250/j}=\Delta^{2500}+\mathfrak C$ where
\[
\mathfrak C=\begin{cases}
3\Delta^{2300}E_4^{600}+
\Delta^{2275}E_4^{675}+
\Delta^{2260}E_4^{720}+
2\Delta^{2255}E_4^{735}+
\Delta^{2252}E_4^{744}+
2\Delta^{2251}E_4^{747}
&\text{if }\quad j=748,749,\\
3\Delta^{2300}E_4^{600}+
\Delta^{2275}E_4^{675}+
\Delta^{2260}E_4^{720}+
2\Delta^{2255}E_4^{735}+
\Delta^{2252}E_4^{744}&\text{if }\quad j=746,747,\\
3\Delta^{2300}E_4^{600}+
\Delta^{2275}E_4^{675}+
\Delta^{2260}E_4^{720}+
2\Delta^{2255}E_4^{735}&\text{if }\quad 736\leq j\leq745,\\
3\Delta^{2300}E_4^{600}+
\Delta^{2275}E_4^{675}+
\Delta^{2260}E_4^{720}&\text{if }\quad 726\leq j\leq735,\\
3\Delta^{2300}E_4^{600}+
\Delta^{2275}E_4^{675}&\text{if }\quad 676\leq j\leq725,\\
3\Delta^{2300}E_4^{600}&\text{if }\quad 626\leq j\leq675.
\end{cases}
\]
In each of the six cases,
the term $\mathfrak C$ raises the $E_4$-divisibility
of $L_2f_{1250/j}$ in
$M_*(2)_{\mathbb Z/5}$
as required by
condition (C4), while keeping
the $q$-order of $f_{1250/j}$
sufficiently large as required
by condition (C2).
\end{exmp}
\begin{rem}
In addition to the modular forms
in Equations \eqref{powersofD} and
\eqref{FTFTN},
Behrens and Laures computed
$f_{25/5,2}=\Delta^{50}$, where $\beta_{25/5,2}$
is the first order 25 element one encounters
in the 5-primary divided beta family.
Computation of the modular forms $f_{i/j,k}$
for $k>1$ will be the subject of a future paper.
\end{rem}

This paper is structured as follows.
In Section \ref{Divided} we recall
how to enumerate the elements of order $p$
in the $p$-primary divided beta family.
In Section \ref{ModularForms} we list results
from the theory of modular forms relevant
to our computations.  Section \ref{at5}
is the technical heart of the paper,
where we compute at the prime 5 and prove
Theorem \ref{main_5}.
In Section \ref{atotherprimes}
we exhibit our computational methods
at the primes 7, 11, 13, and 677,
and for arbitrary primes $p\geq5$ we
conjecture which of the modular forms arising from Behrens' theorem are simply powers
of the Ramanujan delta function.
\section{The divided beta family}
\label{Divided}
In this section we show how to
enumerate the order $p$ elements
$\beta_{i/j}$
of the $p$-primary divided beta family.
Given $p$, let $a_n$ be defined
as in Section \ref{Intro}.
If we write $i=rp^n$
where $(r,p)=1$, then
\cite[Theorem~2.6]{MRW} implies that the
entire $p$-primary
divided beta family comprises the elements
$\beta_{i/j,k}$ for ordered triples of positive
integers $(i,j,k)$ subject to the following
rules:
\begin{enumerate}[i)]
\item if $r=1$, then $1\leq j\leq p^n$,
\item $p^{k-1}|j\leq a_{n-k+1}$,
\item if $p^k|j$, then $j>a_{n-k}$.
\end{enumerate}
The order $p$ divided beta family elements $\beta_{i/j}$
are therefore characterized as follows.
\begin{lem}\label{betas}
The divided beta family elements of
order $p$ of the form $\beta_{p^n/j}$
are
\[
\{\beta_{p^n/j}:1\leq j\leq p^n\text{ and }j\neq p,2p,\ldots,
a_{n-2}p\}.
\]
and those of the form $\beta_{rp^n/j}$
where $r>1$ and $(r,p)=1$ are
\[
\{\beta_{rp^n/j}:1\leq j\leq a_n\text{ and }j\neq p,2p,\ldots,
a_{n-2}p\}.
\]
\end{lem}
\begin{proof}
Consider first the case $r=1$.
Rule i) allows $j$ to range between
1 and $p^n$.  Since $k=1$,
rule ii) does not apply
any additional constraints.
Rule iii) says that $j$ cannot
be both a multiple of $p$ and
$\leq a_{n-1}=p^{n-1}+p^{n-2}-1$.
Therefore, we must insist that
\[
j\neq p,2p,\ldots,p^{n-1}-p^{n-2}-5=a_{n-2}p.
\]

In the case $r>1$, rule i)
does not apply and rule ii)
allows $j$ to range between
1 and $a_n$.  Rule iii)
disallows the same values of $j$
as in the case $r=1$.
\end{proof}
\begin{exmp}\label{exmp_i}
Given $p$, the element $\beta_i=\beta_{i/1,1}$ is a divided
beta family element of order $p$ for any $i\geq1$.
\end{exmp}
\begin{exmp}\label{exmp_ii}
Suppose $p=5$.  The divided beta family elements
of the form $\beta_{5r/j}$ are
\[
\beta_{5r/5},\quad
\beta_{5r/4},\quad
\beta_{5r/3},\quad
\beta_{5r/2},\quad
\beta_{5r/1}=\beta_{5r}.
\]
Those of the form
$\beta_{25/j}$ are
\[
\beta_{25/25},\quad \beta_{25/24},
\quad\ldots,\quad\beta_{25/6},\quad\beta_{25/4},
\quad\ldots,\quad\beta_{25/1}=\beta_{25}
\]
while those of the form $\beta_{25r/j}$ with $r>1$
are
\[
\beta_{25r/29},\quad \beta_{25r/28},\quad\ldots,\quad
\beta_{25r/6},\quad\beta_{25r/4},\quad
\ldots,\quad\beta_{25r/1}=\beta_{25r}.
\]
In particular,
as noted in Section \ref{Intro},
$\beta_{50/29}$ is a 5-primary
divided beta family element, while $\beta_{25/29}$
is not.
\end{exmp}
\section{Modular forms}\label{ModularForms}

In this section we record
facts from the theory of modular forms
required for our computations.
Much of this can be found in
\cite{Lang:Mod} unless otherwise noted.

If $t\geq4$ is an even integer
and $q=e^{2\pi iz}$, the
{\em weight $t$ Eisenstein series}
$E_t\in M_t$ is given by the formula
\[
E_t(z)=\dfrac12\sum_{(m,n)=1}\dfrac1{(mz+n)^t}
=1-\dfrac{2t}{B_t}\sum_{n=1}^{\infty}
\sigma_{t-1}(n)q^n
\]
where $B_t$ is the $t$-th Bernoulli number
and $\sigma_{t-1}(n)$ is the sum of the $(t-1)$st
powers of the divisors of $n$.
The {\em Ramanujan Delta function}
$\Delta$ is a modular form of weight 12
given by the formula
\[
\Delta(z)=\dfrac{E_4^3(z)-E_6^2(z)}{1728}
=\sum_{n=1}^{\infty}\tau(n)q^n=q+\cdots.
\]
The graded (by weight) ring $M_*$ of
all modular forms over $\mathbb Z$
for the full modular group is,
by a result of Deligne \cite[Proposition~6.1]{Del:Ell},
\[
M_*=\dfrac{\mathbb Z[E_4,E_6,\Delta]}{1728\Delta=E_4^3-E_6^2}.
\]
\begin{prop}{\cite[Theorem~10.4.3]{Lang:Mod}}\label{basis}
If $t\equiv0\mod 4$, then
$\{\Delta^aE_4^b:0<a,b\in\mathbb Z, 12a+4b=t\}$
is a $\mathbb Z$-basis for $M_t$.
\end{prop}

Let $M_*(2)[1/2]$ denote the
graded ring of modular forms
for $\Gamma_0(2)$ with 2 inverted.
The structure of this ring
is well-known (see, e.g.,
\cite[Appendix~I]{HBJ}).
\begin{prop}\label{leveltwo}
There exist modular forms
$\delta\in M_2(2)[1/2]$
and $\varepsilon\in M_4(2)[1/2]$
with
\begin{align*}
\delta(q)&=4^{-1}+6q+6q^2+
24q^3+6q^4+36q^5+24q^6+\cdots,\\
\varepsilon(q)&=16^{-1}-q+7q^2-28q^3
+71q^4-126q^5+196q^6-\cdots,
\end{align*}
and such that $M_*(2)[1/2]
=\mathbb Z[1/2][\delta,\varepsilon]$.
\end{prop}
Given an integer $N\geq1$,
we define two linear operators
on modular forms:
the first is
\begin{align*}
\iota_N:M_*&\to M_*(N)\\
f&\mapsto f
\end{align*}
obtained by regarding $f\in M_*$
as a modular form for $\Gamma_0(N)$;
the second is the {\em Verschiebung}
\begin{align*}
V_N:M_*&\to M_*(N)\\
f(q)&\mapsto f(q^N)
\end{align*}
which satisfies $V_N(fg)=V_N(f)V_N(g)$
since $q\mapsto q^N$ is a ring
endomorphism of $\mathbb Z[[q]]$.
In particular, we will often write
$V_Nf^m$ for $1\leq m\in\mathbb Z$, which is unambiguous since
$V_N(f^m)=(V_Nf)^m$.
The linear operator
appearing in condition (C4) of Behrens'
theorem is
\[
L_N=V_N-\iota_N.
\]
We will not distinguish between
$f$ and $\iota_Nf$ and so
we will write $L_Nf=V_Nf-f$.

It will be
convenient to define
\[
\mu=\delta^2-\varepsilon\in M_4(2)[1/2]
\]
and direct computations
in $M_*(2)[1/2]$
yield the identities
\begin{align}
E_4&=64\mu+16\
\label{E4identity},\\
V_2E_4&=4\mu+16\varepsilon
\label{vE4identity},\\
\Delta&=64\mu\varepsilon^2\label{deltaidentity},\\
V_2\Delta&=\mu^2\varepsilon\label{vdeltaidentity}.
\end{align}

Let $M_t(N)_{\mathbb Z/p}$ be
the weight $t$ modular forms
for $\Gamma_0(N)$ defined over
$\mathbb Z/p$.  As long as
$N$ is invertible in $\mathbb Z/p$,
\[
M_t(N)_{\mathbb Z/p}=M_t(N)\otimes
\mathbb Z/p
\]
(see \cite[Equation~(1.1)]{Beh:Cong}).
We will not draw a notational distinction
between $f\in M_t(N)$ and its mod $p$
reduction $f\in M_t(N)_{\mathbb Z/p}$
as the meaning will always be clear from
the context.

The following result of Serre
is a generalization of
the well known congruence
$E_{p-1}(q)\equiv 1\mod p$, and
dictates exactly when congruences
between modular forms
of different weights modulo
$p$ can occur (see \cite[Theorem~10.2]{Beh:Cong}
or \cite[Corollary~4.4.2]{Katz:padic}).
\begin{prop}[Serre]\label{Serre}
Let $f_1\in M_{t_1}(N)_{\mathbb Z/p}$
and
$f_2\in M_{t_2}(N)_{\mathbb Z/p}$
with $t_1<t_2$.  Then
\[
f_1(q)=f_2(q)\in\mathbb Z/p[[q]]
\]
(that is, $f_1\equiv f_2\mod p$)
if and only if
$t_1\equiv t_2\mod (p-1)$
and $f_2=E_{p-1}^{\frac{t_2-t_1}
	{p-1}}f_1$.
\end{prop}

Finally,
the following rigidity result
implies that checking condition
(C4) for a given $f_{i/j}$
often reduces to a computation
with $\Gamma_0(\ell)$ modular forms
at a {\em single} prime $\ell\neq p$
(see \cite[Theorem~1.5]{Beh:Cong}).
\begin{prop}[Behrens]\label{rigidity}
	If the prime $\ell_0$ is a topological
	generator of $\mathbb Z_p^{\times}$
	and $f$ is a modular form
	of weight $t\equiv (p-1)$
	satisfying conditions (C1) through (C3),
	as well as condition (C4) for
	$\ell=\ell_0$, then $f$ satisfies
	condition (C4) for all primes
	$\ell\neq p$.
\end{prop}

\section{Computations at the prime 5}
\label{at5}
In this section we fix $p=5$
and prove Theorem \ref{main_5}.
The first goal is to work toward a proof
that the modular forms identified as
$f_{i/j}$ in Theorem \ref{main_5} satisfy
conditions (C1), (C2), and (C3) of Behrens'
theorem.
\begin{prop}\label{setup}
Suppose $f\in M_t$ with $t\equiv0\mod 12$.
\begin{enumerate}[(a)]
\item There exist integers $c_0,\ldots,c_{t/12}$
such that
\[
f=c_0\Delta^{t/12}+c_1\Delta^{(t/12)-1}E_4^3+\cdots+	c_{(t/12)-1}\Delta E_4^{(t/4)-3}+c_{t/12}E_4^{t/4}.
\]
\item If $c_0\neq0\mod 5$ then $f(q)\not\equiv0\mod 5$
and $f$ is not
divisible by $E_4$ in $(M_*)_{\mathbb Z/5}$.
\item If $c_m\neq0$ and
$c_{m+1},c_{m+2},\ldots,c_{t/12}=0$
for some $m$, $0\leq m\leq t/12$, then
\[
\operatorname{ord}_qf=\dfrac t{12}-m.
\]
\end{enumerate}
\end{prop}
\begin{proof}
Part (a) is a special case of Proposition
\ref{basis}.  Part (b) follows from the fact
that the integral basis for $M_t$ given by
Proposition \ref{basis} remains a basis
for the vector space $(M_t)_{\mathbb Z/5}$
(see, e.g., Section 1 of \cite{Jochnowitz}).
Since $\Delta(q)=q+\cdots$ and $E_4(q)=1+\cdots$,
the value of $\operatorname{ord}_q(f)$ is the smallest power of
$\Delta$ appearing in the expansion of $f$ given
in Part (a), and Part (c) follows.
\end{proof}
\begin{cor}\label{C1C3}
The modular forms identified as $f_{i/j}$
in Theorem \ref{main_5}
satisfy conditions (C1) and (C3) of Behrens' theorem.
\end{cor}
\begin{proof}
The modular form $f_{i/j}$ has weight
$24i$ and so has an expansion
as in Proposition \ref{setup}(a).
The formulas given in
Theorem \ref{main_5} show that each such expansion
has leading coefficient $c_0=1$ by construction.
Thus $f_{i/j}(q)\neq0\mod 5$ by Proposition \ref{setup}(b),
and so $f_{i/j}$ satisfies condition (C1).
Condition (C3) is also satisfied by
Propositions \ref{setup}(b) and \ref{Serre}.
\end{proof}
\begin{lem}\label{qineq}
At the prime 5, condition (C2) is equivalent
to $\operatorname{ord}_qf_{i/j}>2i-\dfrac j3$.
\end{lem}
\begin{proof}
At the prime 5, condition (C2) says that either
$\operatorname{ord}_qf_{i/j}>(24i-4j)/12$ or $\operatorname{ord}_qf_{i/j}=(24i-4j-2)/12$.
But $24i-4j-2$ is never divisible by 12 for integer values
of $i$ and $j$, so condition (C2) reduces to the inequality
only.
\end{proof}
\begin{prop}\label{C2ONLY}
The modular forms $f_{i/j}$ as identified
in Theorem 1.1 all satisfy condition (C2) of Behrens' theorem.
\end{prop}
\begin{proof}
Suppose $i=r\cdot 5^n$ with $(r,5)=1$.  Consider first the case $j\leq p^n$,
so that $f_{r\cdot5^n/j}=\Delta^{2r\cdot5^n}$.  By
Lemma \ref{qineq}, verifying condition (C2)
is equivalent to verifying the inequality
\begin{equation}\label{checkineq}
\operatorname{ord}_q\Delta^{2r\cdot5^n}>2r\cdot5^n-\dfrac j3.
\end{equation}
but $\operatorname{ord}_q\Delta^{2r\cdot5^n}=2r\cdot5^n$
by Proposition \ref{setup}(c),
so \eqref{checkineq} clearly holds for $j\geq1$.

Next, suppose $j>p^n$.  Let $u$ be the
positive integer between 1 and $n-1$ such that
\[
5^n+5^{n-1}-5^{n-u}+1\leq j\leq
5^n+5^{n-1}-5^{n-u-1}.
\]
Assume $j>5^n+5^{n-1}-5^{n-u}+2\cdot5^{n-u-1}$.
By the formulas given in Theorem \ref{main_5},
we must verify condition (C2) for
\[
f_{r\cdot5^n/j}=\Delta^{2r\cdot5^n}+\sum_{m=0}^{u-1}
(C_{m,n,r}+D_{m,n,r}),
\]
a form whose $q$-order is the power
of $\Delta$ occurring in $D_{u-1,n,r}$ by Proposition
\ref{setup}(c).
Therefore, by Lemma \ref{qineq}, we must verify
the inequality
\begin{equation}\label{checkineq2}
8\cdot5^{n-1}+5^{n-u-1}+2(r-1)5^n>2r\cdot5^n-\dfrac j3.
\end{equation}
It suffices to verify \eqref{checkineq2} for the
smallest possible value of $j$, which in this case
is \[j=5^n+5^{n-1}-5^{n-u}+2\cdot5^{n-u-1}+1.\]
At this value of $j$, \eqref{checkineq2} becomes
\[
8\cdot5^{n-1}+5^{n-u-1}+2(r-1)5^n>2r\cdot5^n-
\dfrac{5^n+5^{n-1}-5^{n-u}+2\cdot5^{n-u-1}+1}3
\]
which is equivalent to $1/3>0$. So \eqref{checkineq2}
holds.

Continuing with the case $j>p^n$, we now assume
$j\leq5^n+5^{n-1}-5^{n-u}+2\cdot5^{n-u-1}$.
By the formulas given in Theorem \ref{main_5}, we must verify
condition (C2) for
\[
f_{r\cdot5^n/j}=\Delta^{2r\cdot5^n}+\sum_{m=0}^{u-2}
(C_{m,n,r}+D_{m,n,r})+C_{u-1,n,r}
\]
whose $q$-order is the power of $\Delta$ occurring
in $C_{u-1,n,r}$ by Proposition \ref{setup}(c).  Therefore,
by Lemma \ref{qineq},
the inequality we must verify is
\begin{equation}\label{checkineq3}
8\cdot5^{n-1}+2\cdot5^{n-u-1}+2(r-1)5^n>2r\cdot5^n-\dfrac j3.
\end{equation}
The smallest possible value of $j$ is now
$j=5^n+5^{n-1}-5^{n-u}+1$, so it suffices to verify
\[
8\cdot5^{n-1}+2\cdot5^{n-u-1}+2(r-1)5^n>2r\cdot5^n-
\dfrac{5^n+5^{n-1}-5^{n-u}+1}3
\]
which is equivalent to
\[
\dfrac{5^{n-u-1}+1}3>0.
\]
So \eqref{checkineq3} holds, and we have shown
condition (C2) is satisfied in all cases.
\end{proof}
\begin{lem}\label{rigidlem}
If a modular form $f_{i/j}$
identified in Theorem \ref{main_5} satisfies
conditions (C1) through (C3), then it satisfies
condition (C4) if and only if $L_2f_{i/j}$ is divisible by
$E_4^j$ in $M_*(2)_{\mathbb Z/5}$.
\end{lem}
\begin{proof}
Since 2 is a topological generator of $\mathbb Z_5^{\times}$,
checking $f_{i/j}$ satisfies condition (C4) is equivalent to showing there exists $g\in M_{24i-4j}(2)$ such
that $(L_2f_{i/j})(q)\equiv g(q)\mod 5$
by Proposition \ref{rigidity}.  The lemma then follows
from Proposition \ref{Serre}.
\end{proof}

The remaining goal of this section is
to complete the proof of Theorem \ref{main_5}
by showing that the modular forms $f_{i/j}$
as identified in the theorem satisfy condition (C4).
To begin, we establish a
method for computing the $E_4$-divisibility
required by Lemma \ref{rigidlem}.
\begin{prop}\label{method}
If $f\in M_t$, then $L_2f\in M_t(2)_{\mathbb Z/5}$
is expressible as a homogeneous element
of $\mathbb Z/5[\mu,\varepsilon]$ of degree $t/4$,
and if $y=4\mu/\varepsilon+1$, then
\begin{equation}\label{inhomog}
L_2f=\varepsilon^{t/4}P(y)
\end{equation}
for an inhomogeneous polynomial $P(y)\in\mathbb Z/5[y]$.
Moreover, $L_2f$ is divisible by $E_4^j$ in
$M_*(2)_{\mathbb Z/5}$ if and only if $P(y)=O(y^j)$
(that is, $P(y)$ is divisible by $y^j$)
in $\mathbb Z/5[y]$.
\end{prop}
\begin{proof}
By Proposition \ref{basis}, $f\in M_t$
is a $\mathbb Z$-linear combination of terms
of the form $\Delta^a E_4^b$ for non-negative integers
$a$, $b$ with $12a+4b=t$.
From the basic properties of $L_N$ outlined
in Section \ref{Intro}, as well as
Equations \eqref{E4identity}-\eqref{vdeltaidentity},
it follows that
\[
L_2(\Delta^aE_4^b)=V_2\Delta^aV_2E_4^b-\Delta^aE_4^b=
(\mu^2\varepsilon)^a(4\mu+16\varepsilon)^b-(64\mu\varepsilon^2)^a(64\mu+16\varepsilon)^b
\in M_t(2)
\]
which is a homogeneous polynomial in $\mu$ and $\varepsilon$
over the integers of degree $3a+b=t/4$.  Passing
to $M_t(2)_{\mathbb Z/5}=M_t(2)\otimes\mathbb Z/5$
yields an expression of $L_2(\Delta^aE_4^b)$
as a homogeneous polynomial of the same degree
over $\mathbb Z/5$.  Thus, $L_2f\in M_t(2)_{\mathbb Z/5}$ is a
sum of such homogeneous polynomials by the linearity
of $L_2$.  If we put $x=\mu/\varepsilon$, then $L_2f=\varepsilon^{t/4}P(x)$
where $P(x)\in\mathbb Z/5[x]$, and making the change
of variable $x=4y+1$ (so that $y=4x+1$) yields
Equation \eqref{inhomog}.

By Equation \eqref{E4identity}, $L_2f$ is divisible by $E_4^j$
in $M_*(2)_{\mathbb Z/5}$ if and only if the corresponding homogeneous
element of $\mathbb Z/5[\mu,\varepsilon]$ is divisible by
$(4\mu+\varepsilon)^j$.  But this is equivalent to
$P(y)$ being divisible by $y^j$ since
\[
(4\mu+\varepsilon)^j=\varepsilon^j(4x+1)^j=\varepsilon^j y^j.
\]
\end{proof}
The next four lemmas produce equations
of the form \eqref{inhomog} for various
modular forms $f$.
\begin{lem}\label{L2TERM} For $0\leq a,b\in\mathbb Z$,
$L_2(\Delta^a E_4^b)=\varepsilon^{3a+b}y^b
((4y+1)^{2a}-(-1)^a(4y+1)^a)$
in $M_*(2)_{\mathbb Z/5}$.
\end{lem}
\begin{proof}
Equation \eqref{vE4identity} implies that
$V_2E_4=E_4$ in $M_*(2)_{\mathbb Z/5}$.  Therefore,
using the notation from the proof
of Proposition \ref{method},
\begin{align*}
L_2(\Delta^a E_4^b)&=V_2\Delta^aV_2E_4^b-\Delta^aE_4^b\\
&=E_4^b(V_2\Delta^a-\Delta^a)\\
&=(4\mu+\varepsilon)^b(\mu^{2a}\varepsilon^a-(-1)^a\mu^a\varepsilon^{2a})\\
&=\varepsilon^{3a+b}(4x+1)^b(x^{2a}-(-1)^ax^a)\\
&=\varepsilon^{3a+b}y^b((4y-1)^{2a}-(-1)^a(4y+1)^a)
\end{align*}
in $M_*(2)_{\mathbb Z/5}$.
\end{proof}
\begin{lem}\label{L2ONDELTA}
For integers $r\geq1$ and $n\geq0$,
$L_2\Delta^{2r\cdot5^n}=\varepsilon^{6r\cdot5^n}
(3ry^{5^n}+O(y^{a_n}))$ in $M_*(2)_{\mathbb Z/5}$.
\end{lem}
\begin{proof}
By Lemma \ref{L2TERM},
\begin{align*}
L_2(\Delta^{2r\cdot5^n})&=\varepsilon^{6\cdot5^n}
((4y+1)^{4r\cdot5^n}-(4y+1)^{2r\cdot5^n})\\
&=\varepsilon^{6\cdot5^n}
((4y^{5^n}+1)^{4r}-(4y^{5^n}+1)^{2r})\\
&=\varepsilon^{6\cdot5^n}(1+4r\cdot4y^{5^n}+O(y^{2\cdot5^n})
-(1+2r\cdot4y^{5^n}+O(y^{2\cdot5^n})))\\
&=\varepsilon^{6\cdot5^n}(3ry^{5^n}+O(y^{a_n})).
\end{align*}
\end{proof}
\begin{lem}\label{CDEXTREME}
For integers $n\geq2$ and $r\geq1$,
\begin{align*}
L_2C_{0,n,r}&=r\varepsilon^{6r\cdot5^n}(2y^{5^n}
+4y^{27\cdot5^{n-2}}+4y^{28\cdot5^{n-2}}
+3y^{29\cdot5^{n-2}}+O(y^{a_n})),\\
L_2D_{0,n,r}&=r\varepsilon^{6r\cdot5^n}(
y^{27\cdot5^{n-2}}+y^{28\cdot5^{n-2}}
+3y^{29\cdot5^{n-2}}
+O(y^{a_n})),\quad\text{and}\\
L_2D_{n-2,n,r}&=r\varepsilon^{6r\cdot5^n}
(2y^{6\cdot5^{n-1}-3}
+2y^{6\cdot5^{n-1}-2}
+O(y^{a_n})).
\end{align*}
in $M_*(2)_{\mathbb Z/5}$.
\end{lem}
\begin{proof}
To begin, we compute $L_2C_{0,n,r}$.
By Lemma \ref{L2TERM},
\begin{align}\label{C01}
\begin{split}
L_2C_{0,n,r}&=4rL_2(
\Delta^{42\cdot5^{n-2}+2(r-1)5^n}E_4^{24\cdot5^{n-2}}
)\\
&=4r\varepsilon^{6r\cdot5^n}y^{24\cdot5^{n-2}}
((4y+1)^{5^{n-2}(100r-16)}-(4y+1)^{5^{n-2}(50r-8)})\\
&=4r\varepsilon^{6r\cdot5^n}y^{24\cdot5^{n-2}}
((4y^{5^{n-2}}+1)^{25(4r-1)+5+4}-
(4y^{5^{n-2}}+1)^{25(2r-1)+5(3)+2}).
\end{split}
\end{align}
In $\mathbb Z/5[y]$,
\begin{align*}
(4y^{5^{n-2}}+1)^{25(4r-1)+5+4}&=
(4y^{5^n}+1)^{4r-1}(4y^{5^{n-1}}+1)
(4y^{5^{n-2}}+1)^4\\
&=(4y^{5^n}+1)^{4r-1}(4y^{5^{n-1}}+1)
(1+y^{5^{n-2}}+y^{2\cdot5^{n-2}}+y^{3\cdot5^{n-2}}+
y^{4\cdot5^{n-2}})\\
&=(4y^{5^n}+1)^{4r-1}
(1+y^{5^{n-2}}+y^{2\cdot5^{n-2}}+y^{3\cdot5^{n-2}}+
y^{4\cdot5^{n-2}}+4y^{5^{n-1}}+O(y^{6\cdot5^{n-2}}))\\
&=1+y^{5^{n-2}}+y^{2\cdot5^{n-2}}+y^{3\cdot5^{n-2}}+
y^{4\cdot5^{n-2}}+4y^{5^{n-1}}+O(y^{6\cdot5^{n-2}})
\end{align*}
and
\begin{align*}
(4y^{5^{n-2}}+1)^{25(2r-1)+5(3)+2}&=
(4y^{5^n}+1)^{2r-1}(4y^{5^{n-1}}+1)^3(4y^{5^{n-2}}+1)^2\\
&=(4y^{5^n}+1)^{2r-1}(1+2y^{5^{n-1}}+3y^{2\cdot5^{n-1}}
+4y^{3\cdot5^{n-1}})
(1+3y^{5^{n-2}}+y^{2\cdot5^{n-2}})\\
&=(4y^{5^n}+1)^{2r-1}(1+3y^{5^{n-2}}+y^{2\cdot5^{n-2}}
+2y^{5^{n-1}}+O(y^{6\cdot5^{n-2}}))\\
&=1+3y^{5^{n-2}}+y^{2\cdot5^{n-2}}
+2y^{5^{n-1}}+O(y^{6\cdot5^{n-2}})
\end{align*}
which, when combined with \eqref{C01}, yield
\begin{align*}
L_2C_{0,n,r}&=4r\varepsilon^{6r\cdot5^n}y^{24\cdot5^{n-2}}
(3y^{5^{n-2}}+y^{3\cdot5^{n-2}}+y^{4\cdot5^{n-2}}
+2y^{5^{n-1}}+O(y^{6\cdot5^{n-2}}))\\
&=r\varepsilon^{6r\cdot5^n}(2y^{5^n}
+4y^{27\cdot5^{n-2}}+4y^{28\cdot5^{n-2}}
+3y^{29\cdot5^{n-2}}+O(y^{a_n}))
\end{align*}
in $M_*(2)_{\mathbb Z/5}$.

Next, we compute $L_2D_{0,n,r}$.  By Lemma \ref{L2TERM},
\begin{align}\label{D01}
\begin{split}
L_2D_{0,n,r}&=3rL_2(\Delta^{41\cdot5^{n-2}+2(r-1)5^n}
E_4^{27\cdot5^{n-2}})\\
&=3r\varepsilon^{6r\cdot5^n}y^{27\cdot5^{n-2}}
((4y+1)^{5^{n-2}(100r-18)}+(4y+1)^{5^{n-2}(50r-9)})\\
&=3r\varepsilon^{6r\cdot5^n}y^{27\cdot5^{n-2}}
((4y^{5^{n-2}}+1)^{25(4r-1)+5+2}
+(4y^{5^{n-2}}+1)^{25(2r-1)+5(3)+1}).
\end{split}
\end{align}
In $\mathbb Z/5[y]$,
\begin{align*}
(4y^{5^{n-2}}+1)^{25(4r-1)+5+2}&=(4y^{5^n}+1)^{4r-1}
(4y^{5^{n-1}}+1)(4y^{5^{n-2}}+1)^2\\
&=(4y^{5^n}+1)^{4r-1}(4y^{5^{n-1}}+1)
(1+3y^{5^{n-2}}+y^{2\cdot5^{n-2}})\\
&=1+3y^{5^{n-2}}+y^{2\cdot5^{n-2}}+O(y^{3\cdot5^{n-2}})
\end{align*}
and
\begin{align*}
(4y^{5^{n-2}}+1)^{25(2r-1)+5(3)+1}&=(4y^{5^n}+1)^{2r-1}
(4y^{5^{n-1}}+1)^3(4y^{5^{n-2}}+1)\\
&=1+4y^{5^{n-2}}+O(y^{3\cdot5^{n-2}})
\end{align*}
which, when combined with \eqref{D01}, yield
\begin{align*}
L_2D_{0,n,r}&=3r\varepsilon^{6\cdot5^n}y^{27\cdot5^{n-2}}
(2+2y^{5^{n-2}}+y^{2\cdot5^{n-2}}+O(y^{3\cdot5^{n-2}}))\\
&=r\varepsilon^{6r\cdot5^n}(
y^{27\cdot5^{n-2}}+y^{28\cdot5^{n-2}}+3y^{29\cdot5^{n-2}}
+O(y^{a_n}))
\end{align*}
in $M_*(2)_{\mathbb Z/5}$.

Finally, we compute $L_2D_{n-2,n,r}$.
By Lemma \ref{L2TERM},
\begin{align*}
L_2D_{n-2,n,r}&=rL_2(\Delta^{8\cdot5^{n-1}+1+2(r-1)5^n}
E_4^{6\cdot5^{n-1}-3})\\
&=r\varepsilon^{6r\cdot5^n}y^{6\cdot5^{n-1}-3}
((4y+1)^{5^{n-1}(20r-4)+2}+(4y+1)^{5^{n-1}(10r-2)+1})\\
&=r\varepsilon^{6r\cdot5^n}y^{6\cdot5^{n-1}-3}
((4y^{5^{n-1}}+1)^{20r-4}(4y+1)^2
+(4y^{5^{n-1}}+1)^{10r-2}(4y+1))\\
&=r\varepsilon^{6r\cdot5^n}y^{6\cdot5^{n-1}-3}(2+2y+O(y^2))\\
&=r\varepsilon^{6r\cdot5^n}(2y^{6\cdot5^{n-1}-3}
+2y^{6\cdot5^{n-1}-2}+O(y^{a_n}))
\end{align*}
in $M_*(2)_{\mathbb Z/5}$.
\end{proof}
\begin{lem}\label{CDGENERAL}
For $1\leq m\leq n-2$,
\begin{equation*}
L_2C_{m,n,r}=r\varepsilon^{6r\cdot5^n}
(4y^{6\cdot5^{n-1}-5^{n-m-1}}
+3y^{6\cdot5^{n-1}-3\cdot5^{n-m-2}}
+3y^{6\cdot5^{n-1}-2\cdot5^{n-m-2}}
+O(y^{a_n}))
\end{equation*}
and for $1\leq m\leq n-3$,
\begin{equation*}
L_2D_{m,n,r}=r\varepsilon^{6r\cdot5^n}
(2y^{6\cdot5^{n-1}-3\cdot5^{n-m-2}}
+2y^{6\cdot5^{n-1}-2\cdot5^{n-m-2}}
+y^{6\cdot5^{n-1}-5^{n-m-2}}
+O(y^{a_n}))
\end{equation*}
in $M_*(2)_{\mathbb Z/5}$.
\end{lem}
\begin{proof}
First, we compute $L_2C_{m,n,r}$
for $1\leq m\leq n-2$.
By Lemma \ref{L2TERM},
\begin{align}\label{CM1}
\begin{split}
L_2C_{m,n,r}&=3rL_2(
\Delta^{8\cdot5^{n-1}
	+2\cdot5^{n-m-2}+2(r-1)5^n}
E_4^{6\cdot5^{n-1}-6\cdot5^{n-m-2}}
)\\
&=3r\varepsilon^{6\cdot5^n}
y^{6\cdot5^{n-1}-6\cdot5^{n-m-2}}
(
(4y+1)^{16\cdot5^{n-1}+4\cdot5^{n-m-2}
+4(r-1)5^n}
\\
&\hskip1.7truein-(4y+1)^{8\cdot5^{n-1}
+2\cdot5^{n-m-2}+2(r-1)5^n}).
\end{split}
\end{align}
In $\mathbb Z/5[y]$,
\begin{align*}
(4y+1)^{16\cdot5^{n-1}+4\cdot5^{n-m-2}
	+4(r-1)5^n}&=
(4y+1)^{(4r-1)5^n+5^{n-1}+
	4\cdot5^{n-m-2}}\\
&=(4y^{5^n}+1)^{4r-1}
(4y^{5^{n-1}}+1)(4y^{5^{n-m-2}}+1)^4\\
&=1+y^{5^{n-m-2}}+y^{2\cdot5^{n-m-2}}
+y^{3\cdot5^{n-m-2}}
+y^{4\cdot5^{n-m-2}}
+O(y^{6\cdot5^{n-m-2}})
\end{align*}
and
\begin{align*}
(4y+1)^{8\cdot5^{n-1}+2\cdot5^{n-m-2}
+2(r-1)5^n}&=
(4y+1)^{(2r-1)5^n+3\cdot5^{n-1}
+2\cdot5^{n-m-2}}\\
&=(4y^{5^n}+1)^{2r-1}
(4y^{5^{n-1}}+1)^3
(4y^{5^{n-m-2}}+1)^2\\
&=1+3y^{5^{n-m-2}}+y^{2\cdot5^{n-m-2}}
+O(y^{6\cdot5^{n-m-2}})
\end{align*}
which, when combined with \eqref{CM1}, yield
\begin{align*}
L_2C_{m,n,r}&=3r\varepsilon^{6\cdot5^n}
y^{6\cdot5^{n-1}-6\cdot5^{n-m-2}}
(3y^{5^{n-m-2}}+y^{3\cdot5^{n-m-2}}
+y^{4\cdot5^{n-m-2}}+O(y^{6\cdot5^{n-m-2}}))\\
&=r\varepsilon^{6\cdot5^n}(4y^{6\cdot5^{n-1}-5^{n-m-1}}
+3y^{6\cdot5^{n-1}-3\cdot5^{n-m-2}}
+3y^{6\cdot5^{n-1}-2\cdot5^{n-m-2}}
+O(y^{a_n}))
\end{align*}
in $M_*(2)_{\mathbb Z/5}$.

Next, we compute $L_2D_{m,n,r}$
for $1\leq m\leq n-3$.
By Lemma \ref{L2TERM},
\begin{align}\label{DM1}
\begin{split}
L_2D_{m,n,r}&=rL_2(
\Delta^{8\cdot5^{n-1}
	+\cdot5^{n-m-2}+2(r-1)5^n}
E_4^{6\cdot5^{n-1}-3\cdot5^{n-m-2}}
)\\
&=r\varepsilon^{6\cdot5^n}
y^{6\cdot5^{n-1}-3\cdot5^{n-m-2}}
(
(4y+1)^{16\cdot5^{n-1}+2\cdot5^{n-m-2}
+4(r-1)5^n}\\
&\hskip1.7truein
+(4y+1)^{8\cdot5^{n-1}+5^{n-m-2}
+2(r-1)5^n}).
\end{split}
\end{align}
In $\mathbb Z/5[y]$,
\begin{align*}
(4y+1)^{16\cdot5^{n-1}+2\cdot5^{n-m-2}
	+4(r-1)5^n}&=
(4y+1)^{(4r-1)5^n+5^{n-1}
	+2\cdot5^{n-m-2}}\\
&=(4y^{5^n}+1)^{4r-1}(4y^{5^{n-1}}+1)
(4y^{5^{n-m-2}}+1)^2\\
&=1+3y^{5^{n-m-2}}+y^{2\cdot5^{n-m-2}}
+O(y^{3\cdot5^{n-m-2}})
\end{align*}
and
\begin{align*}
(4y+1)^{8\cdot5^{n-1}+5^{n-m-2}
+2(r-1)5^n}&=
(4y+1)^{(2r-1)5^n+3\cdot5^{n-1}
+5^{n-m-2}}\\
&=(4y^{5^n}+1)^{2r-1}(4y^{5^{n-1}}+1)^3
(4y^{5^{n-m-2}}+1)\\
&=1+4y^{5^{n-m-2}}+O(y^{3\cdot5^{n-m-2}})
\end{align*}
which, when combined with \eqref{DM1}, yield
\begin{align*}
L_2D_{m,n,r}&=r\varepsilon^{6\cdot5^n}
y^{6\cdot5^{n-1}-3\cdot5^{n-m-2}}
(2+2y^{5^{n-m-2}}+y^{2\cdot5^{n-m-2}}+O(y^{3\cdot5^{n-m-2}}))\\
&=r\varepsilon^{6r\cdot5^n}
(2y^{6\cdot5^{n-1}-3\cdot5^{n-m-2}}
+2y^{6\cdot5^{n-1}-2\cdot5^{n-m-2}}
+y^{6\cdot5^{n-1}-5^{n-m-2}}
+O(y^{a_n}))
\end{align*}
in $M_*(2)_{\mathbb Z/5}$.  Note that the assumption $m\leq n-3$ is essential,
because if $m=n-2$,
\[
6\cdot5^{n-1}-5^{n-m-2}=6\cdot5^{n-1}-1=a_n
\]
and so the term $y^{6\cdot5^{n-1}-5^{n-m-2}}
=y^{a_n}$ would
not appear.  This is why the computation
of $L_2D_{n-2,n,r}$ is handled separately in Lemma \ref{CDEXTREME}.
\end{proof}
\begin{thm}\label{divisibility}
The modular forms $f_{r\cdot 5^n/j}$
as identified in Theorem \ref{main_5}
satisfy condition (C4) of Behrens' theorem.
\end{thm}
\begin{proof}
By Corollary \ref{C1C3}, Proposition \ref{C2ONLY},
and Lemma \ref{rigidlem},
it suffices
to show that $L_2f_{r\cdot 5^n/j}$ is divisible by
$E_4^j$ in $M_*(2)_{\mathbb Z/5}$.

Consider first the case $j\leq 5^n$, so that
$f_{r\cdot 5^n/j}=\Delta^{2r\cdot 5^n}$.
Lemma \ref{L2ONDELTA} implies that
\[
L_2\Delta^{2r\cdot 5^n}=\varepsilon^{6\cdot5^n}(3ry^{5^n}+O(y^{a_n}))
\inM_*(2)_{\mathbb Z/5}
\]
showing divisibility by $E_4^{5^n}$
and verifying condition (C4) in this case.

Next, suppose $j>5^n$, and let $u$ be the positive integer
between 1 and $n-1$ such that
\[
5^n+5^{n-1}-5^{n-u}+1\leq j\leq 5^n+5^{n-1}-5^{n-u-1}.
\]   Assume first
that $j>5^n+5^{n-1}-5^{n-u}+2\cdot5^{n-u-1}$,
so that the modular form in question is
\[
f_{r\cdot5^n/j}=\Delta^{2r\cdot5^n}+
\sum_{m=0}^{u-1}(C_{m,n,r}+D_{m,n,r}).
\]
To show that $L_2f_{r\cdot5^n/j}$
is divisible by $E_4^j$
in $M_*(2)_{\mathbb Z/5}$, it suffices to show divisibility
by $E_4^{5^n+5^{n-1}-5^{n-u-1}}$
since $5^n+5^{n-1}-5^{n-u-1}$ is the largest
possible $j$-value in this case.
By Lemmas \ref{L2ONDELTA} and \ref{CDEXTREME},
\begin{equation}\label{L2ONSTART}
L_2(\Delta^{2r\cdot5^n}+C_{0,n,r}+D_{0,n,r})=
\varepsilon^{6r\cdot5^n}
(ry^{29\cdot5^{n-2}}+O(y^{a_n}))
\end{equation}
in $M_*(2)_{\mathbb Z/5}$.
Lemmas \ref{CDEXTREME} and \ref{CDGENERAL}
imply
\begin{equation}\label{L2ONCDNM2}
L_2(C_{n-2,n,r}+D_{n-2,n,r})=\varepsilon^{6r\cdot5^n}
(4ry^{6\cdot5^{n-1}-5}+O(y^{a_n}))\inM_*(2)_{\mathbb Z/5}
\end{equation}
and for $1\leq m\leq n-3$,
Lemma \ref{CDGENERAL} implies
\begin{equation}\label{L2ONCD}
L_2(C_{m,n,r}+D_{m,n,r})=\varepsilon^{6r\cdot5^n}
(4ry^{6\cdot5^{n-1}-5^{n-m-1}}
+ry^{6\cdot5^{n-1}-5^{n-m-2}}+O(y^{a_n}))
\inM_*(2)_{\mathbb Z/5}.
\end{equation}
Therefore, for $1\leq u\leq n-2$,
\begin{align}\label{L2ONCOMBO}
\begin{split}
L_2f_{r\cdot5^n/j}&=\varepsilon^{6r\cdot5^n}
\left(
ry^{29\cdot5^{n-2}}+\sum_{m=1}^{u-1}
(4ry^{6\cdot5^{n-1}-5^{n-m-1}}
+ry^{6\cdot5^{n-1}-5^{n-m-2}})+O(y^{a_n})
\right)\\
&=\varepsilon^{6r\cdot5^n}(ry^{6\cdot5^{n-1}-5^{n-u-1}}
+O(y^{a_n}))\inM_*(2)_{\mathbb Z/5}
\end{split}
\end{align}
by Equations \eqref{L2ONSTART} and \eqref{L2ONCD},
showing divisibility by $E_4^{5^n+5^{n-1}-5^{n-u-1}}$.
For the case $u=n-1$, we obtain
\[
L_2f_{r\cdot5^n/j}=\varepsilon^{6r\cdot5^n}
(ry^{6\cdot5^{n-1}-5}
+4ry^{6\cdot5^{n-1}-5}+O(y^{a_n}))
=\varepsilon^{6r\cdot5^n}\cdot O(y^{a_n})
\]
from Equations \eqref{L2ONCDNM2} and \eqref{L2ONCOMBO},
showing divisibility
by $E_4^{5^n+5^{n-1}-5^{n-(n-1)-1}}
=E_4^{a_n}$.  Thus, for
$j>5^n+5^{n-1}-5^{n-u}+2\cdot5^{n-u-1}$,
condition (C4) is verified.

Continuing with the case $j>5^n$, assume
now that $j\leq
5^n+5^{n-1}-5^{n-u}+2\cdot5^{n-u-1}$,
so that the modular form in question is
\[
f_{r\cdot5^n/j}=\Delta^{2r\cdot5^n}+
\sum_{m=0}^{u-2}(C_{m,n,r}+D_{m,n,r})
+C_{u-1,n,r}.
\]
In this case, it suffices to show
divisibility by
$E_4^{5^n+5^{n-1}-5^{n-u}+2\cdot5^{n-u-1}}$
since $5^n+5^{n-1}-5^{n-u}+2\cdot5^{n-u-1}$
is the largest possible value of $j$.
For $u=1$, Lemmas \ref{L2ONDELTA}
and \ref{CDEXTREME} together imply
\begin{align*}
L_2f_{r\cdot5^n/j}&=L_2(\Delta^{2r\cdot5^n}
+C_{0,n,r})\\
&=\varepsilon^{6r\cdot5^n}(3ry^{5^n}+2ry^{5^n}
+4ry^{27\cdot5^{n-2}}
+4ry^{28\cdot5^{n-2}}
+3ry^{29\cdot5^{n-2}}
+O(y^{a_n}))\\
&=\varepsilon^{6r\cdot5^n}(
4ry^{27\cdot5^{n-2}}
+4ry^{28\cdot5^{n-2}}
+3ry^{29\cdot5^{n-2}}
+O(y^{a_n}))\inM_*(2)_{\mathbb Z/5}
\end{align*}
showing divisibility by
$E_4^{5^n+5^{n-1}-5^{n-1}+2\cdot5^{n-2}}
=E_4^{27\cdot5^{n-2}}$.
For $2\leq u\leq n-1$, we obtain
\begin{align*}
L_2f_{r\cdot5^n/j}&=\varepsilon^{6r\cdot5^n}
\bigg(ry^{29\cdot 5^{n-2}}
+\sum_{m=1}^{u-2}(4ry^{6\cdot5^{n-1}-5^{n-m-1}}
+ry^{6\cdot5^{n-1}-5^{n-m-2}})\\
&\hskip1.2truein
+4ry^{6\cdot5^{n-1}-5^{n-u}}
+3ry^{6\cdot5^{n-1}-3\cdot5^{n-u-1}}
+3ry^{6\cdot5^{n-1}-2\cdot5^{n-u-1}}
+O(y^{a_n})
\bigg)\\
&=\varepsilon^{6r\cdot5^n}
(3ry^{6\cdot5^{n-1}-3\cdot5^{n-u-1}}
+3ry^{6\cdot5^{n-1}-2\cdot5^{n-u-1}}
+O(y^{a_n}))\inM_*(2)_{\mathbb Z/5}
\end{align*}
from
Equations \eqref{L2ONSTART},
\eqref{L2ONCDNM2}, and
\eqref{L2ONCD},
showing divisibility by
\[
E_4^{5^n+5^{n-1}-5^{n-u}+2\cdot5^{n-u-1}}=
E_4^{6\cdot5^{n-1}-3\cdot5^{n-u-1}}.
\]
Thus, condition (C4) is verified for
this last remaining case.
\end{proof}

Theorem \ref{main_5} follows from
Corollary \ref{C1C3}, Proposition \ref{C2ONLY},
and Theorem \ref{divisibility}.

\section{Computations at other primes}
\label{atotherprimes}
\subsection{The prime 7}

In the case $p=7$ we are hunting for
$f_{i/j}\in M_{48i}$ associated to each
order 7 generator $\beta_{i/j}$.
The demand on $q$-order in condition (C2) says
\[
\operatorname{ord}_qf_{i/j}>\dfrac{48i-6j}{12}\quad\text{or}\quad
\operatorname{ord}_qf_{i/j}=\dfrac{48i-6j-2}{12}
\]
which is equivalent to $\operatorname{ord}_qf_{i/j}>4i-\dfrac12j$.
In particular, if $j=1$, the inequality
becomes
\begin{equation}\label{qorder7}
\operatorname{ord}_qf_{i}\geq 4i.
\end{equation}
When combined with parts (a) and (c) of
Proposition \ref{setup}, the inequality
\eqref{qorder7} forces
\[
f_i=c\Delta^{4i}
\]
for some integer $c$ prime to 7.
Taking $c=1$ and recalling that $\beta_i$
is a 7-primary divided beta family element
for all integers $i\geq1$ (see Example \ref{exmp_i})
yields the following theorem.
\begin{thm}\label{thm7}
For any integer $i\geq1$,
the 7-primary divided beta family element $\beta_i$
has corresponding modular form $f_i=\Delta^{4i}$.
\end{thm}
\begin{rem}
Proposition \ref{rigidity} does not apply
in the case $\ell_0=2$ and $p=7$
since 2 is not a topological generator
of $\mathbb Z_7^{\times}$.  Further computations at
the prime 7 could be done by working with
$\Gamma_0(3)$ modular forms instead,
since $\ell=3$
{\em is} a topological generator of $\mathbb Z_7^{\times}$.
\end{rem}

\subsection{The prime 11}\label{theprime11} We shall compute
$f_{i/j}\in M_{120i}$ at the prime 11 in
the case $1\leq j\leq 11^n$, where $i=r\cdot11^n$
with $(r,11)=1$.  Since 2 is a
topological generator of $\mathbb Z_{11}^{\times}$,
Proposition \ref{rigidity} applies, and so it suffices to compute
with $\Gamma_0(2)$ modular forms.  In $M_*(2)[1/2]$,
\begin{align*}
E_{10}&=-32768\delta\mu^2-4096\delta\mu\varepsilon+1024\delta\varepsilon^2,\\
V_2E_{10}&=-32\delta\mu^2+128\delta\mu\varepsilon+1024\varepsilon^2
\end{align*}
and so in $M_*(2)_{\mathbb Z/11}$,
\begin{equation}\label{E10at11}
E_{10}=V_2E_{10}=\delta\mu^2+7\delta\mu\varepsilon+\delta\varepsilon^2
=\delta(\mu+3\varepsilon)(\mu+4\varepsilon).
\end{equation}
Equations \eqref{deltaidentity} and \eqref{vdeltaidentity}
imply the identities
\begin{align}
\Delta&=9\mu\varepsilon^2,\label{deltaat11}\\
V_2\Delta&=\mu^2\varepsilon\label{vdeltaat11}
\end{align}
in $M_*(2)_{\mathbb Z/11}$.
\begin{prop}\label{prop11}
If $f\in M_t$, then $L_2f\in M_t(2)_{\mathbb Z/11}$
is expressible as a homogeneous element of
$\mathbb Z/11[\mu,\varepsilon]$ of degree $t/4$, and
if $x=\mu/\varepsilon$, then
\begin{equation}\label{lemma11equation}
L_2f=\varepsilon^{t/4}P(x)
\end{equation}
for an inhomogeneous polynomial $P(x)\in\mathbb Z/11[x]$.
Moreover, if $P(x)$ is divisible by
$(x+1)^j(x+3)^j(x+4)^j$ in $\mathbb Z/11[x]$,
then $L_2f$ is divisible by $E_{10}^j$ in
$M_*(2)_{\mathbb Z/11}$.
\end{prop}
\begin{proof}
This proposition is analogous to Proposition \ref{method},
and the proof of \eqref{lemma11equation} is similar
to the proof of \eqref{inhomog}.

If $P(x)$ is divisible by
$(x+1)^j(x+3)^j(x+4)^j$ in $\mathbb Z/11[x]$,
then the corresponding homogeneous element
of $\mathbb Z/11[\mu,\varepsilon]$ is divisible
by $(\mu+\varepsilon)^j(\mu+3\varepsilon)^j(\mu+4\varepsilon)^j$.
This in turn implies divisibility by $E_{10}^j$
since
\[
(\mu+\varepsilon)^j(\mu+3\varepsilon)^j(\mu+4\varepsilon)^j=\delta^{j}E_{10}^j
\]
by Equation \eqref{E10at11} and the identity
$\delta^2=\mu+\varepsilon$.
\end{proof}
\begin{thm}\label{thm11}
Given an 11-primary divided beta family element
$\beta_{r\cdot11^n/j}$ with $(r,11)=1$ and $1\leq j\leq11^n$,
the corresponding modular form is
$f_{r\cdot11^n/j}=\Delta^{10r\cdot11^n}\in M_{120r\cdot11^n}$.
\end{thm}
\begin{proof}
The modular form $\Delta^{10r\cdot11^n}$
has nonzero Fourier expansion modulo 11, and
its $q$-order is
\[
\operatorname{ord}_q\Delta^{10r\cdot11^n}=10r\cdot11^n
>\dfrac{120r\cdot11^n-10j}{12}
\]
for $j\geq1$.  Any power of $\Delta=E_4^3+10E_6^2$
is not divisible by $E_{10}=E_4E_6$
in $(M_*)_{\mathbb Z/11}$.  Thus, conditions (C1), (C2),
and (C3) are satisfied.

Using the notation from Proposition \ref{prop11}, Equations
\eqref{deltaat11} and
\eqref{vdeltaat11} imply
\begin{align}\label{11calculation}
\begin{split}
L_2\Delta^{10r\cdot11^n}&=V_2\Delta^{10r\cdot11^n}
-\Delta^{10r\cdot11^n}\\
&=(\mu^{20r}\varepsilon^{10r}-9^{10r}\mu^{10r}\varepsilon^{10r})^{11^n}\\
&=\varepsilon^{30r\cdot11^n}x^{10r\cdot11^n}(x^{10r}-1)^{11^n}\\
&=\varepsilon^{30r\cdot11^n}x^{10r\cdot11^n}(x^{10}-1)^{11^n}
(x^{10(r-1)}+x^{10(r-2)}+\cdots+x^{10}+1)^{11^n}.
\end{split}
\end{align}
Moreover, in $\mathbb Z/11[x]$,
\[
x^{10}-1=(x+1)(x+2)(x+3)(x+4)(x+5)(x+6)(x+7)(x+8)(x+9)(x+10)
\]
and so Equation \eqref{11calculation} has the form
$L_2\Delta^{10r\cdot11^n}=\varepsilon^{30r\cdot11^n}P(x)$
where $P(x)$ is divisible by
\[
(x+1)^{11^n}(x+3)^{11^n}(x+4)^{11^n}.
\]
By Proposition \ref{prop11}, this implies
$L_2\Delta^{10r\cdot11^n}$ is divisible by $E_{10}^{11^n}$
(and hence by $E_{10}^j$ for $1\leq j\leq11^n$)
in $M_*(2)_{\mathbb Z/11}$.
This verifies condition (C4) in the case $\ell=2$ by
Proposition \ref{Serre}.
\end{proof}

\subsection{The primes 13 and 677}
We shall establish analogs of Theorem \ref{thm11}
at the primes 13 and 677.  The proofs will follow
the structure of Subsection \ref{theprime11}
but will be made more concise.
Since 2 is a topological generator of
$\mathbb Z_{13}^{\times}$ and $\mathbb Z_{677}^{\times}$,
it suffices in both cases to compute with $\Gamma_0(2)$
modular forms when checking condition (C4).
\begin{thm}\label{thm13}
Given a 13-primary divided beta family element
$\beta_{r\cdot13^n/j}$ with $(r,13)=1$
and $1\leq j\leq13^n$, the corresponding modular
form is $f_{r\cdot13^n/j}=\Delta^{14r\cdot13^n}\in
M_{168r\cdot13^n}$.
\end{thm}
\begin{proof}
The modular form $\Delta^{14r\cdot13^n}$
has nonzero Fourier expansion modulo 11,
and its $q$-order is
\[
\operatorname{ord}_q\Delta^{14r\cdot13^n}=14r\cdot13^n>\dfrac{168r\cdot13^n-12j}{12}
\]
for $j\geq1$.  Any power of $\Delta=12E_4^3+E_6^2$
is not divisible by $E_{12}=6E_4^3+8E_6^2$
in $(M_*)_{\mathbb Z/13}$.
Thus, conditions (C1), (C2),
and (C3) are satisfied.

In $M_*(2)_{\mathbb Z/13}$,
\begin{align}
\Delta&=12\mu\varepsilon^2,\label{deltaat13}\\
V_2\Delta&=\mu^2\varepsilon,\label{vdeltaat13}
\end{align}
and
\begin{equation}\label{E12at13}
E_{12}=V_2E_{12}=12\mu^3+9\mu^2\varepsilon+4\mu\varepsilon^2+\varepsilon^3
=12(\mu+12\varepsilon)(\mu^2+5\mu\varepsilon+\varepsilon^2).
\end{equation}
Using the notation from Proposition \ref{prop11}, Equations \eqref{deltaat13} and
\eqref{vdeltaat13} imply
\begin{align}\label{13calculation}
\begin{split}
L_2\Delta^{14r\cdot13^n}&=V_2\Delta^{14r\cdot13^n}
-\Delta^{14r\cdot13^n}\\
&=(\mu^{28r}\varepsilon^{14r}-\mu^{14r}\varepsilon^{28r})^{13^n}\\
&=\varepsilon^{42r\cdot13^n}x^{14r\cdot13^n}(x^{14r}-1)^{13^n}\\
&=\varepsilon^{42r\cdot13^n}x^{14r\cdot13^n}(x^{14}-1)^{13^n}
(x^{14(r-1)}+x^{14(r-2)}+\cdots+x^{14}+1)^{13n}.
\end{split}
\end{align}
Moreover, in $\mathbb Z/13[x]$,
\[
x^{14}-1=(x+1)(x+12)
(x^2+3x+1)(x^2+5x+1)(x^2+6x+1)(x^2+7x+1)(x^2+8x+1)
(x^2+10x+1)
\]
and so Equation \eqref{13calculation}
has the form $L_2\Delta^{14r\cdot13^n}
=\varepsilon^{42r\cdot13^n}P(x)$ where $P(x)$
is divisible by
\[
(x+12)^{13^n}(x^2+5x+1)^{13^n}.
\]
Using Equation \eqref{E12at13}
and arguing as in the proof of Proposition
\ref{prop11} shows this is equivalent to
$L_2\Delta^{14r\cdot13^n}$ being divisible by
$E_{12}^{13^n}$ (and hence by $E_{12}^j$
for $1\leq j\leq13^n$)
in $M_*(2)_{\mathbb Z/13}$.
This verifies condition (C4) in the case $\ell=2$ by
Proposition \ref{Serre}.
\end{proof}
\begin{thm}\label{thm677}
Given a 677-primary divided beta family element
$\beta_{r\cdot677^n/j}$ with $(r,677)=1$
and $1\leq j\leq677^n$, the corresponding modular
form is $f_{r\cdot677^n/j}=\Delta^{38194r\cdot677^n}\in
M_{458328r\cdot677^n}$.
\end{thm}
\begin{proof}
The modular form $\Delta^{38194r\cdot677^n}$
has nonzero Fourier expansion modulo 677,
and its $q$-order is
\[
\operatorname{ord}_q\Delta^{38194r\cdot677^n}=38194r\cdot677^n
>\dfrac{458328r\cdot677^n-676j}{12}
\]
for $j\geq1$.  Any power of $\Delta$
is not divisible by
\[
E_{676}=66\Delta^{56}E_4+654\Delta^{55}E_4^4+\cdots+60\Delta E_4^{166}
+E_4^{169}
\]
in $(M_*)_{\mathbb Z/677}$.
Thus, conditions (C1), (C2),
and (C3) are satisfied.

In $M_*(2)_{\mathbb Z/677}$,
\begin{align}
	\Delta&=64\mu\varepsilon^2,\label{deltaat677}\\
	V_2\Delta&=\mu^2\varepsilon,\label{vdeltaat677}
\end{align}
and
\begin{align}\label{E676at677}
\begin{split}
&E_{676}=V_2E_{676}\\
&=676\mu^{169}+127\mu^{168}\varepsilon+236\mu^{167}\varepsilon^{2}
+375\mu^{166}\varepsilon^{3}+522\mu^{165}\varepsilon^{4}+222\mu^{164}\varepsilon^{5}
+232\mu^{163}\varepsilon^{6}+195\mu^{162}\varepsilon^{7}\\
&+220\mu^{161}\varepsilon^{8}+22\mu^{160}\varepsilon^{9}+461\mu^{159}\varepsilon^{10}
+582\mu^{158}\varepsilon^{11}+541\mu^{157}\varepsilon^{12}+283\mu^{156}\varepsilon^{13}
+577\mu^{155}\varepsilon^{14}\\
&+598\mu^{154}\varepsilon^{15}+263\mu^{153}\varepsilon^{16}+361\mu^{152}\varepsilon^{17}
+577\mu^{151}\varepsilon^{18}+540\mu^{150}\varepsilon^{19}+90\mu^{149}\varepsilon^{20}
+222\mu^{148}\varepsilon^{21}\\
&+248\mu^{147}\varepsilon^{22}+164\mu^{146}\varepsilon^{23}+494\mu^{145}\varepsilon^{24}
+361\mu^{144}\varepsilon^{25}+107\mu^{143}\varepsilon^{26}+404\mu^{142}\varepsilon^{27}
+469\mu^{141}\varepsilon^{28}\\
&+265\mu^{140}\varepsilon^{29}+21\mu^{139}\varepsilon^{30}+4\mu^{138}\varepsilon^{31}
+317\mu^{137}\varepsilon^{32}+369\mu^{136}\varepsilon^{33}+189\mu^{135}\varepsilon^{34}
+283\mu^{134}\varepsilon^{35}\\
&+490\mu^{133}\varepsilon^{36}+543\mu^{132}\varepsilon^{37}+81\mu^{131}\varepsilon^{38}
+372\mu^{130}\varepsilon^{39}+302\mu^{129}\varepsilon^{40}+401\mu^{128}\varepsilon^{41}
+293\mu^{127}\varepsilon^{42}\\
&+199\mu^{126}\varepsilon^{43}+532\mu^{125}\varepsilon^{44}+49\mu^{124}\varepsilon^{45}
+431\mu^{123}\varepsilon^{46}+127\mu^{122}\varepsilon^{47}+208\mu^{121}\varepsilon^{48}
+596\mu^{120}\varepsilon^{49}\\
&+277\mu^{119}\varepsilon^{50}+222\mu^{118}\varepsilon^{51}+325\mu^{117}\varepsilon^{52}
+97\mu^{116}\varepsilon^{53}+599\mu^{115}\varepsilon^{54}+576\mu^{114}\varepsilon^{55}
+169\mu^{113}\varepsilon^{56}\\
&+152\mu^{112}\varepsilon^{57}+528\mu^{111}\varepsilon^{58}+273\mu^{110}\varepsilon^{59}
+380\mu^{109}\varepsilon^{60}+353\mu^{108}\varepsilon^{61}+428\mu^{107}\varepsilon^{62}
+248\mu^{106}\varepsilon^{63}\\
&+478\mu^{105}\varepsilon^{64}+327\mu^{104}\varepsilon^{65}+529\mu^{103}\varepsilon^{66}
+262\mu^{102}\varepsilon^{67}+426\mu^{101}\varepsilon^{68}+94\mu^{100}\varepsilon^{69}
+347\mu^{99}\varepsilon^{70}\\
&+474\mu^{98}\varepsilon^{71}+59\mu^{97}\varepsilon^{72}+210\mu^{96}\varepsilon^{73}
+240\mu^{95}\varepsilon^{74}+653\mu^{94}\varepsilon^{75}+228\mu^{93}\varepsilon^{76}
+218\mu^{92}\varepsilon^{77}\\
&+262\mu^{91}\varepsilon^{78}+518\mu^{90}\varepsilon^{79}+508\mu^{89}\varepsilon^{80}
+284\mu^{88}\varepsilon^{81}+97\mu^{87}\varepsilon^{82}+606\mu^{86}\varepsilon^{83}
+127\mu^{85}\varepsilon^{84}\\
&+550\mu^{84}\varepsilon^{85}+71\mu^{83}\varepsilon^{86}+580\mu^{82}\varepsilon^{87}
+393\mu^{81}\varepsilon^{88}+169\mu^{80}\varepsilon^{89}+159\mu^{79}\varepsilon^{90}
+415\mu^{78}\varepsilon^{91}\\
&+459\mu^{77}\varepsilon^{92}+449\mu^{76}\varepsilon^{93}+24\mu^{75}\varepsilon^{94}
+437\mu^{74}\varepsilon^{95}+467\mu^{73}\varepsilon^{96}+618\mu^{72}\varepsilon^{97}
+203\mu^{71}\varepsilon^{98}\\
&+330\mu^{70}\varepsilon^{99}+583\mu^{69}\varepsilon^{100}+251\mu^{68}\varepsilon^{101}
+415\mu^{67}\varepsilon^{102}+148\mu^{66}\varepsilon^{103}+350\mu^{65}\varepsilon^{104}
+199\mu^{64}\varepsilon^{105}\\
&+429\mu^{63}\varepsilon^{106}+249\mu^{62}\varepsilon^{107}+324\mu^{61}\varepsilon^{108}
+297\mu^{60}\varepsilon^{109}+404\mu^{59}\varepsilon^{110}+149\mu^{58}\varepsilon^{111}
+525\mu^{57}\varepsilon^{112}\\
&+508\mu^{56}\varepsilon^{113}+101\mu^{55}\varepsilon^{114}+78\mu^{54}\varepsilon^{115}
+580\mu^{53}\varepsilon^{116}+352\mu^{52}\varepsilon^{117}+455\mu^{51}\varepsilon^{118}
+400\mu^{50}\varepsilon^{119}\\
&+81\mu^{49}\varepsilon^{120}+469\mu^{48}\varepsilon^{121}+550\mu^{47}\varepsilon^{122}
+246\mu^{46}\varepsilon^{123}+628\mu^{45}\varepsilon^{124}+145\mu^{44}\varepsilon^{125}
+478\mu^{43}\varepsilon^{126}\\
&+384\mu^{42}\varepsilon^{127}+276\mu^{41}\varepsilon^{128}+375\mu^{40}\varepsilon^{129}
+305\mu^{39}\varepsilon^{130}+596\mu^{38}\varepsilon^{131}+134\mu^{37}\varepsilon^{132}
+187\mu^{36}\varepsilon^{133}\\
&+394\mu^{35}\varepsilon^{134}+488\mu^{34}\varepsilon^{135}+308\mu^{33}\varepsilon^{136}
+360\mu^{32}\varepsilon^{137}+673\mu^{31}\varepsilon^{138}+656\mu^{30}\varepsilon^{139}
+412\mu^{29}\varepsilon^{140}\\
&+208\mu^{28}\varepsilon^{141}+273\mu^{27}\varepsilon^{142}+570\mu^{26}\varepsilon^{143}
+316\mu^{25}\varepsilon^{144}+183\mu^{24}\varepsilon^{145}+513\mu^{23}\varepsilon^{146}
+429\mu^{22}\varepsilon^{147}\\
&+455\mu^{21}\varepsilon^{148}+587\mu^{20}\varepsilon^{149}+137\mu^{19}\varepsilon^{150}
+100\mu^{18}\varepsilon^{151}+316\mu^{17}\varepsilon^{152}+414\mu^{16}\varepsilon^{153}
+79\mu^{15}\varepsilon^{154}\\
&+100\mu^{14}\varepsilon^{155}+394\mu^{13}\varepsilon^{156}+136\mu^{12}\varepsilon^{157}
+95\mu^{11}\varepsilon^{158}+216\mu^{10}\varepsilon^{159}+655\mu^{9}\varepsilon^{160}
+457\mu^{8}\varepsilon^{161}\\
&+482\mu^{7}\varepsilon^{162}+445\mu^{6}\varepsilon^{163}+455\mu^{5}\varepsilon^{164}
+155\mu^{4}\varepsilon^{165}+302\mu^{3}\varepsilon^{166}+441\mu^{2}\varepsilon^{167}
+550\mu\varepsilon^{168}+\varepsilon^{169}.
\end{split}
\end{align}
Using the notation from Proposition \ref{prop11}, Equations \eqref{deltaat677} and
\eqref{vdeltaat677}
imply
\begin{align}\label{677calculation}
\begin{split}
&L_2\Delta^{38194r\cdot677^n}\\
&=V_2\Delta^{38194r\cdot677^n}
-\Delta^{38194r\cdot677^n}\\
&=(\mu^{76388r}\varepsilon^{38194r}-\mu^{38194r}\varepsilon^{76388r})^{677^n}\\
&=\varepsilon^{114582r\cdot677^n}x^{38194r\cdot677^n}(x^{38194r}-1)^{677^n}\\
&=\varepsilon^{114582r\cdot677^n}x^{38194r\cdot677^n}(x^{38194}-1)^{677^n}
(x^{38194(r-1)}+x^{38194(r-2)}+\cdots+x^{38194}+1)^{677n}.
\end{split}
\end{align}
Moreover, in $\mathbb Z/677[x]$, one can check
that the inhomogeneous polynomial
\[
676x^{169}+127x^{168}+\cdots+550x+1
\]
corresponding to the homogeneous polynomial
in $\mu$ and $\varepsilon$ in Equation
\eqref{E676at677} divides $x^{38194}-1$.
As in the cases $p=11$ and $p=13$, Equation
\eqref{677calculation} implies this is equivalent to
$L_2\Delta^{38194r\cdot677^n}$ being divisible by
$E_{676}^{677^n}$ (and hence by $E_{676}^j$
for $1\leq j\leq677^n$)
in $M_*(2)_{\mathbb Z/677}$.
This verifies condition (C4) in the case $\ell=2$ by
Proposition \ref{Serre}.
\end{proof}

\subsection{A conjecture}
Theorems \ref{main_5}, \ref{thm7},
\ref{thm11},
\ref{thm13}, and
\ref{thm677} all give evidence for
the following conjecture.
\begin{conj}\label{conj}
Let $p\geq5$ be a prime.  If $1\leq i\in\mathbb Z$
and $i=rp^n$ with $(r,p)=1$, then
\[
f_{i/j}=\Delta^{i(p^2-1)/12}
\]
for $1\leq j\leq p^n$,
and $f_{i/j}\neq\Delta^{i(p^2-1)}$
for all other values of $j$
allowed by Lemma \ref{betas}.
\end{conj}
In theory, one possible approach to
Conjecture \ref{conj}
would be to obtain a closed
formula for the Eisenstein series $E_{p-1}
\in M_{p-1}(2)_{\mathbb Z/p}$
as a polynomial in $\mu$ and $\varepsilon$
(and $\delta$ if $p\equiv 3\mod 4$).
Specific instances of this include Equations
\eqref{E10at11}, \eqref{E12at13}, and \eqref{E676at677},
each obtainable by employing the recursive formula
\[
(t-3)(2t-1)(2t+1)\dfrac{B_{2t}}{(2t)!}
E_{2t}=-3\sum_{a+b=t}
\dfrac{(2a-1)(2b-1)B_{2a}B_{2b}}{(2a)!(2b)!}
E_{2a}
E_{2b}
\]
with $t=(p-1)/2$ and then reducing modulo $p$ for $p=11$, 13, and 677,
respectively.
Another approach
would be to find a more conceptual reason why
$E_{p-1}^{p^n}$ should always divide
$L_{\ell}\Delta^{i(p^2-1)/12}$
in $M_*(\ell)_{\mathbb Z/p}$.

The next logical step beyond Conjecture \ref{conj}
is to generalize Theorem \ref{main_5}
by computing all modular forms $f_{i/j}$
at arbitrary primes $p\geq7$.  We are hopeful
that the ``correction terms'' used in this paper
at the prime 5 somehow have natural $p$-primary analogs.
In particular, one can first try to generalize
Equation \eqref{FTFTN} by producing modular
forms $f_{p^2/a_2}$ satisfying conditions (C1) through
(C4) for $i=p^2$ and $j=a_2$.

\subsection*{Acknowledgments}  The author would like to
thank Amanda Folsom, Tom Shemanske, and Mark Behrens
for helpful correspondence.

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