Let $a(n)$ and $b(n)$ be define by the following;


$E\_6/E\_4 = 1 - 744q + 159768q^2 - 36866976q^3 + 8507424792q^4 - 1963211493744q^5 + \cdots = \Sigma a(n)q^n,$
$E\_8/E\_6 = 1 + 984q + 574488q^2 + 307081056q^3 + 164453203992q^4 + 88062998451984q^5 + \cdots = \Sigma b(n)q^n.$


(Please see A288261( <https://oeis.org/A288261> ) and A288840 ( <https://oeis.org/A288840> ).)


Are $\frac{a(5^m \* n) - a(n)}{3000}$ and $\frac{b(5^m \* n) - b(n)}{3000}$ integers for all $m, n >= 0$?


(For example)


$\frac{a(5) - a(1)}{3000} = \frac{-1963211493744 - (-744)}{3000} = -654403831$. 


$\frac{b(5) - b(1)}{3000} = \frac{88062998451984 - 984}{3000} = 29354332817$.



Properties of coefficients in expansion of $E_6/E_4$ and $E_8/E_6$
The congruence is true for all $m,n$, and **Manyama**'s calculations for
$n \leq 500$ are more than enough to prove it. Similar congruences
hold for the coefficients of other quotients of modular forms; e.g. the
$q^n$ and $q^{5n}$ coefficients of $E\_6/E\_8$ are also congruent $\bmod 3000$
for all $n$, while those of $E\_4/E\_6$ are also congruent $\bmod 120$.


As **David Loeffler** noted already in a comment, once we know the
$m=1$ congruences
$$
a(5n) \equiv a(n) \bmod 3000,
\qquad
b(5n) \equiv b(n) \bmod 3000
$$
for all $n$, the congruences for $m \geq 2$ follows by induction.
So we need only prove it for $m=1$.


The key is the following observation: for any power series
$F(q) = \sum\_n c\_n q^n$ and any integer $r>0$, the series
$F^{[r]}(q) := \sum\_n c\_{rn} q^n$ equals $r^{-1} \sum\_{q\_1^r = q} F(q\_1)$,
the sum extending over the $r$ complex roots $q\_1$ of $q$.


In our setting, $F$ is a meromorphic modular form for
$\Gamma(1) = {\rm SL\_2}({\bf Z})$,
so $F^{[r]}$ is a meromorphic modular form of the same weight for the
congruence subgroup $\Gamma\_0(r)$ --- indeed it is a linear combination of
$F$ with the image of $F$ under the Hecke operator $T\_r$.
Thus the same is true of $F^{[r]} - F$.
So, to show that all the coefficients of $F^{[r]} - F$ are multiples of $M$,
it is enough to recognize $(F^{[r]} - F) / M$ as a meromorphic modular form
with integer coefficients. This is essentially routine with a package
such as **gp** once we account for the poles of $F$ and $F^{[r]}$.


Both $F = E\_6/E\_4$ and $F = E\_8/E\_6$ are meromorphic forms of weight $2$,
and the desired congruence has $r=5$ and $M=3000$. The modular curve
$X\_0(5)$ associated to $\Gamma\_0(5)$ is rational, parametrized by the
"Hauptmodul"
$$
h = (\eta\_1/\eta\_5)^6
 = \frac1q \prod\_{k=1}^\infty \left(\frac{1-q^k}{1-q^{5k}} \right)^{\!6}
 = q^{-1} - 6 + 9q + 10q^2 - 30q^3 + 6q^4 - 25q^5 \cdots;
$$
and weight-2 meromorphic forms are rational functions of $h$ times the
Eisenstein series
$$
e\_2 = -q \frac{dh/dq}{h}
 = 1 + 6q + 18q^2 + 24q^3 + 42q^4 + 6q^5 + 72q^6 + 48q^7 + \cdots.
$$


Now if $F = E\_6/E\_4$ then $F$ has simple poles at the zeros of $j=0$,
which are the $\Gamma$ orbit of the cube root of unity
$\rho = (-1 + \sqrt{-3})/2$; so $F^{[5]}$ has simple poles at the
$\Gamma$ orbits of $(\rho+a)/5$, and it is known that those are
the points where $j$ is a root of some quadratic polynomial, namely
 $X^2 + 654403829760 X + 5 \cdot 101376^3 =: Q(X)$.
So $j \, Q(j) (F^{[5]}-F) / e\_2$ is a modular function on $X\_0(5)$
with no poles except at the cusps (the zeros of $e\_2$ turn out to
cancel automatically); since $F^{[5]}-F$ is regular at the cusps,
multiplying by a cubic in $j$ yields a linear combination of
$\sum\_{t=-15}^3 \alpha\_t h^t$, which we can determine by computing the
first $19$ coefficients of $j Q(j) (F^{[5]}-F) / e\_2$
(we actually computed twice as many coefficients for a sanity check).
We find that the integers $\alpha\_n$ are all multiples of $3000$,
which proves the congruence $a(5n) \equiv a(n) \bmod 3000$ because
$h$, and thus also each $h^t$ with $t \in \bf Z$, has integer coefficients
and leading coefficient $1$.


The case $F = E\_8/E\_6$ is treated similarly; here the simple poles are
at roots of $j-1728$, which are the $\Gamma$ orbit of $i = \sqrt{-1}$,
and $F^{[5]}$ has simple poles at the $\Gamma$ orbits of $(i+a)/5$,
where $j$ is either $1728$ again (for $a \equiv \pm 2 \bmod 5$) or
a root of the quadratic $Q\_1(X) := X^2 - 44031499226496X - 6635376^3$.
Again we calculate that
$$
(j-1728) Q\_1(j) (F^{[5]}-F) / e\_2 = \sum\_{t=-15}^3 \beta\_t h^t
$$
with each $\beta\_t \in 3000 \bf Z$, which proves the congruence
$b(5n) \equiv b(n) \bmod 3000$.


