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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)

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

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

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  • $\begingroup$ How many examples were checked? $\endgroup$ – Alexey Ustinov Jun 24 '17 at 16:14
  • $\begingroup$ @Alexey Ustinov: For m = 1, 500 examples were checked. And for m = 2, 100 examples were checked. $\endgroup$ – TOM Jun 24 '17 at 16:20
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    $\begingroup$ The statement for all $m, n$ follows from the statement for $m = 1$ and all $n$, so we can forget about higher $m$. Presumably there is a $U_5$-eigenform lurking here somewhere. $\endgroup$ – David Loeffler Jun 25 '17 at 8:06
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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 $b(5n) \equiv b(n) \bmod 3000$.

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