Let D in Z/3[[x]] be sum ((a\_n)(x^n)) where the sum runs over all n prime to 6 and a\_n is the mod 3 reduction of the number of ideals of norm n in the ring of integers of Q(root(-3)). (So D=x+2(x^7)+2(x^13)+2(x^19)+x^25+(higher degree terms).)


There are formal Hecke operators T\_p: Z/3[[x]]-->Z/3[[x]] for all primes p other than 3. Clearly the T\_p with p=2 mod 3 annihilate D. Experimentally I find that they annihilate D^7,
D^61 and D^547 as well.


Question: If k=(1+3^m)/4 with m odd, do the T\_p with p=2 mod 3 annihilate D^k?


Remark: D is the reduction of the expansion of the modular form (eta(2z))^12, but it's not clear that this helps. 


EDIT: If I've made no mistake, Noam's formula can be tweaked to show that D^k=V-U where U and V are the mod 3 reductions of the theta series attached to the binary quadratic forms 36xx+6xy+kyy and 36xx+30yy+(k+6)yy. This raises the question of which linear combos of theta series of primitive binary quadratic forms lie in the space spanned by the powers of D. I think similar questions have been answered by Nicolas and Serre (ell=2, level=1) and perhaps by Bellaiche and or Serre (ell=3, level=1). 



Are these powers of a characteristic 3 power series annihilated by certain Hecke operators?
Yes, and $D \equiv \eta(2z)^{12}$ certainly helps: it means
$$
D^{(1+3^m)/4} \equiv \eta(2z)^{3+3^{m+1}} \equiv \eta(6z) \, \eta(2\cdot 3^{m+1}z).
$$
Now $\eta = \sum\_a \chi(a) q^{a^2/24}$ where $a$ runs over integers
congruent to $1 \bmod 6$ and $\chi$ is the even Dirichlet character $\bmod\,12$.
Since $m$ is odd, say $m=2n+1$, we have
$$
\eta(6z) \, \eta(2\cdot 3^{m+1}z) = 
\mathop{\sum\sum}\_{a,b \equiv 1 \bmod 6} \chi(ab) q^{(a^2 + 3^{2n+1}b^2)/4}.
$$
All the exponents are odd, so our form is in $\ker T\_2$.
If $p \equiv 5 \bmod 6$ then $(a^2 + 3^{2n+1}b^2)/4$
is a multiple of $p$ *iff* $a,b \equiv 0 \bmod p$, whence
$T\_p$ multiplies our form by $p^{w-1} + 1$ where $w$ is the weight.
Since $w = (1+3^m)/4$ is odd and $p \equiv 2 \bmod 3$, we have
$p^{w-1}+1 = 0$ in ${\bf Z}/3{\bf Z}$ and we're done.


