Define A(0), A(1), A(2) ... in Z/3[[x]] as follows. For n in N let s=3^(2n+1). Then A(n) is sum ((a_k)(x^k)) were a_k is the mod 3 reduction of the number of representations of k by the principal positive binary quadratic form of discriminant -s, and the sum runs over all k prime to 3.
Example 1__ When n=0, we take the form to be uu+uv+vv. The number of representations of any non-zero k by this form is a multiple of 6, and so A(0)=0.
Example 2__ -A(1) is the mod 3 reduction of (the expansion of) the usual weight 12 cusp form for the full modular group.
Experiment suggests a recursion for the A(n) which would allow one to write each A(n), n>0, as a polynomial of degree (s-3)/24 in A=A(1). Explicitly I ask if the following holds:
(*)___ A(n+2)=((A^3s)+(A^2s)+1)(A(n+1))-(A^2s)(A(n))
If (*) holds, then for example:
A(2)=(A^10)+(A^7)+A
A(3)-A(2) is sum(A^k), k in {91,88,82,64,61}
A(4)-A(3) is sum(A^k), k in {820,817,811,793,790,739,736,730,577,574,568,550,547}
I've experimentally verified the above 3 identities, and I'm sure that modular forms could be used to settle the truth of (*) for any fixed n.