Recursions for some binary theta series in characteristic 3 Define $A(0), A(1), A(2) \dots$ in ${\bf Z}/3[[x]]$ as follows. For $n$ in $\bf N$ let $s=3^{2n+1}$. Then $A(n) = \sum a_kx^k$ where $a_k$ 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 $u^2+uv+v^2$. 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:
$A(2)=A^{10}+A^7+A$
$A(3)-A(2) = \sum A^k$, $k$ in $\{91,88,82,64,61\}$
$A(4)-A(3) = \sum A^k$, $k$ 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 $n$.
EDIT: As far as the weaker question of finding a proof that each $A(n)$ is a polynomial in $A$,
there is a possibly relevant article by Skoruppa. (arXiv:0807.4694v2---Reduction mod $\ell$ of theta series of level $\ell^n$). But his main result is restricted to characteristic >3. I quote the result:
It is proved that the theta series of an even lattice whose level is a power of a prime $\ell$ is congruent modulo $\ell$ to an elliptic modular form of level one.
(Note that every elliptic modular form over $\bf Q$ of level 1 has as its modulo 3 reduction a polynomial in $A$. So one would want to extend Skoruppa's result to $\ell=3$).
EDIT: It seems one can use an earlier result of Serre, Theorem 5.4 of "Divisibilite ..." appearing in L'Enseignement Mathematique (1976) to quickly show that the A(n) are polynomials in A. Namely let m=(s+1)/4, and consider the quaternary form uu+uv+mvv+ww+wt+tt.
It's classical that the theta series attached to this form is a weight 2 modular form for Gamma_0 (N), where N is a power of 3. The Serre theorem just cited shows that the mod 3 reduction of such a modular form is the reduction of a modular form for Gamma(1), and so is a Z/3 linear combination of powers of delta= -A.
Now the reduction of the theta series for ww+wt+tt is 1, so the reduction of the theta series for uu+uv+mvv is also a Z/3 linear combination of powers of A. It follows that A(n) is a Z/3 linear combination of A, A^4, A^7, A^10,...
 A: We establish the recursion for all $n$ by writing
the rank-2 theta series $A(n)$ in terms of the rank-1 thetas
$$
S(q) := \sum_{m \in \bf Z} q^{m^2} = 1 + 2q + 2q^4 + 2q^9 + \cdots,
$$ $$
T(q) := \sum_{m \in \bf Z} q^{(m+\frac12)^2}
= 2q^{1/4} + 2q^{9/4} + 2q^{25/4} + \cdots.
$$
The lattice corresponding to the principal positive binary
form of discriminant $-s = -3^{2n+1}$ is the union of the
rectangular lattice ${\bf Z} \oplus {\bf Z}\langle s \rangle$
and its translate by $(1/2, 1/2)$.  The quadratic form
is a multiple of $3$ iff the ${\bf Z}$ or ${\bf Z} + \frac12$
term is a multiple of $3$.  Hence
$$
A(n) = (S(q)-S(q^9)) \, S(q^s) + (T(q)-T(q^9)) \, T(q^s).
$$
Because $9$ and $s$ are powers of $3$, an equivalent formula
in characerstic $3$ is
$$
A(n) = (S-S^9) S^s + (T-T^9) T^s
$$
where $S=S(q)$, $T=T(q)$.  We claim that the recursion
is already satisfied by $A_1(n) := S^s$ and $A_2(n) := T^s$
separately, from which it will follow by linearity for
$A(n) = (S-S^9) A_1(n) + (T-T^9) A_2(n)$.  For both $i=1$ and $i=2$
each side of the recursion
$$
A_i(n+2) = (A^{3s} + A^{2s} + 1) A(n+1) - A^{2s} A(n)
$$
is the ($3^{2n}$)-th power of its $n=0$ case
$$
A_i(2) = (A^9 + A^6 + 1) A_i(1) - A^6 A_i(0),
$$
so we need only verify this last identity for both $i$.
Now $S$ and $T$ are related by $S^4 + T^4 = 1$, because
in characteristic zero $S^4 + T^4$ is the theta series of
the $D_4$ lattice, whose automorphism group contains a
$3$-cycle that acts freely on nonzero vectors.  [Check:
$A(0)$ vanishes because it equals
$$
S^4-S^{12} + T^4-T^{12} = (S^4+T^4) - (S^4+T^4)^3 = 1 - 1^3 = 0.]
$$
Thus
$$
A(1) = S^{28} - S^{36} + T^{28} - T^{36} 
 = S^{28} - S^{36} + (1-S^4)^7 - (1-S^4)^9,
$$
which comes to $S^{24} - S^{16} + S^{12} - S^4$,
and by symmetry also
$A(1) = T^{24} - T^{16} + T^{12} - T^4$.
Then $A_i(2) = (A^9 + A^6 + 1) A_i(1) - A^6 A_i(0)$
is just an identity in $({\bf Z}/3{\bf Z})[S]$ or
$({\bf Z}/3{\bf Z})[T]$, which we verify by direct computation
to complete the proof.
