A sudden inspiration struck!

Let B be A(q^3). Then the following identity, * , holds: (AB+A+B)^4 = AB. To see this, note that A is the mod 2 reduction of the expansion of the weight 12 cusp form delta(z) + delta(13z) for Gamma_0(13). So B is the reduction of a weight 12 cusp form for Gamma_0 (39). Then both sides of * come from modular forms for Gamma_0 (39) of weight 96. Since the Sturm bound in this weight and level is
(96/12)*39=312, it suffices to show that the expansions appearing in * agree through q^312; this is quickly verified.

For n non-negative, let d_n be T_3 (b_n). One calculates that d_1 = d_5 = 0, d_3 = c_1 + c_3, and d_7 = c_5 + c_7. The standard argument used by Nicolas and Serre uses * above to show first that d_(n+4) is (Ad_(n+1) + (A^4)d_n)/(A^4 + 1), and then that d_(n+8) is ((A^2)(d_(n+2)) + (A^8)d_n)/(A^8+1). Since multiplication by A^2/(A^2 + 1) stabilizes the space spanned by the c_n, so does
multiplication by 1/(A^2 + 1), A^2/(A^8 + 1), and A^8/(A^8 + 1). An induction then shows that all the d_n, n odd, lie in this space.

Remarks:

I had earlier looked for a relation between A and B and by a general method found a symmetric one of total degree 176 of the form (1+B^64)A^112 + a sum of monomials, each of degree < 112 in A, is equal to 0. This led to an impenetrable recursion. I imagined that my relation might be reducible, but couldn't factor it on the computer, and it took me a while to find * .

Suppose instead B is A(q^p) with p a prime other than 2 or 13. Is there a nice relation between A and B? I looked briefly at p=7. The general method I used in in 3. then gives a similar symmetric relation of total degree 352, with 64 and 112 replaced by 128 and 224. But I'm sure that there's something simpler, giving a useful recursion for the T_7 (b_n). I wonder what the general result is.

EDIT(10/16)

- The following experimental results aren't directly relevant to the original question, but they do address 4. above, and indicate that there are
identities much like * when B is A(q^p) rather than simply A(q^3).

Let fp in Z/2[x,y] be the unique irreducible polynomial with fp(A(q),A(q^p)) = 0, so that f3 is x^4*y^4+x^4+y^4+x*y. I've calculated fp for p=3,5,7,11,17,19 and discovered the following in each case:

a. fp(x,y) = fp(y,x)

b. The degree of fp in x is p+1

c. The coefficient of x^(p+1) in fp is (y+1)^k for some k. (In the six cases, k is 4,4,4,8,12,8 respectively).

d. fp(1,y) is (y+1)^k for some k. (In the six cases k is 1,3,5,6,9,14 respectively)

Things do get messier as p grows--f17 and f19 are sums of 46 and 48 monomials. It's reminiscent of the modular equation for the j-invariant, and an approach to showing that a,b,c,d hold in general might be to compare the mod 2 reduction of
the uniformizer for the genus 0 curve attached to Gamma_0 (13) with A.