Let l=2m+1 be prime. In my previous MO question, "What are the polynomial relations between these characteristic 2 thetas?", I defined a subring of Z/2[[x]] as follows:

The subring, S, is generated by [1],...,[m] where [i] is the sum of the x^(n^2), n running over all integers congruent to i mod l.

QUESTION...... Let F=x+x^9+x^25+x^49+...,G=F(x^l), and H=G(x^l). Are G and H in S?

The answer is yes when l=3,5 or 7. When l=7, if we set a=[1],b=[2] and c=[3], we have the curious identities H=(abc)^3*(abc+ba^3+cb^3+ac^3), and G=(abc)^2+a^7+b^7+c^7+H.

Remark 1... Kevin Buzzard explained to me that one can decide whether an explicitly given identity such as the ones we've displayed holds by using the theory of characteristic 2 modular forms and computer calculation. But how does one produce these putative identities?

Remark 2... For all l one can show in an elementary way that H is in the field of fractions of S. In fact if a=[i], b=[2i] and c=[4i], then H is the quotient of a^8(a^8+b^2) by b^4+c. Furthermore for l at most 13, H is in S. (One shows that the quotient lies in S, by combining the "quintic relations" of my MO question cited earlier with Groebner basis computer calculations.)

I'll sketch an argument giving the l=7 identities. Let C be the curve in affine 3-space defined by the ideal of quintic relations. C has 3 linear branches at the origin and 3 linear branches at each of the seven points (r,r^4,r^9) with r^7=1. Passing to projective 3-space we find that (the Zariski closure of) C has 14 simple points at infinity. The formula for H as a quotient shows that H has zeros of order 49 at the branches at the origin, simple zeros at the branches at the other singular points, and poles of order 12 at infinity. This leads to the identity for H. To get the identity for G one notes that (GH)+(GH)^2+(G+H)^8=0--see my MO question, "What's known about the reduction...?" It follows from this that if G is in the field of fractions of S then G+H has zeros of order 7 at the branches at the origin, of order 3 at the branches at the other singular points, and poles of order 6 at infinity. This suggests that G+H=(abc)^2+a^7+b^7+c^7. To verify this we set J=(abc)^2+a^7+b^7+c^7+H, and use Groebner basis computer calculations to show that JH+(JH)^2+(J+H)^8=0; it then follows that J=G.

EDIT: I think I can now show that when l=11, G is NOT in the field of fractions of S, even though H is in S. I'll make this an answer once I'm surer of it.

EDIT #2: My supposed counterexample when l=11 is incorrect; G like H is in S. I had the wrong modular equation of degree 11 relating G and H. Once I found the correct equation, in Cayley's article, I was able to argue as in the case l=7.

FINAL(?) EDIT: As I've shown in my answer, G and H are indeed always in S. And I've produced a simple conjectural explicit formula for G+H that holds for l<1500. Whether there is anything comparably simple for H isn't clear. At any rate here are formulas for H when l<24. I write C(a,b,c) for the sum of the [ra]*[rb]*[rc] where r runs from 1 through (l-1)/2;
more generally (a,b,c) can be replaced by any multi-set. P is the product of the [r] where
r runs from 1 through (l-1)/2. The identity when l=17 is striking.

l=3.......... [1]^9 +[1]^12

l=5.......... P^5 +P^6

l=7.......... (P^3)(P+C(1,1,1,2))

l=11.........(P^2)(C(1,1,3)+C(1,1,2,4))

l=13.........P*(P+[1][2][3][5]+[1][4][5][6]+[2][3][4][6]+C(1,1,2,2,2,5))

l=17.........P*([1][2][4][8]+[3][5][6][7])

l=19.........P*([2][3][5]+[4][6][9]+[1][7][8]+C(3,3,2,4))

l=23.........P*(C(1,2,3,3)+C(1,2,4,5)+C(1,4,4,6)+C(1,2,2,5,9))