Define b(n) by b(0)= 0, b(3n)= 9b(n), b(3n+1)= 9b(n) + 1, b(3n+2)= 9b(n) + 3. (this sequence does not show up in o.e.i.s but the similar Moser-de Bruijn sequence A000695 appears in many situations, one being the theory of mod 2 modular forms).

One seems to encounter the b(n) when studying mod 3 modular forms of level Gamma_0 (N) for various small N. But what follows is largely empirical though supported by overwhelming evidence. I would welcome (but do not expect, particularly in the mysterious level 5 case) an explanation.

LEVEL 1

Let F in Z/3[[q]] be the mod 3 reduction of the level 1 weight 12 cusp form. Let V be the subspace spanned by the F^k with (k,3)= 1. The formal Hecke operator T2: Z/3[[q]]–> Z/3[[q]] stabilizes V. (V is the space of level 1 mod 3 modular forms killed by U3). Let K be the kernel of T2: V–> V, and s(0) < s(1) < s(2) ... the degrees (in F) of the (non-zero) elements of K arranged in order of increasing size.

CONJECTURE 1: s(n)= 9b(n) + 1.

This has been computer verified for n < 54; in particular s(53)= 9019. Here's how one makes the calculations. Let A(n) in Z/3[F], n prime to 3, be T2(F^n). The A(n) satisfy the degree 9 recursion A(n+9)= 2(F^9)A(n) + (F^3)A(n+3), with initial conditions A(1)= 0, A(2)= F, A(4)= F^2, A(5)= F^4, A(7)= F^5, A(8)= F^7 + F^4. This permits the rapid calculation of the A(n) and of those n for which A(n) is a Z/3-linear combination of A(k) for k < n. These n are the degrees of the elements of K.

LEVEL 5

Let v in Z/3[[q]] be the mod 3 reduction of the weight 4 cusp form of level Gamma_0 (5).Then v(1) = v + v^3 and v(2)= v^2 – v^3 are killed by U3. Let V be the subspace of Z/3[t] spanned by the t^k, k prime to 3. Let i: V–> Z/3[[q]] be the Z/3-linear imbedding taking t^(3k+1) to (v^3k)v(1) and t^(3k+2) to (v^3k)v(2). The identification of i(V) with V gives a degree function (degree in t) on i(V). It can be shown that the Tp, when p is other than 3 or 5, stabilize i(V). (i(V) is the space of mod 3 modular forms of level Gamma_0 (5) killed by U3 and fixed by the mod 3 Fricke involution W5). Let K be the kernel of (T2 + I): i(V)–> i(V). Let t(0) < t(1) < t(2) .. be the degrees of the elements of K arranged in order of increasing size.

CONJECTURE 2: t(2n)= 27b(n) + 1 or 27b(2n) + 2 according as b(n) is even or odd. t(2n+1)= 27b(n) + 11 or 27b(n) + 10 according as b(n) is even or odd.

CONJECTURE 3: The conjecture still holds when K is replaced by the kernel of T7.

CONJECTURE 4: The conjecture still holds when K is replaced by the kernel of T11 + I

All three of these conjectures hold for n < 27. In particular all three t(53) are 7381. The recursions used to establish these results are messy; they are of degrees 12, 24, and 39 respectively.

Remark–There is a similar computer supported conjecture in level 11 for the kernel of (T7 _ I) acting on a space analogous to i(V). But now v is the reduction of the weight 2 cusp form of level Gamma_0 (11), v(1)= v + v^3 and v(2)= v^2 + v^3 + v^6.