Dear Robert, You ask: Apparently if k>4 and congruent to 14 mod 16 there are 2 eigenforms of tame level 1, weight k and 2-adic slope 6, i.e., d(k,6)=2. Gouvea-Mazur predict no congruences between these pairs of forms. I suspect there are. One knows d(k,3)=1 if k\con 0 mod 4 (in Emerton's range) and I think this means that the corresponding eigeforms should be congruent mod 16. Could you check this? WEIGHT 14: First, I found two (new) eigenforms f = q - 64*q^2 - 1836*q^3 + 4096*q^4 + 3990*q^5 + 117504*q^6 - 433432*q^7 ... and g = q + 64*q^2 + 1236*q^3 + 4096*q^4 - 57450*q^5 + 79104*q^6 + 64232*q^7 ... in S_{14}(Gamma_0(2)) that have slope 6. These should span your space with d(14,6) = 2. These two forms are congruent *to each other *modulo* 2^7. WEIGHT 30: Next, we investigate k=30 and tame level 1. The characteristic polynomial of T_2 on S_30(Gamma_0(1)) is x^2-8640*x-454569984; the slopes of the 2-adic Newton polygon are 6 and 6, so there are two 2-adic forms in S_30(Gamma_0(1);Q2bar) that have slope 6. Here's what one of the conjugate eigenforms in S_30(1) looks like: h= q + a*q^2 + (-552*a - 99180)*q^3 + (8640*a - 82300928)*q^4 + (116000*a - 9240014250)*q^5 + ... where, of course, a satisfies a^2-8640*a-454569984=0. CONGRUENCES?: To check your congruence question, let's try to compute the valuation of h-f at 2: h-f = (a + 64)*q^2 + (-552*a - 97344)*q^3 + (8640*a - 82305024)*q^4 + (116000*a - 9240018240)*q^5 + ... Computationally it is very easy to compute the characteristic polynomials of the coefficients of the above q-expansion, and then find the slopes of the roots. We get p charpoly(a_p(h-f)) slopes of roots ---------------------------------------------------------------- 2 x^2 - 8768*x - 454012928 6,6 3 x^2 + 4963968*x - 138035555758080 6,6 5 x^2 + 17477796480*x + 70000527489971097600 6,6 So I think it is *highly* likely that each Q2-bar conjugate of h is congruent to f modulo 2^6, just as you suspected. Best regards, William