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Oct 10, 2015 at 14:06 comment added paul Monsky I now have a proof, using mod 2 modular forms of levels 3 and 5, that if the c(n) sum to zero, then so do the corresponding b(n). But I still have no clue as how to answer my original questions, either by elementary or modular form methods. For this I'd want to show for the first question that the image of the kernel of I+U_3 in N2/N1 is large (and a corresponding result in level 5 for the variation question).
Oct 2, 2015 at 20:14 comment added paul Monsky no, put it this way--the image in N2/N1 of the kernel is epsilon*(N2/N1) where epsilon is an element of the (completed) shallow Hecke algebra whose square kills N2/N1. (In fact epsilon can be taken to be a power series in T_7, T_13, and T_5--see my note). Similar things should go on for the recursion connected with level 5.
Oct 2, 2015 at 18:11 comment added paul Monsky I have to leave now; on return I'll finish this last comment.
Oct 2, 2015 at 18:00 comment added paul Monsky I have a precise conjecture about the relations between the c(n), but it isn't easily expressed in the above language, The real question is "what is the kernel of the Hecke operator I+U_3" on the space M(odd) of odd mod 2 modular forms of level 3?" The identity that you just verified indicates that this kernel lies in a subspace N2 of M(odd) that I define in my arXiv note 1508.7523. In fact I believe that the kernel is spanned by
Oct 2, 2015 at 16:07 comment added Peter Mueller Indeed, for $n\le1000$ this holds for both questions. However, there are many more vanishing sums among the $b(n)$'s than among the $c(n)$'s. So while the $b(n)$'s are easier (and have a not too complicated explicit expression), these vanishing sums don't seem to give a hint for the $c(n)$'s.
Oct 2, 2015 at 1:05 comment added paul Monsky Thanks, Peter. I have further conjectures related to these recursions, that you might like to provide evidence for (or refute). Let b(n) be defined by b(n+4)=b(n+3)+zb(n), b(0)=b(1)=0, b(2)=b(3)=1. I believe that if some c(n) in the first question sum to 0, then so do the corresponding b(n). Similarly in the second question, but now b(n+6)=b(n+5)+zb(n), while b(0)=b(1)=b(2)=0, b(3)=b(4)=b(5)=1.
Oct 1, 2015 at 22:31 history edited Peter Mueller CC BY-SA 3.0
finally got how to type program code
Oct 1, 2015 at 22:24 history answered Peter Mueller CC BY-SA 3.0