On Aug 1, 11:10 pm, "D. S. McNeil" <[email protected]> wrote:
> > Thanks for this.  There still seems to be a manual step in going from,
> > say,
> > s1 = 2*(1,8) - (1,9)
> > to
> >  s1 = 2*b[1] - b[2]
>
> I may be misunderstanding you.  Are you saying you want to enter the line
>
>     s1 = 2*(1,8)-(1,9)
>
> verbatim and have it work?  

No, s1 will be the result of a calculation on sums of modular symbols
over various residue classes.

Sorry if I was not expressing myself clearly enough.

Jack

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