We have theory that tells us that for curves of analytic rank 0 or 1,
Sha is finite and while BSD is not completely proved for such curves,
the formula it claims for #Sha is certainly rational -- and Sage does
compute this exactly.  But for analytic rank >1 the "analytic Sha" is
just the value predicted by BSD which is not known theoretically to be
rational and can only be computed as a floating point approximation,
for which it is most honest to leave as such.

John Cremona

On 16 October 2013 10:39, Georgi Guninski <gunin...@guninski.com> wrote:
> Why E.sha().an() = 0 and E.sha().an_numerical() = 1.0 ?
>
> On 5.11 on linux and on cloud.sagemath.com:
>
> sage: E=EllipticCurve(QQ,[0, 0, 1, -79, 
> 342]);E.sha().an(),E.sha().an_numerical()
> (0, 1.00000000000000)
> #^ why different and 0 is integer ?
> sage: E.sha().bound()
> (0, 0)
>
> sage: type(E.sha().an())
> <type 'sage.rings.integer.Integer'>
> sage: E.rank(),E.analytic_rank()
> (5, 5)
>
> sage: E.conductor()
> 19047851
>
> E.sha().an_padic(3) # stopped this at about 7G of RAM
>
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