Re: Question on finding a Riemann Zeta function zero for high values of s

Hmm.. I am encountering problems after patching and recompiling.

I am unable to do the 10^21 value that Odlyzko gave as

s = 1/2 +  1370919909931995308226.68016095*I

the gp pari program hangs.

If you go over to https://www.lmfdb.org/zeros/zeta/?limit=100&N=103800788260

and pick the first zeta(s) = 0 in that row starting with 103800788260 
entry which is given as

zeta zero count   imag value

103800788260    30610045972.0390910630030666727984229474273

I don't get zero ??? (using 57 bits precision)

> ? t  = 1/2 + 30610045972.0390910630030666727984229474273000000000000000*I
> ? zeta(t)
> %80 = -0.000645385413125445700656816505388581204917603751967671567187 
> + 0.00141432995627737373722021733088168156756968182126816038178*I
> ? s = 1/2 + 30610045972.0389751581198913418102142029585467738192498830*I
> ? zeta(s)
> %82 = -4.7188538541968517077111529282072387417 E-46 + 
> 1.03058368130580333982284683408181013424 E-45*I
so t and s are off in the 0.0000x place in the decimal expansion.

So which one is correct? s ? or t ? or neither?

For the next entry or the 103800788261 zero number of the Riemann zeta 
function for complex 1/2 + x*I

> s = 1/2 + 30610045972.3301196133311775056695486332490000000000000000*I
but GP Pari gives
> ? t = 1/2 + 30610045972.3301823075035849673628150748546420868088900184*I
So which one is correct? s or t or none?

This needs some careful investigation.

Randall

On 10/22/25 14:51, Cohen Henri wrote:
>
> Sorry, should have replied to list. In integrand_h0, replace line 309
>
> p1=gmul(expIxy(pmd,gsqr(zn),prec), by
>
> p1 = gmul(pmd, mulcxI(gsqr(zn))); p1r = greal(p1);
>  if (gcmpgs(mpabs(p1r), 2*prec2nbits(prec)) > 0) p1 = p1r;
>  p1 = gmul(gexp(p1, prec),
>
> and declare p1r as a GEN. This is not good, but works here. It is to 
> avoid an
>
> overflow in gsincos when taking the exponential of A+I*B where B is 
> huge and
>
> A is huge negative or positive. The condition I wrote is ridiculous, 
> but I don't know what
>
> to put else.
>
> Henri
>