- --- In primenumbers@y..., d.broadhurst@o... wrote:
> Sorry about that. I had uploaded an old file.

OK, I'll delete mine.

> Have now remedied that, with your 2 gems at the top:-)

> > David...could you also update hd1k.zip with the 2 new factors and

Sure. Delete the files you've uploaded and I'll upload the updated

> > Base10.zip with all 3 new factors?

> Sorry, my file management skills are minimal.

> It's already a minor miracle that I managed to

> get the previous version more or less OK.

> Please Greg, will you take over as Base10 DB manager?

versions.

Greg - Thank you, Greg and David.

My program shows that 15793939750*R(13860)/R(10)+1 is prime,

however I have little confidence in this result:-)

I hope that the result is as same as David's.

The coefficients of cubic equation t3X^3 + t2X^2 + t1X + t0 = 0 are

t3 = 9499179946...9373462290 1400 digits

t2 = 2983442091...4520303837 2765 digits

t1 = -5786178299...9234821312 4165 digits

t0 = -8297037200...6079883289 2764 digits

The equation has three real roots r1,r2,r3.

r1=-1.433940810...*10^{-1401}

integral part of r2 246804372...4038770926 ( 1382 digits)

fractional part of r2 0.9565089129....

integral part of r3 -246804372...8946383526 ( 1382 digits)

fractional part of r2 0.5941852755....

Nearest integers of r1,r2 and r3 are not root of the equation.

# It was not easy to compute roots of the cubic equation.

Best regards.

Satoshi Tomabechi - David Broadhurst wrote:
> Have you coded N+1?

Yes, of course.

Furthermore I coded modified KP

with additional square tests by you.

http://groups.yahoo.com/group/primeform/message/2142

But I have not coded combined KP.

http://groups.yahoo.com/group/primenumbers/message/1743

> Have you coded 10/3*Max(F1,F2)+Min(F1,F2)>1 ?

Sorry, I don't know such condition.

Please tell me reference or pointer to reference.

Satoshi Tomabechi - Satoshi TOMABECHI wrote:
> > Have you coded 10/3*Max(F1,F2)+Min(F1,F2)>1 ?

By F1 and F2 I meant the fractions of N-1 and N+1.

> Sorry, I don't know such condition.

When F1>F2 this condition is what I used for Fib(25561),

but it also works for F2>F1. Hence the Max and Min.

David