## Re: Two factors of R(13860)

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• ... OK, I ll delete mine. ... Sure. Delete the files you ve uploaded and I ll upload the updated versions. Greg
Message 1 of 16 , Oct 2, 2001
> Have now remedied that, with your 2 gems at the top:-)

OK, I'll delete mine.

> > David...could you also update hd1k.zip with the 2 new factors and
> > 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
• ... Sure thing, Boss! That folder is now totally Broadhurst free. Thanks for taking over! David
Message 2 of 16 , Oct 2, 2001
Greg Childers commanded:
> Delete the files you've uploaded
> and I'll upload the updated versions.
Sure thing, Boss! That folder is now
Thanks for taking over!
David
• 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
Message 3 of 16 , Oct 2, 2001
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
• ... 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
Message 4 of 16 , Oct 2, 2001
> 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.

> 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
• ... By F1 and F2 I meant the fractions of N-1 and N+1. When F1 F2 this condition is what I used for Fib(25561), but it also works for F2 F1. Hence the Max and
Message 5 of 16 , Oct 2, 2001
Satoshi TOMABECHI wrote:
> > Have you coded 10/3*Max(F1,F2)+Min(F1,F2)>1 ?
> Sorry, I don't know such condition.
By F1 and F2 I meant the fractions of N-1 and N+1.
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
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