[PrimeNumbers] Re: Two factors of R(13860)
- 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.
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.
- David Broadhurst wrote:
> Have you coded N+1?Yes, of course.
Furthermore I coded modified KP
with additional square tests by you.
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 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.