- Hi to all,

The new Wilson-Mills SG primality test has an important

corollary. A new world record for the existing SG (Sophie Germain)

WR holders Underbakke, Jobling and Gallot.

Namely, the largest Sophie Germain prime p is also the largest known

prime factor of a factorial closed form, possibly of any non trivial

closed form. This closed form is (2p)! + 1.

Reminder: The Wilson-Mills primality test for Sophie Germain's is

if q and p are prime and q = 2*p +1 then q divides (2p)! + 1.

i.e. if 2*p +1 (=q) is a factor of (2p)! + 1 then 2*p + 1 is

prime.

Once again, thanks to Henri Lifchitz, the existing record holder who

had a test of

q divides 3^p 1. But (2p)! + 1 is a larger number

than 3^p 1 !

Stare at this long enough and you will see that the `salient'

feature of the Wilson-Mills test is that the Sophie Germain primes

are `nestled' inside the test as (q,p). All you have to do then is

substitute a real SG pair for (q,p) to obtain a factor ( or `lever' a

factor ) from the closed form (2p)! + 1. Of course, a factor q

can also be `levered' from the form 3^p 1 via a Sophie Germain

(q,p) which is Henri's test. An interesting question is now, what

are the Sophie Germain 'primality test forms' f(p).

I.e. q factors f(p). We now have 2. 3^p -1 and (2p)! + 1.

So to conclude: The largest p of a Sophie Germain pair is

109433307 * 2^66452 1 (from the Prime Pages) and so the largest

known closed factorial form number with the largest known prime

factor is

(2* 109433307* 2^66452 2)! + 1 with factor

109433307*2^66452 1

Regards,

Paul Mills,

Kenilworth,

England. > Yes, I see it now,

I'm not sure that you can clearly define "irreducible in the logical

> (p + 1)^p == p + 1 mod p

> (p^p + 1)^p == p^p + 1 mod p etc

>

> So we have a strange object, an open `closed' form. The closure

> here being only in the logical order, and semi-open in the algebraic

> order (extended via the theorem). I did say the theory starts.

>

>

> Of course you can only use 1 theorem once in any form, no repeated

> applications etc. Forms must be irreducible in the logical as well

> as algebraic order. Once again the point of this is that we can

order". Or what you mean by using 1 theorem only once. After all,

I'm sure that most of the methods we use actually use certain

theorems meny times over: Example

Theorem: for any integers x and y, x*y = y*x.

Does that mean they are discounted?

To my mind, every theorem is reducible in a logical sense, except

of course, the axioms of the logical system...

Yours, Mike H...

> now produce ever larger numbers of which we know the factor.

Michael Hartley : Michael.Hartley@...

> There is an error in my previous posting, q = 3*p + 1 is difficult

> for q and p odd prime. If q = k*p +m in a Generalised Sophie

> Germain then k and m have opposite parity

>

> regards,

>

> Paul Mills

>

>

>

>

>

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> The Prime Pages : http://www.primepages.org

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>

Head, Department of Information Technology,

Sepang Institute of Technology

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"For centuries, people thought the moon was made of green cheese.

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