- In a message dated 02/01/2002 08:31:16 GMT Standard Time,

d.broadhurst@... writes:

> > L(2^(p-1))mod 2^p -1 = 0 or, [CASE1]

What's going on here is a special case of the following.

> > L(2^(p-1) -1) mod 2^p -1 = 0 [CASE2]

>

> p CASE

> [ 2, CASE1]

> [ 3, CASE1]

> [ 5, CASE2]

> [ 7, CASE1]

> [13, CASE2]

> [17, CASE2]

> [19, CASE1]

>

> interesting....

>

Let q be any prime = 3 mod 4. (This includes any of the form 2^p-1.)

Then q divides L(m), where

m = (q-1)/2 if q = +-1 mod 5,

and m = (q+1)/2 if q = +-2 mod 5.

[See e.g. Ribenboim's "New Book of Prime Number Records", Chap 2 Sec IV for a

comprehensive treatment of all this.]

For the Mersennes, this gives:-

p q=2^p-1 q mod 5 m=(q+-1)/2 CASE

2 3 -2 2 CASE1

3 7 2 4 CASE1

5 31 1 15 CASE2

7 127 2 64 CASE1

11 2047=23*89 [not prime] --

13 8191 1 4095 CASE2

17 131071 1 65535 CASE2

19 524287 2 262144 CASE1

Numbers q = 3 mod 4 which divide L(m) are not, however, NECESSARILY prime,

only "Lucas pseudoprimes"; so we don't have a new primality test here,

unfortunately.

Mike Oakes - mikeoakes2@a... wrote:

05/01/2002> d.broadhurst@o... writes:

Mersenne with (prime or composite) p <= 4500. [I'm sure you'll

>

> > Thanks for the clarification, Mike.

> > But Shane's question was not answered by either of us:

> > is there a Lucas mersenne pseudoprime,

> > M(p)=2^p-1 with prime p and composite M(p)

> > and one of the two tests satisfied?

> > It's a nice question!

> >

> I have obtained a partial answer to this question: there is no such

appreciate the number of CPU cycles that went into this result,

David -- L(2^4499) is quite big! -- did I compute it? -- "Were

there but world enough and time", as the poet put it...]> However, I expect there actually to be infinitely many such

Mersennes, for the following reason:- about one in 100,000 integers

are "Lucas pseudoprimes" (as defined in my previous posting), the

first being 15251=101*151, the 23rd> being 1970299=199*9901 - and no, they don't ALL have just 2 primes

factors:-). So, if there is nothing special about the form (2^n-1)

(a big IF), and if the density of these pseudoprimes doesn't

decrease too much with increasing size (another big IF), then one

would expect about 1 in 100,000 Mersennes to be Lucas pseudoprimes.> Anyone up to shedding light on these IFs, and/or extending the

search range

> above p=4500?

Hello Mike, You had sent me a program for this, could you send it

> Mike Oakes

again?

I am wondering if PRP/Newpen can be verified first by:

L(2^n-1) mod (k*2^n +/-1)=0

Then if positive execute PRP, and finally proth.

Does the 1/100,000 probability still hold?

What do you think about this variation?