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## Re: Lucas mersenne ?

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• Yes Shane, yours was indeed a nice observation. The classic Lucas-Lehmer test for Mersennes uses the Lucas sequence with p=2, q=-2, d=p^2-4*q=12 (Ribenboim
Message 1 of 12 , Jan 2, 2002
Yes Shane, yours was indeed a nice observation.

The classic Lucas-Lehmer test for Mersennes
uses the Lucas sequence with
p=2, q=-2, d=p^2-4*q=12 (Ribenboim p.92).

You can probably use the same batch of theorems
from Ribenboim to prove the necessity and sufficiency
of your more complicated two-case test with
p=1, q=-1, d=p^2-q=5 (i.e. the rabbit case).

Happy new year

David
• In a message dated 02/01/2002 08:31:16 GMT Standard Time, ... What s going on here is a special case of the following. Let q be any prime = 3 mod 4. (This
Message 2 of 12 , Jan 5, 2002
In a message dated 02/01/2002 08:31:16 GMT Standard Time,

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

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

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
• 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
Message 3 of 12 , Jan 5, 2002
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!
David
• ... So, then I most ask the most obvious question ? Using the Fibonacci sequence: Case 1: F(2^p) mod 2^p -1 = 0 Case 2: F(2^p -2) mod 2^p -1 = 0 F(4)=3 C1
Message 4 of 12 , Jan 5, 2002
> 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!
> David

So, then I most ask the most obvious question ?
Using the Fibonacci sequence:

Case 1: F(2^p) mod 2^p -1 = 0
Case 2: F(2^p -2) mod 2^p -1 = 0

F(4)=3 C1
F(8)=7 C1
F(30)=31 C2
F(128)=127 C1

F(8190,8192)=8191 ?
F( ... ?

The cases so far between Lucas and Fibonacci are identical.
Is the Golden String involved ? (ie.1011010110110...)

Maybe I've gone to far !
Shane F.
• ... I think not, Shane. It seems to me that you have stayed in the same place, since F(2*n)=L(n)*F(n). David Broadhurst
Message 5 of 12 , Jan 5, 2002
Shane Findley wrote:
> Maybe I've gone to far !
I think not, Shane.
It seems to me that you have stayed in the same place,
since F(2*n)=L(n)*F(n).
• One more, issue. Case 2, seems to have 2^p +1 /3 as a prime divisor too ?
Message 6 of 12 , Jan 5, 2002
One more, issue.
Case 2, seems to have 2^p +1 /3 as a prime divisor too ?
• ... I guess what you mean is:- if L(2^(p-1)-1) mod (2^p-1) = 0 [CASE2} then L(2^(p-1)-1) mod (2^p+1)/3 = 0. Well, as per my previous communication, write q =
Message 7 of 12 , Jan 6, 2002
Shane wrote:
> One more, issue.
> Case 2, seems to have 2^p +1 /3 as a prime divisor too ?

I guess what you mean is:-
if L(2^(p-1)-1) mod (2^p-1) = 0 [CASE2}
then L(2^(p-1)-1) mod (2^p+1)/3 = 0.

Well, as per my previous communication, write q = 2^p-1.

If CASE2, then q mod 5 = +-1, and q divides L(m) with m=(q-1)/2 = 2^(p-1)-1.
But q mod 5 = +1 => (2^p+1) mod 5 = 3 => (2^p+1)/3 mod 5 = +1,
and q mod 5 = -1 => (2^p) mod 5 = 0 which is impossible.
So in CASE2, (2^p+1)/3 mod 5 = 1,
so (2^p+1)/3 divides L(n) with n=((2^p+1)/3-1)/2 = (2^(p-1)-1)/3.
But any prime dividing L(n) also divides L(3*n)=L(2^(p-1)-1).
Q.E.D.

Mike Oakes
• In a message dated 05/01/2002 12:42:45 GMT Standard Time, ... I have obtained a partial answer to this question: there is no such Mersenne with (prime or
Message 8 of 12 , Jan 8, 2002
In a message dated 05/01/2002 12:42:45 GMT Standard Time,

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

Mike Oakes
• mikeoakes2@a... wrote: 05/01/2002 ... Mersenne with (prime or composite) p
Message 9 of 12 , Sep 30, 2002
mikeoakes2@a... wrote:
05/01/2002
>
> > 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
Mersenne with (prime or composite) p <= 4500. [I'm sure you'll
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?
> Mike Oakes

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