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Old 2002-11-09, 10:27   #1
TTn
 

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Default Mersenne composite using fibonacci

Let p= 1 mod 4

If Mp, does not divide F(Mp-1), then Mp is composite.


For example.
2^13466917 divides F(2^13466917 -2)
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Old 2002-11-09, 10:30   #2
TTn
 

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Oops,
2^13466917 -1 divides F(2^13466917 -2)

And so may be prime!
sure enough it is.
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Old 2002-11-09, 10:41   #3
TTn
 

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What would take longer?

(a)The execution of the Lucas-Lehmer test on 2^13466917 -1
or,
(b)Dividing F(2^13466917 -2) by 2^13466917 -1, make sure it is not compostie
?
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Old 2002-11-12, 18:07   #4
svempasnake
 
Aug 2002

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Default Mersenne composite using fibonacci

Quote:
Originally Posted by TTn
What would take longer?

(a)The execution of the Lucas-Lehmer test on 2^13466917 -1
or,
(b)Dividing F(2^13466917 -2) by 2^13466917 -1, make sure it is not compostie
?
I hope (a), but I think (b). My fantasy fails when trying to figure out any way of dealing with that number. F(2^13466917 -2) just looks like an astronomically long number to me. :(
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Old 2002-11-21, 10:54   #5
TTn
 

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Well, Fibonacci numbers are smaller than the currently used Lehmer test numbers, but the algorithm lay undiscovered.

For example Lucas sequence 2,1, 3 ,4, 7 ,11,18,29...
L(2^n) =
3, then 3^2 -2 =7, then 7^2-2=47, and so on, just like Lehmer test but with a smaller starting number of three.



I found more !


Let p be a prime>7 satisfying the following conditions:
1. p= 2,4(mod 5)
2. 2^[p+1] -3, is also prime
Then (2^[p+1]-3) | F(2^p-1)

Let p be a prime>5 satisfying the following conditions:
1. p = 4 (mod5)
2. 2^[p+1]-1 is also prime
Then(2^[p+1]-1) | L(2^p-1)



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Old 2002-11-23, 03:54   #6
toferc
 
Aug 2002

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Never mind.
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