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 2002-11-09, 10:27 #1 TTn   13·523 Posts 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)
 2002-11-09, 10:30 #2 TTn   3×52×61 Posts Oops, 2^13466917 -1 divides F(2^13466917 -2) And so may be prime! sure enough it is.
 2002-11-09, 10:41 #3 TTn   2×32×52×17 Posts 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 ?
2002-11-12, 18:07   #4
svempasnake

Aug 2002

3×7 Posts
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. :(

 2002-11-21, 10:54 #5 TTn   19·433 Posts 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)
 2002-11-23, 03:54 #6 toferc   Aug 2002 2·3·5 Posts Never mind.

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