20021109, 10:27  #1 
2^{4}×3×53 Posts 
Mersenne composite using fibonacci
Let p= 1 mod 4
If Mp, does not divide F(Mp1), then Mp is composite. For example. 2^13466917 divides F(2^13466917 2) 
20021109, 10:30  #2 
5·1,451 Posts 
Oops,
2^13466917 1 divides F(2^13466917 2) And so may be prime! sure enough it is. 
20021109, 10:41  #3 
2^{2}·3·811 Posts 
What would take longer?
(a)The execution of the LucasLehmer test on 2^13466917 1 or, (b)Dividing F(2^13466917 2) by 2^13466917 1, make sure it is not compostie ? 
20021112, 18:07  #4  
Aug 2002
3·7 Posts 
Mersenne composite using fibonacci
Quote:


20021121, 10:54  #5 
1110_{8} 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 7^22=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^p1) 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^p1) 
20021123, 03:54  #6 
Aug 2002
30_{10} Posts 
Never mind.

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