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Old 2004-10-06, 16:40   #1
jocelynl
 
Sep 2002

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Default factors of lucas sequences

Is there a project that looks for factors of the lucas sequence used in prime95
s(n)=s(n-1)^2-2 where s(0)=4 but not using modulo
The numbers get large rapidly. How far has it been factored.

Joss
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Old 2004-10-06, 16:51   #2
R.D. Silverman
 
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Quote:
Originally Posted by jocelynl
Is there a project that looks for factors of the lucas sequence used in prime95
s(n)=s(n-1)^2-2 where s(0)=4 but not using modulo
The numbers get large rapidly. How far has it been factored.

Joss
IMHO, we already have enough factoring projects. We don't need
another. Furthermore, this sequence grows doubly exponentially.
Factoring beyond the first few terms would be hopeless.
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Old 2010-09-23, 21:32   #3
science_man_88
 
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well if I can figure right bob Mersenne numbers are a lucas sequence that I never remember hearing.

if you use U(3,2) you get:

U0 = 0
U1 = 1
U2 = (3*1)-(2*0)=3-0 = 3
U3 = (3*3)-(2*1)=9-2 = 7
U4 = (3*7)-(2*3)=21-6 = 15

....

well i think you get the point.

I checked V that go along with them aren't the exponents though precprime() in pari for each member gives an exponent that's known up to 127.

Last fiddled with by science_man_88 on 2010-09-23 at 21:34
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