lucas-lehmer theorem
 

Hello,
there are various generalization of the lucas-lehmer test,for example one is in [url]http://mathworld.wolfram.com/Lucas-LehmerTest.html[/url] for N+1.
I think that i've found another generalization to the theorem and proved the sufficent part.
but it seems that the necessary part is beyond my mathematical skill.so,suppose that the theorem that i'm speak about it is not know,my question is:
do i have to spent months(years?) to gain the right amount of math knowledge and try to complete the proof,and maybe let another person discover the same thing, or it's better to release an incomplete theorem and let the necessary part to someone else?

[QUOTE=Robot2357]
do i have to spent months(years?) to gain the right amount of math knowledge and try to complete the proof,and maybe let another person discover the same thing, or it's better to release an incomplete theorem and let the necessary part to someone else?[/QUOTE]

Mathematics is a collaborative endeavor. Post what you know here and people will discuss it.

John

[QUOTE=Robot2357]Hello,
there are various generalization of the lucas-lehmer test, ...
I think that i've found another generalization to the theorem and proved the sufficent part.[/QUOTE]It is not widely known that the LLT can be used to prove the primality of other numbers than Mersenne numbers. Moreover, many people think that only numbers N for which a factorization of N+1 is known (like Mersenne numbers) can be proved prime by the LLT. As an example, the LLT can also be used for Fermat numbers, see [URL=http://tony.reix.free.fr/Mersenne/PrimalityTest2FermatNumbers.pdf]my paper[/URL].
It is the same for the Pépin's test. It can be used for proving the primality of a kind of Generalized Fermat Numbers, see [URL=http://tony.reix.free.fr/Mersenne/PrimalityTest2FermatNumbers.pdf]Saouter[/URL] .
Let describe your findings here. Better Mathematicians than me will help you to understand if it is new or not.
Tony

new one
 

S(n)=2*S(n-1)^2 -1 where S(0)=2

[QUOTE=hoca;94349]S(n)=2*S(n-1)^2 -1 where S(0)=2[/QUOTE]

A.k.a. [url=http://www.mersenneforum.org/showthread.php?t=6800]"LLT in slight disguise."[/url]

[QUOTE=hoca;94349]S(n)=2*S(n-1)^2 -1 where S(0)=2[/QUOTE]


Don't cross post. I already answered you in another thread.

[URL="http://numbertheoryapplets.weebly.com/generalization-of-lucas-lehmer-riesel-test.html"]Generalization of Lucas-Lehmer-Riesel Test[/URL]