20100711, 13:29  #1 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
lucas lehmer outstretch
I was told to post on the forum instead of annoying CRGreathouse. so here I am
basically i wan't to find a way to easily connect certain p values with s in a new way. s^{2}2 =k*2^{p}1 (s^{2}2)/k + 1 = 2^{p} p=log_{2}((s^{2}2)/k + 1) this works for s(n1) can we figure out a generalisation for any amount of n change ? if so can we apply it easily and can anyone get the log_{2}() part reduced more ? 
20100711, 13:34  #2 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}×1,531 Posts 

20100711, 14:03  #3 
Aug 2006
3^{2}·5·7·19 Posts 
There's an arithmetic error in your post.
I don't understand your goal here. What does "for any amount of n change" mean? What does "get the log_{2}() part reduced more" mean? 
20100711, 16:08  #4  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
sorry log_{2}(2^p) Last fiddled with by science_man_88 on 20100711 at 16:09 

20100711, 16:14  #5 
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Jul 2009
Dumbassville
8384_{10} Posts 
s^{2}2 =k*(2^{p}1) =correction
(s^{2}2)/k =2^{p}1 ((s^{2}2)/k)+1 =2^{p} log_{2}(((s^{2}2)/k)+1) =p Last fiddled with by science_man_88 on 20100711 at 16:15 
20100711, 16:20  #6 
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Jul 2009
Dumbassville
10000011000000_{2} Posts 
the n change part means for any s within the sequence (so for 2^51 instead of using the 5th one in the sequence; n=4 as n=0 for the first; I could try and use a lower number.) the reason a wanted the log_{2}(((s22)/k)+1) reduced is to get a lower comparison of s and p. like p*k isn't that big compared with k*2^p1 same idea in reducing the log.

20100713, 00:16  #7  
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Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


20100714, 12:35  #8 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
log_{2}(((s^4  4*s^2 + 2)/k)+1) = p is the next one up. I can continue this but the equations get longer and without getting a lower fraction to go with it works out to log(Mersenne number(p) +1) which isn't very helpful.

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