20060513, 06:22  #1  
Feb 2005
404_{8} Posts 
Penultimate LucasLehmer step
Following the email by PeterLawrence Montgomery dated 11 Dec 2003 (quoted below).
1) There are results about some other starting values in Bas Jansen's thesis "Mersenne primes and class field theory" available in PDF file pages 18551858 (5760) 2) Now we have 43 Mersenne primes known. Is it possible to get the sign of a[p2]/2^((p+1)/2) mod Mp for them from the existing databases (without redoing LucasLehmer iterations)? Quote:


20060513, 13:09  #2  
Feb 2004
France
1110100101_{2} Posts 
Quote:
This document is'nt crystal clear for me, probably because it is only a summary (and because I'm an amateur ...). Is that all clear for you ? Thanks, Tony 

20060513, 13:20  #3  
Feb 2005
100000100_{2} Posts 
Quote:
Quote:


20060513, 13:23  #4 
Feb 2004
France
3×311 Posts 
I know about another guy doing a PhD thesis on the same kind of subject. I've forgotten his (complex) name. I think I remember his thesis is planned to end soon. I'll try to find his name in my emails ...
T. 
20060513, 13:28  #5  
Feb 2005
2^{2}×5×13 Posts 
Quote:


20060513, 13:55  #6  
Feb 2004
France
3×311 Posts 
Quote:
T. 

20060524, 23:31  #7 
Einyen
Dec 2003
Denmark
2×17×101 Posts 
This is as far as I got with just C++ and GMP library: From M2M32: 15 times s[p3]=+2^((p+1)/2) 16 times s[p3]=2^((p+1)/2) We need prime95 to be configured to save last iteration for the rest of them. Last fiddled with by ATH on 20060524 at 23:35 
20070514, 11:55  #8 
Einyen
Dec 2003
Denmark
3434_{10} Posts 
I went back and finished the list with prime95:
M2: p=3 s[p3]== 2^((p+1)/2) (mod Mp) M3: p=5 s[p3]== 2^((p+1)/2) (mod Mp) M4: p=7 s[p3]== 2^((p+1)/2) (mod Mp). M5: p=13 s[p3]== 2^((p+1)/2) (mod Mp). M6: p=17 s[p3]== 2^((p+1)/2) (mod Mp). M7: p=19 s[p3]== 2^((p+1)/2) (mod Mp). M8: p=31 s[p3]== 2^((p+1)/2) (mod Mp). M9: p=61 s[p3]== 2^((p+1)/2) (mod Mp). M10: p=89 s[p3]== 2^((p+1)/2) (mod Mp). M11: p=107 s[p3]== 2^((p+1)/2) (mod Mp). M12: p=127 s[p3]== 2^((p+1)/2) (mod Mp). M13: p=521 s[p3]== 2^((p+1)/2) (mod Mp). M14: p=607 s[p3]== 2^((p+1)/2) (mod Mp). M15: p=1279 s[p3]== 2^((p+1)/2) (mod Mp). M16: p=2203 s[p3]== 2^((p+1)/2) (mod Mp). M17: p=2281 s[p3]== 2^((p+1)/2) (mod Mp). M18: p=3217 s[p3]== 2^((p+1)/2) (mod Mp). M19: p=4253 s[p3]== 2^((p+1)/2) (mod Mp). M20: p=4423 s[p3]== 2^((p+1)/2) (mod Mp). M21: p=9689 s[p3]== 2^((p+1)/2) (mod Mp). M22: p=9941 s[p3]== 2^((p+1)/2) (mod Mp). M23: p=11213 s[p3]== 2^((p+1)/2) (mod Mp). M24: p=19937 s[p3]== 2^((p+1)/2) (mod Mp). M25: p=21701 s[p3]== 2^((p+1)/2) (mod Mp). M26: p=23209 s[p3]== 2^((p+1)/2) (mod Mp). M27: p=44497 s[p3]== 2^((p+1)/2) (mod Mp). M28: p=86243 s[p3]== 2^((p+1)/2) (mod Mp). M29: p=110503 s[p3]== 2^((p+1)/2) (mod Mp). M30: p=132049 s[p3]== 2^((p+1)/2) (mod Mp). M31: p=216091 s[p3]== 2^((p+1)/2) (mod Mp). M32: p=756839 s[p3]== 2^((p+1)/2) (mod Mp). M33: p=859433 s[p3]== 2^((p+1)/2) (mod Mp). M34: p=1257787 s[p3]== 2^((p+1)/2) (mod Mp). M35: p=1398269 s[p3]== 2^((p+1)/2) (mod Mp). M36: p=2976221 s[p3]== 2^((p+1)/2) (mod Mp). M37: p=3021377 s[p3]== 2^((p+1)/2) (mod Mp). M38: p=6972593 s[p3]== 2^((p+1)/2) (mod Mp). M39: p=13466917 s[p3]== 2^((p+1)/2) (mod Mp). M40?:p=20996011 s[p3]== 2^((p+1)/2) (mod Mp). M41?:p=24036583 s[p3]== 2^((p+1)/2) (mod Mp). M42?:p=25964951 s[p3]== 2^((p+1)/2) (mod Mp). M43?:p=30402457 s[p3]== 2^((p+1)/2) (mod Mp). M44?:p=32582657 s[p3]== 2^((p+1)/2) (mod Mp). So 22+ and 21 so far. Here is residues from the last iterations from prime95 for each exponent: results.txt I have intermediate savefiles from M32M44 if anyone needs for anything else. Edit: Prime95 gave some wierd residues for p<521 so I calculated them myself and pasted into the Prime95 log. Last fiddled with by ATH on 20070514 at 12:14 
20070514, 17:28  #9 
∂^{2}ω=0
Sep 2002
República de California
5×2,351 Posts 
Nice work, ATH. No correlation with the usual suspects (congruence of exponent mod 4 and 6), and no apparent correlation with regular/irregularprime status of the exponent  the roughly equal fraction of + signs would also seem to rule the latter out, even in the absence of an exponentbyexponent comparison. (Not that we expect any correlation based on that to begin with, but just by way of kicking the tires, as it were.)
Seems pretty random to me... 
20070514, 22:51  #10 
Feb 2005
104_{16} Posts 
Thanks, ATH.
Please submit more terms to http://www.research.att.com/~njas/sequences/A123271 
20070515, 00:07  #11 
Einyen
Dec 2003
Denmark
D6A_{16} Posts 
I'm not sure how to submit terms, can you do it? The missing terms are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 Btw it doesn't say that the sequence starts for n=2 or is that what "OFFSET 2,1" means? Last fiddled with by ATH on 20070515 at 00:07 
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