Hi, I've read the "new Mersenne Divisors conj." thread, and I propose the following test for Sophie Germain primes > 3 that does not require 2 different cases...
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djbroadhurst
Jan 19, 2010 5:55 pm
... Looks good. But you can speed it up by a factor of 2 by working with Mod(3/4,2*p+1) JL(p)=Mod(3/4,2*p+1)^p==1; ...
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j_chrtn
Jan 19, 2010 6:56 pm
... Right. 3/4 appears in the proof. Do you know if this test is something well known? I ask this because I've never seen it before. J-L...
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djbroadhurst
Jan 20, 2010 12:38 pm
... It's not well known because it's quite unnecessary. The foolproof "unified Sophie" test that works in *all* cases is much simpler. If p is prime, then q =...
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j_chrtn
Jan 20, 2010 8:34 pm
... Unnecessary but at least correct. ... This is closed to Henri Lifchitz's test 3^p = 1 (mod q) but your's also handles p=2 and p=3. ... "There Is More Than...
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bhelmes_1
Jan 23, 2010 2:14 pm
A beautifull day Results for Primes mod some numbers up to 10^14 is ready http://beablue.selfip.net/devalco/table_of_primes.htm I checked the results to P mod...
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Andrey Kulsha
andrey_601
Jan 23, 2010 3:34 pm
... 10^15 or 10^16. http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind1001&L=nmbrthry&T=0&X=14ADB57FE44944E3D4&P=327 Best regards, Andrey...
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Peter Lesala
plesala@...
Jan 27, 2010 6:47 am
Hi, It is high time I deliver what I promised; still keeping my fingers crossed that I am not duplicating something already known. At the time of writing the...
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Bill Bouris
leavemsg1
Jan 27, 2010 3:02 pm
Pete, I did it by hand; sorry for the intrusion, last time. the counter-example for 46; 45 is not enough! notice how the columns are formed; that's how I'm...
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djbroadhurst
Jan 27, 2010 4:14 pm
... It's easy to prove that N = 2^2203 - 2^499 - 1 is prime since we know the complete factorization of N + 1 = 2^499*(2^1704 - 1) which contains no prime with...
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Peter Lesala
plesala@...
Jan 27, 2010 4:41 pm
In the previous message I should have included the line involving 2^7 -1, which has primes 2^7 -1+ 2^2 and 2^7 -1+ 2^6. This line is very important because n =...
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djbroadhurst
Jan 27, 2010 6:20 pm
... Here's an easy proof where one doesn't even need to look at the Cunningham tables: pfgw -tp -f300 -q"2^521+2^160-1" 2^521+2^160-1 is prime!...
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djbroadhurst
Jan 27, 2010 11:12 pm
... This "only a PRP" entry from Peter ... was clearly misplaced. With b/a > 1/3, the primality proof requires merely the knowledge that 2 is a prime: pfgw -tp...
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Sren
srentospace
Jan 28, 2010 10:25 pm
I mean I have read, that the formula n*n-n+41 is the formula giving most primes of such kind of formulas. Is that right?...
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Sren
srentospace
Jan 28, 2010 10:29 pm
I mean I have read, that the formula n*n-n+41 is the formula giving most primes of such kind of formulas. Is that right?...
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Bob Gilson
bobgillson
Jan 29, 2010 10:10 am
Isn't it amazing that a Swedish pop group can have so much influence! ________________________________ From: Bill Bouris <leavemsg1@...> To: Peter...
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Jens Kruse Andersen
jkand71
Jan 29, 2010 1:05 pm
... It depends what is meant by "most primes" and "such kind of formulas". n^2-n+41 gives distinct primes for n = 1 to 40. This is the largest known number of...
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miltbrown@...
Jan 31, 2010 1:09 pm
... Euler discovered this with p=41. It is the values on the height of the number triangle: ... ... ... ... ... ... 45 46 47 48 49 42 43 44 41 This is similar...
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Peter Lesala
plesala@...
Jan 31, 2010 2:26 pm
Thanks a lot David. The reference you gave is exactly what i have been looking for. Peter. ... From: djbroadhurst To: primenumbers@yahoogroups.com Sent:...
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Bill Bouris
leavemsg1
Jan 31, 2010 8:47 pm
Proth theorem extended(but not by much, but w/out any sacrifice): Let Q= k*2^n +1, where n=> 3 is an odd natural number & k<= 2^n+1. If for some 'a',...
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William
fatbobforman
Feb 1, 2010 2:50 am
Like I believe that the primes (except 2,3,5)are all of the form 30x + y where y in {7,11,13,17,19,23,29,31} and I am working on a search algorithm for what I...
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Bill Bouris
leavemsg1
Feb 1, 2010 2:19 pm
Group, I believe that if 'N' passes the 2-PRP test that either 3, 5, or 7 divides 'N-1', one or all of them, or 'N' will be divisible by 3, 5, or 7 and be...
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Paul Leyland
xilmanuk
Feb 1, 2010 2:39 pm
Aargh! I misunderstood the example and posted complete claptrap. My apologies. To make amends: all Fermat numbers pass the 2-PRP test. Many composite ...
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Bill Bouris
leavemsg1
Feb 1, 2010 2:45 pm
forwarded... ... From: Bill Bouris <leavemsg1@...> To: Paul Leyland <paul@...> Sent: Mon, February 1, 2010 8:42:15 AM Subject: Re: {Spam?}...
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Paul Leyland
xilmanuk
Feb 1, 2010 2:54 pm
If you used your elderly computer to search for known results rather than for computations you would have discovered ...
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djbroadhurst
Feb 1, 2010 5:49 pm
... n = 1097343303233 is not prime n is 2-PRP n is 3-PRP n is coprime to 3*5*7 n-1 is coprime to 3*5*7 There are 95 such counterexamples in ...
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Bill Bouris
leavemsg1
Feb 1, 2010 7:00 pm
Here's how I want to define the magnum 357 test! take N; if 'N' passes the 2-PRP test, then... if 3, 5, or 7 divides 'N', then 'N' is composite... and we are...
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Jack Brennen
jbrennen
Feb 1, 2010 7:33 pm
Try N = 25326001. Try N = 3215031751. Then fix the test to handle those correctly. :)...
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djbroadhurst
Feb 2, 2010 12:27 am
... It takes less than 2 seconds to prove that there are at least 124880 counterexamples with N < 10^15. See: ...
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djbroadhurst
Feb 2, 2010 3:35 pm
... In less than 20 seconds, I found 993305 counterexamples in tables from William Galway and Richard Pinch. There are 319226 numbers N < 10^15 that are 2-PSP...
