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Messages 15848 - 15877 of 19504   Oldest  |  < Older  |  Newer >  |  Newest
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15848
The solution of Goldbach's conjectur 1)The conventional statement: BELOW every even natural number 2n >2 ,there is at least one prime pair( P_i , P_ j ) such...
zaljohar
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Jan 3, 2005
11:21 am
15849
I am happy to announce that ECMNet Client/Server 2.5.6 is finally released. You can find it at http://www.loria.fr/~zimmerma/records/ecmnet.html. At this...
Mark Rodenkirch
mgrogue
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Jan 4, 2005
2:29 am
15850
Define a q-residue modulo n as follows: An integer a is a q-residue modulo n if and only if it exists an integer b such as b^(2^q)=a mod n Obviously if q=1...
hl59126
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Jan 4, 2005
3:08 pm
15851
Is the source include with the release? Rob ... From: Mark Rodenkirch [mailto:mgrogue@...] Sent: Monday, January 03, 2005 8:29 PM To:...
Robert Sitton
rsitton@...
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Jan 5, 2005
6:26 am
15852
Congratulations to all at Seventeen or Bust for finding that 28433*2^7830457+1 is prime. This has 2,357,207 digits and is the fourth largest known prime. There...
Paul Jobling
paul_joblinguk
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Jan 5, 2005
10:57 am
15853
hleleu ... Define a 2^n-residue modulo m as follows: An integer a is a 2^n-residue modulo m if and only if it exists an integer b such as b^(2^n)=a mod m ...
Payam Samidoost
samidoost
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Jan 5, 2005
11:11 am
15854
... Of course....
Mark Rodenkirch
mgrogue
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Jan 5, 2005
12:10 pm
15855
I want to try and get the follow congruencies to combine it with Proth's theorem. Any suggestions? 1) Let N = the number to test a^(2*N) - a^(N-1)*(a^2+1) + 1...
John W. Nicholson
reddwarf2956
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Jan 7, 2005
4:26 am
15856
I've recently discovered primes and find it fashinating, i'm currently trying to find as many primes as possible that follow the pattern of (x^x)+1 = prime. So...
marcus bunny
marcus_bunny
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Jan 7, 2005
5:00 pm
15857
... You have found all of the known examples (for x an integer). Note first that x=1 is a solution. Assume for the remainder of this discussion that x>1. If x...
jbrennen
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Jan 7, 2005
7:34 pm
15858
Hi, ... This pattern can be extended adnauseam Probable primes in x^x +/- 1,2,3,..k x^x - x +/- 1,2,3,..k x^x +/- x^k +/- k x^x +/- x(x-k) +/- 1,2,3,..k .. .. ...
cino hilliard
hillcino368
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Jan 8, 2005
2:50 am
15859
If I'm given a fraction x/y = a_0 + 1/(a_1 + 1/(a_2 + ... + 1/(a_n)...)), what is the fastest way to compute w/z = a_0 + 1/(a_1 + 1/(a_2 + ... + ...
Décio Luiz Gazzoni...
deciogazzoni
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Jan 8, 2005
4:10 am
15860
I'm sure most of us have seen the proof that there is no largest prime by taking one plus the product of any list of primes and showing that either that number...
jbohanon3
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Jan 8, 2005
6:10 am
15861
In a message dated 08/01/2005 06:12:34 GMT Standard Time, ... No. For one thing, the first term of your series 2,3,7,43,... is = +2 mod 5, the second is = -2...
mikeoakes2@...
mikeoakes2
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Jan 8, 2005
9:05 am
15862
Hello! I have created a ms EXCEL macro to create a colorful graphic using a sequential list of prime numbers 2 through 19997 (about 2262 in the list). The...
slarom69
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Jan 8, 2005
9:28 am
15863
... What if we get a squareful number? Suppose, for arguments sake, that 2*3*7+1 is 4693 which is -2 mod 5 as required. 4693 = 13*19^2. If we only use one 19...
Jens Kruse Andersen
jkand71
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Jan 8, 2005
12:57 pm
15864
hi, Consider 1. If p is prime and p+k is prime then p+k divides p^(p+k) + k. 2. the converse If p is prime and p+k is composite and p+k divides p^(p+k) +k,...
cino hilliard
hillcino368
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Jan 9, 2005
8:23 am
15865
This program is a rough test for the solution of Goldbach's conjecture presented in the message titled: final corrected version of trying solving goldbach's...
zaljohar
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Jan 9, 2005
3:21 pm
15866
Concerning the Hardy-Littlewood conjecture, pi(x+y)-pi(x)<=pi(y). latest find pi(x+3243)-pi(x) = 457 = pi(3243) [0] also, large grouping of -1's around 3429 ...
Tom
thoeng
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Jan 10, 2005
3:41 am
15867
Guys, I've been looking at the following equation; R=4^k +/- P*2^(k-1) - 1, where k = 2,3,4 ... and P is a prime number. The two (or four) most interesting...
Cletus Emmanuel
cemmanu
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Jan 10, 2005
8:00 pm
15868
Your equations are a special form of the more general form 2^a +/- b*2^c +/- 1 Phil C. wrote a sieve base-10 of this form for me; you could ask him to write a...
Rob Binnekamp
robdine2004
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Jan 11, 2005
3:10 pm
15869
I'm announcing the discovery of the 38th Carol prime, (2^227493-1)^2- 2 by me this 11th of January. Steven Harvey...
harvey563@...
harvey563
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Jan 11, 2005
7:29 pm
15870
Steven, Congrats! On you new Carol Prime. Now you are on top of the Carol/Kynea top 10 list again... It is good to see all the interest in these numbers and...
Cletus Emmanuel
cemmanu
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Jan 11, 2005
8:46 pm
15871
The actual number is (2^226749-1)^2-2 with 136517 digits. Steven Harvey...
harvey563@...
harvey563
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Jan 11, 2005
8:55 pm
15872
Congratulations to Predrag Minovic for finding a new record Sophie Germain Prime pair. 7068555 · 2121302-1 36523 L100 2005 Sophie Germain (2p+1) 7068555 ·...
David Underbakke
dlunderbakke
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Jan 11, 2005
9:52 pm
15873
Thank you! I think I was lucky to select n=121300 as the exponent. The new record has the second smallest multiplier among top-20 Sohpie Germain primes. I used...
pminovic
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Jan 11, 2005
10:31 pm
15874
1 REM This is a program showing that below every even number y the superficial prime pairs fulfils Goldbach's conjecture:as y increases the number of even...
zaljohar
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Jan 12, 2005
10:48 am
15875
does anybody know where to find such k's for which neither 6k-1 nor 6k+1 is prime or a program that finds such k's. please e-mail me if you have such...
mcnamara_gio
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Jan 13, 2005
1:16 pm
15876
Hi all, What can I use to sieve for N=35*10^k - 1? And, would this form be considered near-repunit? I did not see this form in Multisieve, but I suspect that...
Cletus Emmanuel
cemmanu
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Jan 13, 2005
2:50 pm
15877
Hi Cletus, I don't consider 3499...99 is near-repdigit because Chris K. Caldwell's Top Twenty page (http://primes.utm.edu/top20/page.php?id=15) says "Let all...
KAMADA Makoto
makotokamada
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Jan 13, 2005
3:42 pm
Messages 15848 - 15877 of 19504   Oldest  |  < Older  |  Newer >  |  Newest
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