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primenumbers · Prime numbers and primality testing

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  • Members: 1092
  • Category: Number Theory
  • Founded: Dec 27, 2000
  • Language: English
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Messages 15665 - 15696 of 25093   Oldest  |  < Older  |  Newer >  |  Newest
Messages: Simplify | Expand Author Sort by Date ^
15665 pminovic Send Email Dec 1, 2004
9:53 am
Jean, Thank you for the new version in almost no time! I tried to test Kynea numbers. It's working but it's slower than pfgw. Two examples are given below. Now...
15666 Paul Underwood
paulunderwooduk Send Email
Dec 1, 2004
10:50 am
... The way to speed up searching Carol/Kynea numbers is through modular reduction. When reducing over 2^n+-2^k-1 ( k<70% of n ingeneral ) we can use additions...
15667 Jean Penné
jpyah2001 Send Email
Dec 1, 2004
12:50 pm
Thank you for your tests ! I am not surprised if LLPP4 deterministic test is slower than pfgw PRP one, because the "Computing U0" loop is more time consuming...
15668 pminovic Send Email Dec 1, 2004
1:23 pm
... PRP ... This is true, it takes about 50 minutes to compute U0, I'll append the lresults.txt file tomorrow. ... No! To prove primality of a PRP using "pfgw...
15671 jim_fougeron Send Email Dec 1, 2004
9:42 pm
I have a 2004 addition for CPAP-4 (yes, it was very easy to find) 78006074.883#+k*2004+R is prime for k = [0...3] Using the special crafted R of: ...
15672 Jens Kruse Andersen
jkand71 Send Email
Dec 2, 2004
12:32 am
... How fortunate that the year is divisible by 3#. Don't expect me to wait until 2010 to beat it :-) Congratulations on beating my ... erm ... improving your...
15673 Ken Davis
kraden Send Email
Dec 2, 2004
1:19 am
The following is posted on behalf of "Jean Penne" who sent his reply to "primenumbers-owner" instead of "primenumbers&quot; by mistake. Cheers Ken ... pfgw ... the ...
15674 patience_and_fortitude
patience_and... Send Email
Dec 2, 2004
2:42 pm
Are perfect cubes + 1 ever prime? i.e.: Is any number of the form (a^3)+1 ever prime? ... It seems like there should be eventually... ... Better yet, is any...
15675 mikeoakes2@...
mikeoakes2 Send Email
Dec 2, 2004
2:53 pm
In a message dated 02/12/2004 14:46:19 GMT Standard Time, ... No. If q is any odd number > 2, then a^q + 1 = (a+1) * (a^(q-1) - a^(q-2) + ... +1) and so has...
15676 Ray Chandler
rayjchandler Send Email
Dec 2, 2004
4:39 pm
So after 1^q+1=2 there are no further primes for odd q. -Ray Chandler ________________________________ From: mikeoakes2@... [mailto:mikeoakes2@...] ...
15677 Paul Leyland
xilmanuk Send Email
Dec 2, 2004
4:48 pm
... We can say even more than that. All primes of the form (a^q + 1) are either 2 (i.e. a=1) or of the form (b^{2^n}+1) where b and n are integers. When...
15678 patience_and_fortitude
patience_and... Send Email
Dec 2, 2004
7:29 pm
...and a similar rational holds for (a^q)-1 counterparts. Thanks, Shawn...
15679 Didier van der Straten
didiervander... Send Email
Dec 2, 2004
11:48 pm
Dear interested primers Have appreciated feedbacks about my initial draft. I start disclosing "my" Goldbach partition count evaluation method and formula. ...
15680 gulland68 Send Email Dec 3, 2004
12:06 am
Someone told me that one of either Goldbach or Riemann had been proven unprovable by current knowledge of mathematics. Can someone inform me as to which it...
15681 Jose Ramón Brox
ambroxius Send Email
Dec 3, 2004
12:50 am
My two cents... Original English Text: My love you are no where, and I want you now here Translating and translating back... My love you are no where, and I...
15682 Jim Doyle
ozyjim2004 Send Email
Dec 3, 2004
12:52 am
Hi I'm not sure if this is a new idea but I have not come across it . Its based on the fact that apart from 2 and 5 all primes have a last digit of 1,3,7 or 9....
15683 Robin Garcia
sopadeajo2001 Send Email
Dec 3, 2004
2:45 am
Robin Garcia <sopadeajo2001@...> wrote: Robin Garcia <sopadeajo2001@...> wrote: Would anybody try to speed up the algorith implemented in a simple...
15684 Jud McCranie
judmccr Send Email
Dec 3, 2004
4:02 am
... As far as I know, neither of them has been proven undecidable. But either one of them could conceivably be undecidable....
15685 Jud McCranie
judmccr Send Email
Dec 3, 2004
4:02 am
This is true, isn't it? For every epsilon > 0, let x = 1+epsilon. There is always a prime between x^n and x^(n+1) for all n>m, where m depends on epsilon....
15686 Sren Nielsen
srentospace Send Email
Dec 3, 2004
4:06 pm
Under visualization there are four links, I couldn't open. I know that at least one link (my own) is closed for ever. I have tried to tell it to Chriss...
15687 Chris Caldwell
primemogul Send Email
Dec 3, 2004
10:08 pm
... Perhaps I can; but what page are you talking about? It always helps to be specific. Perhaps you mean: http://primes.utm.edu/links/visualization/ This is...
15688 Jens Kruse Andersen
jkand71 Send Email
Dec 3, 2004
10:26 pm
... I also chose to search an easy improvement with a "special" d: 46313478 * 1201#/1302643 + x498 + 2310n, for n = 0..3 Using the special crafted x498 at the...
15689 cino hilliard
hillcino368 Send Email
Dec 4, 2004
5:13 am
Hi, ... By Ohakm's Razor, the long division (x^n+1)/(x+1) is the best proof that x+1 divides. However We also can also argue as follows using Baconian...
15690 primenumbers@yahoogro... Send Email Dec 4, 2004
9:35 am
Hello, This email message is a notification to let you know that a file has been uploaded to the Files area of the primenumbers group. File :...
15691 Norman Luhn
nluhn Send Email
Dec 4, 2004
11:22 am
Hello members ! The start of a record-breaking run of 170 consecutive integers (4339207185939-4339207186108) that are non-semiprimes. No greater runs less than...
15692 Michael Gian
pastmyprime2 Send Email
Dec 5, 2004
10:19 pm
As I have read, for prime numbers, p, 2^p-1 are called Prime- Exponent Mersenne numbers. These are sub-group of 2^k-1, k positive integers, which are called ...
15693 Décio Luiz Gazzoni...
deciogazzoni Send Email
Dec 6, 2004
1:11 am
... In response to this astonishing (for me at least) claim, I had to try and prove that this is true. I recommend anyone else to try to prove it before ...
15694 David Cleaver
wraythex Send Email
Dec 6, 2004
10:39 am
Hello everyone, I visited the ECMNET page located at: http://www.loria.fr/~zimmerma/records/ecmnet.html and saw that the latest version of ecm available is...
15695 Ignacio Larrosa Ca...
ilarrosa Send Email
Dec 6, 2004
10:55 am
Sunday, December 05, 2004 11:19 PM [GMT+1=CET], ... The sum of the first n odd cubes is S(n) = Sum((2k-1)^3, k, 1, n) = 2n^4 - n^2 = n^2(2n^2 - 1) If n = 2^m,...
15696 Mazzarello Gianni
mzzgv Send Email
Dec 6, 2004
11:55 am
It seems correct, but we can say it in an equivalent way: the set of prime numbers = 2, 5 PLUS all odd numbers > 5, not ending by 5 and not composite. ...
Messages 15665 - 15696 of 25093   Oldest  |  < Older  |  Newer >  |  Newest
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